# 貝氏定理基礎 · 脈衝 R06 本批已通過 Schema、去重、來源快照與獨立答案重算；仍未標記為人工審定。 本題庫已有對應的中英逐步詳解。原題的 candidate 與英文 ai_candidate 狀態保留；AI 覆核、有限計算與 Lean 形式證明分開標示。 batch-20261014-advance-20261102-r06-bayes-theorem-foundations · 2026-10-14 · 100 題 ## 1. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,8],[1,8],[1,8],[5,8]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/4, 1/2, 1/4, 1/8, 1/2, not_causal) Wave5第二層核對=(1/4, 1/2, 1/4, 1/8, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0001/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 2. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,9],[1,9],[1,9],[6,9]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/9, 1/2, 2/9, 1/9, 1/2, not_causal) Wave5第二層核對=(2/9, 1/2, 2/9, 1/9, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0002/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 3. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,10],[1,10],[1,10],[7,10]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/5, 1/2, 1/5, 1/10, 1/2, not_causal) Wave5第二層核對=(1/5, 1/2, 1/5, 1/10, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0003/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 4. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,11],[1,11],[1,11],[8,11]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/11, 1/2, 2/11, 1/11, 1/2, not_causal) Wave5第二層核對=(2/11, 1/2, 2/11, 1/11, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0004/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 5. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,12],[1,12],[1,12],[9,12]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/6, 1/2, 1/6, 1/12, 1/2, not_causal) Wave5第二層核對=(1/6, 1/2, 1/6, 1/12, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0005/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 6. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,13],[1,13],[1,13],[10,13]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/13, 1/2, 2/13, 1/13, 1/2, not_causal) Wave5第二層核對=(2/13, 1/2, 2/13, 1/13, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0006/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 7. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,14],[1,14],[1,14],[11,14]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/7, 1/2, 1/7, 1/14, 1/2, not_causal) Wave5第二層核對=(1/7, 1/2, 1/7, 1/14, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0007/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 8. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,15],[1,15],[1,15],[12,15]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/15, 1/2, 2/15, 1/15, 1/2, not_causal) Wave5第二層核對=(2/15, 1/2, 2/15, 1/15, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0008/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 9. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,16],[1,16],[1,16],[13,16]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/8, 1/2, 1/8, 1/16, 1/2, not_causal) Wave5第二層核對=(1/8, 1/2, 1/8, 1/16, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0009/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 10. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,17],[1,17],[1,17],[14,17]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/17, 1/2, 2/17, 1/17, 1/2, not_causal) Wave5第二層核對=(2/17, 1/2, 2/17, 1/17, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0010/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 11. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,18],[1,18],[1,18],[15,18]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/9, 1/2, 1/9, 1/18, 1/2, not_causal) Wave5第二層核對=(1/9, 1/2, 1/9, 1/18, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0011/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 12. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,19],[1,19],[1,19],[16,19]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/19, 1/2, 2/19, 1/19, 1/2, not_causal) Wave5第二層核對=(2/19, 1/2, 2/19, 1/19, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0012/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 13. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,20],[1,20],[1,20],[17,20]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/10, 1/2, 1/10, 1/20, 1/2, not_causal) Wave5第二層核對=(1/10, 1/2, 1/10, 1/20, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0013/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 14. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,21],[1,21],[1,21],[18,21]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/21, 1/2, 2/21, 1/21, 1/2, not_causal) Wave5第二層核對=(2/21, 1/2, 2/21, 1/21, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0014/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 15. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,22],[1,22],[1,22],[19,22]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/11, 1/2, 1/11, 1/22, 1/2, not_causal) Wave5第二層核對=(1/11, 1/2, 1/11, 1/22, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0015/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 16. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,23],[1,23],[1,23],[20,23]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/23, 1/2, 2/23, 1/23, 1/2, not_causal) Wave5第二層核對=(2/23, 1/2, 2/23, 1/23, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0016/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 17. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,24],[1,24],[1,24],[21,24]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/12, 1/2, 1/12, 1/24, 1/2, not_causal) Wave5第二層核對=(1/12, 1/2, 1/12, 1/24, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0017/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 18. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,25],[1,25],[1,25],[22,25]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/25, 1/2, 2/25, 1/25, 1/2, not_causal) Wave5第二層核對=(2/25, 1/2, 2/25, 1/25, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0018/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 19. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,26],[1,26],[1,26],[23,26]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/13, 1/2, 1/13, 1/26, 1/2, not_causal) Wave5第二層核對=(1/13, 1/2, 1/13, 1/26, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0019/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 20. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,27],[1,27],[1,27],[24,27]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/27, 1/2, 2/27, 1/27, 1/2, not_causal) Wave5第二層核對=(2/27, 1/2, 2/27, 1/27, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0020/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 21. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,28],[1,28],[1,28],[25,28]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/14, 1/2, 1/14, 1/28, 1/2, not_causal) Wave5第二層核對=(1/14, 1/2, 1/14, 1/28, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0021/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 22. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,29],[1,29],[1,29],[26,29]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/29, 1/2, 2/29, 1/29, 1/2, not_causal) Wave5第二層核對=(2/29, 1/2, 2/29, 1/29, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0022/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 23. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,30],[1,30],[1,30],[27,30]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/15, 1/2, 1/15, 1/30, 1/2, not_causal) Wave5第二層核對=(1/15, 1/2, 1/15, 1/30, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0023/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 24. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,31],[1,31],[1,31],[28,31]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(2/31, 1/2, 2/31, 1/31, 1/2, not_causal) Wave5第二層核對=(2/31, 1/2, 2/31, 1/31, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0024/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 25. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-two-event-reversal；資料={"masses":[[1,32],[1,32],[1,32],[29,32]],"hypothesis":[0,1],"evidence":[0,2]}。請依 [prior,likelihood,evidence_mass,joint,posterior,boundary] 欄序回答。 查看答案與解釋(1/16, 1/2, 1/16, 1/32, 1/2, not_causal) Wave5第二層核對=(1/16, 1/2, 1/16, 1/32, 1/2, not_causal)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0025/) 貝氏定理 · qf-mod33-bayes-two-event-reversal · candidate ## 26. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,8],[1,8],[1,8],[5,8]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/4, 1/2, 1/4, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/4, 1/2, 1/4, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0026/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 27. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,9],[1,9],[1,9],[6,9]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 7/9, 1/7, 2/9, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 7/9, 1/7, 2/9, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0027/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 28. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,10],[1,10],[1,10],[7,10]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/5, 1/2, 1/5, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/5, 1/2, 1/5, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0028/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 29. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,11],[1,11],[1,11],[8,11]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 9/11, 1/9, 2/11, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 9/11, 1/9, 2/11, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0029/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 30. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,12],[1,12],[1,12],[9,12]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/6, 1/2, 1/6, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/6, 1/2, 1/6, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0030/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 31. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,13],[1,13],[1,13],[10,13]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 11/13, 1/11, 2/13, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 11/13, 1/11, 2/13, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0031/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 32. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,14],[1,14],[1,14],[11,14]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/7, 1/2, 1/7, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/7, 1/2, 1/7, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0032/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 33. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,15],[1,15],[1,15],[12,15]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 13/15, 1/13, 2/15, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 13/15, 1/13, 2/15, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0033/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 34. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,16],[1,16],[1,16],[13,16]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/8, 1/2, 1/8, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/8, 1/2, 1/8, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0034/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 35. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,17],[1,17],[1,17],[14,17]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 15/17, 1/15, 2/17, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 15/17, 1/15, 2/17, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0035/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 36. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,18],[1,18],[1,18],[15,18]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/9, 1/2, 1/9, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/9, 1/2, 1/9, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0036/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 37. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,19],[1,19],[1,19],[16,19]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 17/19, 1/17, 2/19, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 17/19, 1/17, 2/19, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0037/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 38. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,20],[1,20],[1,20],[17,20]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/10, 1/2, 1/10, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/10, 1/2, 1/10, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0038/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 39. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,21],[1,21],[1,21],[18,21]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 19/21, 1/19, 2/21, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 19/21, 1/19, 2/21, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0039/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 40. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,22],[1,22],[1,22],[19,22]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/11, 1/2, 1/11, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/11, 1/2, 1/11, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0040/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 41. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,23],[1,23],[1,23],[20,23]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 21/23, 1/21, 2/23, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 21/23, 1/21, 2/23, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0041/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 42. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,24],[1,24],[1,24],[21,24]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/12, 1/2, 1/12, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/12, 1/2, 1/12, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0042/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 43. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,25],[1,25],[1,25],[22,25]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 23/25, 1/23, 2/25, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 23/25, 1/23, 2/25, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0043/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 44. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,26],[1,26],[1,26],[23,26]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/13, 1/2, 1/13, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/13, 1/2, 1/13, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0044/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 45. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,27],[1,27],[1,27],[24,27]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 25/27, 1/25, 2/27, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 25/27, 1/25, 2/27, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0045/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 46. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,28],[1,28],[1,28],[25,28]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/14, 1/2, 1/14, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/14, 1/2, 1/14, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0046/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 47. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,29],[1,29],[1,29],[26,29]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 27/29, 1/27, 2/29, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 27/29, 1/27, 2/29, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0047/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 48. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,30],[1,30],[1,30],[27,30]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/15, 1/2, 1/15, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/15, 1/2, 1/15, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0048/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 49. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,31],[1,31],[1,31],[28,31]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":1}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(1, 29/31, 1/29, 2/31, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(1, 29/31, 1/29, 2/31, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0049/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 50. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-partition-posterior；資料={"masses":[[1,32],[1,32],[1,32],[29,32]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2],"target_index":0}。請依 [target_index,prior,likelihood,evidence_mass,posterior,partition_certificate] 欄序回答。 查看答案與解釋(0, 1/16, 1/2, 1/16, 1/2, partition_pairwise_disjoint_exhaustive) Wave5第二層核對=(0, 1/16, 1/2, 1/16, 1/2, partition_pairwise_disjoint_exhaustive)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0050/) 貝氏定理 · qf-mod33-bayes-partition-posterior · candidate ## 51. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,8],[1,8],[1,8],[5,8]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/4, 3/4], [1/2, 1/6], [1/8, 1/8], 1/4, 1/4, 是) Wave5第二層核對=([1/4, 3/4], [1/2, 1/6], [1/8, 1/8], 1/4, 1/4, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0051/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 52. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,9],[1,9],[1,9],[6,9]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/9, 7/9], [1/2, 1/7], [1/9, 1/9], 2/9, 2/9, 是) Wave5第二層核對=([2/9, 7/9], [1/2, 1/7], [1/9, 1/9], 2/9, 2/9, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0052/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 53. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,10],[1,10],[1,10],[7,10]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/5, 4/5], [1/2, 1/8], [1/10, 1/10], 1/5, 1/5, 是) Wave5第二層核對=([1/5, 4/5], [1/2, 1/8], [1/10, 1/10], 1/5, 1/5, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0053/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 54. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,11],[1,11],[1,11],[8,11]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/11, 9/11], [1/2, 1/9], [1/11, 1/11], 2/11, 2/11, 是) Wave5第二層核對=([2/11, 9/11], [1/2, 1/9], [1/11, 1/11], 2/11, 2/11, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0054/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 55. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,12],[1,12],[1,12],[9,12]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/6, 5/6], [1/2, 1/10], [1/12, 1/12], 1/6, 1/6, 是) Wave5第二層核對=([1/6, 5/6], [1/2, 1/10], [1/12, 1/12], 1/6, 1/6, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0055/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 56. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,13],[1,13],[1,13],[10,13]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/13, 11/13], [1/2, 1/11], [1/13, 1/13], 2/13, 2/13, 是) Wave5第二層核對=([2/13, 11/13], [1/2, 1/11], [1/13, 1/13], 2/13, 2/13, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0056/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 57. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,14],[1,14],[1,14],[11,14]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/7, 6/7], [1/2, 1/12], [1/14, 1/14], 1/7, 1/7, 是) Wave5第二層核對=([1/7, 6/7], [1/2, 1/12], [1/14, 1/14], 1/7, 1/7, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0057/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 58. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,15],[1,15],[1,15],[12,15]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/15, 13/15], [1/2, 1/13], [1/15, 1/15], 2/15, 2/15, 是) Wave5第二層核對=([2/15, 13/15], [1/2, 1/13], [1/15, 1/15], 2/15, 2/15, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0058/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 59. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,16],[1,16],[1,16],[13,16]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/8, 7/8], [1/2, 1/14], [1/16, 1/16], 1/8, 1/8, 是) Wave5第二層核對=([1/8, 7/8], [1/2, 1/14], [1/16, 1/16], 1/8, 1/8, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0059/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 60. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,17],[1,17],[1,17],[14,17]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/17, 15/17], [1/2, 1/15], [1/17, 1/17], 2/17, 2/17, 是) Wave5第二層核對=([2/17, 15/17], [1/2, 1/15], [1/17, 1/17], 2/17, 2/17, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0060/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 61. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,18],[1,18],[1,18],[15,18]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/9, 8/9], [1/2, 1/16], [1/18, 1/18], 1/9, 1/9, 是) Wave5第二層核對=([1/9, 8/9], [1/2, 1/16], [1/18, 1/18], 1/9, 1/9, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0061/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 62. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,19],[1,19],[1,19],[16,19]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/19, 17/19], [1/2, 1/17], [1/19, 1/19], 2/19, 2/19, 是) Wave5第二層核對=([2/19, 17/19], [1/2, 1/17], [1/19, 1/19], 2/19, 2/19, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0062/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 63. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,20],[1,20],[1,20],[17,20]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/10, 9/10], [1/2, 1/18], [1/20, 1/20], 1/10, 1/10, 是) Wave5第二層核對=([1/10, 9/10], [1/2, 1/18], [1/20, 1/20], 1/10, 1/10, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0063/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 64. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,21],[1,21],[1,21],[18,21]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/21, 19/21], [1/2, 1/19], [1/21, 1/21], 2/21, 2/21, 是) Wave5第二層核對=([2/21, 19/21], [1/2, 1/19], [1/21, 1/21], 2/21, 2/21, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0064/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 65. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,22],[1,22],[1,22],[19,22]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/11, 10/11], [1/2, 1/20], [1/22, 1/22], 1/11, 1/11, 是) Wave5第二層核對=([1/11, 10/11], [1/2, 1/20], [1/22, 1/22], 1/11, 1/11, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0065/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 66. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,23],[1,23],[1,23],[20,23]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/23, 21/23], [1/2, 1/21], [1/23, 1/23], 2/23, 2/23, 是) Wave5第二層核對=([2/23, 21/23], [1/2, 1/21], [1/23, 1/23], 2/23, 2/23, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0066/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 67. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,24],[1,24],[1,24],[21,24]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/12, 11/12], [1/2, 1/22], [1/24, 1/24], 1/12, 1/12, 是) Wave5第二層核對=([1/12, 11/12], [1/2, 1/22], [1/24, 1/24], 1/12, 1/12, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0067/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 68. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,25],[1,25],[1,25],[22,25]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/25, 23/25], [1/2, 1/23], [1/25, 1/25], 2/25, 2/25, 是) Wave5第二層核對=([2/25, 23/25], [1/2, 1/23], [1/25, 1/25], 2/25, 2/25, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0068/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 69. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,26],[1,26],[1,26],[23,26]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/13, 12/13], [1/2, 1/24], [1/26, 1/26], 1/13, 1/13, 是) Wave5第二層核對=([1/13, 12/13], [1/2, 1/24], [1/26, 1/26], 1/13, 1/13, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0069/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 70. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,27],[1,27],[1,27],[24,27]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/27, 25/27], [1/2, 1/25], [1/27, 1/27], 2/27, 2/27, 是) Wave5第二層核對=([2/27, 25/27], [1/2, 1/25], [1/27, 1/27], 2/27, 2/27, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0070/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 71. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,28],[1,28],[1,28],[25,28]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/14, 13/14], [1/2, 1/26], [1/28, 1/28], 1/14, 1/14, 是) Wave5第二層核對=([1/14, 13/14], [1/2, 1/26], [1/28, 1/28], 1/14, 1/14, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0071/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 72. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,29],[1,29],[1,29],[26,29]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/29, 27/29], [1/2, 1/27], [1/29, 1/29], 2/29, 2/29, 是) Wave5第二層核對=([2/29, 27/29], [1/2, 1/27], [1/29, 1/29], 2/29, 2/29, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0072/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 73. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,30],[1,30],[1,30],[27,30]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/15, 14/15], [1/2, 1/28], [1/30, 1/30], 1/15, 1/15, 是) Wave5第二層核對=([1/15, 14/15], [1/2, 1/28], [1/30, 1/30], 1/15, 1/15, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0073/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 74. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,31],[1,31],[1,31],[28,31]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([2/31, 29/31], [1/2, 1/29], [1/31, 1/31], 2/31, 2/31, 是) Wave5第二層核對=([2/31, 29/31], [1/2, 1/29], [1/31, 1/31], 2/31, 2/31, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0074/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 75. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-total-evidence；資料={"masses":[[1,32],[1,32],[1,32],[29,32]],"hypotheses":[[0,1],[2,3]],"evidence":[0,2]}。請依 [priors,likelihoods,joints,total,evidence_mass,equal] 欄序回答。 查看答案與解釋([1/16, 15/16], [1/2, 1/30], [1/32, 1/32], 1/16, 1/16, 是) Wave5第二層核對=([1/16, 15/16], [1/2, 1/30], [1/32, 1/32], 1/16, 1/16, 是)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0075/) 貝氏定理 · qf-mod33-bayes-total-evidence · candidate ## 76. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,8],[1,8],[1,8],[5,8]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0076/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 77. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,9],[1,9],[1,9],[6,9]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0077/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 78. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,10],[1,10],[1,10],[7,10]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0078/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 79. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,11],[1,11],[1,11],[8,11]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0079/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 80. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,12],[1,12],[1,12],[9,12]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0080/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 81. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,13],[1,13],[1,13],[10,13]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0081/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 82. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,14],[1,14],[1,14],[11,14]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0082/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 83. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,15],[1,15],[1,15],[12,15]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0083/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 84. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,16],[1,16],[1,16],[13,16]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0084/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 85. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,17],[1,17],[1,17],[14,17]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0085/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 86. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,18],[1,18],[1,18],[15,18]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0086/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 87. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,19],[1,19],[1,19],[16,19]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0087/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 88. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,20],[1,20],[1,20],[17,20]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0088/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 89. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,21],[1,21],[1,21],[18,21]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0089/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 90. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,22],[1,22],[1,22],[19,22]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0090/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 91. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,23],[1,23],[1,23],[20,23]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0091/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 92. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,24],[1,24],[1,24],[21,24]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0092/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 93. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,25],[1,25],[1,25],[22,25]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0093/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 94. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,26],[1,26],[1,26],[23,26]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0094/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 95. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,27],[1,27],[1,27],[24,27]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0095/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 96. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,28],[1,28],[1,28],[25,28]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0096/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 97. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,29],[1,29],[1,29],[26,29]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0097/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 98. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,30],[1,30],[1,30],[27,30]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0098/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 99. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,31],[1,31],[1,31],[28,31]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0099/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate ## 100. Bayes定理 完整有限原子law、事件與正conditioning denominator。操作=bayes-zero-denominator-boundary；資料={"masses":[[1,32],[1,32],[1,32],[29,32]],"hypothesis":[0,1],"evidence":[]}。請依 [evidence_mass,conditional_value,defined,boundary] 欄序回答。 查看答案與解釋(0, undefined, 否, conditional_probability_requires_positive_denominator) Wave5第二層核對=(0, undefined, 否, conditional_probability_requires_positive_denominator)；依完整明示資料重算；partition須pairwise disjoint且exhaustive並由total probability正規化；不可漏prior、反轉零分母或把Bayes當causation。 [本題逐步詳解（AI 二次覆核）](https://ocme.evemisslab.com/zh-Hant/questions/q-20261014-advance-20261102-r06-bayes-theorem-0100/) 貝氏定理 · qf-mod33-bayes-zero-denominator-boundary · candidate [原始題目 JSONL](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r06-bayes-theorem-foundations/questions.jsonl) · [驗收報告](https://ocme.evemisslab.com/data/questions/batches/batch-20261014-advance-20261102-r06-bayes-theorem-foundations/validation-report.json)