{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0001","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ffc0c053b82f3e936ca7a68ec88cdacb733435c8fd7d769c2b49044974f9089b","family_id":"qf-conditional-weighted-probability","seed":202610047001,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[],\"B\":[-12,-11,-10,-9],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10,-9],\"denominator\":\"4\",\"operation\":\"weighted_conditional\",\"weights\":[[-12,\"0\"],[-11,\"1\"],[-10,\"2\"],[-9,\"1\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. A=∅, B={-12, -11, -10, -9}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 4. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0002","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"806e86ed94b29ba8c8864455cc1d5c793cc0fb22d8abdfb8d601713101239b3e","family_id":"qf-conditional-weighted-probability","seed":202610047002,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-11,-10,-9,-8],\"B\":[-11,-10,-9,-8],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8],\"denominator\":\"8\",\"operation\":\"weighted_conditional\",\"weights\":[[-11,\"1\"],[-10,\"2\"],[-9,\"3\"],[-8,\"2\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. A={-11, -10, -9, -8}, B={-11, -10, -9, -8}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 8 and the B mass numerator is 8. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0003","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f274a007a97672edf9d227c55999bd07d94f21c1e6a07b976f9b48bfed8db70c","family_id":"qf-conditional-weighted-probability","seed":202610047003,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-10,-9],\"B\":[-9,-8],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8,-7],\"denominator\":\"12\",\"operation\":\"weighted_conditional\",\"weights\":[[-10,\"2\"],[-9,\"3\"],[-8,\"4\"],[-7,\"3\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. A={-10, -9}, B={-9, -8}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"3/7","explanation":"The intersection mass numerator is 3 and the B mass numerator is 7. Their common positive denominator cancels, giving P(A|B)=3/7; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0004","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"48352d713de330e423cb4f59b23cb3fd6ea4d1e174dbbcc55f371b3852b8e45d","family_id":"qf-conditional-weighted-probability","seed":202610047004,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-9],\"B\":[-7,-6],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7,-6],\"denominator\":\"13\",\"operation\":\"weighted_conditional\",\"weights\":[[-9,\"0\"],[-8,\"4\"],[-7,\"5\"],[-6,\"4\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. A={-9}, B={-7, -6}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 9. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0005","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f2b344fd32978bd759aa83cd2012611db1795405d738d59f8c45f0a52a4bf752","family_id":"qf-conditional-weighted-probability","seed":202610047005,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-6,-5],\"B\":[-8,-6,-5],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6,-5],\"denominator\":\"13\",\"operation\":\"weighted_conditional\",\"weights\":[[-8,\"1\"],[-7,\"1\"],[-6,\"6\"],[-5,\"5\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. A={-6, -5}, B={-8, -6, -5}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"11/12","explanation":"The intersection mass numerator is 11 and the B mass numerator is 12. Their common positive denominator cancels, giving P(A|B)=11/12; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0006","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bd00815754067c8094185f6e9042542eb1cc65fa69bbfdd376a968df58b24b72","family_id":"qf-conditional-weighted-probability","seed":202610047006,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[],\"B\":[-7,-6,-5,-4],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5,-4],\"denominator\":\"6\",\"operation\":\"weighted_conditional\",\"weights\":[[-7,\"1\"],[-6,\"1\"],[-5,\"1\"],[-4,\"3\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. A=∅, B={-7, -6, -5, -4}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 12. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0007","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc0047e49343ae3384a1e8ce01cddd6762db1d3fdded885398f82a68cfad6f44","family_id":"qf-conditional-weighted-probability","seed":202610047007,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-6,-5,-4,-3],\"B\":[-6,-5,-4,-3],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4,-3],\"denominator\":\"13\",\"operation\":\"weighted_conditional\",\"weights\":[[-6,\"0\"],[-5,\"3\"],[-4,\"3\"],[-3,\"7\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. A={-6, -5, -4, -3}, B={-6, -5, -4, -3}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 13 and the B mass numerator is 13. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0008","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a496465bff080888a190460dfe233bf8cd584bbdaa31dedca2047e7b32e78268","family_id":"qf-conditional-weighted-probability","seed":202610047008,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-5,-4],\"B\":[-4,-3],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3,-2],\"denominator\":\"17\",\"operation\":\"weighted_conditional\",\"weights\":[[-5,\"1\"],[-4,\"4\"],[-3,\"4\"],[-2,\"8\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. A={-5, -4}, B={-4, -3}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/2","explanation":"The intersection mass numerator is 4 and the B mass numerator is 8. Their common positive denominator cancels, giving P(A|B)=1/2; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0009","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ebfe79c92834d235528ec65970fce5706d19f0412720e95c63481836d8b3353a","family_id":"qf-conditional-weighted-probability","seed":202610047009,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-4],\"B\":[-2,-1],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2,-1],\"denominator\":\"17\",\"operation\":\"weighted_conditional\",\"weights\":[[-4,\"2\"],[-3,\"1\"],[-2,\"5\"],[-1,\"9\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. A={-4}, B={-2, -1}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 14. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0010","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1216858949a6bd6a9f4f092bacd824d94f8ec366386e325bd773e8d98d47403","family_id":"qf-conditional-weighted-probability","seed":202610047010,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-1,0],\"B\":[-3,-1,0],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1,0],\"denominator\":\"9\",\"operation\":\"weighted_conditional\",\"weights\":[[-3,\"0\"],[-2,\"1\"],[-1,\"3\"],[0,\"5\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. A={-1, 0}, B={-3, -1, 0}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 16 and the B mass numerator is 16. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0011","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"284e2819be022e49863c9a7538fac7fd5261e12b17540b42f55d33b6f7886058","family_id":"qf-conditional-weighted-probability","seed":202610047011,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[],\"B\":[-2,-1,0,1],\"F\":\"powerset(S)\",\"S\":[-2,-1,0,1],\"denominator\":\"17\",\"operation\":\"weighted_conditional\",\"weights\":[[-2,\"1\"],[-1,\"3\"],[0,\"2\"],[1,\"11\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. A=∅, B={-2, -1, 0, 1}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 17. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0012","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"91aa695d081a9822812567889fa74041e81a555a472c1fd713f80872d46bdc0f","family_id":"qf-conditional-weighted-probability","seed":202610047012,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[-1,0,1,2],\"B\":[-1,0,1,2],\"F\":\"powerset(S)\",\"S\":[-1,0,1,2],\"denominator\":\"21\",\"operation\":\"weighted_conditional\",\"weights\":[[-1,\"2\"],[0,\"4\"],[1,\"3\"],[2,\"12\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. A={-1, 0, 1, 2}, B={-1, 0, 1, 2}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 21 and the B mass numerator is 21. