{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0001","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6b2dc51087ff737b6a573e510e698c3ad5d26ebaeaf2a52b8c16c3a8362cdab9","family_id":"qf-mod38-markov-qualification","seed":202611074001,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,6],[2,6],[3,6]],\"values\":[0,1,3]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/6, 2/6, 3/6] and aligned total values X=[0, 1, 3]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/6, 1/3, 1/2], [0, 1, 3], Yes, 11/6, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/6, 1/3, 1/2], [0, 1, 3], Yes, 11/6, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0002","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2896c82cb9d3ae1b6cc798f853ce3e7cf7b401fd681997266b7b80b3f3780f34","family_id":"qf-mod38-markov-qualification","seed":202611074002,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,7],[2,7],[4,7]],\"values\":[0,2,4]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/7, 2/7, 4/7] and aligned total values X=[0, 2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/7, 2/7, 4/7], [0, 2, 4], Yes, 20/7, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/7, 2/7, 4/7], [0, 2, 4], Yes, 20/7, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0003","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e1b9f04e035c5f3fdc01f8e9ea6477824f8e24e555ef4d64888e350cc817623b","family_id":"qf-mod38-markov-qualification","seed":202611074003,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,8],[2,8],[5,8]],\"values\":[0,3,5]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/8, 2/8, 5/8] and aligned total values X=[0, 3, 5]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/8, 1/4, 5/8], [0, 3, 5], Yes, 31/8, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/8, 1/4, 5/8], [0, 3, 5], Yes, 31/8, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0004","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"af042f5bb0bf7de774809e3b1fe57f4a63abdc4767e2aa89e28909ba1789fd82","family_id":"qf-mod38-markov-qualification","seed":202611074004,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,9],[2,9],[6,9]],\"values\":[0,4,6]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/9, 2/9, 6/9] and aligned total values X=[0, 4, 6]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/9, 2/9, 2/3], [0, 4, 6], Yes, 44/9, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/9, 2/9, 2/3], [0, 4, 6], Yes, 44/9, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0005","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"decf01bf9c9bd81cc341379699fda63fa0330e37a38f74221a932247219b41db","family_id":"qf-mod38-markov-qualification","seed":202611074005,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,10],[2,10],[7,10]],\"values\":[0,5,7]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/10, 2/10, 7/10] and aligned total values X=[0, 5, 7]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/10, 1/5, 7/10], [0, 5, 7], Yes, 59/10, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/10, 1/5, 7/10], [0, 5, 7], Yes, 59/10, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0006","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d88189cb28b421260df1a9086d3d01252f95e528ea33a83715dc0e80134c939a","family_id":"qf-mod38-markov-qualification","seed":202611074006,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,11],[2,11],[8,11]],\"values\":[0,6,8]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/11, 2/11, 8/11] and aligned total values X=[0, 6, 8]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/11, 2/11, 8/11], [0, 6, 8], Yes, 76/11, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/11, 2/11, 8/11], [0, 6, 8], Yes, 76/11, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0007","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"db3c80d67282f8f7f22c984e08dc50e6d8cf249582077e77ebb0bcec620c9afe","family_id":"qf-mod38-markov-qualification","seed":202611074007,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,12],[2,12],[9,12]],\"values\":[0,7,9]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/12, 2/12, 9/12] and aligned total values X=[0, 7, 9]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/12, 1/6, 3/4], [0, 7, 9], Yes, 95/12, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/12, 1/6, 3/4], [0, 7, 9], Yes, 95/12, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0008","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a2169c091088323d070caca4457d9b747f0193b8b23f69d1228d46a26b0311a","family_id":"qf-mod38-markov-qualification","seed":202611074008,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,13],[2,13],[10,13]],\"values\":[0,8,10]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/13, 2/13, 10/13] and aligned total values X=[0, 8, 10]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/13, 2/13, 10/13], [0, 8, 10], Yes, 116/13, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/13, 2/13, 10/13], [0, 8, 10], Yes, 116/13, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0009","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a2fd65eac825ca82622bb3056a5a3bcbb42970e399fd0eaf5bd121b597fb3594","family_id":"qf-mod38-markov-qualification","seed":202611074009,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,14],[2,14],[11,14]],\"values\":[0,9,11]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/14, 2/14, 11/14] and aligned total values X=[0, 9, 11]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/14, 1/7, 11/14], [0, 9, 11], Yes, 139/14, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/14, 1/7, 11/14], [0, 9, 11], Yes, 139/14, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0010","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"482b1008006a016df7936315861f8c0b531c9e2ce9fb4510a21dc1d4d86bef18","family_id":"qf-mod38-markov-qualification","seed":202611074010,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,15],[2,15],[12,15]],\"values\":[0,10,12]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/15, 2/15, 12/15] and aligned total values X=[0, 10, 12]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/15, 2/15, 4/5], [0, 10, 12], Yes, 164/15, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/15, 2/15, 4/5], [0, 10, 12], Yes, 164/15, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0011","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5a4d0e5fca931668b676add621f378799de5f2682e9095c129485b03f4a056a1","family_id":"qf-mod38-markov-qualification","seed":202611074011,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,16],[2,16],[13,16]],\"values\":[0,11,13]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/16, 2/16, 13/16] and aligned total values X=[0, 11, 13]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/16, 1/8, 13/16], [0, 11, 13], Yes, 191/16, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/16, 1/8, 13/16], [0, 11, 13], Yes, 191/16, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0012","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e128037228fd6ace0bb12bb5fffcd23d8590a760a3a7d09944a172a630f99fb8","family_id":"qf-mod38-markov-qualification","seed":202611074012,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,17],[2,17],[14,17]],\"values\":[0,12,14]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/17, 2/17, 14/17] and aligned total values X=[0, 12, 14]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/17, 2/17, 14/17], [0, 12, 14], Yes, 220/17, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/17, 2/17, 14/17], [0, 12, 14], Yes, 220/17, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0013","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"80d423b70f29268c9e0d5eab2299c8fbb9e8be04fa05d1fd1c60b670e4c0b418","family_id":"qf-mod38-markov-qualification","seed":202611074013,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,18],[2,18],[15,18]],\"values\":[0,13,15]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/18, 2/18, 15/18] and aligned total values X=[0, 13, 15]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/18, 1/9, 5/6], [0, 13, 15], Yes, 251/18, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/18, 1/9, 5/6], [0, 13, 15], Yes, 