{"schema_version":"ocme-question-localization-v0.1","question_id":"q-20261003-r07-probability-0001","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5b00246fe0f4e8e1a25c92c3b7e1110704cc35749eb3f249b95bd3eae7990394","family_id":"qf-probability-uniform-event","seed":202610037001,"semantic_template":"qf-probability-uniform-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-12,-11],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-12, -11} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Explicit equiprobability permits |E|/|S|=0/2=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0002","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"33bc47036169163b875e285d45722db9db3612cd6f873abb38dd704e11793c2e","family_id":"qf-probability-uniform-event","seed":202610037002,"semantic_template":"qf-probability-uniform-event/{\"E\":[-11,-10,-9],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-11, -10, -9}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"Explicit equiprobability permits |E|/|S|=3/3=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0003","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fe778dbfcf6f4210d5e61522eaa7a8759636bbfdc2ee06b73312aa10d8a836c6","family_id":"qf-probability-uniform-event","seed":202610037003,"semantic_template":"qf-probability-uniform-event/{\"E\":[-10],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8,-7],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-10, -9, -8, -7} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-10}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/4","explanation":"Explicit equiprobability permits |E|/|S|=1/4=1/4. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0004","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4a62ca4d88fa7616abbacd1801ecf7b23147de9b8b5e64e8350487b5ec5ce17f","family_id":"qf-probability-uniform-event","seed":202610037004,"semantic_template":"qf-probability-uniform-event/{\"E\":[-5],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7,-6,-5],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-9, -8, -7, -6, -5} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-5}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/5","explanation":"Explicit equiprobability permits |E|/|S|=1/5=1/5. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0005","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f4c72e78a6d0bd1311a723ad7f262502e3b4ce7fdb70afad80917d2ad39aaa3d","family_id":"qf-probability-uniform-event","seed":202610037005,"semantic_template":"qf-probability-uniform-event/{\"E\":[-8,-6,-4],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6,-5,-4,-3],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-8, -7, -6, -5, -4, -3} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-8, -6, -4}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=3/6=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0006","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f373e15f43de7320239ec813945cd4715a260111eae99b1a5849bce81e046371","family_id":"qf-probability-uniform-event","seed":202610037006,"semantic_template":"qf-probability-uniform-event/{\"E\":[-7,-6],\"F\":\"powerset(S)\",\"S\":[-7,-6],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-7, -6} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-7, -6}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"Explicit equiprobability permits |E|/|S|=2/2=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0007","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3f0e81b4b9120666eef4abc064144fbd9e35375cbce0e871206820268032b67f","family_id":"qf-probability-uniform-event","seed":202610037007,"semantic_template":"qf-probability-uniform-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Explicit equiprobability permits |E|/|S|=0/3=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0008","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"840473ff07bbe55b49dc82c5e5b7e9da18017b0b44fdea84a4dc7dd92e7d8686","family_id":"qf-probability-uniform-event","seed":202610037008,"semantic_template":"qf-probability-uniform-event/{\"E\":[-5,-4,-3,-2],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3,-2],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-5, -4, -3, -2} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-5, -4, -3, -2}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"Explicit equiprobability permits |E|/|S|=4/4=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0009","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"09627b740705a00fb024358390dab835416d4f8da86dad79b7e6e6e1be1a23db","family_id":"qf-probability-uniform-event","seed":202610037009,"semantic_template":"qf-probability-uniform-event/{\"E\":[-4],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2,-1,0],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-4, -3, -2, -1, 0} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-4}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/5","explanation":"Explicit equiprobability permits |E|/|S|=1/5=1/5. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0010","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a6323070343f136a1196881c331e71f96292e8da3d86a44afcb24f0af508a30f","family_id":"qf-probability-uniform-event","seed":202610037010,"semantic_template":"qf-probability-uniform-event/{\"E\":[2],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1,0,1,2],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-3, -2, -1, 0, 1, 2} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={2}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/6","explanation":"Explicit equiprobability permits |E|/|S|=1/6=1/6. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0011","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b02e9a67dd1c416a04236f7edd67a3cd6d8ab36bfd3845dbf8ca602a799f989e","family_id":"qf-probability-uniform-event","seed":202610037011,"semantic_template":"qf-probability-uniform-event/{\"E\":[-2],\"F\":\"powerset(S)\",\"S\":[-2,-1],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-2, -1} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-2}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0012","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f63282d37e4167426382551dff1ba9c271d8002767486e90595777eda697aeea","family_id":"qf-probability-uniform-event","seed":202610037012,"semantic_template":"qf-probability-uniform-event/{\"E\":[-1,0],\"F\":\"powerset(S)\",\"S\":[-1,0,1],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={-1, 0}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"2/3","explanation":"Explicit equiprobability permits |E|/|S|=2/3=2/3. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0013","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"86d52cd78b9e1dd4d811d67f18771e37533951fb3eb40e9921ab57ed11547807","family_id":"qf-probability-uniform-event","seed":202610037013,"semantic_template":"qf-probability-uniform-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[0,1,2,3],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={0, 1, 2, 3} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Explicit equiprobability permits |E|/|S|=0/4=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0014","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"150e833565bbca935bddd52cf925cb62f8999f3a95c8ac6b6c5f198d0b15f338","family_id":"qf-probability-uniform-event","seed":202610037014,"semantic_template":"qf-probability-uniform-event/{\"E\":[1,2,3,4,5],\"F\":\"powerset(S)\",\"S\":[1,2,3,4,5],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={1, 2, 3, 4, 5} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={1, 2, 3, 4, 5}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"Explicit equiprobability permits |E|/|S|=5/5=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0015","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"856a5a1ef0aec73dce4a6aabb7e63ae70d12cf61982791295e6af02d8b29712c","family_id":"qf-probability-uniform-event","seed":202610037015,"semantic_template":"qf-probability-uniform-event/{\"E\":[2],\"F\":\"powerset(S)\",\"S\":[2,3,4,5,6,7],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={2, 3, 4, 5, 6, 7} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={2}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/6","explanation":"Explicit equiprobability permits |E|/|S|=1/6=1/6. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0016","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d06ae98e1c6492bb23326425ce5ebe00443df4906f86a277214d684227f1f0a6","family_id":"qf-probability-uniform-event","seed":202610037016,"semantic_template":"qf-probability-uniform-event/{\"E\":[4],\"F\":\"powerset(S)\",\"S\":[3,4],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={3, 4} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={4}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0017","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"f68d5cde6e178382e5e835e66ca927dd824d42b2b42c279562c6940aaa93b2d0","family_id":"qf-probability-uniform-event","seed":202610037017,"semantic_template":"qf-probability-uniform-event/{\"E\":[4,6],\"F\":\"powerset(S)\",\"S\":[4,5,6],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={4, 6}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"2/3","explanation":"Explicit equiprobability permits |E|/|S|=2/3=2/3. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0018","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b7821a32fc372d40a305fb9da5649f8e8e70c88207177ba8e3bbfc168de92e38","family_id":"qf-probability-uniform-event","seed":202610037018,"semantic_template":"qf-probability-uniform-event/{\"E\":[5,6],\"F\":\"powerset(S)\",\"S\":[5,6,7,8],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={5, 6, 7, 8} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={5, 6}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=2/4=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0019","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2708e486721efc7d3e7ff216a172506cbb20ae68625a779dc933c2c758efddb5","family_id":"qf-probability-uniform-event","seed":202610037019,"semantic_template":"qf-probability-uniform-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[6,7,8,9,10],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={6, 7, 8, 9, 10} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Explicit equiprobability permits |E|/|S|=0/5=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0020","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"4cfe5921a6c4b04bcfc52882534fe1aaab77526400a86230f7030c363e5f2821","family_id":"qf-probability-uniform-event","seed":202610037020,"semantic_template":"qf-probability-uniform-event/{\"E\":[7,8,9,10,11,12],\"F\":\"powerset(S)\",\"S\":[7,8,9,10,11,12],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={7, 8, 9, 10, 11, 12} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={7, 8, 9, 10, 11, 12}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/1","explanation":"Explicit equiprobability permits |E|/|S|=6/6=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0021","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ee7283123972e0c898e4ff1a511043e3e5dcb469c78b21b3baa37fac9dcd330b","family_id":"qf-probability-uniform-event","seed":202610037021,"semantic_template":"qf-probability-uniform-event/{\"E\":[8],\"F\":\"powerset(S)\",\"S\":[8,9],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={8, 9} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={8}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=1/2=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0022","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"271a0f4af978f7ee0441483d1380994fe61f2463d2368b5c6f6aba26171545c1","family_id":"qf-probability-uniform-event","seed":202610037022,"semantic_template":"qf-probability-uniform-event/{\"E\":[11],\"F\":\"powerset(S)\",\"S\":[9,10,11],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={11}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/3","explanation":"Explicit equiprobability permits |E|/|S|=1/3=1/3. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0023","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"67b5412c205ad3bf9b4f66f74f27a91f467a06caab6de29e0d5f31829776b337","family_id":"qf-probability-uniform-event","seed":202610037023,"semantic_template":"qf-probability-uniform-event/{\"E\":[10,12],\"F\":\"powerset(S)\",\"S\":[10,11,12,13],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={10, 11, 12, 13} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={10, 12}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"1/2","explanation":"Explicit equiprobability permits |E|/|S|=2/4=1/2. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0024","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e7468363dcf4a4863470bfb6495d6c1519b87d4646f7ef0fc7824e0679ba6794","family_id":"qf-probability-uniform-event","seed":202610037024,"semantic_template":"qf-probability-uniform-event/{\"E\":[11,12],\"F\":\"powerset(S)\",\"S\":[11,12,13,14,15],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={11, 12, 13, 14, 15} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E={11, 12}, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"2/5","explanation":"Explicit equiprobability permits |E|/|S|=2/5=2/5. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0025","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6d126920f2d894bf3557af25bd68de445d9d9b361082f5cd8b36171391975baf","family_id":"qf-probability-uniform-event","seed":202610037025,"semantic_template":"qf-probability-uniform-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[12,13,14,15,16,17],\"equiprobable\":true,\"operation\":\"uniform_event\"}"},"fields":{"title":"Event Probability with Explicit Equiprobability","stem":"The nonempty finite sample space is S={12, 13, 14, 15, 16, 17} and the event family is F=𝒫(S). All elementary outcomes are explicitly equiprobable. For E=∅, find P(E) exactly as a reduced fraction with positive denominator.","answer":{"kind":"expression","value":"0/1","explanation":"Explicit equiprobability permits |E|/|S|=0/6=0/1. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Event 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-r07-probability-0026","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c2c4942a07015dea6a814763c78538905d2c028ab7e45fec9bed31458705bea2","family_id":"qf-probability-weighted-event","seed":202610037026,"semantic_template":"qf-probability-weighted-event/{\"E\":[-10],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10],\"denominator\":\"2\",\"operation\":\"weighted_event\",\"weights\":[[-12,\"1\"],[-11,\"1\"],[-10,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(2), 1/(2), 0/(2)] and E={-10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0027","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b47a4b0eebad2d05a5d14a6f00958dfae3a10194d37a7afab5b08e7cbcb3572","family_id":"qf-probability-weighted-event","seed":202610037027,"semantic_template":"qf-probability-weighted-event/{\"E\":[-11,-10,-9],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9],\"denominator\":\"5\",\"operation\":\"weighted_event\",\"weights\":[[-11,\"2\"],[-10,\"2\"],[-9,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(5), 2/(5), 1/(5)] and E={-11, -10, -9}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0028","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b4c86539642ceb2dc24d8d7d8688257c9e88ba6b4121a7164937730d7b4aebb1","family_id":"qf-probability-weighted-event","seed":202610037028,"semantic_template":"qf-probability-weighted-event/{\"E\":[-10],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8],\"denominator\":\"8\",\"operation\":\"weighted_event\",\"weights\":[[-10,\"3\"],[-9,\"3\"],[-8,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(8), 3/(8), 2/(8)] and E={-10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"3/8","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 3/8. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0029","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a63f5fe08bb31d5e08e2e0f90d3ae6ccfcf77bf93e767983aaf9fe9422bcd155","family_id":"qf-probability-weighted-event","seed":202610037029,"semantic_template":"qf-probability-weighted-event/{\"E\":[-7],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7],\"denominator\":\"2\",\"operation\":\"weighted_event\",\"weights\":[[-9,\"1\"],[-8,\"1\"],[-7,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [4/(8), 4/(8), 0/(8)] and E={-7}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0030","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"35b7b473bdd9a0caf761b65b3c39bcc8847c4c517801fb67f988a3587045be77","family_id":"qf-probability-weighted-event","seed":202610037030,"semantic_template":"qf-probability-weighted-event/{\"E\":[-8,-7],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6],\"denominator\":\"11\",\"operation\":\"weighted_event\",\"weights\":[[-8,\"5\"],[-7,\"5\"],[-6,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [5/(11), 5/(11), 1/(11)] and E={-8, -7}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"10/11","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 10/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0031","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0263938fe424a868e0df7bbaee0b07cee167ce8f3a5c0f1dafe6e5987a7a9394","family_id":"qf-probability-weighted-event","seed":202610037031,"semantic_template":"qf-probability-weighted-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5],\"denominator\":\"9\",\"operation\":\"weighted_event\",\"weights\":[[-7,\"6\"],[-6,\"1\"],[-5,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [6/(9), 1/(9), 2/(9)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0032","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6a3c48e9ae70cce77b321ec8c03680225d93009a8c8e492a391ccbd04799596e","family_id":"qf-probability-weighted-event","seed":202610037032,"semantic_template":"qf-probability-weighted-event/{\"E\":[-6,-5,-4],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4],\"denominator\":\"9\",\"operation\":\"weighted_event\",\"weights\":[[-6,\"7\"],[-5,\"2\"],[-4,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [7/(9), 2/(9), 0/(9)] and E={-6, -5, -4}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0033","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5eec4877106aafc3e5c329b007d8de32a98f1e9400ec2a1665f3aa8d2783aefc","family_id":"qf-probability-weighted-event","seed":202610037033,"semantic_template":"qf-probability-weighted-event/{\"E\":[-5],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3],\"denominator\":\"12\",\"operation\":\"weighted_event\",\"weights\":[[-5,\"8\"],[-4,\"3\"],[-3,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [8/(12), 3/(12), 1/(12)] and E={-5}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"2/3","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 2/3. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0034","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"141d92fdfaf54d56bdaf7bd356287c08621b51e4f0cf2a40b56e438629185e21","family_id":"qf-probability-weighted-event","seed":202610037034,"semantic_template":"qf-probability-weighted-event/{\"E\":[-2],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2],\"denominator\":\"15\",\"operation\":\"weighted_event\",\"weights\":[[-4,\"9\"],[-3,\"4\"],[-2,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [9/(15), 4/(15), 2/(15)] and E={-2}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"2/15","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 2/15. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0035","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c0de2e34dca08bce3f6517ada5267c2e92ef51801da4f874ef6ecfd8770f00aa","family_id":"qf-probability-weighted-event","seed":202610037035,"semantic_template":"qf-probability-weighted-event/{\"E\":[-3,-2],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1],\"denominator\":\"3\",\"operation\":\"weighted_event\",\"weights\":[[-3,\"2\"],[-2,\"1\"],[-1,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [10/(15), 5/(15), 0/(15)] and E={-3, -2}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0036","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9012dbb0d7abd584955222e386983b8e835b8f808fc5fc9608b1f1136b0b00c9","family_id":"qf-probability-weighted-event","seed":202610037036,"semantic_template":"qf-probability-weighted-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-2,-1,0],\"denominator\":\"13\",\"operation\":\"weighted_event\",\"weights\":[[-2,\"11\"],[-1,\"1\"],[0,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the weights are [11/(13), 1/(13), 1/(13)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0037","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6131e4e08ccc925185530a7ed9a0d51a2e58d44fdebe822c6d42eb44dd6f7d1","family_id":"qf-probability-weighted-event","seed":202610037037,"semantic_template":"qf-probability-weighted-event/{\"E\":[-1,0,1],\"F\":\"powerset(S)\",\"S\":[-1,0,1],\"denominator\":\"8\",\"operation\":\"weighted_event\",\"weights\":[[-1,\"6\"],[0,\"1\"],[1,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [12/(16), 2/(16), 2/(16)] and E={-1, 0, 1}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0038","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fabf0c60f0a7e79a4a326b99687637cb1e3c60a580ff4d76a337a114523b81dd","family_id":"qf-probability-weighted-event","seed":202610037038,"semantic_template":"qf-probability-weighted-event/{\"E\":[0],\"F\":\"powerset(S)\",\"S\":[0,1,2],\"denominator\":\"16\",\"operation\":\"weighted_event\",\"weights\":[[0,\"13\"],[1,\"3\"],[2,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [13/(16), 3/(16), 0/(16)] and E={0}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"13/16","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 13/16. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0039","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c13e09884ae729b317c5967877a0c182775f431112ddcc00db2a35f04796607e","family_id":"qf-probability-weighted-event","seed":202610037039,"semantic_template":"qf-probability-weighted-event/{\"E\":[3],\"F\":\"powerset(S)\",\"S\":[1,2,3],\"denominator\":\"19\",\"operation\":\"weighted_event\",\"weights\":[[1,\"14\"],[2,\"4\"],[3,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [14/(19), 4/(19), 1/(19)] and E={3}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/19","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/19. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0040","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5f27d4c03fb0c0644196afbbd9d1a15d28af26b6169ba54d71424b592b481d2e","family_id":"qf-probability-weighted-event","seed":202610037040,"semantic_template":"qf-probability-weighted-event/{\"E\":[2,3],\"F\":\"powerset(S)\",\"S\":[2,3,4],\"denominator\":\"22\",\"operation\":\"weighted_event\",\"weights\":[[2,\"15\"],[3,\"5\"],[4,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [15/(22), 5/(22), 2/(22)] and E={2, 3}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"10/11","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 10/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0041","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"babaa34504ba92b3ec0a2fc9145d8e494b214e5a3e58b3a02532637dc6d43187","family_id":"qf-probability-weighted-event","seed":202610037041,"semantic_template":"qf-probability-weighted-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[3,4,5],\"denominator\":\"17\",\"operation\":\"weighted_event\",\"weights\":[[3,\"16\"],[4,\"1\"],[5,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [16/(17), 1/(17), 0/(17)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0042","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"87f2bbe71c61e302e3356fed161076fbe6a28b0a4ae870f5eb262610b74dbdb8","family_id":"qf-probability-weighted-event","seed":202610037042,"semantic_template":"qf-probability-weighted-event/{\"E\":[4,5,6],\"F\":\"powerset(S)\",\"S\":[4,5,6],\"denominator\":\"20\",\"operation\":\"weighted_event\",\"weights\":[[4,\"17\"],[5,\"2\"],[6,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [17/(20), 2/(20), 1/(20)] and E={4, 5, 6}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0043","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"58ac307e42019b4e0a3b83b33d6d43b903409260e3769bca507cfcba7e40b83b","family_id":"qf-probability-weighted-event","seed":202610037043,"semantic_template":"qf-probability-weighted-event/{\"E\":[5],\"F\":\"powerset(S)\",\"S\":[5,6,7],\"denominator\":\"23\",\"operation\":\"weighted_event\",\"weights\":[[5,\"18\"],[6,\"3\"],[7,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [18/(23), 3/(23), 2/(23)] and E={5}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"18/23","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 18/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0044","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"15d165d87db140872cc76c87e1231b50f7aa566cf1a6ea09529e1979958cee3e","family_id":"qf-probability-weighted-event","seed":202610037044,"semantic_template":"qf-probability-weighted-event/{\"E\":[8],\"F\":\"powerset(S)\",\"S\":[6,7,8],\"denominator\":\"23\",\"operation\":\"weighted_event\",\"weights\":[[6,\"19\"],[7,\"4\"],[8,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [19/(23), 4/(23), 0/(23)] and E={8}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0045","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5690c516c56140fa49780d2308b77e5edf4d08368a7a4fb2b43b9c0cb27b5c2c","family_id":"qf-probability-weighted-event","seed":202610037045,"semantic_template":"qf-probability-weighted-event/{\"E\":[7,8],\"F\":\"powerset(S)\",\"S\":[7,8,9],\"denominator\":\"26\",\"operation\":\"weighted_event\",\"weights\":[[7,\"20\"],[8,\"5\"],[9,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [20/(26), 5/(26), 1/(26)] and E={7, 8}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"25/26","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 25/26. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0046","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"88eaecac27254336e157a6a76587292002d4bd0dabcd3d42beadd458e856f841","family_id":"qf-probability-weighted-event","seed":202610037046,"semantic_template":"qf-probability-weighted-event/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[8,9,10],\"denominator\":\"24\",\"operation\":\"weighted_event\",\"weights\":[[8,\"21\"],[9,\"1\"],[10,\"2\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [21/(24), 1/(24), 2/(24)] and E=∅. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0047","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e6c4b7ada818baf9bcfb398f2db6ad38e0cae733079c533efb24c9609100acd6","family_id":"qf-probability-weighted-event","seed":202610037047,"semantic_template":"qf-probability-weighted-event/{\"E\":[9,10,11],\"F\":\"powerset(S)\",\"S\":[9,10,11],\"denominator\":\"12\",\"operation\":\"weighted_event\",\"weights\":[[9,\"11\"],[10,\"1\"],[11,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the weights are [22/(24), 2/(24), 0/(24)] and E={9, 10, 11}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0048","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6899a648ff53203ed5342f7b29de37c6b6ed03d55b768edb71c03388a1afbc63","family_id":"qf-probability-weighted-event","seed":202610037048,"semantic_template":"qf-probability-weighted-event/{\"E\":[10],\"F\":\"powerset(S)\",\"S\":[10,11,12],\"denominator\":\"27\",\"operation\":\"weighted_event\",\"weights\":[[10,\"23\"],[11,\"3\"],[12,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the weights are [23/(27), 3/(27), 1/(27)] and E={10}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"23/27","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 23/27. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0049","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8bd97066a8ee7bdcbca10fabb7933e558053d6d02a0359bec6cf48ac73b7ee28","family_id":"qf-probability-weighted-event","seed":202610037049,"semantic_template":"qf-probability-weighted-event/{\"E\":[13],\"F\":\"powerset(S)\",\"S\":[11,12,13],\"denominator\":\"15\",\"operation\":\"weighted_event\",\"weights\":[[11,\"12\"],[12,\"2\"],[13,\"1\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the weights are [24/(30), 4/(30), 2/(30)] and E={13}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/15","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/15. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0050","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"656048c542ff54d64a3b45b5c6d750e2debff42ef03ba2ad6762f44bdb1ba22a","family_id":"qf-probability-weighted-event","seed":202610037050,"semantic_template":"qf-probability-weighted-event/{\"E\":[12,13],\"F\":\"powerset(S)\",\"S\":[12,13,14],\"denominator\":\"6\",\"operation\":\"weighted_event\",\"weights\":[[12,\"5\"],[13,\"1\"],[14,\"0\"]]}"},"fields":{"title":"Weighted Event Probability without Equiprobability","stem":"The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the weights are [25/(30), 5/(30), 0/(30)] and E={12, 13}. Find P(E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the event outcomes. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Weighted Event Probability without 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-r07-probability-0051","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"68b2ea5fd484d8760040f3a3e99466fd7a213a052124a2eb84cc07e3f0b4b0b7","family_id":"qf-probability-model-validity","seed":202610037051,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-12,-11,-10],\"denominator\":\"2\",\"operation\":\"model_validity\",\"weights\":[[-12,\"1\"],[-11,\"1\"],[-10,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [1/(2), 1/(2), 0/(2)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 2 and the common denominator is 2. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0052","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"e87d62d01ddb6a0b99228a96b13d04d40c5529e4029c492a714612458f1a0403","family_id":"qf-probability-model-validity","seed":202610037052,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-11,-10,-9],\"denominator\":\"5\",\"operation\":\"model_validity\",\"weights\":[[-11,\"-2\"],[-10,\"2\"],[-9,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-2/(5), 2/(5), 1/(5)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 1 and the common denominator is 5. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0053","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c061793a453167d77feb1a97be4f93ed62ffef30fa04a91f5cb55d4212faed2b","family_id":"qf-probability-model-validity","seed":202610037053,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-10,-9,-8],\"denominator\":\"9\",\"operation\":\"model_validity\",\"weights\":[[-10,\"3\"],[-9,\"3\"],[-8,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [3/(9), 3/(9), 2/(9)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 8 and the common denominator is 9. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0054","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"40a0c4291256fe914d9321f755034f10c954b4481dca7819720dc26051830673","family_id":"qf-probability-model-validity","seed":202610037054,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-9,-8,-7],\"denominator\":\"0\",\"operation\":\"model_validity\",\"weights\":[[-9,\"1\"],[-8,\"1\"],[-7,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [4/(0), 4/(0), 0/(0)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 8 and the common denominator is 0. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0055","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b634293f98ddfc4b8b88b56d8242368f1386ad6c1033d8e9ce411249fa4b99ad","family_id":"qf-probability-model-validity","seed":202610037055,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-8,-7,-6],\"denominator\":\"11\",\"operation\":\"model_validity\",\"weights\":[[-8,\"5\"],[-7,\"5\"],[-6,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [5/(11), 5/(11), 1/(11)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 11 and the common denominator is 11. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0056","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1d8d7aa6a993e2e057d1115b7f753a958a701cf54eb5eda7563fe9db5b6edee1","family_id":"qf-probability-model-validity","seed":202610037056,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-7,-6,-5],\"denominator\":\"9\",\"operation\":\"model_validity\",\"weights\":[[-7,\"-6\"],[-6,\"1\"],[-5,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-6/(9), 1/(9), 2/(9)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to -3 and the common denominator is 9. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0057","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fed965a584e1f5f14d07aad777a695a7a2a63fc36216ce8f68e86e2c88a4d7f","family_id":"qf-probability-model-validity","seed":202610037057,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-6,-5,-4],\"denominator\":\"10\",\"operation\":\"model_validity\",\"weights\":[[-6,\"7\"],[-5,\"2\"],[-4,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [7/(10), 2/(10), 0/(10)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 9 and the common denominator is 10. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0058","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"fa04dcf189753e5c1f69a88111625909d7b11b99adeaf83ee58b785ca3e30c2c","family_id":"qf-probability-model-validity","seed":202610037058,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-5,-4,-3],\"denominator\":\"-12\",\"operation\":\"model_validity\",\"weights\":[[-5,\"8\"],[-4,\"3\"],[-3,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [8/(-12), 3/(-12), 1/(-12)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 12 and the common denominator is -12. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0059","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"74326a9a004bc44c8d93c786eaf04d659dab0cf612b21bb7693af33f806c2a37","family_id":"qf-probability-model-validity","seed":202610037059,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-4,-3,-2],\"denominator\":\"15\",\"operation\":\"model_validity\",\"weights\":[[-4,\"9\"],[-3,\"4\"],[-2,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [9/(15), 4/(15), 2/(15)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 15 and the common denominator is 15. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0060","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5fc6b4c81a52c1c1835842fd54f67283bb681a75882651256e6d81eaa57b01a2","family_id":"qf-probability-model-validity","seed":202610037060,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-3,-2,-1],\"denominator\":\"3\",\"operation\":\"model_validity\",\"weights\":[[-3,\"-2\"],[-2,\"1\"],[-1,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-10/(15), 5/(15), 0/(15)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to -5 and the common denominator is 15. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0061","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d486ff38f06ba34aa263d1e8bef82be5377ff1e4363d5f47f2f94c1c2cee9388","family_id":"qf-probability-model-validity","seed":202610037061,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-2,-1,0],\"denominator\":\"14\",\"operation\":\"model_validity\",\"weights\":[[-2,\"11\"],[-1,\"1\"],[0,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [11/(14), 1/(14), 1/(14)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 13 and the common denominator is 14. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0062","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"234c5e4bc3221d830d8687459754a544bc241589cda45ff898f76fed8d94cfa8","family_id":"qf-probability-model-validity","seed":202610037062,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[-1,0,1],\"denominator\":\"-8\",\"operation\":\"model_validity\",\"weights\":[[-1,\"6\"],[0,\"1\"],[1,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [12/(-16), 2/(-16), 2/(-16)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 16 and the common denominator is -16. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0063","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"359428f521f20c09864064f59fc8a7af4b13e3e1c2abeeb637702169e7ca396a","family_id":"qf-probability-model-validity","seed":202610037063,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[0,1,2],\"denominator\":\"16\",\"operation\":\"model_validity\",\"weights\":[[0,\"13\"],[1,\"3\"],[2,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [13/(16), 3/(16), 0/(16)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 16 and the common denominator is 16. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0064","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"011c03b279909f9adf5e666e6ea661b3a462845e04f4203fd890d08fdee6d022","family_id":"qf-probability-model-validity","seed":202610037064,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[1,2,3],\"denominator\":\"19\",\"operation\":\"model_validity\",\"weights\":[[1,\"-14\"],[2,\"4\"],[3,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-14/(19), 4/(19), 1/(19)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to -9 and the common denominator is 19. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0065","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"23d7e2469f4d1094fe234965024b1c5393beb0a38a9fd79190f848d2dc9cd2d7","family_id":"qf-probability-model-validity","seed":202610037065,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[2,3,4],\"denominator\":\"23\",\"operation\":\"model_validity\",\"weights\":[[2,\"15\"],[3,\"5\"],[4,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [15/(23), 5/(23), 2/(23)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 22 and the common denominator is 23. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0066","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a16cfd9e1db145f295f06bb35283aa579805831998d2b3378949661b11b0110c","family_id":"qf-probability-model-validity","seed":202610037066,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[3,4,5],\"denominator\":\"-17\",\"operation\":\"model_validity\",\"weights\":[[3,\"16\"],[4,\"1\"],[5,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [16/(-17), 1/(-17), 0/(-17)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 17 and the common denominator is -17. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0067","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"9dd21f2c771d71276e10b57094028d380ca77bd620922a002e7ee8e2b6754da0","family_id":"qf-probability-model-validity","seed":202610037067,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[4,5,6],\"denominator\":\"20\",\"operation\":\"model_validity\",\"weights\":[[4,\"17\"],[5,\"2\"],[6,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [17/(20), 2/(20), 1/(20)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 20 and the common denominator is 20. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0068","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a1154cdca0e2ad5b4ee0b8a1ac63aeb4a6d13881ebb8f0d9018b826eb369d1f3","family_id":"qf-probability-model-validity","seed":202610037068,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[5,6,7],\"denominator\":\"23\",\"operation\":\"model_validity\",\"weights\":[[5,\"-18\"],[6,\"3\"],[7,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-18/(23), 3/(23), 2/(23)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to -13 and the common denominator is 23. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0069","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"cc8978ee8a9f697a8cbde37e5c2b906cf89b3f8f46bff0b5d3d0ff4bfee2f21c","family_id":"qf-probability-model-validity","seed":202610037069,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[6,7,8],\"denominator\":\"24\",\"operation\":\"model_validity\",\"weights\":[[6,\"19\"],[7,\"4\"],[8,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [19/(24), 4/(24), 0/(24)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 23 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0070","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"513b5fa1d12698207b28a984f10e6918b1337526cc81f510b055fdb77a1bc973","family_id":"qf-probability-model-validity","seed":202610037070,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[7,8,9],\"denominator\":\"-26\",\"operation\":\"model_validity\",\"weights\":[[7,\"20\"],[8,\"5\"],[9,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [20/(-26), 5/(-26), 1/(-26)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 26 and the common denominator is -26. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0071","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"bcb15b18f33e67641233e6ddd0c4f20823aee02b2866020e08bc813ba1aa5e67","family_id":"qf-probability-model-validity","seed":202610037071,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[8,9,10],\"denominator\":\"24\",\"operation\":\"model_validity\",\"weights\":[[8,\"21\"],[9,\"1\"],[10,\"2\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [21/(24), 1/(24), 2/(24)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 24 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0072","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"8ee457277c9fc202363e53ae39368b0ce5e07401efd40a137c5aa7efe1340c60","family_id":"qf-probability-model-validity","seed":202610037072,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[9,10,11],\"denominator\":\"12\",\"operation\":\"model_validity\",\"weights\":[[9,\"-11\"],[10,\"1\"],[11,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [-22/(24), 2/(24), 0/(24)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to -20 and the common denominator is 24. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0073","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"49a67255a2f75f40af2e383e646dad32a746352eb141b62d299410fa0d1876c1","family_id":"qf-probability-model-validity","seed":202610037073,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[10,11,12],\"denominator\":\"28\",\"operation\":\"model_validity\",\"weights\":[[10,\"23\"],[11,\"3\"],[12,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [23/(28), 3/(28), 1/(28)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 27 and the common denominator is 28. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0074","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"625928b9459e9dd62fcd026f03b0ae73909b7e599aa45c7a8a68049081bd9c58","family_id":"qf-probability-model-validity","seed":202610037074,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[11,12,13],\"denominator\":\"-15\",\"operation\":\"model_validity\",\"weights\":[[11,\"12\"],[12,\"2\"],[13,\"1\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [24/(-30), 4/(-30), 2/(-30)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"No","explanation":"The weight numerators sum to 30 and the common denominator is -30. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is No. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0075","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"5509902f0ee29673cd9913a2f6f33e41b395b7e32dc56047817458dc6bdb4d7f","family_id":"qf-probability-model-validity","seed":202610037075,"semantic_template":"qf-probability-model-validity/{\"F\":\"powerset(S)\",\"S\":[12,13,14],\"denominator\":\"6\",\"operation\":\"model_validity\",\"weights\":[[12,\"5\"],[13,\"1\"],[14,\"0\"]]}"},"fields":{"title":"Validity of a Candidate Finite Probability Model","stem":"The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the proposed weights are [25/(30), 5/(30), 0/(30)]. Do they define a valid probability model with nonnegative weights summing to one?","answer":{"kind":"short_text","value":"Yes","explanation":"The weight numerators sum to 30 and the common denominator is 30. Check a positive denominator, nonnegative weights, and a total of exactly one; the answer is Yes. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Validity of a Candidate Finite Probability 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-r07-probability-0076","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"d93836b98c45c43bc7790540b972c8acc767b73c5c0506e088072eac5383d9f9","family_id":"qf-probability-complement","seed":202610037076,"semantic_template":"qf-probability-complement/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-12,-11,-10],\"denominator\":\"3\",\"operation\":\"complement\",\"weights\":[[-12,\"0\"],[-11,\"1\"],[-10,\"2\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-12, -11, -10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(3), 1/(3), 2/(3)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0077","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c34b17d027b598fd66d3b4146d2430cea26c1a840ab2c067a9d94f25a2ac2319","family_id":"qf-probability-complement","seed":202610037077,"semantic_template":"qf-probability-complement/{\"E\":[-11,-10,-9],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9],\"denominator\":\"6\",\"operation\":\"complement\",\"weights\":[[-11,\"1\"],[-10,\"2\"],[-9,\"3\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-11, -10, -9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(6), 2/(6), 3/(6)] and E={-11, -10, -9}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0078","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"febb37c40b6c4e2383b3182caeb32db7d38939332a2d8081dfc631a93b2b3f7d","family_id":"qf-probability-complement","seed":202610037078,"semantic_template":"qf-probability-complement/{\"E\":[-10],\"F\":\"powerset(S)\",\"S\":[-10,-9,-8],\"denominator\":\"9\",\"operation\":\"complement\",\"weights\":[[-10,\"2\"],[-9,\"3\"],[-8,\"4\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-10, -9, -8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(9), 3/(9), 4/(9)] and E={-10}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"7/9","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/9. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0079","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3e93a824e07582ea542e8fd5785ef58bce2bd113a80d22982e3055d5a9427803","family_id":"qf-probability-complement","seed":202610037079,"semantic_template":"qf-probability-complement/{\"E\":[-7],\"F\":\"powerset(S)\",\"S\":[-9,-8,-7],\"denominator\":\"12\",\"operation\":\"complement\",\"weights\":[[-9,\"3\"],[-8,\"4\"],[-7,\"5\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-9, -8, -7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(12), 4/(12), 5/(12)] and E={-7}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"7/12","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/12. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0080","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c3dad56395e95f2163500460b310032e6e65e9c5ce55b99e50dcfbda48bb9515","family_id":"qf-probability-complement","seed":202610037080,"semantic_template":"qf-probability-complement/{\"E\":[-8,-7],\"F\":\"powerset(S)\",\"S\":[-8,-7,-6],\"denominator\":\"11\",\"operation\":\"complement\",\"weights\":[[-8,\"0\"],[-7,\"5\"],[-6,\"6\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-8, -7, -6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(11), 5/(11), 6/(11)] and E={-8, -7}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"6/11","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 6/11. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0081","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"aabb3fd69e4ef90d680c1498eebb3506deb6dae4672fef66ace03c85c0d87f31","family_id":"qf-probability-complement","seed":202610037081,"semantic_template":"qf-probability-complement/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-7,-6,-5],\"denominator\":\"9\",\"operation\":\"complement\",\"weights\":[[-7,\"1\"],[-6,\"1\"],[-5,\"7\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-7, -6, -5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(9), 1/(9), 7/(9)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0082","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"c998764c4c0084dbf8197c2fa599103e740ba47329626cc5a238cb631190ad78","family_id":"qf-probability-complement","seed":202610037082,"semantic_template":"qf-probability-complement/{\"E\":[-6,-5,-4],\"F\":\"powerset(S)\",\"S\":[-6,-5,-4],\"denominator\":\"6\",\"operation\":\"complement\",\"weights\":[[-6,\"1\"],[-5,\"1\"],[-4,\"4\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-6, -5, -4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(12), 2/(12), 8/(12)] and E={-6, -5, -4}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0083","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2bdba829f2bed566f2a68d7ae128422d4fc98c7fde797f95de1b7b5198e093e4","family_id":"qf-probability-complement","seed":202610037083,"semantic_template":"qf-probability-complement/{\"E\":[-5],\"F\":\"powerset(S)\",\"S\":[-5,-4,-3],\"denominator\":\"5\",\"operation\":\"complement\",\"weights\":[[-5,\"1\"],[-4,\"1\"],[-3,\"3\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-5, -4, -3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(15), 3/(15), 9/(15)] and E={-5}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"4/5","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 4/5. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0084","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2b3bee2e276b662676259622215f6d9c140cd2088a3533d2536e16d1f8d43bd5","family_id":"qf-probability-complement","seed":202610037084,"semantic_template":"qf-probability-complement/{\"E\":[-2],\"F\":\"powerset(S)\",\"S\":[-4,-3,-2],\"denominator\":\"7\",\"operation\":\"complement\",\"weights\":[[-4,\"0\"],[-3,\"2\"],[-2,\"5\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-4, -3, -2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(14), 4/(14), 10/(14)] and E={-2}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"2/7","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 2/7. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0085","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"1fcbd4003810592712c90594d15c7f66cf7e70e8358a30524af5c97e4c851750","family_id":"qf-probability-complement","seed":202610037085,"semantic_template":"qf-probability-complement/{\"E\":[-3,-2],\"F\":\"powerset(S)\",\"S\":[-3,-2,-1],\"denominator\":\"17\",\"operation\":\"complement\",\"weights\":[[-3,\"1\"],[-2,\"5\"],[-1,\"11\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-3, -2, -1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(17), 5/(17), 11/(17)] and E={-3, -2}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"11/17","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 11/17. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0086","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"a588d093ec8127c26473bd8ec862b300d6b26aecc878e566bd36e659bdffcb4b","family_id":"qf-probability-complement","seed":202610037086,"semantic_template":"qf-probability-complement/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[-2,-1,0],\"denominator\":\"15\",\"operation\":\"complement\",\"weights\":[[-2,\"2\"],[-1,\"1\"],[0,\"12\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-2, -1, 0} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(15), 1/(15), 12/(15)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0087","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0b522c285c988b52ce4b902ec3d3bd055855a4c6a871af6beefd5cfa9e7550d9","family_id":"qf-probability-complement","seed":202610037087,"semantic_template":"qf-probability-complement/{\"E\":[-1,0,1],\"F\":\"powerset(S)\",\"S\":[-1,0,1],\"denominator\":\"18\",\"operation\":\"complement\",\"weights\":[[-1,\"3\"],[0,\"2\"],[1,\"13\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={-1, 0, 1} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(18), 2/(18), 13/(18)] and E={-1, 0, 1}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0088","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3ed69a13342e54c30d85bf33aa4cb1b111b9301b08da755ebed7f01cf1b1ba21","family_id":"qf-probability-complement","seed":202610037088,"semantic_template":"qf-probability-complement/{\"E\":[0],\"F\":\"powerset(S)\",\"S\":[0,1,2],\"denominator\":\"17\",\"operation\":\"complement\",\"weights\":[[0,\"0\"],[1,\"3\"],[2,\"14\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={0, 1, 2} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(17), 3/(17), 14/(17)] and E={0}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0089","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"3033d950be8979abd6b1171444f6720835ce54ca4786b5a9416a6dc294088940","family_id":"qf-probability-complement","seed":202610037089,"semantic_template":"qf-probability-complement/{\"E\":[3],\"F\":\"powerset(S)\",\"S\":[1,2,3],\"denominator\":\"20\",\"operation\":\"complement\",\"weights\":[[1,\"1\"],[2,\"4\"],[3,\"15\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={1, 2, 3} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(20), 4/(20), 15/(20)] and E={3}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/4","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/4. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0090","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"2c042c152ccbe94eb8f6a3453756e6e6305b8f014d52c686824d141eb21603f2","family_id":"qf-probability-complement","seed":202610037090,"semantic_template":"qf-probability-complement/{\"E\":[2,3],\"F\":\"powerset(S)\",\"S\":[2,3,4],\"denominator\":\"23\",\"operation\":\"complement\",\"weights\":[[2,\"2\"],[3,\"5\"],[4,\"16\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={2, 3, 4} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(23), 5/(23), 16/(23)] and E={2, 3}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"16/23","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 16/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0091","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"509049b2dcf898ed09d072ccd9d58070609694853e3bdc500c285c3bd8080f05","family_id":"qf-probability-complement","seed":202610037091,"semantic_template":"qf-probability-complement/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[3,4,5],\"denominator\":\"21\",\"operation\":\"complement\",\"weights\":[[3,\"3\"],[4,\"1\"],[5,\"17\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={3, 4, 5} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(21), 1/(21), 17/(21)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0092","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"b9cff3a37f2706e2a63272cf8ef179ad723f86118c504cd14b65ee9e9d0c5e39","family_id":"qf-probability-complement","seed":202610037092,"semantic_template":"qf-probability-complement/{\"E\":[4,5,6],\"F\":\"powerset(S)\",\"S\":[4,5,6],\"denominator\":\"10\",\"operation\":\"complement\",\"weights\":[[4,\"0\"],[5,\"1\"],[6,\"9\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={4, 5, 6} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(20), 2/(20), 18/(20)] and E={4, 5, 6}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0093","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"499fb8ecd7ef7f5b8b6382af39ea1a5974382d928f51c0a4b27b96357e0098c0","family_id":"qf-probability-complement","seed":202610037093,"semantic_template":"qf-probability-complement/{\"E\":[5],\"F\":\"powerset(S)\",\"S\":[5,6,7],\"denominator\":\"23\",\"operation\":\"complement\",\"weights\":[[5,\"1\"],[6,\"3\"],[7,\"19\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={5, 6, 7} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(23), 3/(23), 19/(23)] and E={5}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"22/23","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 22/23. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0094","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"6e949b71c917b6bf3c1f2ac8b36a0339c9569237c7ade9d8f80b3bf6fd2f0ba4","family_id":"qf-probability-complement","seed":202610037094,"semantic_template":"qf-probability-complement/{\"E\":[8],\"F\":\"powerset(S)\",\"S\":[6,7,8],\"denominator\":\"13\",\"operation\":\"complement\",\"weights\":[[6,\"1\"],[7,\"2\"],[8,\"10\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={6, 7, 8} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(26), 4/(26), 20/(26)] and E={8}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"3/13","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 3/13. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0095","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"ef087ce60e81e142aaec706a2648ebfb921e759d617c8024abe9dcf6f4f4ef29","family_id":"qf-probability-complement","seed":202610037095,"semantic_template":"qf-probability-complement/{\"E\":[7,8],\"F\":\"powerset(S)\",\"S\":[7,8,9],\"denominator\":\"29\",\"operation\":\"complement\",\"weights\":[[7,\"3\"],[8,\"5\"],[9,\"21\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={7, 8, 9} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(29), 5/(29), 21/(29)] and E={7, 8}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"21/29","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 21/29. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0096","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"303a2585e5f939b3b4084ef2f4cf3640d14d37e5aa403fefa041087e1db8ef08","family_id":"qf-probability-complement","seed":202610037096,"semantic_template":"qf-probability-complement/{\"E\":[],\"F\":\"powerset(S)\",\"S\":[8,9,10],\"denominator\":\"23\",\"operation\":\"complement\",\"weights\":[[8,\"0\"],[9,\"1\"],[10,\"22\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={8, 9, 10} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(23), 1/(23), 22/(23)] and E=∅. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"1/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 1/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0097","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"25aab09df0bc6ae2e1c61c585673e244b931615824f1221447f63ee984fde8f7","family_id":"qf-probability-complement","seed":202610037097,"semantic_template":"qf-probability-complement/{\"E\":[9,10,11],\"F\":\"powerset(S)\",\"S\":[9,10,11],\"denominator\":\"26\",\"operation\":\"complement\",\"weights\":[[9,\"1\"],[10,\"2\"],[11,\"23\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={9, 10, 11} and the event family is F=𝒫(S). In the displayed order of S, the weights are [1/(26), 2/(26), 23/(26)] and E={9, 10, 11}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"0/1","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 0/1. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0098","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"feed75df7a39249f856c5bfc12e6538eef92bedab2bdc228c327c299d3178e6b","family_id":"qf-probability-complement","seed":202610037098,"semantic_template":"qf-probability-complement/{\"E\":[10],\"F\":\"powerset(S)\",\"S\":[10,11,12],\"denominator\":\"29\",\"operation\":\"complement\",\"weights\":[[10,\"2\"],[11,\"3\"],[12,\"24\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={10, 11, 12} and the event family is F=𝒫(S). In the displayed order of S, the weights are [2/(29), 3/(29), 24/(29)] and E={10}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"27/29","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 27/29. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0099","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"0682cd1310b975c2342f5c1e3f908993836f03b84ffca5ac8db0e03cfc858008","family_id":"qf-probability-complement","seed":202610037099,"semantic_template":"qf-probability-complement/{\"E\":[13],\"F\":\"powerset(S)\",\"S\":[11,12,13],\"denominator\":\"32\",\"operation\":\"complement\",\"weights\":[[11,\"3\"],[12,\"4\"],[13,\"25\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={11, 12, 13} and the event family is F=𝒫(S). In the displayed order of S, the weights are [3/(32), 4/(32), 25/(32)] and E={13}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"7/32","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 7/32. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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-r07-probability-0100","batch_id":"batch-20261003-r07-probability-foundations","locale":"en","translation_state":"ai_candidate","source":{"question_sha256":"81446b406ed68400873ca23411c1fdd86944f426447ce657cf957be16efd1b8c","family_id":"qf-probability-complement","seed":202610037100,"semantic_template":"qf-probability-complement/{\"E\":[12,13],\"F\":\"powerset(S)\",\"S\":[12,13,14],\"denominator\":\"31\",\"operation\":\"complement\",\"weights\":[[12,\"0\"],[13,\"5\"],[14,\"26\"]]}"},"fields":{"title":"Complement Probability in a Weighted Model","stem":"The nonempty finite sample space is S={12, 13, 14} and the event family is F=𝒫(S). In the displayed order of S, the weights are [0/(31), 5/(31), 26/(31)] and E={12, 13}. Find P(S∖E) exactly as a reduced fraction with positive denominator; do not assume equiprobability.","answer":{"kind":"expression","value":"26/31","explanation":"Check nonnegative weights summing to one, then sum the weights of the complement outcomes in the same S. The exact probability is 26/31. A nonempty event may have zero probability. Here 𝒫 denotes the full powerset and P denotes the probability law."},"skills":["Complement Probability in a 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."}}