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0013","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fbf81ac061ade848ae090f8fc1da82da9829126e51c57153983df820fc2e8f89","family_id":"qf-conditional-weighted-probability","seed":202610047013,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[0,1],\"B\":[1,2],\"F\":\"powerset(S)\",\"S\":[0,1,2,3],\"denominator\":\"18\",\"operation\":\"weighted_conditional\",\"weights\":[[0,\"0\"],[1,\"1\"],[2,\"4\"],[3,\"13\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. A={0, 1}, B={1, 2}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/5","explanation":"The intersection mass numerator is 1 and the B mass numerator is 5. Their common positive denominator cancels, giving P(A|B)=1/5; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0014","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bf9d8b11a1400f3583cc64fefa0bcf07f5cbcd4474aff273210308359921bc5e","family_id":"qf-conditional-weighted-probability","seed":202610047014,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[1],\"B\":[3,4],\"F\":\"powerset(S)\",\"S\":[1,2,3,4],\"denominator\":\"22\",\"operation\":\"weighted_conditional\",\"weights\":[[1,\"1\"],[2,\"2\"],[3,\"5\"],[4,\"14\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. A={1}, B={3, 4}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 19. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0015","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"02ba7fa6464c84688a2e67f4e8211ecb0f325ac296d08a97d0f09049ea1d2a48","family_id":"qf-conditional-weighted-probability","seed":202610047015,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[4,5],\"B\":[2,4,5],\"F\":\"powerset(S)\",\"S\":[2,3,4,5],\"denominator\":\"26\",\"operation\":\"weighted_conditional\",\"weights\":[[2,\"2\"],[3,\"3\"],[4,\"6\"],[5,\"15\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. A={4, 5}, B={2, 4, 5}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"21/23","explanation":"The intersection mass numerator is 21 and the B mass numerator is 23. Their common positive denominator cancels, giving P(A|B)=21/23; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0016","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5aaf20f1a26ed091fa64199a3699ca1c2e6bcc43af8315c94c5226f05e7d9422","family_id":"qf-conditional-weighted-probability","seed":202610047016,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[],\"B\":[3,4,5,6],\"F\":\"powerset(S)\",\"S\":[3,4,5,6],\"denominator\":\"11\",\"operation\":\"weighted_conditional\",\"weights\":[[3,\"0\"],[4,\"2\"],[5,\"1\"],[6,\"8\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. A=∅, B={3, 4, 5, 6}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 22. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0017","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f33e57747a07c4dddbd551e2ebe6080d068fd340569a5034e4a48e335ef4e39f","family_id":"qf-conditional-weighted-probability","seed":202610047017,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[4,5,6,7],\"B\":[4,5,6,7],\"F\":\"powerset(S)\",\"S\":[4,5,6,7],\"denominator\":\"22\",\"operation\":\"weighted_conditional\",\"weights\":[[4,\"1\"],[5,\"1\"],[6,\"3\"],[7,\"17\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. A={4, 5, 6, 7}, B={4, 5, 6, 7}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 22 and the B mass numerator is 22. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0018","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9c83f0f0e71bb47bcd675abdb1d500691858fa33d69513612eb403216a29c07f","family_id":"qf-conditional-weighted-probability","seed":202610047018,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[5,6],\"B\":[6,7],\"F\":\"powerset(S)\",\"S\":[5,6,7,8],\"denominator\":\"13\",\"operation\":\"weighted_conditional\",\"weights\":[[5,\"1\"],[6,\"1\"],[7,\"2\"],[8,\"9\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. A={5, 6}, B={6, 7}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/3","explanation":"The intersection mass numerator is 2 and the B mass numerator is 6. Their common positive denominator cancels, giving P(A|B)=1/3; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0019","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c9451ac44aac71897818330448741e3fe4d82b14d8495b0914ff44c8477b86a9","family_id":"qf-conditional-weighted-probability","seed":202610047019,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[6],\"B\":[8,9],\"F\":\"powerset(S)\",\"S\":[6,7,8,9],\"denominator\":\"27\",\"operation\":\"weighted_conditional\",\"weights\":[[6,\"0\"],[7,\"3\"],[8,\"5\"],[9,\"19\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. A={6}, B={8, 9}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 24. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0020","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3de1daf673bc0636512149f16cfc0c1979e4e71edc2b7f4b8f44bb386c100282","family_id":"qf-conditional-weighted-probability","seed":202610047020,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[9,10],\"B\":[7,9,10],\"F\":\"powerset(S)\",\"S\":[7,8,9,10],\"denominator\":\"31\",\"operation\":\"weighted_conditional\",\"weights\":[[7,\"1\"],[8,\"4\"],[9,\"6\"],[10,\"20\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. A={9, 10}, B={7, 9, 10}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"26/27","explanation":"The intersection mass numerator is 26 and the B mass numerator is 27. Their common positive denominator cancels, giving P(A|B)=26/27; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0021","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2efaa64e3be3fdf72baeaeb773df5ce9e62607e34b49ce34a3db329bbc6d5379","family_id":"qf-conditional-weighted-probability","seed":202610047021,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[],\"B\":[8,9,10,11],\"F\":\"powerset(S)\",\"S\":[8,9,10,11],\"denominator\":\"26\",\"operation\":\"weighted_conditional\",\"weights\":[[8,\"2\"],[9,\"1\"],[10,\"2\"],[11,\"21\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. A=∅, B={8, 9, 10, 11}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 26. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0022","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e0535c02594d8248c9a9a93b8da9c5ae7eac8a2978e89906c4500fe1b0d95b47","family_id":"qf-conditional-weighted-probability","seed":202610047022,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[9,10,11,12],\"B\":[9,10,11,12],\"F\":\"powerset(S)\",\"S\":[9,10,11,12],\"denominator\":\"27\",\"operation\":\"weighted_conditional\",\"weights\":[[9,\"0\"],[10,\"2\"],[11,\"3\"],[12,\"22\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. A={9, 10, 11, 12}, B={9, 10, 11, 12}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 27 and the B mass numerator is 27. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0023","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8e086ce79723c5cfecd6649e69f5baf8e91a2bf1749f531593238823625f0dbb","family_id":"qf-conditional-weighted-probability","seed":202610047023,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[10,11],\"B\":[11,12],\"F\":\"powerset(S)\",\"S\":[10,11,12,13],\"denominator\":\"31\",\"operation\":\"weighted_conditional\",\"weights\":[[10,\"1\"],[11,\"3\"],[12,\"4\"],[13,\"23\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. A={10, 11}, B={11, 12}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"3/7","explanation":"The intersection mass numerator is 3 and the B mass numerator is 7. Their common positive denominator cancels, giving P(A|B)=3/7; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0024","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f7b9a8aa41a8dbb832cbc9e9a51613136ec098673a057a3b51b73f6e432ab2d1","family_id":"qf-conditional-weighted-probability","seed":202610047024,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[11],\"B\":[13,14],\"F\":\"powerset(S)\",\"S\":[11,12,13,14],\"denominator\":\"35\",\"operation\":\"weighted_conditional\",\"weights\":[[11,\"2\"],[12,\"4\"],[13,\"5\"],[14,\"24\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. A={11}, B={13, 