{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261014-advance-20261102-r06-bayes-theorem-0014",
  "pilot_number": "P250-0114",
  "number": 114,
  "family_id": "qf-mod33-bayes-two-event-reversal",
  "title": {
    "zh-Hant": "第 114 題：Bayes定理",
    "en": "Question 114: Prior and Likelihood Must Both Enter a Bayes Reversal"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261014-advance-20261102-r06-bayes-theorem-0014",
    "batch_id": "batch-20261014-advance-20261102-r06-bayes-theorem-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-bayes-theorem"
    ],
    "title_zh": "Bayes定理",
    "stem_zh": "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。",
    "answer": {
      "kind": "expression",
      "value": "(2/21, 1/2, 2/21, 1/21, 1/2, not_causal)",
      "explanation_zh": "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。"
    },
    "difficulty": {
      "band": "undergraduate_core",
      "level": 2,
      "skills": [
        "Bayes定理"
      ]
    },
    "variant": {
      "family_id": "qf-mod33-bayes-two-event-reversal",
      "seed": 202611026014,
      "parameters": {
        "masses": [
          [
            1,
            21
          ],
          [
            1,
            21
          ],
          [
            1,
            21
          ],
          [
            18,
            21
          ]
        ],
        "hypothesis": [
          0,
          1
        ],
        "evidence": [
          0,
          2
        ],
        "domain": "Exact_finite_atomic_Bayes",
        "operation": "bayes-two-event-reversal",
        "semantic_template": "qf-mod33-bayes-two-event-reversal/{\"evidence\":[0,2],\"hypothesis\":[0,1],\"masses\":[[1,21],[1,21],[1,21],[18,21]]}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_finite_or_symbolic_model",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication",
        "producer_json_schema"
      ],
      "warning_zh": "候選須由主端獨立重算；不構成人工審定、任意維basis theorem、bounded-to-global kernel/image、一般compactness、因果推論、language equivalence、Lean或形式Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "model_unknown",
      "generated_at": "2026-10-14",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261014-advance-20261102-r06-bayes-theorem-0014",
    "batch_id": "batch-20261014-advance-20261102-r06-bayes-theorem-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "86b0a06e2067997c3ffffe80bc191d02e3284e58411f6603b9a988aaecea13df",
      "family_id": "qf-mod33-bayes-two-event-reversal",
      "seed": 202611026014,
      "semantic_template": "qf-mod33-bayes-two-event-reversal/{\"evidence\":[0,2],\"hypothesis\":[0,1],\"masses\":[[1,21],[1,21],[1,21],[18,21]]}"
    },
    "fields": {
      "title": "Prior and Likelihood Must Both Enter a Bayes Reversal",
      "stem": "Use the same complete finite atomic probability law with aligned nonnegative exact masses=[1/21, 1/21, 1/21, 18/21], summing to 1, and canonical event evidence=[0, 2]. Event entries are increasing distinct outcome indices from 0 through 3. Every actual conditional probability requires a strictly positive denominator; a dedicated zero-denominator audit records undefined instead of dividing. Priors, likelihoods, intersections and evidence mass all come from this one law. Keep reduced exact fractions with positive denominators, omitting /1 for integers. Bayes conditioning does not imply causation. Hypothesis=[0, 1] must have positive prior, and evidence must have positive mass. Give (prior,likelihood,evidence_mass,joint,posterior,not_causal).",
      "answer": {
        "kind": "expression",
        "value": "(2/21, 1/2, 2/21, 1/21, 1/2, not_causal)",
        "explanation": "Compute P(H), P(E|H), P(E), and the same joint P(H∩E), then form P(H|E)=joint/evidence_mass. Likelihood and posterior need not be equal; dropping the prior or reversing a zero denominator is invalid. The exact result, retaining the requested order, is (2/21, 1/2, 2/21, 1/21, 1/2, not_causal)."
      },
      "skills": [
        "Prior and Likelihood Must Both Enter a Bayes Reversal"
      ],
      "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review."