251/18, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0014","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ec7fb417278d113362ecb5723eb312e83b1de1ba8b537dd06c6b5d7aef10065a","family_id":"qf-mod38-markov-qualification","seed":202611074014,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,19],[2,19],[16,19]],\"values\":[0,14,16]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/19, 2/19, 16/19] and aligned total values X=[0, 14, 16]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/19, 2/19, 16/19], [0, 14, 16], Yes, 284/19, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/19, 2/19, 16/19], [0, 14, 16], Yes, 284/19, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0015","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aa0add468e757ba888b1a139ba415befffbff207774e50de645b330e98b752ba","family_id":"qf-mod38-markov-qualification","seed":202611074015,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,20],[2,20],[17,20]],\"values\":[0,15,17]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/20, 2/20, 17/20] and aligned total values X=[0, 15, 17]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/20, 1/10, 17/20], [0, 15, 17], Yes, 319/20, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/20, 1/10, 17/20], [0, 15, 17], Yes, 319/20, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0016","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5e65a5e5b12cad7ac5a7c221bc3deeabaa03d8b858280b679846f1de94f6735b","family_id":"qf-mod38-markov-qualification","seed":202611074016,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,21],[2,21],[18,21]],\"values\":[0,16,18]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/21, 2/21, 18/21] and aligned total values X=[0, 16, 18]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/21, 2/21, 6/7], [0, 16, 18], Yes, 356/21, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/21, 2/21, 6/7], [0, 16, 18], Yes, 356/21, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0017","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2a96a0f44403fbe8af145eb10657f53d90273bfbcefa3cc456319c12b329c57","family_id":"qf-mod38-markov-qualification","seed":202611074017,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,22],[2,22],[19,22]],\"values\":[0,17,19]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/22, 2/22, 19/22] and aligned total values X=[0, 17, 19]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/22, 1/11, 19/22], [0, 17, 19], Yes, 395/22, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/22, 1/11, 19/22], [0, 17, 19], Yes, 395/22, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0018","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e12f59349433a7429f4a1cdf0be28d87ccb5db8b3b43932444c9240bdeb07de","family_id":"qf-mod38-markov-qualification","seed":202611074018,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,23],[2,23],[20,23]],\"values\":[0,18,20]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/23, 2/23, 20/23] and aligned total values X=[0, 18, 20]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/23, 2/23, 20/23], [0, 18, 20], Yes, 436/23, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/23, 2/23, 20/23], [0, 18, 20], Yes, 436/23, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0019","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cdfd556958043bdc9e0ba5d8221b19cf4a9e5f1e09b8a4d36c0d8e3c54069dd","family_id":"qf-mod38-markov-qualification","seed":202611074019,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,24],[2,24],[21,24]],\"values\":[0,19,21]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/24, 2/24, 21/24] and aligned total values X=[0, 19, 21]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/24, 1/12, 7/8], [0, 19, 21], Yes, 479/24, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/24, 1/12, 7/8], [0, 19, 21], Yes, 479/24, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0020","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4336410e1964961be0c1b6bad0a405b35669f4f6c81dfefcc6e296accbadd691","family_id":"qf-mod38-markov-qualification","seed":202611074020,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,25],[2,25],[22,25]],\"values\":[0,20,22]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/25, 2/25, 22/25] and aligned total values X=[0, 20, 22]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/25, 2/25, 22/25], [0, 20, 22], Yes, 524/25, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/25, 2/25, 22/25], [0, 20, 22], Yes, 524/25, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0021","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"69c4aa1633790d6b93bc4259a84e6aea93ceb9865bf45f101c7bad41c6bee178","family_id":"qf-mod38-markov-qualification","seed":202611074021,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,26],[2,26],[23,26]],\"values\":[0,21,23]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/26, 2/26, 23/26] and aligned total values X=[0, 21, 23]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/26, 1/13, 23/26], [0, 21, 23], Yes, 571/26, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/26, 1/13, 23/26], [0, 21, 23], Yes, 571/26, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0022","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"28f9c6ecfccdf670b10ba921a0cdb80e31ea9da49ac0f9f2e7e05c608b6b532e","family_id":"qf-mod38-markov-qualification","seed":202611074022,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,27],[2,27],[24,27]],\"values\":[0,22,24]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/27, 2/27, 24/27] and aligned total values X=[0, 22, 24]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/27, 2/27, 8/9], [0, 22, 24], Yes, 620/27, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/27, 2/27, 8/9], [0, 22, 24], Yes, 620/27, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0023","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0cad64281d36c5c47f7b007bcb130e93c252826a55c8a4e61e6cf8d9c5c3b4f5","family_id":"qf-mod38-markov-qualification","seed":202611074023,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,28],[2,28],[25,28]],\"values\":[0,23,25]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/28, 2/28, 25/28] and aligned total values X=[0, 23, 25]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/28, 1/14, 25/28], [0, 23, 25], Yes, 671/28, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/28, 1/14, 25/28], [0, 23, 25], Yes, 671/28, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0024","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c1e37d586b00559930b5ab70e5fe460735269a2cdb4531994a8785ad2f511401","family_id":"qf-mod38-markov-qualification","seed":202611074024,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,29],[2,29],[26,29]],\"values\":[0,24,26]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/29, 2/29, 26/29] and aligned total values X=[0, 24, 26]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/29, 2/29, 26/29], [0, 24, 26], Yes, 724/29, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/29, 2/29, 26/29], [0, 24, 26], Yes, 724/29, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0025","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d85c998ac9ca8d046e0f68f7a92c1a2b60ec7c276104cda383f083037863fd72","family_id":"qf-mod38-markov-qualification","seed":202611074025,"semantic_template":"qf-mod38-markov-qualification/{\"masses\":[[1,30],[2,30],[27,30]],\"values\":[0,25,27]}"},"fields":{"title":"Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity","stem":"Use complete normalized finite law with exact masses=[1/30, 2/30, 27/30] and aligned total values X=[0, 25, 27]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Give (masses, values, all_values_nonnegative, expectation, source_Markov_qualification, certificate).","answer":{"kind":"expression","value":"([1/30, 1/15, 9/10], [0, 25, 27], Yes, 779/30, Yes, finite_expectation_and_nonnegative_almost_sure)","explanation":"Return normalized masses, values, the every-entry nonnegativity flag and exact expectation. The source qualification repeats that pointwise flag, a sufficient condition for almost-sure nonnegativity. Null-mass-only negative ambiguities are rejected by the English adapter for source review, not converted into a false theorem claim. The exact result, retaining the requested order, is ([1/30, 1/15, 9/10], [0, 25, 27], Yes, 779/30, Yes, finite_expectation_and_nonnegative_almost_sure)."