14}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"The intersection mass numerator is 0 and the B mass numerator is 29. Their common positive denominator cancels, giving P(A|B)=0/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0025","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d123635c5c66a1b635f2224dd7c4d8f2272d961aab5d4100563917aeacdc6a35","family_id":"qf-conditional-weighted-probability","seed":202610047025,"semantic_template":"qf-conditional-weighted-probability/{\"A\":[14,15],\"B\":[12,14,15],\"F\":\"powerset(S)\",\"S\":[12,13,14,15],\"denominator\":\"32\",\"operation\":\"weighted_conditional\",\"weights\":[[12,\"0\"],[13,\"1\"],[14,\"6\"],[15,\"25\"]]}"},"fields":{"title":"Conditional Probability in a Finite Weighted Model","stem":"The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. A={14, 15}, B={12, 14, 15}, and P(B)>0. Compute P(A|B) exactly as a reduced fraction with positive denominator, without assuming equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"The intersection mass numerator is 31 and the B mass numerator is 31. Their common positive denominator cancels, giving P(A|B)=1/1; this is not the product P(A)P(B)."},"skills":["Conditional Probability in a Finite Weighted 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-20261003-advance-20261004-r07-conditional-probability-0026","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8a75d2da24bc942d3b25b9c8b81966ea68801245312b2c50b7d0a33a40d6d2eb","family_id":"qf-conditional-uniform-probability","seed":202610047026,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[],\"B\":[-12,-11,-10,-9],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10,-9],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={-12, -11, -10, -9} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0027","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"51cfd6470b419cdfbebff78054969548ba01aa8fde7596909d5729dba261ea43","family_id":"qf-conditional-uniform-probability","seed":202610047027,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-11,-10,-9,-8,-7],\"B\":[-11,-10],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8,-7],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-11, -10, -9, -8, -7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-11, -10, -9, -8, -7} and B={-11, -10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0028","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"678af7de4afb6372fc4df6da99aad81c7646579a08e0bf1e1b23b7307735942c","family_id":"qf-conditional-uniform-probability","seed":202610047028,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-10],\"B\":[-6,-5],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8,-7,-6,-5],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-10, -9, -8, -7, -6, -5}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-10} and B={-6, -5} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0029","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82ffd7ad32a89bc7edbd28f4ff58521ce7df1ef58ab185d1982d57994454f2dd","family_id":"qf-conditional-uniform-probability","seed":202610047029,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-9,-8,-7],\"B\":[-9,-8,-7,-6],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7,-6],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-9, -8, -7} and B={-9, -8, -7, -6} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"3/4","explanation":"|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0030","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b39291190f9020a24ca7e8d7a9e614e2c168cdee95140df7d0091d6e14090ab","family_id":"qf-conditional-uniform-probability","seed":202610047030,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-8,-7],\"B\":[-8,-7],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6,-5,-4],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-8, -7, -6, -5, -4}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-8, -7} and B={-8, -7} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0031","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a441a07f708bcec69ce744143f4b263320d5f0176ea1cac6a7df150ec68b6a20","family_id":"qf-conditional-uniform-probability","seed":202610047031,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-7,-5,-3],\"B\":[-3,-2],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5,-4,-3,-2],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-7, -6, -5, -4, -3, -2}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-7, -5, -3} and B={-3, -2} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0032","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"10124d090a9211c41cbaf250014ba3d1fa2b03dfedb59770ca4a6b4eec0d4cf5","family_id":"qf-conditional-uniform-probability","seed":202610047032,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[],\"B\":[-6,-5,-4,-3],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4,-3],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={-6, -5, -4, -3} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0033","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"522a263290b1a75d9e7943f22558e44264bd0ee913d95c06922de37cd4fa988d","family_id":"qf-conditional-uniform-probability","seed":202610047033,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-5,-4,-3,-2,-1],\"B\":[-5,-4],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3,-2,-1],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-5, -4, -3, -2, -1}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-5, -4, -3, -2, -1} and B={-5, -4} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0034","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87876855d778cea306f5bc1cd1815dbac02f97e8cc1616646d09899f72cf0a1b","family_id":"qf-conditional-uniform-probability","seed":202610047034,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-4],\"B\":[0,1],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2,-1,0,1],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-4, -3, -2, -1, 0, 1}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-4} and B={0, 1} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0035","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a853edd444c14572dd6ca10f0b47d04c38cf1c0ef68b622a0b20706cbd733087","family_id":"qf-conditional-uniform-probability","seed":202610047035,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-3,-2,-1],\"B\":[-3,-2,-1,0],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1,0],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-3, -2, -1} and B={-3, -2, -1, 0} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"3/4","explanation":"|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0036","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8b154dcdbb961f5484f92834af9f004985ca56d0ee12d07ed98d7fcde9170e2a","family_id":"qf-conditional-uniform-probability","seed":202610047036,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-2,-1],\"B\":[-2,-1],\"F\":\"powerset(S)\",\"S\":[-2,-1,0,1,2],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-2, -1, 0, 1, 2}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-2, -1} and B={-2, -1} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0037","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"209751227018a30c763d872c7fd213f9fc528182779d67f1e9c7557d01c117a5","family_id":"qf-conditional-uniform-probability","seed":202610047037,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[-1,1,3],\"B\":[3,4],\"F\":\"powerset(S)\",\"S\":[-1,0,1,2,3,4],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-1, 0, 1, 2, 3, 4}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-1, 1, 3} and B={3, 4} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0038","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3287d6cb46b2006a826ac359952a1be710c8019dba33f3dbc23c9859ac97ff81","family_id":"qf-conditional-uniform-probability","seed":202610047038,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[],\"B\":[0,1,2,3],\"F\":\"powerset(S)\",\"S\":[0,1,2,3],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={0, 