    }
  },
  "givens": [
    {
      "id": "G001",
      "text": {
        "zh-Hant": "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。",
        "en": "Use the same complete finite atomic probability law with aligned nonnegative exact masses=[1/21, 1/21, 1/21, 18/21], summing to 1, and canonical event evidence=[0, 2]. Event entries are increasing distinct outcome indices from 0 through 3. Every actual conditional probability requires a strictly positive denominator; a dedicated zero-denominator audit records undefined instead of dividing. Priors, likelihoods, intersections and evidence mass all come from this one law. Keep reduced exact fractions with positive denominators, omitting /1 for integers. Bayes conditioning does not imply causation. Hypothesis=[0, 1] must have positive prior, and evidence must have positive mass. Give (prior,likelihood,evidence_mass,joint,posterior,not_causal)."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "P(Ω) := sum(1/21, 1/21, 1/21, 6/7)",
      "after": "P(Ω) = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "先加總全部原子質量，確認這是一個機率分布。",
        "en": "Sum all atom masses first to verify normalization."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(Ω)，不以小數近似替代。",
          "en": "Recompute P(Ω) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/21",
          "1/21",
          "1/21",
          "6/7"
        ],
        "output": "1",
        "label": "P(Ω)"
      }
    },
    {
      "id": "S002",
      "uses": [
        "G001",
        "S001"
      ],
      "before": "H=[0, 1]; E=[0, 2]",
      "after": "H∩E=[0]",
      "relation": "implies",
      "rule_id": "event_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "同時屬於兩事件的樣本點才可貢獻交集機率。",
        "en": "Only sample points in both events contribute to their joint probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "不是把兩事件機率直接相乘；尚未給獨立性。",
          "en": "Do not multiply the event probabilities: independence has not been assumed."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "P(H) := sum(1/21, 1/21)",
      "after": "P(H) = 2/21",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐點累加 H 的質量。",
        "en": "Add the masses of the points in H."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(H)，不以小數近似替代。",
          "en": "Recompute P(H) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/21",
          "1/21"
        ],
        "output": "2/21",
        "label": "P(H)"
      }
    },
    {
      "id": "S004",
      "uses": [
        "G001",
        "S003"
      ],
      "before": "P(E) := sum(1/21, 1/21)",
      "after": "P(E) = 2/21",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "逐點累加 E 的質量，作為 posterior 的條件分母。",
        "en": "Add the masses in E; this is the posterior's conditioning denominator."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(E)，不以小數近似替代。",
          "en": "Recompute P(E) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/21",
          "1/21"
        ],
        "output": "2/21",
        "label": "P(E)"
      }
    },
    {
      "id": "S005",
      "uses": [
        "G001",
        "S004"
      ],
      "before": "P(H∩E) := sum(1/21)",
      "after": "P(H∩E) = 1/21",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "交集只累加一次，不重複計數。",
        "en": "Sum the intersection once, without double-counting."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(H∩E)，不以小數近似替代。",
          "en": "Recompute P(H∩E) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "sum",
        "inputs": [
          "1/21"
        ],
        "output": "1/21",
        "label": "P(H∩E)"
      }
    },
    {
      "id": "S006",
      "uses": [
        "G001",
        "S005"
      ],
      "before": "P(E|H) := div(1/21, 2/21)",
      "after": "P(E|H) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "P(H)=2/21>0，所以可用交集機率除以 P(H)。",
        "en": "Since P(H)=2/21>0, divide the joint probability by P(H)."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(E|H)，不以小數近似替代。",
          "en": "Recompute P(E|H) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/21",
          "2/21"
        ],
        "output": "1/2",
        "label": "P(E|H)"
      }
    },
    {
      "id": "S007",
      "uses": [
        "G001",
        "S006"
      ],