},"skills":["Keep the Source Pointwise Check Distinct from Almost-Sure Nonnegativity"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0026","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c13bca17086b05b16aba0c8cf3496b9d67e976e6022dfa0323453729437a484","family_id":"qf-mod38-markov-tail-probability","seed":202611074026,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,6],[2,6],[3,6]],\"threshold\":2,\"values\":[0,1,3]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/6, 2/6, 3/6] and aligned total values X=[0, 1, 3]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=2. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(2, 1/2, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (2, 1/2, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0027","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"63c1699d48c5423147d9745ac6350edb07bdf90448805c447bf96425e940c2e3","family_id":"qf-mod38-markov-tail-probability","seed":202611074027,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,7],[2,7],[4,7]],\"threshold\":3,\"values\":[0,2,4]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/7, 2/7, 4/7] and aligned total values X=[0, 2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=3. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(3, 4/7, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (3, 4/7, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0028","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a870539a4a5209e5459410ae7700d2856d1c5dc5a1ba35af3c662b41f9b24492","family_id":"qf-mod38-markov-tail-probability","seed":202611074028,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,8],[2,8],[5,8]],\"threshold\":4,\"values\":[0,3,5]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/8, 2/8, 5/8] and aligned total values X=[0, 3, 5]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=4. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(4, 5/8, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (4, 5/8, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0029","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96191091c00d3b0e762484c2d89aa1a6c426244c9b502bcacdd08b74b37ba8f1","family_id":"qf-mod38-markov-tail-probability","seed":202611074029,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,9],[2,9],[6,9]],\"threshold\":5,\"values\":[0,4,6]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/9, 2/9, 6/9] and aligned total values X=[0, 4, 6]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=5. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(5, 2/3, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (5, 2/3, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0030","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88e24e0dfa7a34e4fc6cb0f173269e8c3bc134a626ccc1605a5e8aed94d4af70","family_id":"qf-mod38-markov-tail-probability","seed":202611074030,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,10],[2,10],[7,10]],\"threshold\":6,\"values\":[0,5,7]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/10, 2/10, 7/10] and aligned total values X=[0, 5, 7]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=6. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(6, 7/10, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (6, 7/10, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0031","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cad832367741899f90dc32e19eeecf87da7e3991a74f2f03813b382cbad3ece","family_id":"qf-mod38-markov-tail-probability","seed":202611074031,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,11],[2,11],[8,11]],\"threshold\":7,\"values\":[0,6,8]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/11, 2/11, 8/11] and aligned total values X=[0, 6, 8]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=7. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(7, 8/11, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (7, 8/11, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0032","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2dfbb999ed781aabe7d18894fb4d3ec33f5bb6158d9aefd43433cc09d6e27c91","family_id":"qf-mod38-markov-tail-probability","seed":202611074032,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,12],[2,12],[9,12]],\"threshold\":8,\"values\":[0,7,9]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/12, 2/12, 9/12] and aligned total values X=[0, 7, 9]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=8. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(8, 3/4, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (8, 3/4, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0033","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1a86445978534414fc9d85b4ea6caa0de06a8412279409262cd4124bad022158","family_id":"qf-mod38-markov-tail-probability","seed":202611074033,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,13],[2,13],[10,13]],\"threshold\":9,\"values\":[0,8,10]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/13, 2/13, 10/13] and aligned total values X=[0, 8, 10]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=9. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(9, 10/13, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (9, 10/13, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0034","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5335488d244a8b80fbdbaa8c5dc357f974de60d4675dd0024782052df1edc8e3","family_id":"qf-mod38-markov-tail-probability","seed":202611074034,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,14],[2,14],[11,14]],\"threshold\":10,\"values\":[0,9,11]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/14, 2/14, 11/14] and aligned total values X=[0, 9, 11]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=10. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(10, 11/14, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (10, 11/14, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0035","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5011a01dd51e622538862b30fa693d834480ed18d96d86349ac7fd98c3e56b62","family_id":"qf-mod38-markov-tail-probability","seed":202611074035,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,15],[2,15],[12,15]],\"threshold\":11,\"values\":[0,10,12]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/15, 2/15, 12/15] and aligned total values X=[0, 10, 12]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=11. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(11, 4/5, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (11, 4/5, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0036","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9b53bbc1cacf36cc1798dd32e569cc4ebb3ed468dcce3b80bd7327cd84fa462f","family_id":"qf-mod38-markov-tail-probability","seed":202611074036,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,16],[2,16],[13,16]],\"threshold\":12,\"values\":[0,11,13]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/16, 2/16, 13/16] and aligned total values X=[0, 11, 13]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=12. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(12, 13/16, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (12, 13/16, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0037","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d8409eebe975d4a11862311f94daa2c32276c4b41056363fc342e40dc02afafc","family_id":"qf-mod38-markov-tail-probability","seed":202611074037,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,17],[2,17],[14,17]],\"threshold\":13,\"values\":[0,12,14]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/17, 2/17, 14/17] and aligned total values X=[0, 12, 14]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=13. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(13, 14/17, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (13, 14/17, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0038","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"73a61213aeff479f58a12f4b71cb0fd272bddfd25919275c9ac04f67b8831cb5","family_id":"qf-mod38-markov-tail-probability","seed":202611074038,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,18],[2,18],[15,18]],\"threshold\":14,\"values\":[0,13,15]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/18, 2/18, 15/18] and aligned total values X=[0, 13, 15]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=14. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(14, 5/6, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (14, 5/6, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0039","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c0b0183e5563c971208e4d191ec352575bcd7a2925db393379c92dcced24f43","family_id":"qf-mod38-markov-tail-probability","seed":202611074039,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,19],[2,19],[16,19]],\"threshold\":15,\"values\":[0,14,16]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/19, 2/19, 16/19] and aligned total values X=[0, 14, 16]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=15. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(15, 16/19, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (15, 16/19, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0040","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"99f7e1dec3a598c901ae2e734d04fb2a3dec33c0303de0513240859974e9fce2","family_id":"qf-mod38-markov-tail-probability","seed":202611074040,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,20],[2,20],[17,20]],\"threshold\":16,\"values\":[0,15,17]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/20, 2/20, 17/20] and aligned total values X=[0, 15, 17]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=16. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(16, 17/20, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (16, 17/20, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0041","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8d769f1b7690bfad6bced35538b3f6874dc1c955a4c0737311d002ef0bb0000","family_id":"qf-mod38-markov-tail-probability","seed":202611074041,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,21],[2,21],[18,21]],\"threshold\":17,\"values\":[0,16,18]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/21, 2/21, 18/21] and aligned total values X=[0, 16, 18]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=17. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(17, 6/7, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (17, 6/7, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0042","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f9e43f094aae656156a1ba9104ba7ce63a89248c96fce2c684e435e1ca1a902a","family_id":"qf-mod38-markov-tail-probability","seed":202611074042,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,22],[2,22],[19,22]],\"threshold\":18,\"values\":[0,17,19]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/22, 2/22, 19/22] and aligned total values X=[0, 17, 19]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=18. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(18, 19/22, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (18, 19/22, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0043","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad8d9534a69058508f86eba5216b12a3b87bca1d9b919710b0bca8bb0eb47c54","family_id":"qf-mod38-markov-tail-probability","seed":202611074043,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,23],[2,23],[20,23]],\"threshold\":19,\"values\":[0,18,20]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/23, 2/23, 20/23] and aligned total values X=[0, 18, 20]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=19. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(19, 20/23, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (19, 20/23, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0044","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dda13b9b782f7eef3a0f3b45766aa4b502d85f9d564aed2ce928c4f367663149","family_id":"qf-mod38-markov-tail-probability","seed":202611074044,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,24],[2,24],[21,24]],\"threshold\":20,\"values\":[0,19,21]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/24, 2/24, 21/24] and aligned total values X=[0, 19, 21]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=20. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(20, 7/8, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (20, 7/8, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0045","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9f42ebdbf92cefb1adf1032042a8ff940fd1ffb54aa3af3e170b7c3f9a37f205","family_id":"qf-mod38-markov-tail-probability","seed":202611074045,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,25],[2,25],[22,25]],\"threshold\":21,\"values\":[0,20,22]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/25, 2/25, 22/25] and aligned total values X=[0, 20, 22]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=21. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(21, 22/25, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (21, 22/25, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0046","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e296cbd08c53d33bc1b6a6636306a0870f207ac23f89b5bdfda0b30bd9b4ce21","family_id":"qf-mod38-markov-tail-probability","seed":202611074046,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,26],[2,26],[23,26]],\"threshold\":22,\"values\":[0,21,23]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/26, 2/26, 23/26] and aligned total values X=[0, 21, 23]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=22. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(22, 23/26, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (22, 23/26, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0047","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6db9d8d43aa279c486ade2c28a2079f7acc8d5fc84688d4217a51f33dd099901","family_id":"qf-mod38-markov-tail-probability","seed":202611074047,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,27],[2,27],[24,27]],\"threshold\":23,\"values\":[0,22,24]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/27, 2/27, 24/27] and aligned total values X=[0, 22, 24]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=23. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(23, 8/9, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (23, 8/9, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0048","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0c0a5b35c7a90a4f0e194303db211467ef5d3081f2b5554cd656481073197f58","family_id":"qf-mod38-markov-tail-probability","seed":202611074048,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,28],[2,28],[25,28]],\"threshold\":24,\"values\":[0,23,25]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/28, 2/28, 25/28] and aligned total values X=[0, 23, 25]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=24. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(24, 25/28, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (24, 25/28, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0049","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9362047066e0079c977b37dd6875dc84e68e8b341c2cac90d88cde0c92c0639f","family_id":"qf-mod38-markov-tail-probability","seed":202611074049,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,29],[2,29],[26,29]],\"threshold\":25,\"values\":[0,24,26]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/29, 2/29, 26/29] and aligned total values X=[0, 24, 26]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=25. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(25, 26/29, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (25, 26/29, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0050","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"845b1de07c8edd75126b36977183a722f1089b75395a1917007b3a8c50d74963","family_id":"qf-mod38-markov-tail-probability","seed":202611074050,"semantic_template":"qf-mod38-markov-tail-probability/{\"masses\":[[1,30],[2,30],[27,30]],\"threshold\":26,\"values\":[0,25,27]}"},"fields":{"title":"The Exact Tail Is a Sum over the Full Threshold Event","stem":"Use complete normalized finite law with exact masses=[1/30, 2/30, 27/30] and aligned total values X=[0, 25, 27]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=26. Give (threshold, tail_probability, certificate).","answer":{"kind":"expression","value":"(26, 9/10, exact_complete_finite_tail)","explanation":"Add exactly the masses at outcomes with X≥threshold, including equality. This event probability can be computed for a signed variable as well; calculating it alone does not assert Markov applicability. The exact result, retaining the requested order, is (26, 9/10, exact_complete_finite_tail)."