1, 2, 3} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0039","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6fdb18c76fabd419ed3aef6836f367d59950e50d6493f9a5cfdb2e95b8a260e8","family_id":"qf-conditional-uniform-probability","seed":202610047039,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[1,2,3,4,5],\"B\":[1,2],\"F\":\"powerset(S)\",\"S\":[1,2,3,4,5],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={1, 2, 3, 4, 5}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={1, 2, 3, 4, 5} and B={1, 2} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0040","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7caee3d8fd9dd651d7b3ccc2f130c904cd3e5719215fb53890b5ade89d47a702","family_id":"qf-conditional-uniform-probability","seed":202610047040,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[2],\"B\":[6,7],\"F\":\"powerset(S)\",\"S\":[2,3,4,5,6,7],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={2, 3, 4, 5, 6, 7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={2} and B={6, 7} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0041","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"90e2b7a41631413885aaed70507784d08810452d44646c4cbfa6e7a541060e81","family_id":"qf-conditional-uniform-probability","seed":202610047041,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[3,4,5],\"B\":[3,4,5,6],\"F\":\"powerset(S)\",\"S\":[3,4,5,6],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={3, 4, 5} and B={3, 4, 5, 6} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"3/4","explanation":"|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0042","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba0a89552931b4db0cb58d7d7605ec095cdc19793ba97dded23bde08110d587f","family_id":"qf-conditional-uniform-probability","seed":202610047042,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[4,5],\"B\":[4,5],\"F\":\"powerset(S)\",\"S\":[4,5,6,7,8],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={4, 5, 6, 7, 8}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={4, 5} and B={4, 5} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0043","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d2114d6958e6b46e0d69fd2723cc560c524e216059ae8f0c0d67b035212d042c","family_id":"qf-conditional-uniform-probability","seed":202610047043,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[5,7,9],\"B\":[9,10],\"F\":\"powerset(S)\",\"S\":[5,6,7,8,9,10],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={5, 6, 7, 8, 9, 10}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={5, 7, 9} and B={9, 10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0044","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7131dddbce0335f80e7e5dac6457cd8f90a2e46ea00d720e32dadd584f23addd","family_id":"qf-conditional-uniform-probability","seed":202610047044,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[],\"B\":[6,7,8,9],\"F\":\"powerset(S)\",\"S\":[6,7,8,9],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={6, 7, 8, 9} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0045","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7a99528e924b718041bd81fea15711d788f7041d572d49c5e1da8a1e26fec211","family_id":"qf-conditional-uniform-probability","seed":202610047045,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[7,8,9,10,11],\"B\":[7,8],\"F\":\"powerset(S)\",\"S\":[7,8,9,10,11],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={7, 8, 9, 10, 11}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={7, 8, 9, 10, 11} and B={7, 8} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0046","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"71775946d78d99479c1b580689f4d57086fca8249a16fdc1805115b00e2612b3","family_id":"qf-conditional-uniform-probability","seed":202610047046,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[8],\"B\":[12,13],\"F\":\"powerset(S)\",\"S\":[8,9,10,11,12,13],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={8, 9, 10, 11, 12, 13}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={8} and B={12, 13} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0047","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4ab05832de422fa3634e7b0c391bf594a00f8c342b42cc84c83a34bd6977b7a6","family_id":"qf-conditional-uniform-probability","seed":202610047047,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[9,10,11],\"B\":[9,10,11,12],\"F\":\"powerset(S)\",\"S\":[9,10,11,12],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={9, 10, 11} and B={9, 10, 11, 12} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"3/4","explanation":"|A∩B|=3 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=3/4. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0048","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b4ec50332d3ac92d4edfb7cb366f3d266c261c805721532f4bb36fba9c483f3","family_id":"qf-conditional-uniform-probability","seed":202610047048,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[10,11],\"B\":[10,11],\"F\":\"powerset(S)\",\"S\":[10,11,12,13,14],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={10, 11, 12, 13, 14}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={10, 11} and B={10, 11} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0049","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4019f5e5c9d860a9b1d4155bc595fae0adfff3908006b6609f743d358da97ca7","family_id":"qf-conditional-uniform-probability","seed":202610047049,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[11,13,15],\"B\":[15,16],\"F\":\"powerset(S)\",\"S\":[11,12,13,14,15,16],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={11, 12, 13, 14, 15, 16}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={11, 13, 15} and B={15, 16} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"|A∩B|=1 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0050","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"16c2775a1bb5bb2df06b6afa99c3db93a693f4824b4529a763cd8fd715def3fd","family_id":"qf-conditional-uniform-probability","seed":202610047050,"semantic_template":"qf-conditional-uniform-probability/{\"A\":[],\"B\":[12,13,14,15],\"F\":\"powerset(S)\",\"S\":[12,13,14,15],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"},"fields":{"title":"Conditional Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A=∅ and B={12, 13, 14, 15} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"|A∩B|=0 and |B|=4. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Conditional Probability with Explicit Equiprobability"],"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-20261003-advance-20261004-r07-conditional-probability-0051","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"7c27d4849aea500d52e3b58c31c1ccf780809b83a7de04fe4727f7ad1020073d","family_id":"qf-conditional-event-validity","seed":202610047051,"semantic_template":"qf-conditional-event-validity/{\"B\":[-12],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10,-9],\"denominator\":\"4\",\"operation\":\"condition_validity\",\"weights\":[[-12,\"0\"],[-11,\"1\"],[-10,\"2\"],[-9,\"1\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. Does B={-12} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 4. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0052","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b8244bc2ea27b81d40e0905372d518120661b4c97520ee7520246d11de3fc8ce","family_id":"qf-conditional-event-validity","seed":202610047052,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8],\"denominator\":\"8\",\"operation\":\"condition_validity\",\"weights\":[[-11,\"1\"],[-10,\"2\"],[-9,\"3\"],[-8,\"2\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 8. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0053","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5d42e30818e75ad23e946621eae6f94ba5d6499f24b1033040f3627ec8b4a520","family_id":"qf-conditional-event-validity","seed":202610047053,"semantic_template":"qf-conditional-event-validity/{\"B\":[-9],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8,-7],\"denominator\":\"12\",\"operation\":\"condition_validity\",\"weights\":[[-10,\"2\"],[-9,\"3\"],[-8,\"4\"],[-7,\"3\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. Does B={-9} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 12. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0054","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ceec9c1b3b7e2ad1c667d3bc6614d68cd524f2214f83e4645014f34e52a6865c","family_id":"qf-conditional-event-validity","seed":202610047054,"semantic_template":"qf-conditional-event-validity/{\"B\":[-9,-8,-7,-6],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7,-6],\"denominator\":\"13\",\"operation\":\"condition_validity\",\"weights\":[[-9,\"0\"],[-8,\"4\"],[-7,\"5\"],[-6,\"4\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. Does B={-9, -8, -7, -6} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 13, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0055","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33c1a08dffec843c793f2e5558705a26c52b3b6c6609588dcca383a1693135b5","family_id":"qf-conditional-event-validity","seed":202610047055,"semantic_template":"qf-conditional-event-validity/{\"B\":[-8],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6,-5],\"denominator\":\"13\",\"operation\":\"condition_validity\",\"weights\":[[-8,\"1\"],[-7,\"1\"],[-6,\"6\"],[-5,\"5\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. Does B={-8} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 1, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0056","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e521a7b16d0778229add07af3866ed7d3455447451cdecf3361821e3129be80d","family_id":"qf-conditional-event-validity","seed":202610047056,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5,-4],\"denominator\":\"6\",\"operation\":\"condition_validity\",\"weights\":[[-7,\"1\"],[-6,\"1\"],[-5,\"1\"],[-4,\"3\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 12. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0057","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"310f01241c135ec36f6021d2a1c395967f45814d2ba88b8cf0400a0df27691f9","family_id":"qf-conditional-event-validity","seed":202610047057,"semantic_template":"qf-conditional-event-validity/{\"B\":[-5],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4,-3],\"denominator\":\"13\",\"operation\":\"condition_validity\",\"weights\":[[-6,\"0\"],[-5,\"3\"],[-4,\"3\"],[-3,\"7\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. Does B={-5} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 13. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0058","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d494486d571432a6bef429244feb9544b1fa68f063e2ed26b9be19d1e3ce0d82","family_id":"qf-conditional-event-validity","seed":202610047058,"semantic_template":"qf-conditional-event-validity/{\"B\":[-5,-4,-3,-2],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3,-2],\"denominator\":\"17\",\"operation\":\"condition_validity\",\"weights\":[[-5,\"1\"],[-4,\"4\"],[-3,\"4\"],[-2,\"8\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. Does B={-5, -4, -3, -2} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 17, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0059","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2a3745f43148b27391cc206d46beec4173560922321174ac1d7be637ea1df62f","family_id":"qf-conditional-event-validity","seed":202610047059,"semantic_template":"qf-conditional-event-validity/{\"B\":[-4],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2,-1],\"denominator\":\"17\",\"operation\":\"condition_validity\",\"weights\":[[-4,\"2\"],[-3,\"1\"],[-2,\"5\"],[-1,\"9\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. Does B={-4} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 2, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0060","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e64a4381fbc9270653cc0d9e08cb2978d95b0876eeccd3b17ae8e8e0492a4584","family_id":"qf-conditional-event-validity","seed":202610047060,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1,0],\"denominator\":\"9\",\"operation\":\"condition_validity\",\"weights\":[[-3,\"0\"],[-2,\"1\"],[-1,\"3\"],[0,\"5\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 18. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0061","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"08f0d54390746ba85e9f19dfdf50adb7e0d08de819d05702000d983977aeed22","family_id":"qf-conditional-event-validity","seed":202610047061,"semantic_template":"qf-conditional-event-validity/{\"B\":[-1],\"F\":\"powerset(S)\",\"S\":[-2,-1,0,1],\"denominator\":\"17\",\"operation\":\"condition_validity\",\"weights\":[[-2,\"1\"],[-1,\"3\"],[0,\"2\"],[1,\"11\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. Does B={-1} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 17. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0062","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"290ed11bc3dd9c08667ab32d2374f641c6bdb45a6d1cf10f32d933ce6ae2d334","family_id":"qf-conditional-event-validity","seed":202610047062,"semantic_template":"qf-conditional-event-validity/{\"B\":[-1,0,1,2],\"F\":\"powerset(S)\",\"S\":[-1,0,1,2],\"denominator\":\"21\",\"operation\":\"condition_validity\",\"weights\":[[-1,\"2\"],[0,\"4\"],[1,\"3\"],[2,\"12\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. Does B={-1, 0, 1, 2} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 21, with positive common denominator 21. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0063","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d3aa0761aa6ec498f68313c659629ef1acc1184870f5c84f5e6f367bb21d565a","family_id":"qf-conditional-event-validity","seed":202610047063,"semantic_template":"qf-conditional-event-validity/{\"B\":[0],\"F\":\"powerset(S)\",\"S\":[0,1,2,3],\"denominator\":\"18\",\"operation\":\"condition_validity\",\"weights\":[[0,\"0\"],[1,\"1\"],[2,\"4\"],[3,\"13\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. Does B={0} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 18. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0064","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1e5eb5cdd593629ece878598f818f23c2d94c9136b66d70a97573b5f2fabb0b9","family_id":"qf-conditional-event-validity","seed":202610047064,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[1,2,3,4],\"denominator\":\"22\",\"operation\":\"condition_validity\",\"weights\":[[1,\"1\"],[2,\"2\"],[3,\"5\"],[4,\"14\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 22. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0065","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c12c98b12003caac023c587a07d443cd40a0f74438150cac92c2eef61c3ad214","family_id":"qf-conditional-event-validity","seed":202610047065,"semantic_template":"qf-conditional-event-validity/{\"B\":[3],\"F\":\"powerset(S)\",\"S\":[2,3,4,5],\"denominator\":\"26\",\"operation\":\"condition_validity\",\"weights\":[[2,\"2\"],[3,\"3\"],[4,\"6\"],[5,\"15\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. Does B={3} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 26. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0066","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9cbe1d04c48c341cc0a72277e4f4bba833a4e15efc8168737be995ba0e124d71","family_id":"qf-conditional-event-validity","seed":202610047066,"semantic_template":"qf-conditional-event-validity/{\"B\":[3,4,5,6],\"F\":\"powerset(S)\",\"S\":[3,4,5,6],\"denominator\":\"11\",\"operation\":\"condition_validity\",\"weights\":[[3,\"0\"],[4,\"2\"],[5,\"1\"],[6,\"8\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. Does B={3, 4, 5, 6} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 22, with positive common denominator 22. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0067","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4e34df14fe701917f9a6d40e666d8238c5e60db9cc15345c0dac1c712866e2fd","family_id":"qf-conditional-event-validity","seed":202610047067,"semantic_template":"qf-conditional-event-validity/{\"B\":[4],\"F\":\"powerset(S)\",\"S\":[4,5,6,7],\"denominator\":\"22\",\"operation\":\"condition_validity\",\"weights\":[[4,\"1\"],[5,\"1\"],[6,\"3\"],[7,\"17\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. Does B={4} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 1, with positive common denominator 22. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0068","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e0ad0bffd3b4026dd7392ccab4a9c72c34bcca57af5af1408c657e21fa810b5","family_id":"qf-conditional-event-validity","seed":202610047068,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[5,6,7,8],\"denominator\":\"13\",\"operation\":\"condition_validity\",\"weights\":[[5,\"1\"],[6,\"1\"],[7,\"2\"],[8,\"9\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 26. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0069","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4521c0d8d05fc8e31835dd59b73e47bcafc3006bf31cc78cde1c908d0df0252f","family_id":"qf-conditional-event-validity","seed":202610047069,"semantic_template":"qf-conditional-event-validity/{\"B\":[7],\"F\":\"powerset(S)\",\"S\":[6,7,8,9],\"denominator\":\"27\",\"operation\":\"condition_validity\",\"weights\":[[6,\"0\"],[7,\"3\"],[8,\"5\"],[9,\"19\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. Does B={7} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 27. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0070","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb6a3768784150b75342f105f077a93462ac2c20c7ec9b61503d15052e4038a3","family_id":"qf-conditional-event-validity","seed":202610047070,"semantic_template":"qf-conditional-event-validity/{\"B\":[7,8,9,10],\"F\":\"powerset(S)\",\"S\":[7,8,9,10],\"denominator\":\"31\",\"operation\":\"condition_validity\",\"weights\":[[7,\"1\"],[8,\"4\"],[9,\"6\"],[10,\"20\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. Does B={7, 8, 9, 10} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 31, with positive common denominator 31. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0071","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"404b0f28cc7cbe1446d94cf78390b6c6292c223744cac1a7b6437353ea41fd8f","family_id":"qf-conditional-event-validity","seed":202610047071,"semantic_template":"qf-conditional-event-validity/{\"B\":[8],\"F\":\"powerset(S)\",\"S\":[8,9,10,11],\"denominator\":\"26\",\"operation\":\"condition_validity\",\"weights\":[[8,\"2\"],[9,\"1\"],[10,\"2\"],[11,\"21\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. Does B={8} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 2, with positive common denominator 26. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0072","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"513f90eda92381157a8cf84e90e580757871274df78095c63a682cae4deb6ab4","family_id":"qf-conditional-event-validity","seed":202610047072,"semantic_template":"qf-conditional-event-validity/{\"B\":[],\"F\":\"powerset(S)\",\"S\":[9,10,11,12],\"denominator\":\"27\",\"operation\":\"condition_validity\",\"weights\":[[9,\"0\"],[10,\"2\"],[11,\"3\"],[12,\"22\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. Does B=∅ meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 27. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0073","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a2ae4bcbaac2441319f168d3218024bbffb1b2e57a22c950801ff6c22dc5ad3","family_id":"qf-conditional-event-validity","seed":202610047073,"semantic_template":"qf-conditional-event-validity/{\"B\":[11],\"F\":\"powerset(S)\",\"S\":[10,11,12,13],\"denominator\":\"31\",\"operation\":\"condition_validity\",\"weights\":[[10,\"1\"],[11,\"3\"],[12,\"4\"],[13,\"23\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. Does B={11} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 3, with positive common denominator 31. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0074","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"05ec5754c6861f0cf7b8ad9f797c5c0ba7d489461572ea90cfcbc8d30dc27be4","family_id":"qf-conditional-event-validity","seed":202610047074,"semantic_template":"qf-conditional-event-validity/{\"B\":[11,12,13,14],\"F\":\"powerset(S)\",\"S\":[11,12,13,14],\"denominator\":\"35\",\"operation\":\"condition_validity\",\"weights\":[[11,\"2\"],[12,\"4\"],[13,\"5\"],[14,\"24\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. Does B={11, 12, 13, 14} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"Yes","explanation":"The B mass numerator is 35, with positive common denominator 35. It must be strictly positive, giving Yes. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0075","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6dc4619c483afd79cd10405ab0da15d0175767f356f0b479c3d396b318addd12","family_id":"qf-conditional-event-validity","seed":202610047075,"semantic_template":"qf-conditional-event-validity/{\"B\":[12],\"F\":\"powerset(S)\",\"S\":[12,13,14,15],\"denominator\":\"32\",\"operation\":\"condition_validity\",\"weights\":[[12,\"0\"],[13,\"1\"],[14,\"6\"],[15,\"25\"]]}"},"fields":{"title":"Positive-Probability Eligibility of a Conditioning Event","stem":"The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. Does B={12} meet the positive-probability requirement for defining P(A|B)?","answer":{"kind":"short_text","value":"No","explanation":"The B mass numerator is 0, with positive common denominator 32. It must be strictly positive, giving No. A nonempty event of zero mass is still ineligible; undefined conditioning is not assigned a zero answer."},"skills":["Positive-Probability Eligibility of a Conditioning Event"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0076","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ba8cc53ebc21230ccd485bbdb13f4e2d85029ad30805b460789e41095e6215a0","family_id":"qf-conditional-direction-comparison","seed":202610047076,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-11],\"B\":[-10],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10,-9],\"denominator\":\"4\",\"operation\":\"compare_direction\",\"weights\":[[-12,\"0\"],[-11,\"1\"],[-10,\"2\"],[-9,\"1\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-12, -11, -10, -9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(4), 1/(4), 2/(4), 1/(4)], summing exactly to one. A={-11}, B={-10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0077","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4b317a4179aaff9061b7df07d11354eec0f565708f0639c13ddbd1a728822ab1","family_id":"qf-conditional-direction-comparison","seed":202610047077,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-10,-9],\"B\":[-9,-8],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8],\"denominator\":\"8\",\"operation\":\"compare_direction\",\"weights\":[[-11,\"1\"],[-10,\"2\"],[-9,\"3\"],[-8,\"2\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-11, -10, -9, -8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(8), 2/(8), 3/(8), 2/(8)], summing exactly to one. A={-10, -9}, B={-9, -8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=3/5 and P(B|A)=3/5, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0078","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"31ee25e1c84bb7e4c412a3d5aaa5e0404ea83108095f28731ddbe9128dfcc84b","family_id":"qf-conditional-direction-comparison","seed":202610047078,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-9,-8],\"B\":[-9,-8],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8,-7],\"denominator\":\"12\",\"operation\":\"compare_direction\",\"weights\":[[-10,\"2\"],[-9,\"3\"],[-8,\"4\"],[-7,\"3\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-10, -9, -8, -7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 3/(12), 4/(12), 3/(12)], summing exactly to one. A={-9, -8}, B={-9, -8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0079","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bddfd731b6872eed239761230e03411b8bf8f949b788b279b3d1b5410bf2666e","family_id":"qf-conditional-direction-comparison","seed":202610047079,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-9,-8,-7,-6],\"B\":[-8,-7],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7,-6],\"denominator\":\"13\",\"operation\":\"compare_direction\",\"weights\":[[-9,\"0\"],[-8,\"4\"],[-7,\"5\"],[-6,\"4\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-9, -8, -7, -6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 