{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261004-advance-20261005-r06-expectation-0001","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0725021add88284ab4493917ee501656e264aa5462e3084655c2a1aefde4be6b","family_id":"qf-expectation-weighted","seed":202610056001,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-10\",\"1/2\"],[\"-8\",\"1/2\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-9/1","explanation":"The weighted numerator is -54 and the common denominator is 6, giving E[X]=-9/1. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0002","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"955ed7235d8af8a80dbb5a11526fe77728bb5ce28918d4cf5610c13cf8fccc55","family_id":"qf-expectation-weighted","seed":202610056002,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-7\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-11/2","explanation":"The weighted numerator is -44 and the common denominator is 8, giving E[X]=-11/2. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0003","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0f8ff3c4af8adf71a76457e2e24d9c381dd176c5389a8cea3f1cab8590e8ea31","family_id":"qf-expectation-weighted","seed":202610056003,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-8\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-6/1","explanation":"The weighted numerator is -60 and the common denominator is 10, giving E[X]=-6/1. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0004","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bb4b8c525edd0969b1687a0d67d4d6df6bd73eb2de31995545c67621c9b817e3","family_id":"qf-expectation-weighted","seed":202610056004,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-9\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-13/2","explanation":"The weighted numerator is -78 and the common denominator is 12, giving E[X]=-13/2. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0005","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2274e84cac02a4defc9a62eb1ae538d355f24c8ed4c58aea11f2cb26d2e19f35","family_id":"qf-expectation-weighted","seed":202610056005,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-6\",\"7/10\"],[\"0\",\"3/10\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-21/5","explanation":"The weighted numerator is -42 and the common denominator is 10, giving E[X]=-21/5. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0006","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6acdec5bfcc69d33374c37090a72b9cc9a280aa3499a094be4e9e1077d15dc25","family_id":"qf-expectation-weighted","seed":202610056006,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-11\",\"2/3\"],[\"-4\",\"1/3\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-26/3","explanation":"The weighted numerator is -104 and the common denominator is 12, giving E[X]=-26/3. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0007","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3004ebf8b43dc2c37a2863d44253d3416191617e976c8bc0ef6a4e30e18acf38","family_id":"qf-expectation-weighted","seed":202610056007,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-4\",\"9/14\"],[\"4\",\"5/14\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-8/7","explanation":"The weighted numerator is -16 and the common denominator is 14, giving E[X]=-8/7. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0008","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ef7d11b8688f520e6c82f324c0f40fa6bfc5b2d1124e916636825213adb32a3","family_id":"qf-expectation-weighted","seed":202610056008,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-13\",\"5/8\"],[\"-4\",\"3/8\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-77/8","explanation":"The weighted numerator is -154 and the common denominator is 16, giving E[X]=-77/8. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0009","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e53b1254d0c5c2ec026d0a8b011c19dcf8e5e7ba213b63bead691455b030a57e","family_id":"qf-expectation-weighted","seed":202610056009,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-2\",\"11/14\"],[\"8\",\"3/14\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"1/7","explanation":"The weighted numerator is 2 and the common denominator is 14, giving E[X]=1/7. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0010","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a61feb82c4a55b579948c05a775e5b0d715ea10828cb24c6a93cf4933c649f4","family_id":"qf-expectation-weighted","seed":202610056010,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-15\",\"3/4\"],[\"-4\",\"1/4\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-49/4","explanation":"The weighted numerator is -196 and the common denominator is 16, giving E[X]=-49/4. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0011","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a5426dad09ac68aa834f239943647f1379a2a3077e5ea495a6922817104f05c0","family_id":"qf-expectation-weighted","seed":202610056011,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"10/3","explanation":"The weighted numerator is 60 and the common denominator is 18, giving E[X]=10/3. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0012","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"068fdb069489e55741c8f79c77592a82df3a150de4cc7de3824e4f0f5d6cad00","family_id":"qf-expectation-weighted","seed":202610056012,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-17\",\"7/10\"],[\"-4\",\"3/10\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-131/10","explanation":"The weighted numerator is -262 and the common denominator is 20, giving E[X]=-131/10. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0013","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a829bdacade989b57d6e4f4721b3df6087920fbbb7b60520a166f25a84ef0da","family_id":"qf-expectation-weighted","seed":202610056013,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"2\",\"5/6\"],[\"16\",\"1/6\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"13/3","explanation":"The weighted numerator is 78 and the common denominator is 18, giving E[X]=13/3. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0014","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"057c83a613d4318425950b8ae3777327fbae09c6e82d54f582fe9999d84719e4","family_id":"qf-expectation-weighted","seed":202610056014,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-19\",\"4/5\"],[\"-4\",\"1/5\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-16/1","explanation":"The weighted numerator is -320 and the common denominator is 20, giving E[X]=-16/1. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0015","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03bfa3f1284bda518c276593117cee1e12f4cc279ae854c244ca8af5912580ee","family_id":"qf-expectation-weighted","seed":202610056015,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"4\",\"17/22\"],[\"20\",\"5/22\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"84/11","explanation":"The weighted numerator is 168 and the common denominator is 22, giving E[X]=84/11. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0016","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f3df3951e00942ae79a5f1bb24c68d5f5bb8f415af5ceb6b551cb040872619f2","family_id":"qf-expectation-weighted","seed":202610056016,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-21\",\"3/4\"],[\"-4\",\"1/4\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-67/4","explanation":"The weighted numerator is -402 and the common denominator is 24, giving E[X]=-67/4. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0017","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"216b75ff23e12ae0c5a478158b368f5e6321dd00149c223bee47f8abbbdd13b4","family_id":"qf-expectation-weighted","seed":202610056017,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"6\",\"19/22\"],[\"24\",\"3/22\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"93/11","explanation":"The weighted numerator is 186 and the common denominator is 22, giving E[X]=93/11. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0018","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d448d441c127ea47ff235f87615d25397413292e3d271cf22d8f604b71bfd9d1","family_id":"qf-expectation-weighted","seed":202610056018,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-23\",\"5/6\"],[\"-4\",\"1/6\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-119/6","explanation":"The weighted numerator is -476 and the common denominator is 24, giving E[X]=-119/6. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0019","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"727b5fa160df7cc556a75c86582a10305e9f9809932b302f51097bbd01aa08ae","family_id":"qf-expectation-weighted","seed":202610056019,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"8\",\"21/26\"],[\"28\",\"5/26\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"154/13","explanation":"The weighted numerator is 308 and the common denominator is 26, giving E[X]=154/13. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0020","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fdbec96d62fd2070b12139d0a6006b6c448f246dfd886e7282041e3771a0610","family_id":"qf-expectation-weighted","seed":202610056020,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-25\",\"11/14\"],[\"-4\",\"3/14\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-41/2","explanation":"The weighted numerator is -574 and the common denominator is 28, giving E[X]=-41/2. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0021","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7b8777cee94fa4b89094e4012bb0d753ca76d2eddda2c05692e03fb1cedab6e5","family_id":"qf-expectation-weighted","seed":202610056021,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"10\",\"23/26\"],[\"32\",\"3/26\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"163/13","explanation":"The weighted numerator is 326 and the common denominator is 26, giving E[X]=163/13. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0022","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"989fc246d9c974563167f56d2fa2395b3265a0704baf088daf99b961000a3a8c","family_id":"qf-expectation-weighted","seed":202610056022,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-27\",\"6/7\"],[\"-4\",\"1/7\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-166/7","explanation":"The weighted numerator is -664 and the common denominator is 28, giving E[X]=-166/7. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0023","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f275fe87e50620ed62ff0416c381ddbd13e563a9269bc6dabdb6e9bb84d52991","family_id":"qf-expectation-weighted","seed":202610056023,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"12\",\"5/6\"],[\"36\",\"1/6\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"16/1","explanation":"The weighted numerator is 480 and the common denominator is 30, giving E[X]=16/1. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0024","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b342f53cb129e12d15af334c137a9a6d1ba5280a50609cdab16aeb0313cfc37","family_id":"qf-expectation-weighted","seed":202610056024,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"-29\",\"13/16\"],[\"-4\",\"3/16\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"-389/16","explanation":"The weighted numerator is -778 and the common denominator is 32, giving E[X]=-389/16. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0025","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"da17431dd833d621eab1b19c19d3b4db4d0ba9a00429f55e32f0a00964377cd3","family_id":"qf-expectation-weighted","seed":202610056025,"semantic_template":"qf-expectation-weighted/{\"F\":\"powerset(S)\",\"law\":[[\"14\",\"9/10\"],[\"40\",\"1/10\"]],\"operation\":\"expectation\"}"},"fields":{"title":"Expectation in a Nonuniform Finite Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Compute E[X] exactly as a reduced positive-denominator fraction; do not replace it with the ordinary average of distinct values.","answer":{"kind":"expression","value":"83/5","explanation":"The weighted numerator is 498 and the common denominator is 30, giving E[X]=83/5. Expectation may be negative or unattained and is not a sample-mean estimate."},"skills":["Expectation in a Nonuniform Finite Model"],"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-20261004-advance-20261005-r06-expectation-0026","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9af33e90d9da12dee5874c298784ffea0cfd56a03194512837f6d525e4be9402","family_id":"qf-expectation-affine","seed":202610056026,"semantic_template":"qf-expectation-affine/{\"a\":0,\"b\":-5,\"normalized_model\":true,\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. Let Z=(0)X+(-5). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-5/1","explanation":"Form Z at each outcome, then weight and sum to obtain -5/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0027","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9a46edf84b1e5bb3ae701108ca296e0366905e9b6d1aa7000b9cd1691f3bf344","family_id":"qf-expectation-affine","seed":202610056027,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":-4,\"law\":[[\"-7\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. Let Z=(-2)X+(-4). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"7/1","explanation":"Form Z at each outcome, then weight and sum to obtain 7/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0028","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0dd6340b9cacd8725ed8310ed43d7e30c08d5fe2aed52f8869b2df14779f9abf","family_id":"qf-expectation-affine","seed":202610056028,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":-3,\"law\":[[\"-8\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. Let Z=(3)X+(-3). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-21/1","explanation":"Form Z at each outcome, then weight and sum to obtain -21/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0029","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"29c8213e49a8ad8adbe70cab5c897736015c3d65f3d50500d374d9b7354a90ca","family_id":"qf-expectation-affine","seed":202610056029,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":-2,\"law\":[[\"-9\",\"1/2\"],[\"-4\",\"1/2\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. Let Z=(-4)X+(-2). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"24/1","explanation":"Form Z at each outcome, then weight and sum to obtain 24/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0030","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"64d98c756dae63af9c211993b6e410c58f4e6b580fcdf8c739bc984b4183cc6c","family_id":"qf-expectation-affine","seed":202610056030,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":-1,\"law\":[[\"-6\",\"7/10\"],[\"0\",\"3/10\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. Let Z=(1)X+(-1). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-26/5","explanation":"Form Z at each outcome, then weight and sum to obtain -26/5, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0031","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e954eef8da4447147c2eb378807e98e26b5873365c6445d47911e47afc5755d7","family_id":"qf-expectation-affine","seed":202610056031,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":0,\"law\":[[\"-11\",\"2/3\"],[\"-4\",\"1/3\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. Let Z=(-2)X+(0). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"52/3","explanation":"Form Z at each outcome, then weight and sum to obtain 52/3, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0032","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c8370831c534b7ad2d47c357098f518bea46dbb5bce424d66fb098006d5cfb4c","family_id":"qf-expectation-affine","seed":202610056032,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":1,\"law\":[[\"-4\",\"9/14\"],[\"4\",\"5/14\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. Let Z=(3)X+(1). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-17/7","explanation":"Form Z at each outcome, then weight and sum to obtain -17/7, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0033","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"310bd7d23a8ad07f1664caade62d4b7f621298d71f4772c1346c5e48ea1f7ca7","family_id":"qf-expectation-affine","seed":202610056033,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":2,\"law\":[[\"-13\",\"5/8\"],[\"-4\",\"3/8\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. Let Z=(-4)X+(2). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"81/2","explanation":"Form Z at each outcome, then weight and sum to obtain 81/2, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0034","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"124efcffaefc4d05d431e05e8714bc1e2f0a776e6c239aff0acad12176b12e04","family_id":"qf-expectation-affine","seed":202610056034,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":3,\"law\":[[\"-2\",\"11/14\"],[\"8\",\"3/14\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. Let Z=(1)X+(3). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"22/7","explanation":"Form Z at each outcome, then weight and sum to obtain 22/7, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0035","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"52f8d91131f694d6787037a1df191499e13c47d1fbd9f12497d5d19cefe9442e","family_id":"qf-expectation-affine","seed":202610056035,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":4,\"law\":[[\"-15\",\"3/4\"],[\"-4\",\"1/4\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. Let Z=(-2)X+(4). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"57/2","explanation":"Form Z at each outcome, then weight and sum to obtain 57/2, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0036","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31a006537407ca189501ab4c5b1b13b0a18104ad6cc004c555fba80258a8fb63","family_id":"qf-expectation-affine","seed":202610056036,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":5,\"law\":[[\"0\",\"13/18\"],[\"12\",\"5/18\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. Let Z=(3)X+(5). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"15/1","explanation":"Form Z at each outcome, then weight and sum to obtain 15/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0037","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1cbba70e14b21866fcd8022fcad385633d549af2d5f5c7ef59158f6dbffe4ae8","family_id":"qf-expectation-affine","seed":202610056037,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":6,\"law\":[[\"-17\",\"7/10\"],[\"-4\",\"3/10\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. Let Z=(-4)X+(6). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"292/5","explanation":"Form Z at each outcome, then weight and sum to obtain 292/5, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0038","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"07386bf079e19ae361f8d7f0cfbc661673c9e3d9535b81098cbd65a028269930","family_id":"qf-expectation-affine","seed":202610056038,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":7,\"law\":[[\"2\",\"5/6\"],[\"16\",\"1/6\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. Let Z=(1)X+(7). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"34/3","explanation":"Form Z at each outcome, then weight and sum to obtain 34/3, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0039","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e5d8342fed51ce8e0b822e4aa392c273f4fb6786261916a030597eab0d1ba155","family_id":"qf-expectation-affine","seed":202610056039,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":8,\"law\":[[\"-19\",\"4/5\"],[\"-4\",\"1/5\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. Let Z=(-2)X+(8). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"40/1","explanation":"Form Z at each outcome, then weight and sum to obtain 40/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0040","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe3a5d03d9fb77cb91349567b8ffb95b1e45165140a6f65787de5c9fc9c6e51a","family_id":"qf-expectation-affine","seed":202610056040,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":9,\"law\":[[\"4\",\"17/22\"],[\"20\",\"5/22\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. Let Z=(3)X+(9). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"351/11","explanation":"Form Z at each outcome, then weight and sum to obtain 351/11, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0041","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9e7b983ba5ddacd015e774ddcccde53af484efd6bc91d4bcd3e9d16760d99c4a","family_id":"qf-expectation-affine","seed":202610056041,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":10,\"law\":[[\"-21\",\"3/4\"],[\"-4\",\"1/4\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. Let Z=(-4)X+(10). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"77/1","explanation":"Form Z at each outcome, then weight and sum to obtain 77/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0042","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ed20a380bb89f51180b560602e9195b2d22b23a796e18d300367a0d8c2a5c8fb","family_id":"qf-expectation-affine","seed":202610056042,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":11,\"law\":[[\"6\",\"19/22\"],[\"24\",\"3/22\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. Let Z=(1)X+(11). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"214/11","explanation":"Form Z at each outcome, then weight and sum to obtain 214/11, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0043","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"df76d4bad70d7975837bd34565b69e8499a6de8137f86076d8c319700b67dcba","family_id":"qf-expectation-affine","seed":202610056043,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":12,\"law\":[[\"-23\",\"5/6\"],[\"-4\",\"1/6\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. Let Z=(-2)X+(12). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"155/3","explanation":"Form Z at each outcome, then weight and sum to obtain 155/3, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0044","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d67a0f00486db4cfe098db8c323098a6fd1b760ba06fbf660b9b2e034e43f214","family_id":"qf-expectation-affine","seed":202610056044,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":13,\"law\":[[\"8\",\"21/26\"],[\"28\",\"5/26\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. Let Z=(3)X+(13). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"631/13","explanation":"Form Z at each outcome, then weight and sum to obtain 631/13, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0045","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8137f054548a672c726462b9ea24f3d3dac64ec566552744f8ff3a079d7f302b","family_id":"qf-expectation-affine","seed":202610056045,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":14,\"law\":[[\"-25\",\"11/14\"],[\"-4\",\"3/14\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. Let Z=(-4)X+(14). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"96/1","explanation":"Form Z at each outcome, then weight and sum to obtain 96/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0046","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"955de9b1628721aea01f714bfa1f1b28c7c5fbd401705eb562fc8e794f1076bf","family_id":"qf-expectation-affine","seed":202610056046,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":15,\"law\":[[\"10\",\"23/26\"],[\"32\",\"3/26\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. Let Z=(1)X+(15). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"358/13","explanation":"Form Z at each outcome, then weight and sum to obtain 358/13, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0047","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6973725c7a6f0c4c4b342d5c0e7b3367dc96a5547aacec1a1286f3d859651a0b","family_id":"qf-expectation-affine","seed":202610056047,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-2,\"b\":16,\"law\":[[\"-27\",\"6/7\"],[\"-4\",\"1/7\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. Let Z=(-2)X+(16). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"444/7","explanation":"Form Z at each outcome, then weight and sum to obtain 444/7, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0048","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dd005d83f89c65b54039d28c147e9f88c37acd49de214e79be6068f6157a184","family_id":"qf-expectation-affine","seed":202610056048,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":3,\"b\":17,\"law\":[[\"12\",\"5/6\"],[\"36\",\"1/6\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. Let Z=(3)X+(17). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"65/1","explanation":"Form Z at each outcome, then weight and sum to obtain 65/1, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0049","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ddcfdf0a0ce10c89edf848eb0bd4c6232850404a1299ed3400a48b65f0b3a4f9","family_id":"qf-expectation-affine","seed":202610056049,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":-4,\"b\":18,\"law\":[[\"-29\",\"13/16\"],[\"-4\",\"3/16\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. Let Z=(-4)X+(18). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"461/4","explanation":"Form Z at each outcome, then weight and sum to obtain 461/4, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0050","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8dd99c272edbb262a3e068922bca796ef143d4af7c8a95ae2a3e9436c14986e4","family_id":"qf-expectation-affine","seed":202610056050,"semantic_template":"qf-expectation-affine/{\"F\":\"powerset(S)\",\"a\":1,\"b\":19,\"law\":[[\"14\",\"9/10\"],[\"40\",\"1/10\"]],\"operation\":\"affine_expectation\"}"},"fields":{"title":"Expectation of an Affine Transformation","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. Let Z=(1)X+(19). Find E[Z] exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"178/5","explanation":"Form Z at each outcome, then weight and sum to obtain 178/5, equivalently aE[X]+b. At a=0 only the constant b remains."},"skills":["Expectation of an Affine Transformation"],"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-20261004-advance-20261005-r06-expectation-0051","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c9c2b18a60ccacbb36e45c8d7fb1911a84b53d3a89ad4de0fe30e0268f20812","family_id":"qf-expectation-same-model-sum","seed":202610056051,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-10\",\"-20\",\"1/2\"],[\"-8\",\"-16\",\"1/2\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -10), (1, -10), (2, -8), (3, 999)}. In the displayed order of S, the nonnegative probability weights are [1/(6), 2/(6), 3/(6), 0/(6)], summing exactly to one. On the same S, Y has table {(0, -20), (1, -20), (2, -16), (3, 1998)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-27/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -27/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0052","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b2c085ac997ff7a5070ea72174e78c77c04632f6b881bfb08d30738fab2a836d","family_id":"qf-expectation-same-model-sum","seed":202610056052,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-7\",\"-13\",\"1/2\"],[\"-4\",\"-7\",\"1/2\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -7), (1, -7), (2, -4), (3, 1000)}. In the displayed order of S, the nonnegative probability weights are [1/(8), 3/(8), 4/(8), 0/(8)], summing exactly to one. On the same S, Y has table {(0, -13), (1, -13), (2, -7), (3, 2001)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-31/2","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -31/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0053","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9027b96827a9ee25d648d4f6fb1a45d3e7b86bae8b4906c87d449fc726e52112","family_id":"qf-expectation-same-model-sum","seed":202610056053,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-8\",\"-14\",\"1/2\"],[\"-4\",\"-6\",\"1/2\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -8), (1, -8), (2, -4), (3, 1001)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 4/(10), 5/(10), 0/(10)], summing exactly to one. On the same S, Y has table {(0, -14), (1, -14), (2, -6), (3, 2004)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-16/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -16/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0054","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"357812b2150a358735c74fc9afa5a8768cae118370bc685e38e3182a6543f142","family_id":"qf-expectation-same-model-sum","seed":202610056054,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-9\",\"-15\",\"1/2\"],[\"-4\",\"-5\",\"1/2\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -9), (1, -9), (2, -4), (3, 1002)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 5/(12), 6/(12), 0/(12)], summing exactly to one. On the same S, Y has table {(0, -15), (1, -15), (2, -5), (3, 2007)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-33/2","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -33/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0055","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1525a64ab3b9c3e6ead0f7edaad97bdbfb90e3f4d7db94fac12507c8b4c87ae0","family_id":"qf-expectation-same-model-sum","seed":202610056055,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-6\",\"-8\",\"7/10\"],[\"0\",\"4\",\"3/10\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -6), (1, -6), (2, 0), (3, 1003)}. In the displayed order of S, the nonnegative probability weights are [1/(10), 6/(10), 3/(10), 0/(10)], summing exactly to one. On the same S, Y has table {(0, -8), (1, -8), (2, 4), (3, 2010)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-43/5","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -43/5. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0056","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b6bf60ad76acba973391953bfc3ed92906cb310df64f4dfc3b52b98bdda81930","family_id":"qf-expectation-same-model-sum","seed":202610056056,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-11\",\"-22\",\"2/3\"],[\"-4\",\"-8\",\"1/3\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -11), (1, -11), (2, -4), (3, 1004)}. In the displayed order of S, the nonnegative probability weights are [1/(12), 7/(12), 4/(12), 0/(12)], summing exactly to one. On the same S, Y has table {(0, -22), (1, -22), (2, -8), (3, 2008)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-26/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -26/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0057","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3a683a79440adf2a015edeba68b9eb3410f65f896288efcbb8a354c2d081c0b2","family_id":"qf-expectation-same-model-sum","seed":202610056057,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-4\",\"-7\",\"9/14\"],[\"4\",\"9\",\"5/14\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -4), (1, -4), (2, 4), (3, 1005)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 8/(14), 5/(14), 0/(14)], summing exactly to one. On the same S, Y has table {(0, -7), (1, -7), (2, 9), (3, 2011)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-17/7","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -17/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0058","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"554d55e20c9a13393b3146ac3aa12b9b7ab3ddb5a7940e0d06bae8c1a2b98d44","family_id":"qf-expectation-same-model-sum","seed":202610056058,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-13\",\"-24\",\"5/8\"],[\"-4\",\"-6\",\"3/8\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -13), (1, -13), (2, -4), (3, 1006)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 9/(16), 6/(16), 0/(16)], summing exactly to one. On the same S, Y has table {(0, -24), (1, -24), (2, -6), (3, 2014)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-215/8","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -215/8. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0059","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a014f46e28ed78fcf09543236ce4891e2df7ffda341486d6207589201184431a","family_id":"qf-expectation-same-model-sum","seed":202610056059,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-2\",\"-1\",\"11/14\"],[\"8\",\"19\",\"3/14\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -2), (1, -2), (2, 8), (3, 1007)}. In the displayed order of S, the nonnegative probability weights are [1/(14), 10/(14), 3/(14), 0/(14)], summing exactly to one. On the same S, Y has table {(0, -1), (1, -1), (2, 19), (3, 2017)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"24/7","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 24/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0060","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"96f0d5034e29a092571568593d1d4d9b624a095ce58c90d95a98f3ffa8a83773","family_id":"qf-expectation-same-model-sum","seed":202610056060,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-15\",\"-26\",\"3/4\"],[\"-4\",\"-4\",\"1/4\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -15), (1, -15), (2, -4), (3, 1008)}. In the displayed order of S, the nonnegative probability weights are [1/(16), 11/(16), 4/(16), 0/(16)], summing exactly to one. On the same S, Y has table {(0, -26), (1, -26), (2, -4), (3, 2020)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-131/4","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -131/4. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0061","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5ccc3cd1e544eb42801254d854ff305eaba3ec36e54f7d628ec99a29f1e1b222","family_id":"qf-expectation-same-model-sum","seed":202610056061,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"0\",\"0\",\"13/18\"],[\"12\",\"24\",\"5/18\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 0), (1, 0), (2, 12), (3, 1009)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 12/(18), 5/(18), 0/(18)], summing exactly to one. On the same S, Y has table {(0, 0), (1, 0), (2, 24), (3, 2018)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"10/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 10/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0062","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"75298b1af335251c637073b1ad238129559664d2ca75bdce08817d498d6a48b6","family_id":"qf-expectation-same-model-sum","seed":202610056062,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-17\",\"-33\",\"7/10\"],[\"-4\",\"-7\",\"3/10\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -17), (1, -17), (2, -4), (3, 1010)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 13/(20), 6/(20), 0/(20)], summing exactly to one. On the same S, Y has table {(0, -33), (1, -33), (2, -7), (3, 2021)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-383/10","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -383/10. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0063","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e822e873704d76962513438922f7d561789b448913f530834447529060f28583","family_id":"qf-expectation-same-model-sum","seed":202610056063,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"2\",\"6\",\"5/6\"],[\"16\",\"34\",\"1/6\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 2), (1, 2), (2, 16), (3, 1011)}. In the displayed order of S, the nonnegative probability weights are [1/(18), 14/(18), 3/(18), 0/(18)], summing exactly to one. On the same S, Y has table {(0, 6), (1, 6), (2, 34), (3, 2024)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"15/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 15/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0064","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"862853998c818e750a368e4d8c29e44c00e650b3c74ee40aedda010e082bee0b","family_id":"qf-expectation-same-model-sum","seed":202610056064,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-19\",\"-35\",\"4/5\"],[\"-4\",\"-5\",\"1/5\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -19), (1, -19), (2, -4), (3, 1012)}. In the displayed order of S, the nonnegative probability weights are [1/(20), 15/(20), 4/(20), 0/(20)], summing exactly to one. On the same S, Y has table {(0, -35), (1, -35), (2, -5), (3, 2027)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-45/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -45/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0065","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcd16dd8acef4d3a079eaa5fb6e06829a7bea1c0982926af414c736b146ebce1","family_id":"qf-expectation-same-model-sum","seed":202610056065,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"4\",\"12\",\"17/22\"],[\"20\",\"44\",\"5/22\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 4), (1, 4), (2, 20), (3, 1013)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 16/(22), 5/(22), 0/(22)], summing exactly to one. On the same S, Y has table {(0, 12), (1, 12), (2, 44), (3, 2030)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"296/11","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 296/11. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0066","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2ecdbc30e4e043b4a12a997d46fd12a9d74ed90b6cb77676c9b2049909ef9c46","family_id":"qf-expectation-same-model-sum","seed":202610056066,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-21\",\"-42\",\"3/4\"],[\"-4\",\"-8\",\"1/4\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -21), (1, -21), (2, -4), (3, 1014)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 17/(24), 6/(24), 0/(24)], summing exactly to one. On the same S, Y has table {(0, -42), (1, -42), (2, -8), (3, 2028)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-201/4","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -201/4. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0067","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2657d3f4f9a2a01e1cdf054864b7b05bba0506c360240537d4699a0ada6e2089","family_id":"qf-expectation-same-model-sum","seed":202610056067,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"6\",\"13\",\"19/22\"],[\"24\",\"49\",\"3/22\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 6), (1, 6), (2, 24), (3, 1015)}. In the displayed order of S, the nonnegative probability weights are [1/(22), 18/(22), 3/(22), 0/(22)], summing exactly to one. On the same S, Y has table {(0, 13), (1, 13), (2, 49), (3, 2031)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"290/11","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 290/11. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0068","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"55dfa7ecee82ebe89bcda41801d5bea94eecf6687bf2f69cf486c61c38aa6aaa","family_id":"qf-expectation-same-model-sum","seed":202610056068,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-23\",\"-44\",\"5/6\"],[\"-4\",\"-6\",\"1/6\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -23), (1, -23), (2, -4), (3, 1016)}. In the displayed order of S, the nonnegative probability weights are [1/(24), 19/(24), 4/(24), 0/(24)], summing exactly to one. On the same S, Y has table {(0, -44), (1, -44), (2, -6), (3, 2034)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-115/2","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -115/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0069","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"59d7d9829f74bf61a3ef80cc7ef6648782851af17d476d05da0b12971016c244","family_id":"qf-expectation-same-model-sum","seed":202610056069,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"8\",\"19\",\"21/26\"],[\"28\",\"59\",\"5/26\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 8), (1, 8), (2, 28), (3, 1017)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 20/(26), 5/(26), 0/(26)], summing exactly to one. On the same S, Y has table {(0, 19), (1, 19), (2, 59), (3, 2037)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"501/13","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 501/13. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0070","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"591f44498f0c9e3ae3bf6928eb2a71d53b77c0d492a55b3cbfecf2a7fd719162","family_id":"qf-expectation-same-model-sum","seed":202610056070,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-25\",\"-46\",\"11/14\"],[\"-4\",\"-4\",\"3/14\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -25), (1, -25), (2, -4), (3, 1018)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 21/(28), 6/(28), 0/(28)], summing exactly to one. On the same S, Y has table {(0, -46), (1, -46), (2, -4), (3, 2040)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-115/2","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -115/2. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0071","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3c27ecc6c6f9cd932956ddab56c970582c4bc0e1d7a089e77d65415cb7c2f32e","family_id":"qf-expectation-same-model-sum","seed":202610056071,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"10\",\"20\",\"23/26\"],[\"32\",\"64\",\"3/26\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 10), (1, 10), (2, 32), (3, 1019)}. In the displayed order of S, the nonnegative probability weights are [1/(26), 22/(26), 3/(26), 0/(26)], summing exactly to one. On the same S, Y has table {(0, 20), (1, 20), (2, 64), (3, 2038)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"489/13","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 489/13. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0072","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"11d1981d5058c10a486df41a3d82365abc90db9a202e060457d70702b6b82100","family_id":"qf-expectation-same-model-sum","seed":202610056072,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-27\",\"-53\",\"6/7\"],[\"-4\",\"-7\",\"1/7\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -27), (1, -27), (2, -4), (3, 1020)}. In the displayed order of S, the nonnegative probability weights are [1/(28), 23/(28), 4/(28), 0/(28)], summing exactly to one. On the same S, Y has table {(0, -53), (1, -53), (2, -7), (3, 2041)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-491/7","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -491/7. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0073","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0ca96e96c8dee98e84f786a7149598ac13c0acfa6faf4cbd10394eb6890471d8","family_id":"qf-expectation-same-model-sum","seed":202610056073,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"12\",\"26\",\"5/6\"],[\"36\",\"74\",\"1/6\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 12), (1, 12), (2, 36), (3, 1021)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 24/(30), 5/(30), 0/(30)], summing exactly to one. On the same S, Y has table {(0, 26), (1, 26), (2, 74), (3, 2044)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"50/1","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 50/1. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0074","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b3397ebd6f0485ce9ba0fead52164785396d368edb4e28a60ba6076e4402d68c","family_id":"qf-expectation-same-model-sum","seed":202610056074,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"-29\",\"-55\",\"13/16\"],[\"-4\",\"-5\",\"3/16\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, -29), (1, -29), (2, -4), (3, 1022)}. In the displayed order of S, the nonnegative probability weights are [1/(32), 25/(32), 6/(32), 0/(32)], summing exactly to one. On the same S, Y has table {(0, -55), (1, -55), (2, -5), (3, 2047)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"-1119/16","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving -1119/16. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0075","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ba77bc2b11dba8bf852f9f71539c0e8e505a13aa029c3f27e332175234f5c5d","family_id":"qf-expectation-same-model-sum","seed":202610056075,"semantic_template":"qf-expectation-same-model-sum/{\"F\":\"powerset(S)\",\"law\":[[\"14\",\"32\",\"9/10\"],[\"40\",\"84\",\"1/10\"]],\"operation\":\"sum_expectation\"}"},"fields":{"title":"Expectation of a Sum on the Same Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). The total function X:S→R has table {(0, 14), (1, 14), (2, 40), (3, 1023)}. In the displayed order of S, the nonnegative probability weights are [1/(30), 26/(30), 3/(30), 0/(30)], summing exactly to one. On the same S, Y has table {(0, 32), (1, 32), (2, 84), (3, 2050)}. Compute E[X+Y] exactly without introducing a separate independent model.","answer":{"kind":"expression","value":"269/5","explanation":"At each identical outcome, add X and Y first and then use the declared weights, giving 269/5. Linearity E[X+Y]=E[X]+E[Y] does not require independence and does not imply a product-expectation formula."},"skills":["Expectation of a Sum on the Same Model"],"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-20261004-advance-20261005-r06-expectation-0076","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"76c50539266fb1cd704c7deddee4c819690c679895302445d557ee1e8587434e","family_id":"qf-expectation-constant","seed":202610056076,"semantic_template":"qf-expectation-constant/{\"c\":-12,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-12 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-12/1","explanation":"The constant multiplies the weight total one, giving E[X]=-12/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0077","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"186729062ee6455f321a542a8b332a6f497a9ebd1608d1674be21e1dc6a0c20f","family_id":"qf-expectation-constant","seed":202610056077,"semantic_template":"qf-expectation-constant/{\"c\":-11,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-11 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-11/1","explanation":"The constant multiplies the weight total one, giving E[X]=-11/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0078","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fbe09843cb4551545d02f26c88f2f5967c1bb5a623c5b319769f7934ec4921b","family_id":"qf-expectation-constant","seed":202610056078,"semantic_template":"qf-expectation-constant/{\"c\":-10,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-10 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-10/1","explanation":"The constant multiplies the weight total one, giving E[X]=-10/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0079","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5c332cd74c3469766215aba76ac6ecc2f4765c756e7eb142695c65e99cc5c079","family_id":"qf-expectation-constant","seed":202610056079,"semantic_template":"qf-expectation-constant/{\"c\":-9,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-9 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-9/1","explanation":"The constant multiplies the weight total one, giving E[X]=-9/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0080","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"867ee9f73ee829a22a7825061d7ce48d0aa4625e06bdf1c828c1031dd9ad1334","family_id":"qf-expectation-constant","seed":202610056080,"semantic_template":"qf-expectation-constant/{\"c\":-8,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-8 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-8/1","explanation":"The constant multiplies the weight total one, giving E[X]=-8/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0081","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"00bc8e225b2764821e42b26417849233c09bb9a40e0e1bd19ebf1890eca9db1e","family_id":"qf-expectation-constant","seed":202610056081,"semantic_template":"qf-expectation-constant/{\"c\":-7,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-7 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-7/1","explanation":"The constant multiplies the weight total one, giving E[X]=-7/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0082","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"624591d95fb69aa15baad2124a4d9faa6bdd90c0aed1ada19d17106d781fdd24","family_id":"qf-expectation-constant","seed":202610056082,"semantic_template":"qf-expectation-constant/{\"c\":-6,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-6 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-6/1","explanation":"The constant multiplies the weight total one, giving E[X]=-6/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0083","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1235effe5986eecbe1867a493d6f5dac873a8783d4d3c9a3f3d0f98b23d34a24","family_id":"qf-expectation-constant","seed":202610056083,"semantic_template":"qf-expectation-constant/{\"c\":-5,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-5 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-5/1","explanation":"The constant multiplies the weight total one, giving E[X]=-5/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0084","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3b72dc9606ad024187c90bc45983ecaa454f5d17b223ed8575404c21b86c0992","family_id":"qf-expectation-constant","seed":202610056084,"semantic_template":"qf-expectation-constant/{\"c\":-4,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-4 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-4/1","explanation":"The constant multiplies the weight total one, giving E[X]=-4/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0085","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cf96660185844157354962c626697626d24084995dd4fdcfc248dff8a82df116","family_id":"qf-expectation-constant","seed":202610056085,"semantic_template":"qf-expectation-constant/{\"c\":-3,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-3 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-3/1","explanation":"The constant multiplies the weight total one, giving E[X]=-3/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0086","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"03178a9da9215b91287eec7fcb904d39b3f71caf73a1d2869bb03a632453bcfd","family_id":"qf-expectation-constant","seed":202610056086,"semantic_template":"qf-expectation-constant/{\"c\":-2,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-2 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-2/1","explanation":"The constant multiplies the weight total one, giving E[X]=-2/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0087","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f975d4f3b8ee6b999776495e7de8b6495a99d14b46236b2f5087b0e94dfc153","family_id":"qf-expectation-constant","seed":202610056087,"semantic_template":"qf-expectation-constant/{\"c\":-1,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=-1 