},"skills":["The Exact Tail Is a Sum over the Full Threshold Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0051","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c4554949b0b8513ceb74653c84d908b5b0266ea8bb7f9685db65a25c5c80c5b7","family_id":"qf-mod38-markov-bound-comparison","seed":202611074051,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,6],[2,6],[3,6]],\"threshold\":2,\"values\":[0,1,3]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/6, 2/6, 3/6] and aligned total values X=[0, 1, 3]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=2>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(1/2, 11/6, 11/12, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (1/2, 11/6, 11/12, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0052","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f0057a39630c0576c8e2b260152d215e16ae92d5cfd18e4a1d579b2bb9412945","family_id":"qf-mod38-markov-bound-comparison","seed":202611074052,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,7],[2,7],[4,7]],\"threshold\":3,\"values\":[0,2,4]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/7, 2/7, 4/7] and aligned total values X=[0, 2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=3>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(4/7, 20/7, 20/21, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (4/7, 20/7, 20/21, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0053","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa849c793d26d66215cb703998ac221b56934d175b5238723d13b293dad58ca9","family_id":"qf-mod38-markov-bound-comparison","seed":202611074053,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,8],[2,8],[5,8]],\"threshold\":4,\"values\":[0,3,5]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/8, 2/8, 5/8] and aligned total values X=[0, 3, 5]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=4>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(5/8, 31/8, 31/32, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (5/8, 31/8, 31/32, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0054","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81efc48756ecc5497199a05252a69a2a4c34dfee20e4f35fe88c39ffa750902f","family_id":"qf-mod38-markov-bound-comparison","seed":202611074054,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,9],[2,9],[6,9]],\"threshold\":5,\"values\":[0,4,6]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/9, 2/9, 6/9] and aligned total values X=[0, 4, 6]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=5>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(2/3, 44/9, 44/45, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (2/3, 44/9, 44/45, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0055","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14dd5d526560e969bf51ae6c605794f28b93e7eba0b53e3d5583b6f4f009d6e0","family_id":"qf-mod38-markov-bound-comparison","seed":202611074055,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,10],[2,10],[7,10]],\"threshold\":6,\"values\":[0,5,7]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/10, 2/10, 7/10] and aligned total values X=[0, 5, 7]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=6>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(7/10, 59/10, 59/60, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (7/10, 59/10, 59/60, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0056","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"500cddc7da94473d69ef434289cea424101c8bf9ca1ba1551de0792ce3cf84b3","family_id":"qf-mod38-markov-bound-comparison","seed":202611074056,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,11],[2,11],[8,11]],\"threshold\":7,\"values\":[0,6,8]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/11, 2/11, 8/11] and aligned total values X=[0, 6, 8]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=7>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(8/11, 76/11, 76/77, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (8/11, 76/11, 76/77, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0057","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"549aea49617133b358c9de117e0e2f8cf93c801d55f909d01874a31e53f205b3","family_id":"qf-mod38-markov-bound-comparison","seed":202611074057,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,12],[2,12],[9,12]],\"threshold\":8,\"values\":[0,7,9]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/12, 2/12, 9/12] and aligned total values X=[0, 7, 9]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=8>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(3/4, 95/12, 95/96, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (3/4, 95/12, 95/96, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0058","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca59099230092e7d56db8e7edddecce153f20efade0754fa302a6b7992a5ebe2","family_id":"qf-mod38-markov-bound-comparison","seed":202611074058,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,13],[2,13],[10,13]],\"threshold\":9,\"values\":[0,8,10]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/13, 2/13, 10/13] and aligned total values X=[0, 8, 10]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=9>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(10/13, 116/13, 116/117, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (10/13, 116/13, 116/117, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0059","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe289df78e73a16aa79399d36f7086c7cf6f78a09351a5b17a5913f9d227875a","family_id":"qf-mod38-markov-bound-comparison","seed":202611074059,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,14],[2,14],[11,14]],\"threshold\":10,\"values\":[0,9,11]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/14, 2/14, 11/14] and aligned total values X=[0, 9, 11]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=10>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(11/14, 139/14, 139/140, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (11/14, 139/14, 139/140, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0060","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"36c3029c99e251c408b24f329694b47d5121c80ca11d54ad0d651c48c53fe34a","family_id":"qf-mod38-markov-bound-comparison","seed":202611074060,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,15],[2,15],[12,15]],\"threshold\":11,\"values\":[0,10,12]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/15, 2/15, 12/15] and aligned total values X=[0, 10, 12]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=11>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(4/5, 164/15, 164/165, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (4/5, 164/15, 164/165, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0061","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55bb77ff65c88712b6a77aedfac28dd20f0e5e9b063e6cdb3dc1b6353363815c","family_id":"qf-mod38-markov-bound-comparison","seed":202611074061,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,16],[2,16],[13,16]],\"threshold\":12,\"values\":[0,11,13]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/16, 2/16, 13/16] and aligned total values X=[0, 11, 13]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=12>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(13/16, 191/16, 191/192, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (13/16, 191/16, 191/192, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0062","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"423b512a7af9d2ed566748b25e655a7f9e1b48da87eb01f4cccb96710d0c3f9b","family_id":"qf-mod38-markov-bound-comparison","seed":202611074062,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,17],[2,17],[14,17]],\"threshold\":13,\"values\":[0,12,14]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/17, 2/17, 14/17] and aligned total values X=[0, 12, 14]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=13>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(14/17, 220/17, 220/221, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (14/17, 220/17, 220/221, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0063","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"967935955d1e671065e9b73c0f848e905edbfc489631cf3dcba234c64850d474","family_id":"qf-mod38-markov-bound-comparison","seed":202611074063,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,18],[2,18],[15,18]],\"threshold\":14,\"values\":[0,13,15]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/18, 