{
  "schema_version": "ocme-derivation-reader-v1",
  "guide_id": "derivation-mko-conditional-probability-01",
  "mko_id": "mko-conditional-probability",
  "title": {
    "zh-Hant": "有限模型的條件機率：正分母、兩個方向與零機率條件",
    "en": "Conditional probability in finite models: positive denominators, the two directions and null conditioning events"
  },
  "contract": {
    "schema_version": "ocme-derivation-guide-v0.1",
    "guide_id": "derivation-mko-conditional-probability-01",
    "source": {
      "mko_id": "mko-conditional-probability",
      "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
      "sha256": "7b0f4824aa1f3e4c64ea306c5bed04843566b8e4bbdf96d94dbc1367dbb68419",
      "english_sha256": "d0ca772e54d88ba2c19452fb83b8dbc37d4f064fe26d89d017fe591f2913621b"
    },
    "mode": "worked_example",
    "audience_prerequisites": [
      "prob_axioms",
      "set_operations_def",
      "subset_def",
      "frac_rules",
      "numeric_evaluation",
      "weight_model"
    ],
    "goal": {
      "zh-Hant": "在 S={1,2,3,4,5,6}、權重 w=(1,2,3,2,4,0)/12 的有限模型中，設 A={2,3}、B={3,4,5,6}、Z={6}。驗證模型合法，計算 P(A|B) 與 P(B|A)，與等可能計數比 |A∩B|/|B| 及乘積 P(A)P(B) 比較，並判定 P(A|Z) 是否有定義。",
      "en": "In the finite model on S={1,2,3,4,5,6} with weights w=(1,2,3,2,4,0)/12, let A={2,3}, B={3,4,5,6} and Z={6}. Validate the model, compute P(A|B) and P(B|A), compare them with the equiprobable count ratio |A∩B|/|B| and with the product P(A)P(B), and decide whether P(A|Z) is defined."
    },
    "givens": [
      {
        "id": "G1",
        "statement": {
          "zh-Hant": "S={1,2,3,4,5,6}，事件族 F=𝒫(S)，權重 w(1)=1/12、w(2)=2/12、w(3)=3/12、w(4)=2/12、w(5)=4/12、w(6)=0；P(E)=Σ_{s∈E} w(s)。此模型未宣告為等可能。",
          "en": "S={1,2,3,4,5,6}, event family F=𝒫(S), weights w(1)=1/12, w(2)=2/12, w(3)=3/12, w(4)=2/12, w(5)=4/12, w(6)=0; P(E)=Σ_{s∈E} w(s). The model is not declared equiprobable."
        }
      },
      {
        "id": "G2",
        "statement": {
          "zh-Hant": "事件 A={2,3}、B={3,4,5,6}、Z={6}。",
          "en": "Events A={2,3}, B={3,4,5,6}, Z={6}."
        }
      },
      {
        "id": "G3",
        "statement": {
          "zh-Hant": "待辦：驗證模型與事件資格；求 P(A|B)、P(B|A)；比較 |A∩B|/|B| 與 P(A)P(B)；判定 P(A|Z)。",
          "en": "To do: validate the model and the events; find P(A|B) and P(B|A); compare with |A∩B|/|B| and P(A)P(B); decide P(A|Z)."
        }
      }
    ],
    "steps": [
      {
        "id": "S01",
        "uses": [
          "G1",
          "weight_model"
        ],
        "before_tex": "w=\\left(\\tfrac{1}{12},\\tfrac{2}{12},\\tfrac{3}{12},\\tfrac{2}{12},\\tfrac{4}{12},0\\right)",
        "after_tex": "w(s)\\ge 0,\\quad \\sum_{s\\in S}w(s)=\\tfrac{12}{12}=1",
        "relation": "implies",
        "rule": "weight_model",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 假設 a1 要求模型合法：每個權重是非負有理數，總和為 1。逐項：1/12、2/12、3/12、2/12、4/12、0 都 ≥0；總和 (1+2+3+2+4+0)/12=12/12=1（N1）。所以這是合法的有限權重模型。w(6)=0 是允許的：樣本點可以存在但權重為零。",
          "en": "MKO assumption a1 requires a valid model: every weight is a nonnegative rational and the total is 1. Item by item, 1/12, 2/12, 3/12, 2/12, 4/12 and 0 are all ≥0; the total is (1+2+3+2+4+0)/12=12/12=1 (N1). So this is a valid finite weighted model. w(6)=0 is allowed: a sample point may exist with weight zero."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "weight_model",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "兩個條件（非負、總和 1）都必須核對；之後所有機率都依賴這一步。",
            "en": "Both conditions (nonnegativity, total 1) must be checked; every later probability depends on this step."