      "before": "P(E|H)·P(H) := mul(1/2, 2/21)",
      "after": "P(E|H)·P(H) = 1/21",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "把條件機率乘回條件事件的機率，應恢復交集機率。",
        "en": "Multiply the conditional probability by the conditioning event's probability to recover the joint."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(E|H)·P(H)，不以小數近似替代。",
          "en": "Recompute P(E|H)·P(H) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "2/21"
        ],
        "output": "1/21",
        "label": "P(E|H)·P(H)"
      }
    },
    {
      "id": "S008",
      "uses": [
        "G001",
        "S007"
      ],
      "before": "P(H|E) := div(1/21, 2/21)",
      "after": "P(H|E) = 1/2",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "反轉條件時，分母改為 P(E)=2/21>0。",
        "en": "When reversing the condition, the denominator is P(E)=2/21>0."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(H|E)，不以小數近似替代。",
          "en": "Recompute P(H|E) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "1/21",
          "2/21"
        ],
        "output": "1/2",
        "label": "P(H|E)"
      }
    },
    {
      "id": "S009",
      "uses": [
        "G001",
        "S008"
      ],
      "before": "P(H|E)=1/2; P(H∩E)=P(E|H)P(H)",
      "after": "P(H|E)=P(E|H)P(H)/P(E)=1/2",
      "relation": "implies",
      "rule_id": "bayes_substitution",
      "conditions": [
        "P(E)>0",
        "P(H)>0"
      ],
      "explanation": {
        "zh-Hant": "將已證的交集乘法式代入 posterior 定義，即得到 Bayes 公式在本題的完整代值。",
        "en": "Substitute the established joint-product identity into the posterior definition to obtain Bayes' formula for this instance."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "條件機率反轉不是把兩個符號交換後保留原分母。",
          "en": "Reversing a conditional does not mean swapping labels while keeping the old denominator."
        }
      },
      "arithmetic": null
    },
    {
      "id": "S010",
      "uses": [
        "G001",
        "S009"
      ],
      "before": "P(H|E)·P(E) := mul(1/2, 2/21)",
      "after": "P(H|E)·P(E) = 1/21",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "再乘回 P(E)，須回到同一交集機率。",
        "en": "Multiply back by P(E) to recover the same joint probability."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 P(H|E)·P(E)，不以小數近似替代。",
          "en": "Recompute P(H|E)·P(E) with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "mul",
        "inputs": [
          "1/2",
          "2/21"
        ],
        "output": "1/21",
        "label": "P(H|E)·P(E)"
      }
    },
    {
      "id": "S011",
      "uses": [
        "G001",
        "S010"
      ],
      "before": "probability identity",
      "after": "no causal direction has been established",
      "relation": "implies",
      "rule_id": "probability_not_causality",
      "conditions": [],
      "explanation": {
        "zh-Hant": "這是同一分布內的條件機率恆等式，沒有由此證明因果方向。",
        "en": "This is a conditional-probability identity in one law, not evidence of a causal direction."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "若條件事件質量為 0，須保留未定義而不是計算 0/0。",
          "en": "With zero conditioning mass the quotient is undefined, not a numerical 0/0 answer."
        }
      },
      "arithmetic": null
    }
  ],
  "math_model": {
    "masses": [
      "1/21",
      "1/21",
      "1/21",
      "6/7"
    ],
    "H": [
      0,
      1
    ],
    "E": [
      0,
      2
    ],
    "intersection": [
      0
    ],
    "prior": "2/21",
    "evidence": "2/21",
    "joint": "1/21",
    "likelihood": "1/2",
    "posterior": "1/2"
  },
  "conclusion": {
    "canonical_value": "(2/21, 1/2, 2/21, 1/21, 1/2, not_causal)",
    "explanation": {
      "zh-Hant": "P(H|E)=1/2；完整答案亦保留 prior、likelihood、evidence 與 joint。",
      "en": "P(H|E)=1/2; the complete answer also retains the prior, likelihood, evidence and joint."
    }
  },
  "definition_note": {
    "zh-Hant": "使用同一完整有限機率分布，兩個條件事件質量都嚴格為正。",
    "en": "Use one complete finite probability law; both conditioning events have strictly positive mass."
  },
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    "formalization": "not_formalized",
    "Lean": "NOT_RUN",
    "Canonical_truth_promotion": false,
    "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged."
  }
}