2/18, 15/18] and aligned total values X=[0, 13, 15]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=14>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(5/6, 251/18, 251/252, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (5/6, 251/18, 251/252, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0064","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bea65a159225b93cfe201efd84978b75d70245488cdb37a39626b9315afecc1d","family_id":"qf-mod38-markov-bound-comparison","seed":202611074064,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,19],[2,19],[16,19]],\"threshold\":15,\"values\":[0,14,16]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/19, 2/19, 16/19] and aligned total values X=[0, 14, 16]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=15>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(16/19, 284/19, 284/285, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (16/19, 284/19, 284/285, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0065","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9a177ca0acd4ae89faa971068f2554e633cb821a55982f5802cc22ddf9f4e2d","family_id":"qf-mod38-markov-bound-comparison","seed":202611074065,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,20],[2,20],[17,20]],\"threshold\":16,\"values\":[0,15,17]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/20, 2/20, 17/20] and aligned total values X=[0, 15, 17]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=16>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(17/20, 319/20, 319/320, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (17/20, 319/20, 319/320, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0066","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2af5139b523cf027fd0ce8dcb29ef325b4e437cbd78b497e2e498032961f9858","family_id":"qf-mod38-markov-bound-comparison","seed":202611074066,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,21],[2,21],[18,21]],\"threshold\":17,\"values\":[0,16,18]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/21, 2/21, 18/21] and aligned total values X=[0, 16, 18]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=17>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(6/7, 356/21, 356/357, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (6/7, 356/21, 356/357, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0067","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"777dfacbe29c48dc3564c5653c6032cda8e8b90eb2ec95640e575264f97e4400","family_id":"qf-mod38-markov-bound-comparison","seed":202611074067,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,22],[2,22],[19,22]],\"threshold\":18,\"values\":[0,17,19]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/22, 2/22, 19/22] and aligned total values X=[0, 17, 19]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=18>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(19/22, 395/22, 395/396, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (19/22, 395/22, 395/396, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0068","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4d8240ce6e1c203d0c0d3b01a3dd0baa580dc96b394f95f05e12dab459a2613","family_id":"qf-mod38-markov-bound-comparison","seed":202611074068,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,23],[2,23],[20,23]],\"threshold\":19,\"values\":[0,18,20]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/23, 2/23, 20/23] and aligned total values X=[0, 18, 20]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=19>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(20/23, 436/23, 436/437, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (20/23, 436/23, 436/437, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0069","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90ebccbf5159c0f05cc0c18486b9d9b64a65d022dc08f6a8aa8a7c9083d02adc","family_id":"qf-mod38-markov-bound-comparison","seed":202611074069,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,24],[2,24],[21,24]],\"threshold\":20,\"values\":[0,19,21]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/24, 2/24, 21/24] and aligned total values X=[0, 19, 21]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=20>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(7/8, 479/24, 479/480, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (7/8, 479/24, 479/480, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0070","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7bc079b864a371985a860af52fab8f28b5776efc6531511af6f15970040b31e","family_id":"qf-mod38-markov-bound-comparison","seed":202611074070,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,25],[2,25],[22,25]],\"threshold\":21,\"values\":[0,20,22]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/25, 2/25, 22/25] and aligned total values X=[0, 20, 22]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=21>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(22/25, 524/25, 524/525, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (22/25, 524/25, 524/525, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0071","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b28e6d76b3fdc0b775a330fa43998debc02a022ecb9db4a62e5c14506b803700","family_id":"qf-mod38-markov-bound-comparison","seed":202611074071,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,26],[2,26],[23,26]],\"threshold\":22,\"values\":[0,21,23]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/26, 2/26, 23/26] and aligned total values X=[0, 21, 23]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=22>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(23/26, 571/26, 571/572, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (23/26, 571/26, 571/572, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0072","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"07112af85fde5794b171e09db756347869d17021595001fffead9c0cd0d2d8e8","family_id":"qf-mod38-markov-bound-comparison","seed":202611074072,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,27],[2,27],[24,27]],\"threshold\":23,\"values\":[0,22,24]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/27, 2/27, 24/27] and aligned total values X=[0, 22, 24]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=23>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(8/9, 620/27, 620/621, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (8/9, 620/27, 620/621, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0073","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb459bb778210b77392ffdd105c7f88f32cfc310f78fadcbc09986f97a70d70c","family_id":"qf-mod38-markov-bound-comparison","seed":202611074073,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,28],[2,28],[25,28]],\"threshold\":24,\"values\":[0,23,25]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/28, 2/28, 25/28] and aligned total values X=[0, 23, 25]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=24>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(25/28, 671/28, 671/672, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (25/28, 671/28, 671/672, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0074","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f3496f160518ebc3875cdbee497807ac91bfbbe2b8b76a6447ad081faf5b5c6","family_id":"qf-mod38-markov-bound-comparison","seed":202611074074,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,29],[2,29],[26,29]],\"threshold\":25,\"values\":[0,24,26]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/29, 2/29, 26/29] and aligned total values X=[0, 24, 26]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=25>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(26/29, 724/29, 724/725, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (26/29, 724/29, 724/725, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0075","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"30d78f4ce447db968c8ba3a09bf67c5e976b757b3a2a4d815e1d9d3ed89053b1","family_id":"qf-mod38-markov-bound-comparison","seed":202611074075,"semantic_template":"qf-mod38-markov-bound-comparison/{\"masses\":[[1,30],[2,30],[27,30]],\"threshold\":26,\"values\":[0,25,27]}"},"fields":{"title":"A Markov Upper Bound Is Not the Tail Probability","stem":"Use complete normalized finite law with exact masses=[1/30, 2/30, 27/30] and aligned total values X=[0, 25, 27]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. For this theorem exercise all table values are nonnegative and threshold=26>0. Retain the raw upper bound without clipping it to1. Give (tail_probability, expectation, raw_upper_bound, bound_holds, certificate).","answer":{"kind":"expression","value":"(9/10, 779/30, 779/780, Yes, upper_bound_not_tail_equality)","explanation":"Compute the actual tail, E[X] and E[X]/threshold separately, then check tail≤bound. The upper bound can exceed1 or be strictly larger than the tail; neither makes it an invalid probability value because it is not the probability itself. The exact result, retaining the requested order, is (9/10, 779/30, 779/780, Yes, upper_bound_not_tail_equality)."