4/(13), 5/(13), 4/(13)], summing exactly to one. A={-9, -8, -7, -6}, B={-8, -7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/1 and P(B|A)=9/13, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0080","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"550b57d0c27b4cb8276753bb0967f12f938bd59e10d6972433327cf95f754dc0","family_id":"qf-conditional-direction-comparison","seed":202610047080,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-7,-5],\"B\":[-8,-7,-6,-5],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6,-5],\"denominator\":\"13\",\"operation\":\"compare_direction\",\"weights\":[[-8,\"1\"],[-7,\"1\"],[-6,\"6\"],[-5,\"5\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-8, -7, -6, -5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(13), 1/(13), 6/(13), 5/(13)], summing exactly to one. A={-7, -5}, B={-8, -7, -6, -5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=6/13 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0081","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e10b5b5d06457d516224c02a26a1cb75a4a66d961623175387832a275c7fdc30","family_id":"qf-conditional-direction-comparison","seed":202610047081,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-6],\"B\":[-5],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5,-4],\"denominator\":\"6\",\"operation\":\"compare_direction\",\"weights\":[[-7,\"1\"],[-6,\"1\"],[-5,\"1\"],[-4,\"3\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-7, -6, -5, -4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(12), 2/(12), 2/(12), 6/(12)], summing exactly to one. A={-6}, B={-5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0082","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5dbfe31bb58498dbd31c2f7318d4efcf5d6263a70f3dd16490e9852cebfe0dbd","family_id":"qf-conditional-direction-comparison","seed":202610047082,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-5,-4],\"B\":[-4,-3],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4,-3],\"denominator\":\"13\",\"operation\":\"compare_direction\",\"weights\":[[-6,\"0\"],[-5,\"3\"],[-4,\"3\"],[-3,\"7\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-6, -5, -4, -3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(13), 3/(13), 3/(13), 7/(13)], summing exactly to one. A={-5, -4}, B={-4, -3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=3/10 and P(B|A)=1/2, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0083","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b1632a1c7bb5c1f2226b845e704991cbed0f13fd095918d45fb596f19368caad","family_id":"qf-conditional-direction-comparison","seed":202610047083,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-4,-3],\"B\":[-4,-3],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3,-2],\"denominator\":\"17\",\"operation\":\"compare_direction\",\"weights\":[[-5,\"1\"],[-4,\"4\"],[-3,\"4\"],[-2,\"8\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-5, -4, -3, -2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 4/(17), 4/(17), 8/(17)], summing exactly to one. A={-4, -3}, B={-4, -3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0084","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fb766006936d3d9136c17ae24ef551c18b1b494fdb081a6f3e37c8f6da5e0923","family_id":"qf-conditional-direction-comparison","seed":202610047084,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-4,-3,-2,-1],\"B\":[-3,-2],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2,-1],\"denominator\":\"17\",\"operation\":\"compare_direction\",\"weights\":[[-4,\"2\"],[-3,\"1\"],[-2,\"5\"],[-1,\"9\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-4, -3, -2, -1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(17), 1/(17), 5/(17), 9/(17)], summing exactly to one. A={-4, -3, -2, -1}, B={-3, -2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/1 and P(B|A)=6/17, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0085","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"47ac038df3296ffb4740ea9cbc7b2f6ece493bafab4e6646b8f14b95462faa84","family_id":"qf-conditional-direction-comparison","seed":202610047085,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-2,0],\"B\":[-3,-2,-1,0],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1,0],\"denominator\":\"9\",\"operation\":\"compare_direction\",\"weights\":[[-3,\"0\"],[-2,\"1\"],[-1,\"3\"],[0,\"5\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-3, -2, -1, 0}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 2/(18), 6/(18), 10/(18)], summing exactly to one. A={-2, 0}, B={-3, -2, -1, 0}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=2/3 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0086","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fd1dcde8569d33cb8e2f7c4919332d38a2332b97ad9de7f732e5202966706790","family_id":"qf-conditional-direction-comparison","seed":202610047086,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[-1],\"B\":[0],\"F\":\"powerset(S)\",\"S\":[-2,-1,0,1],\"denominator\":\"17\",\"operation\":\"compare_direction\",\"weights\":[[-2,\"1\"],[-1,\"3\"],[0,\"2\"],[1,\"11\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-2, -1, 0, 1}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(17), 3/(17), 2/(17), 11/(17)], summing exactly to one. A={-1}, B={0}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0087","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e59d7ea004cc067cfa088fe9fa692b37fc7161c4d385d1c94fd303ee56f44ae","family_id":"qf-conditional-direction-comparison","seed":202610047087,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[0,1],\"B\":[1,2],\"F\":\"powerset(S)\",\"S\":[-1,0,1,2],\"denominator\":\"21\",\"operation\":\"compare_direction\",\"weights\":[[-1,\"2\"],[0,\"4\"],[1,\"3\"],[2,\"12\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={-1, 0, 1, 2}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(21), 4/(21), 3/(21), 12/(21)], summing exactly to one. A={0, 1}, B={1, 2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/5 and P(B|A)=3/7, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0088","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d555a0682dfb13d79c5a8a31065798a0d3985c82e707e244eef0568c1364f7be","family_id":"qf-conditional-direction-comparison","seed":202610047088,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[1,2],\"B\":[1,2],\"F\":\"powerset(S)\",\"S\":[0,1,2,3],\"denominator\":\"18\",\"operation\":\"compare_direction\",\"weights\":[[0,\"0\"],[1,\"1\"],[2,\"4\"],[3,\"13\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={0, 1, 2, 3}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(18), 1/(18), 4/(18), 13/(18)], summing exactly to one. A={1, 2}, B={1, 2}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0089","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1451ea962b1641d0a277da7aa23faba058d713335e0e6b2d4ccb912aa15d3b97","family_id":"qf-conditional-direction-comparison","seed":202610047089,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[1,2,3,4],\"B\":[2,3],\"F\":\"powerset(S)\",\"S\":[1,2,3,4],\"denominator\":\"22\",\"operation\":\"compare_direction\",\"weights\":[[1,\"1\"],[2,\"2\"],[3,\"5\"],[4,\"14\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={1, 2, 3, 4}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 2/(22), 5/(22), 14/(22)], summing exactly to one. A={1, 2, 3, 4}, B={2, 3}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/1 and P(B|A)=7/22, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0090","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cad1cf439353e1c7dcdbfd20a93b246d0cd5cd3e7245632f47bdd68198991fa9","family_id":"qf-conditional-direction-comparison","seed":202610047090,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[3,5],\"B\":[2,3,4,5],\"F\":\"powerset(S)\",\"S\":[2,3,4,5],\"denominator\":\"26\",\"operation\":\"compare_direction\",\"weights\":[[2,\"2\"],[3,\"3\"],[4,\"6\"],[5,\"15\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={2, 3, 4, 5}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 3/(26), 6/(26), 15/(26)], summing exactly to one. A={3, 5}, B={2, 