          }
        },
        "leaf_basis": "weight_model + finite_sum_rules + numeric_evaluation"
      },
      {
        "id": "S02",
        "uses": [
          "G1",
          "G2",
          "subset_def"
        ],
        "before_tex": "A=\\{2,3\\},\\ B=\\{3,4,5,6\\},\\ Z=\\{6\\}",
        "after_tex": "A,B,Z\\subseteq S",
        "relation": "implies",
        "rule": "subset_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "事件資格：A、B、Z 的每個元素都在 S={1,2,3,4,5,6} 中，所以三者都是 S 的子集合，屬於事件族 𝒫(S)。",
          "en": "Event eligibility: every element of A, B and Z lies in S={1,2,3,4,5,6}, so all three are subsets of S and belong to the event family 𝒫(S)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "subset_def"
          ],
          "explanation": {
            "zh-Hant": "逐元素核對即可；不宣稱反向。",
            "en": "An element-by-element check suffices; no converse is claimed."
          }
        },
        "leaf_basis": "subset_def"
      },
      {
        "id": "S03",
        "uses": [
          "S01",
          "S02",
          "weight_model"
        ],
        "before_tex": "P(B)=w(3)+w(4)+w(5)+w(6)",
        "after_tex": "P(B)=\\tfrac{3+2+4+0}{12}=\\tfrac{9}{12}=\\tfrac{3}{4}",
        "relation": "equivalent",
        "rule": "weight_model",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 第一段：先算 B 中的機率質量。P(B) 是 B 的四個樣本點權重之和：(3+2+4+0)/12=9/12=3/4（N2）。",
          "en": "MKO paragraph one: first compute the probability mass in B. P(B) is the sum of the weights of the four sample points of B: (3+2+4+0)/12=9/12=3/4 (N2)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "weight_model",
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "有理數的精確求值，兩邊是同一個數。",
            "en": "An exact rational evaluation; both sides are the same number."
          }
        },
        "leaf_basis": "weight_model + finite_sum_rules + frac_rules + numeric_evaluation"
      },
      {
        "id": "S04",
        "uses": [
          "S03",
          "mko_cond_prob_def"
        ],
        "before_tex": "P(B)=\\tfrac{3}{4}",
        "after_tex": "P(B)>0,\\ \\text{so}\\ P(A\\mid B)\\ \\text{is defined}",
        "relation": "implies",
        "rule": "mko_cond_prob_def",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "依 MKO，P(A|B) 只在 P(B)>0 時定義。3/4>0，所以可以條件化在 B 上。這個判斷必須在任何除法之前做。",
          "en": "By the MKO, P(A|B) is defined only when P(B)>0. Since 3/4>0, we may condition on B. This check must come before any division."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_cond_prob_def",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要的是 P(B) 嚴格大於零，不是 B 非空（見 S14）。",
            "en": "What is needed is P(B) strictly positive, not merely B nonempty (see S14)."
          }
        },
        "leaf_basis": "mko_cond_prob_def + order_basics"
      },
      {
        "id": "S05",
        "uses": [
          "G2",
          "set_operations_def"
        ],
        "before_tex": "A\\cap B",
        "after_tex": "\\{3\\}",
        "relation": "equivalent",
        "rule": "set_operations_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "x∈A∩B 當且僅當 x∈A 且 x∈B。A 的元素是 2 與 3：2∉B，3∈B。所以 A∩B={3}。",
          "en": "x∈A∩B exactly when x∈A and x∈B. The elements of A are 2 and 3: 2∉B and 3∈B. So A∩B={3}."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "set_operations_def",
            "set_extensionality"
          ],
          "explanation": {
            "zh-Hant": "兩個集合有相同元素，依集合相等兩向成立。",
            "en": "The two sets have the same elements, so by set equality it holds both ways."
          }
        },
        "leaf_basis": "set_operations_def + set_extensionality"
      },
      {
        "id": "S06",
        "uses": [
          "S05",
          "weight_model"
        ],
        "before_tex": "P(A\\cap B)=P(\\{3\\})",
        "after_tex": "w(3)=\\tfrac{3}{12}=\\tfrac{1}{4}",
        "relation": "equivalent",
        "rule": "weight_model",
        "conditions": [],
        "explanation": {
          "zh-Hant": "MKO 第一段的第二步：算 A∩B 中的質量。A∩B={3}，所以 P(A∩B)=w(3)=3/12=1/4（N3）。",
          "en": "The second step of MKO paragraph one: compute the mass in A∩B. A∩B={3}, so P(A∩B)=w(3)=3/12=1/4 (N3)."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "weight_model",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "精確求值，兩向成立。",
            "en": "An exact evaluation, holding both ways."