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"-1/1","explanation":"The constant multiplies the weight total one, giving E[X]=-1/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0088","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7d6c04f5a5fd4b7502092eae00e3750a53e9ed99167bc6f866311df9730bed14","family_id":"qf-expectation-constant","seed":202610056088,"semantic_template":"qf-expectation-constant/{\"c\":0,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=0 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"The constant multiplies the weight total one, giving E[X]=0/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0089","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"de12330a38f41aecee5064f171ac25ce11ac3af825565d9de7902eca8a1d7ea7","family_id":"qf-expectation-constant","seed":202610056089,"semantic_template":"qf-expectation-constant/{\"c\":1,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=1 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"The constant multiplies the weight total one, giving E[X]=1/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0090","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"50165e0f547f603510ef3adb6b3524e383d3a44ce03ca1051faaeadaf8bf14c0","family_id":"qf-expectation-constant","seed":202610056090,"semantic_template":"qf-expectation-constant/{\"c\":2,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=2 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"2/1","explanation":"The constant multiplies the weight total one, giving E[X]=2/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0091","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a54010e8e3794ce1e2bd1716d934718fd9e5b079eee8f24ce168ef9e0e3a27d6","family_id":"qf-expectation-constant","seed":202610056091,"semantic_template":"qf-expectation-constant/{\"c\":3,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=3 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"3/1","explanation":"The constant multiplies the weight total one, giving E[X]=3/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0092","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4f7b3e0856c0e8299f351f25fccfdae1cba1ded239493ab9a9b0bd5a70d195c3","family_id":"qf-expectation-constant","seed":202610056092,"semantic_template":"qf-expectation-constant/{\"c\":4,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=4 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"4/1","explanation":"The constant multiplies the weight total one, giving E[X]=4/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0093","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4692b843087bb9a8691304ae2e0d98dbc8376903d1ad18f798a45b64e7519e7e","family_id":"qf-expectation-constant","seed":202610056093,"semantic_template":"qf-expectation-constant/{\"c\":5,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=5 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"5/1","explanation":"The constant multiplies the weight total one, giving E[X]=5/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0094","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"32cae651ee206966cdef8a3fe574b5d6778ed914accc888ae5c20bffaa51b579","family_id":"qf-expectation-constant","seed":202610056094,"semantic_template":"qf-expectation-constant/{\"c\":6,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=6 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"6/1","explanation":"The constant multiplies the weight total one, giving E[X]=6/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0095","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ad1036bb8f57d1a6c2edb944f4103ddd6dc84d618740481aebc7e4a8a79e1007","family_id":"qf-expectation-constant","seed":202610056095,"semantic_template":"qf-expectation-constant/{\"c\":7,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=7 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"7/1","explanation":"The constant multiplies the weight total one, giving E[X]=7/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0096","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"53531d3e10aaac723d391d335cca575f0488aea189851d87c3144e63cf600af4","family_id":"qf-expectation-constant","seed":202610056096,"semantic_template":"qf-expectation-constant/{\"c\":8,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=8 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"8/1","explanation":"The constant multiplies the weight total one, giving E[X]=8/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0097","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a907586c18a03598c1395a3e1ad539366b10a7440db470e36aced069b85b2b0","family_id":"qf-expectation-constant","seed":202610056097,"semantic_template":"qf-expectation-constant/{\"c\":9,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=9 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"9/1","explanation":"The constant multiplies the weight total one, giving E[X]=9/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0098","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"dd032e97dfe7a25e7d25ce5f54e114fc9f312ea03e585b8943b85d3218c9fc58","family_id":"qf-expectation-constant","seed":202610056098,"semantic_template":"qf-expectation-constant/{\"c\":10,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=10 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"10/1","explanation":"The constant multiplies the weight total one, giving E[X]=10/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0099","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"83250e6c70d247aef48693688ec5df82d2334659b68974faf75b0a030639283b","family_id":"qf-expectation-constant","seed":202610056099,"semantic_template":"qf-expectation-constant/{\"c\":11,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=11 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"11/1","explanation":"The constant multiplies the weight total one, giving E[X]=11/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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-20261004-advance-20261005-r06-expectation-0100","batch_id":"batch-20261004-advance-20261005-r06-expectation-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c37c0962ebb236e5fc2515ca534d589940b8ca30f0e439eb6a8cee80326fe0d2","family_id":"qf-expectation-constant","seed":202610056100,"semantic_template":"qf-expectation-constant/{\"c\":12,\"normalized_model\":true,\"operation\":\"constant_expectation\"}"},"fields":{"title":"Expectation of a Constant in a Normalized Model","stem":"The nonempty finite sample space is S={0, 1}, with F=𝒫(S). In the displayed order of S, the nonnegative probability weights are [1/(4), 3/(4)], summing exactly to one. For X(s)=12 at every outcome, compute E[X] as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"12/1","explanation":"The constant multiplies the weight total one, giving E[X]=12/1. Different normalized weight representations do not change this constant-expectation result."},"skills":["Expectation of a Constant in a Normalized Model"],"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."}}