3, 4, 5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=9/13 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0091","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"20d516d88e03a06247a73552fb8adf4a4cb1458f4b1b384a370df048d3f9feca","family_id":"qf-conditional-direction-comparison","seed":202610047091,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[4],\"B\":[5],\"F\":\"powerset(S)\",\"S\":[3,4,5,6],\"denominator\":\"11\",\"operation\":\"compare_direction\",\"weights\":[[3,\"0\"],[4,\"2\"],[5,\"1\"],[6,\"8\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={3, 4, 5, 6}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(22), 4/(22), 2/(22), 16/(22)], summing exactly to one. A={4}, B={5}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0092","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"82eb2a409ff8da977b7e96090be11b0a2def8272a57e0b9ff6af25c5ce7c31e8","family_id":"qf-conditional-direction-comparison","seed":202610047092,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[5,6],\"B\":[6,7],\"F\":\"powerset(S)\",\"S\":[4,5,6,7],\"denominator\":\"22\",\"operation\":\"compare_direction\",\"weights\":[[4,\"1\"],[5,\"1\"],[6,\"3\"],[7,\"17\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={4, 5, 6, 7}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(22), 1/(22), 3/(22), 17/(22)], summing exactly to one. A={5, 6}, B={6, 7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=3/20 and P(B|A)=3/4, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0093","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c90fa868afb5f6e45130f4b3d757e64562b1800d8aab4717938e7072b04f7726","family_id":"qf-conditional-direction-comparison","seed":202610047093,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[6,7],\"B\":[6,7],\"F\":\"powerset(S)\",\"S\":[5,6,7,8],\"denominator\":\"13\",\"operation\":\"compare_direction\",\"weights\":[[5,\"1\"],[6,\"1\"],[7,\"2\"],[8,\"9\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={5, 6, 7, 8}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 2/(26), 4/(26), 18/(26)], summing exactly to one. A={6, 7}, B={6, 7}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0094","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef22829ca048e4f6c00f50d2d603dde44f32e7ed557e09cd6f4e5200fdb7b69a","family_id":"qf-conditional-direction-comparison","seed":202610047094,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[6,7,8,9],\"B\":[7,8],\"F\":\"powerset(S)\",\"S\":[6,7,8,9],\"denominator\":\"27\",\"operation\":\"compare_direction\",\"weights\":[[6,\"0\"],[7,\"3\"],[8,\"5\"],[9,\"19\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={6, 7, 8, 9}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 3/(27), 5/(27), 19/(27)], summing exactly to one. A={6, 7, 8, 9}, B={7, 8}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/1 and P(B|A)=8/27, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0095","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1f8b821d8626b5c0b6a766f8ced4cffc73186fc151617aec65977b4d3d1c0bcb","family_id":"qf-conditional-direction-comparison","seed":202610047095,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[8,10],\"B\":[7,8,9,10],\"F\":\"powerset(S)\",\"S\":[7,8,9,10],\"denominator\":\"31\",\"operation\":\"compare_direction\",\"weights\":[[7,\"1\"],[8,\"4\"],[9,\"6\"],[10,\"20\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={7, 8, 9, 10}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 4/(31), 6/(31), 20/(31)], summing exactly to one. A={8, 10}, B={7, 8, 9, 10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=24/31 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0096","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2f51553e7d9935fa3c3d5f0bcbcb71447ca8664476f9bc9d37bd43d454bd2153","family_id":"qf-conditional-direction-comparison","seed":202610047096,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[9],\"B\":[10],\"F\":\"powerset(S)\",\"S\":[8,9,10,11],\"denominator\":\"26\",\"operation\":\"compare_direction\",\"weights\":[[8,\"2\"],[9,\"1\"],[10,\"2\"],[11,\"21\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={8, 9, 10, 11}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(26), 1/(26), 2/(26), 21/(26)], summing exactly to one. A={9}, B={10}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=0/1 and P(B|A)=0/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0097","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"61468c646931f17d29f77a128a184a241ed4bf7dbd671a7bba606e001bc3b845","family_id":"qf-conditional-direction-comparison","seed":202610047097,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[10,11],\"B\":[11,12],\"F\":\"powerset(S)\",\"S\":[9,10,11,12],\"denominator\":\"27\",\"operation\":\"compare_direction\",\"weights\":[[9,\"0\"],[10,\"2\"],[11,\"3\"],[12,\"22\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={9, 10, 11, 12}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(27), 2/(27), 3/(27), 22/(27)], summing exactly to one. A={10, 11}, B={11, 12}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=3/25 and P(B|A)=3/5, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0098","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9a109c809278a0ce1d81e1b48a72dd5a88f1f660e95b2c359635310b8ae0b14","family_id":"qf-conditional-direction-comparison","seed":202610047098,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[11,12],\"B\":[11,12],\"F\":\"powerset(S)\",\"S\":[10,11,12,13],\"denominator\":\"31\",\"operation\":\"compare_direction\",\"weights\":[[10,\"1\"],[11,\"3\"],[12,\"4\"],[13,\"23\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={10, 11, 12, 13}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [1/(31), 3/(31), 4/(31), 23/(31)], summing exactly to one. A={11, 12}, B={11, 12}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"Yes","explanation":"P(A|B)=1/1 and P(B|A)=1/1, so the answer is Yes. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0099","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4591c7ff531f60b9a01ab2f99143aff11d4edbe6bce55d32faba9dd577cb8161","family_id":"qf-conditional-direction-comparison","seed":202610047099,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[11,12,13,14],\"B\":[12,13],\"F\":\"powerset(S)\",\"S\":[11,12,13,14],\"denominator\":\"35\",\"operation\":\"compare_direction\",\"weights\":[[11,\"2\"],[12,\"4\"],[13,\"5\"],[14,\"24\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={11, 12, 13, 14}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [2/(35), 4/(35), 5/(35), 24/(35)], summing exactly to one. A={11, 12, 13, 14}, B={12, 13}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=1/1 and P(B|A)=9/35, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}
{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-advance-20261004-r07-conditional-probability-0100","batch_id":"batch-20261003-advance-20261004-r07-conditional-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1c9a7ce3962fc535f469055c9c76f22f2cd00dc7a9016f04c0e82a4d9182e17c","family_id":"qf-conditional-direction-comparison","seed":202610047100,"semantic_template":"qf-conditional-direction-comparison/{\"A\":[13,15],\"B\":[12,13,14,15],\"F\":\"powerset(S)\",\"S\":[12,13,14,15],\"denominator\":\"32\",\"operation\":\"compare_direction\",\"weights\":[[12,\"0\"],[13,\"1\"],[14,\"6\"],[15,\"25\"]]}"},"fields":{"title":"Comparing Two Directions of Conditional Probability","stem":"The nonempty finite sample space is S={12, 13, 14, 15}, with event family F=𝒫(S). In the displayed order of S, the nonnegative normalized weights are [0/(32), 1/(32), 6/(32), 25/(32)], summing exactly to one. A={13, 15}, B={12, 13, 14, 15}, and both P(A) and P(B) are positive. Are P(A|B) and P(B|A) equal?","answer":{"kind":"short_text","value":"No","explanation":"P(A|B)=13/16 and P(B|A)=1/1, so the answer is No. Each direction uses its own positive conditioning denominator. Their numerical comparison does not establish independence."},"skills":["Comparing Two Directions of Conditional Probability"],"verification_warning":"This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."}}