          }
        },
        "leaf_basis": "weight_model + frac_rules + numeric_evaluation"
      },
      {
        "id": "S07",
        "uses": [
          "S04",
          "S06",
          "mko_cond_prob_def"
        ],
        "before_tex": "P(A\\mid B)=\\frac{P(A\\cap B)}{P(B)}",
        "after_tex": "\\frac{1/4}{3/4}=\\frac{1}{3}",
        "relation": "equivalent",
        "rule": "mko_cond_prob_def",
        "conditions": [
          "P(B)=\\tfrac{3}{4}\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "最後相除：P(A|B)=(1/4)/(3/4)=(1/4)·(4/3)=1/3（N4）。除以 3/4 需要 3/4≠0，S04 已確認。",
          "en": "Finally divide: P(A|B)=(1/4)/(3/4)=(1/4)·(4/3)=1/3 (N4). Dividing by 3/4 needs 3/4≠0, confirmed in S04."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "反向：P(A|B)·P(B)=(1/3)·(3/4)=1/4=P(A∩B)（N5）；乘以 3/4 與除以 3/4 互為反運算，前提是 3/4≠0。",
            "en": "Backward: P(A|B)·P(B)=(1/3)·(3/4)=1/4=P(A∩B) (N5); multiplying and dividing by 3/4 undo each other, given 3/4≠0."
          }
        },
        "leaf_basis": "mko_cond_prob_def + frac_rules + mul_inverse_nonzero + numeric_evaluation"
      },
      {
        "id": "S08",
        "uses": [
          "S03",
          "S06",
          "frac_rules"
        ],
        "before_tex": "\\frac{3/12}{9/12}",
        "after_tex": "\\frac{3}{9}=\\frac{1}{3}",
        "relation": "equivalent",
        "rule": "frac_rules",
        "conditions": [
          "12\\neq 0",
          "9\\neq 0"
        ],
        "explanation": {
          "zh-Hant": "MKO 第一段：共同正分母可以約掉。兩個質量都以同一模型的十二分之幾表示：P(A∩B)=3/12、P(B)=9/12；依分數規則 (3/12)/(9/12)=(3·12)/(12·9)=3/9=1/3（N6、N7）。與 S07 一致。能這樣約分，前提是兩者來自同一個已驗證的權重模型，且事件資格已確認。",
          "en": "MKO paragraph one: a common positive denominator may be cancelled. Both masses are expressed in twelfths of the same model: P(A∩B)=3/12 and P(B)=9/12; by the fraction rules (3/12)/(9/12)=(3·12)/(12·9)=3/9=1/3 (N6, N7), agreeing with S07. Cancelling this way presupposes that both come from the same validated weight model and that the events are eligible."
        },
        "reverse_audit": {
          "kind": "inverse_equivalence",
          "required_ids": [
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "分數規則在分母非零時是等式，雙向成立。",
            "en": "The fraction rules are equalities when the denominators are nonzero, so this holds both ways."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S09",
        "uses": [
          "S05",
          "S07",
          "L_split"
        ],
        "before_tex": "A\\cap B\\subseteq B",
        "after_tex": "P(A\\cap B)\\le P(B),\\ \\text{so}\\ 0\\le P(A\\mid B)\\le 1",
        "relation": "theorem_application",
        "rule": "L_split",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "A∩B⊆B，依引理 L_split，P(B)=P(A∩B)+P(B∖(A∩B))≥P(A∩B)。兩邊除以正數 P(B)，得 P(A|B)≤1；分子、分母都非負，得 P(A|B)≥0。本例 1/4≤3/4（N18），1/3 確實在 [0,1] 中。",
          "en": "A∩B⊆B, so by Lemma L_split, P(B)=P(A∩B)+P(B∖(A∩B))≥P(A∩B). Dividing by the positive P(B) gives P(A|B)≤1; numerator and denominator are nonnegative, so P(A|B)≥0. Here 1/4≤3/4 (N18), and 1/3 does lie in [0,1]."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_split",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "除以正數保持不等式方向，所以需要 P(B)>0；引理在 S09.1 證明。",
            "en": "Dividing by a positive number keeps the inequality direction, so P(B)>0 is needed; the lemma is proved in S09.1."