},"skills":["A Markov Upper Bound Is Not the Tail Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0076","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ca1665513bdb3ba44049203f1d03339e09d073f0e2f7945529ae2a4bf7a890d3","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074076,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":2,\"values\":[-1,3]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-1, 3]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=2, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-1, 3], No, 1/2, 1/2, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-1, 3], No, 1/2, 1/2, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0077","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0a37e3262be69c5c3bcf7a725e3d53d17fc45b48e246ecae4c919521e2c80a8c","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074077,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":3,\"values\":[-2,4]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-2, 4]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=3, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-2, 4], No, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-2, 4], No, 1/2, 1/3, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0078","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"13b5a40a59ab0bc11c4904481f92957022a7fdb49cd1ec7ac27059451b5fe050","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074078,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":4,\"values\":[-3,5]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-3, 5]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=4, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-3, 5], No, 1/2, 1/4, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-3, 5], No, 1/2, 1/4, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0079","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a7acfe316a392fa8e0ea4780407d54c773515913727a91c1ac12259c27b8b52e","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074079,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":5,\"values\":[-4,6]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-4, 6]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=5, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-4, 6], No, 1/2, 1/5, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-4, 6], No, 1/2, 1/5, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0080","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"206fbfc64a3bbe6207536461b1ab5840c6acee521367e25693aac12992015026","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074080,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":6,\"values\":[-5,7]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-5, 7]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=6, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-5, 7], No, 1/2, 1/6, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-5, 7], No, 1/2, 1/6, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0081","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"14a244a3c1e2b0b45dbfab7620ddd8879495c6cb6220f037d5b5ab09ec2f046e","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074081,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":7,\"values\":[-6,8]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-6, 8]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=7, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-6, 8], No, 1/2, 1/7, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-6, 8], No, 1/2, 1/7, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0082","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3adcc13a6f22a3afb3eca1c149780794d1b1313cea0706a9598a4bbbabc61f72","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074082,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":8,\"values\":[-7,9]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-7, 9]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=8, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-7, 9], No, 1/2, 1/8, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-7, 9], No, 1/2, 1/8, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0083","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc7405307d34da247ef40f54511dc0d4d4e5b8c8447092c04fe9d6f483091e27","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074083,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":9,\"values\":[-8,10]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-8, 10]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=9, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-8, 10], No, 1/2, 1/9, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-8, 10], No, 1/2, 1/9, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0084","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8f90f7e0a04d74645928051c2efb854853ebdb9a2bf20726dfb6e37d22b376b3","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074084,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":10,\"values\":[-9,11]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-9, 11]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=10, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-9, 11], No, 1/2, 1/10, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-9, 11], No, 1/2, 1/10, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0085","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4144193fd681e7f017c6b3528846b76a5bc0086b1a9b643e4bd34946635b8225","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074085,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":11,\"values\":[-10,12]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-10, 12]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=11, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-10, 12], No, 1/2, 1/11, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-10, 12], No, 1/2, 1/11, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0086","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f735eda19ab0f325a495bb75f1d8d0954299d5a20163b311dbe0dfdf58a87bd","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074086,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":12,\"values\":[-11,13]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-11, 13]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=12, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-11, 13], No, 1/2, 1/12, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-11, 13], No, 1/2, 1/12, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0087","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61a8721f9d687bae3c68bd65f43715858bcfc2e533c7df941a5f8501b97a802d","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074087,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":13,\"values\":[-12,14]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-12, 14]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=13, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-12, 14], No, 1/2, 1/13, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-12, 14], No, 1/2, 1/13, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0088","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b73890b598f925c8b1d8ef0b96b9ad1048e0d4290dae13cf3f8decd4996ad0c","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074088,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":14,\"values\":[-13,15]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-13, 15]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=14, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-13, 15], No, 1/2, 1/14, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-13, 15], No, 1/2, 1/14, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0089","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"726e4ba3094156c059f80b8c1405db6029030e79b30247c403792b0551397d77","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074089,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":15,\"values\":[-14,16]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-14, 16]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=15, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-14, 16], No, 1/2, 1/15, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-14, 16], No, 1/2, 1/15, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0090","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0d344e27eb0114c6568606ebc0c923b6d56faacca0ed739bd9d252166fb8e5b0","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074090,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":16,\"values\":[-15,17]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-15, 17]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=16, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-15, 17], No, 1/2, 1/16, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-15, 17], No, 1/2, 1/16, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0091","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c983dd79b598b3f4b77ac386c5ab8b5a54b067a1b5512fd646b2b8b742cfba5","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074091,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":17,\"values\":[-16,18]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-16, 18]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=17, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-16, 18], No, 1/2, 1/17, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-16, 18], No, 1/2, 1/17, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0092","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"57139fb9313eab1590be9629292205dfbaab007ec715a61d5eb7ef25ac1eac8a","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074092,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":18,\"values\":[-17,19]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-17, 19]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=18, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-17, 19], No, 1/2, 1/18, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-17, 19], No, 1/2, 1/18, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0093","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de267258e51dcda065c54cd6a955ea0f9e103a58b5aff3c25ed0bef0a0c48808","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074093,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":19,\"values\":[-18,20]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-18, 20]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=19, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-18, 20], No, 1/2, 1/19, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-18, 20], No, 1/2, 1/19, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0094","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c5e9913624d3968b546ff02deeb7a03abd748e7691442b5daab6a2ce09889e4","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074094,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":20,\"values\":[-19,21]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-19, 21]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=20, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-19, 21], No, 1/2, 1/20, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-19, 21], No, 1/2, 1/20, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0095","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"922939b36698996fb582eb5ff3b95aad74cf463148691a9911674754b835d15b","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074095,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":21,\"values\":[-20,22]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-20, 22]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=21, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-20, 22], No, 1/2, 1/21, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-20, 22], No, 1/2, 1/21, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0096","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"376da8912815ad203d37f073055bab605a6b075b8ab4b022c87f46b0a0644fd7","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074096,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":22,\"values\":[-21,23]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-21, 23]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=22, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-21, 23], No, 1/2, 1/22, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-21, 23], No, 1/2, 1/22, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0097","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0aea7fd00b02fd070b2f6167a2fddc87e3c7416962e00a149003b3c910cf8955","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074097,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":23,\"values\":[-22,24]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-22, 24]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=23, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-22, 24], No, 1/2, 1/23, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-22, 24], No, 1/2, 1/23, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0098","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"039d1b268f2283fc2974571276005fd71805fb38c8b1979c86bb95cf4173304e","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074098,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":24,\"values\":[-23,25]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-23, 25]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=24, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-23, 25], No, 1/2, 1/24, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-23, 25], No, 1/2, 1/24, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0099","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11fa989a4cea223a15d9d5e0e0500578bc23b42094d112210ebdb3fd3adc0411","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074099,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":25,\"values\":[-24,26]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-24, 26]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=25, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-24, 26], No, 1/2, 1/25, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-24, 26], No, 1/2, 1/25, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261014-advance-20261107-r04-markov-inequality-0100","batch_id":"batch-20261014-advance-20261107-r04-markov-inequality-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d59c9e3b7405369e01a8b3293ef0e82f3e16cadaf2dd29bdd1fb87bd2a2afae8","family_id":"qf-mod38-markov-negative-variable-boundary","seed":202611074100,"semantic_template":"qf-mod38-markov-negative-variable-boundary/{\"masses\":[[1,2],[1,2]],\"threshold\":26,\"values\":[-25,27]}"},"fields":{"title":"A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail","stem":"Use complete normalized finite law with exact masses=[1/2, 1/2] and aligned total values X=[-25, 27]. Each mass is nonnegative and the masses sum to1. All moments here are finite. The prototype checks every value for its pointwise nonnegativity flag; a null-mass negative value is not by itself failure of the theorem's almost-sure premise. Positive threshold=26, with a negative value at positive mass in this boundary case. Give (values, pointwise_Markov_applicable, tail_probability, raw_expectation_over_threshold, boundary).","answer":{"kind":"expression","value":"([-25, 27], No, 1/2, 1/26, Markov_not_applicable_to_signed_variable_without_transformation)","explanation":"Report the value table, No for source qualification, the exact tail and the raw quotient E[X]/threshold. Without the nonnegative premise Markov provides no general guarantee; the numeric inequality may fail or coincidentally hold. Do not turn inapplicability into a proved reverse inequality. The exact result, retaining the requested order, is ([-25, 27], No, 1/2, 1/26, Markov_not_applicable_to_signed_variable_without_transformation)."},"skills":["A Failed Nonnegativity Premise Does Not Force This Particular Inequality to Fail"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}