          }
        },
        "leaf_basis": "weight_model + finite_sum_rules + order_basics + set_operations_def"
      },
      {
        "id": "S09.1",
        "uses": [
          "weight_model",
          "set_operations_def",
          "finite_sum_rules"
        ],
        "before_tex": "E\\subseteq E'\\subseteq S",
        "after_tex": "P(E')=P(E)+P(E'\\setminus E)\\ge P(E)",
        "relation": "implies",
        "rule": "finite_sum_rules",
        "conditions": [
          "w(s)\\ge 0"
        ],
        "explanation": {
          "zh-Hant": "E⊆E′ 時，E′ 的每個元素恰屬於 E 或 E′∖E 之一（集合運算定義），兩者互斥。所以有限和 Σ_{s∈E′} w(s) 拆成 Σ_{s∈E} w(s)+Σ_{s∈E′∖E} w(s)，即 P(E′)=P(E)+P(E′∖E)。每個權重 ≥0，所以 P(E′∖E)≥0，得 P(E)≤P(E′)。這證得引理 L_split。",
          "en": "When E⊆E′, each element of E′ lies in exactly one of E and E′∖E (definition of set operations), and the two are disjoint. So the finite sum Σ_{s∈E′} w(s) splits as Σ_{s∈E} w(s)+Σ_{s∈E′∖E} w(s), that is P(E′)=P(E)+P(E′∖E). Every weight is ≥0, so P(E′∖E)≥0 and P(E)≤P(E′). This proves Lemma L_split."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "weight_model",
            "set_operations_def",
            "finite_sum_rules",
            "order_basics"
          ],
          "explanation": {
            "zh-Hant": "需要互斥拆分與權重非負；若允許負權重，不等式部分就不成立。",
            "en": "It needs the disjoint split and nonnegative weights; with negative weights the inequality part would fail."
          }
        },
        "leaf_basis": "weight_model + set_operations_def + finite_sum_rules + order_basics"
      },
      {
        "id": "S10",
        "uses": [
          "S03",
          "S06",
          "S07",
          "L_split",
          "G2"
        ],
        "before_tex": "B\\setminus(A\\cap B)=A^{c}\\cap B=\\{4,5,6\\}",
        "after_tex": "P(A^{c}\\cap B)=\\tfrac{1}{2},\\ P(A\\mid B)+P(A^{c}\\mid B)=\\tfrac{1}{3}+\\tfrac{2}{3}=1",
        "relation": "implies",
        "rule": "L_split",
        "conditions": [
          "P(B)>0"
        ],
        "explanation": {
          "zh-Hant": "交叉核對：B∖(A∩B)={4,5,6}=Aᶜ∩B，質量 (2+4+0)/12=1/2（N8）。依 L_split，P(A∩B)+P(Aᶜ∩B)=1/4+1/2=3/4=P(B)（N9），吻合。P(Aᶜ|B)=(1/2)/(3/4)=2/3（N10），與 P(A|B) 相加得 1（N11）：在 B 之下，A 與 Aᶜ 的條件機率加總為 1。",
          "en": "Cross-check: B∖(A∩B)={4,5,6}=Aᶜ∩B, with mass (2+4+0)/12=1/2 (N8). By L_split, P(A∩B)+P(Aᶜ∩B)=1/4+1/2=3/4=P(B) (N9), as it should. P(Aᶜ|B)=(1/2)/(3/4)=2/3 (N10), and adding P(A|B) gives 1 (N11): given B, the conditional probabilities of A and Aᶜ sum to 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "L_split",
            "S03",
            "S06",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "這是獨立的一致性檢查，不是 P(A|B) 的推導所必需。",
            "en": "This is an independent consistency check, not needed for deriving P(A|B)."
          }
        },
        "leaf_basis": "weight_model + set_operations_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S11",
        "uses": [
          "S06",
          "G1",
          "G2",
          "mko_cond_prob_def"
        ],
        "before_tex": "P(B\\mid A)=\\frac{P(B\\cap A)}{P(A)}",
        "after_tex": "P(A)=\\tfrac{5}{12}>0,\\quad P(B\\mid A)=\\frac{1/4}{5/12}=\\frac{3}{5}\\neq\\frac{1}{3}",
        "relation": "implies",
        "rule": "mko_cond_prob_def",
        "conditions": [
          "P(A)>0"
        ],
        "explanation": {
          "zh-Hant": "反方向要另外確認分母：P(A)=w(2)+w(3)=(2+3)/12=5/12>0（N12），所以 P(B|A) 有定義。B∩A=A∩B={3}，P(B|A)=(1/4)/(5/12)=(1/4)·(12/5)=3/5（N13）。3/5≠1/3（N14）：兩個方向的條件機率一般不同，因為分母不同。",
          "en": "The other direction needs its own denominator check: P(A)=w(2)+w(3)=(2+3)/12=5/12>0 (N12), so P(B|A) is defined. B∩A=A∩B={3}, and P(B|A)=(1/4)/(5/12)=(1/4)·(12/5)=3/5 (N13). 3/5≠1/3 (N14): conditional probabilities in the two directions generally differ, because their denominators differ."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_cond_prob_def",
            "set_operations_def",
            "frac_rules"
          ],
          "explanation": {
            "zh-Hant": "兩個方向各自需要正分母（MKO 第三段）；交集可交換，但分母不能互換。",
            "en": "Each direction needs its own positive denominator (MKO paragraph three); the intersection is symmetric, but the denominators may not be swapped."
          }
        },
        "leaf_basis": "mko_cond_prob_def + set_operations_def + frac_rules + numeric_evaluation"
      },
      {
        "id": "S12",
        "uses": [
          "S05",
          "S07",
          "G1"
        ],
        "before_tex": "\\frac{|A\\cap B|}{|B|}=\\frac{1}{4}",
        "after_tex": "\\frac{1}{4}\\neq\\frac{1}{3}=P(A\\mid B)",
        "relation": "implies",
        "rule": "mko_cond_prob_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "若誤把模型當成等可能，會用計數比 |A∩B|/|B|=1/4。但本模型權重不相等（例如 w(5)=4/12≠w(6)=0），而且沒有宣告等可能（MKO 假設 a2），所以計數比不是這個模型的條件機率；正確值是 1/3（N15）。",
          "en": "Treating the model as equiprobable by mistake would give the count ratio |A∩B|/|B|=1/4. But the weights here are unequal (for example w(5)=4/12≠w(6)=0) and no equiprobable model was declared (MKO assumption a2), so the count ratio is not this model's conditional probability; the correct value is 1/3 (N15)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_cond_prob_def",
            "weight_model"
          ],
          "explanation": {
            "zh-Hant": "只需一組不相等的權重就說明模型不是等可能；此步只排除錯誤算法。",
            "en": "One pair of unequal weights shows the model is not equiprobable; this step only rules out a wrong method."
          }
        },
        "leaf_basis": "mko_cond_prob_def + weight_model + numeric_evaluation"
      },
      {
        "id": "S13",
        "uses": [
          "S03",
          "S07",
          "S11"
        ],
        "before_tex": "P(A)P(B)=\\tfrac{5}{12}\\cdot\\tfrac{3}{4}",
        "after_tex": "\\tfrac{5}{16}\\neq\\tfrac{1}{3}=P(A\\mid B)",
        "relation": "implies",
        "rule": "frac_rules",
        "conditions": [],
        "explanation": {
          "zh-Hant": "條件機率不是把 P(A) 乘上 P(B)：P(A)P(B)=(5/12)(3/4)=15/48=5/16（N16），而 P(A|B)=1/3，兩者不等（N17）。正確的乘法關係是 P(A∩B)=P(A|B)·P(B)=(1/3)(3/4)=1/4（S07 的反向核對）。",
          "en": "Conditional probability is not P(A) times P(B): P(A)P(B)=(5/12)(3/4)=15/48=5/16 (N16), while P(A|B)=1/3, and the two differ (N17). The correct multiplicative relation is P(A∩B)=P(A|B)·P(B)=(1/3)(3/4)=1/4 (the backward check of S07)."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "frac_rules",
            "numeric_evaluation"
          ],
          "explanation": {
            "zh-Hant": "只需精確比較兩個有理數；本稿不由此推論任何獨立性結論。",
            "en": "Only an exact comparison of two rationals is needed; no conclusion about independence is drawn here."
          }
        },
        "leaf_basis": "frac_rules + numeric_evaluation"
      },
      {
        "id": "S14",
        "uses": [
          "G1",
          "G2",
          "mko_cond_prob_def",
          "mul_inverse_nonzero"
        ],
        "before_tex": "Z=\\{6\\}\\neq\\varnothing,\\ P(Z)=w(6)",
        "after_tex": "P(Z)=0,\\ \\text{so}\\ P(A\\mid Z)\\ \\text{is undefined}",
        "relation": "implies",
        "rule": "mko_cond_prob_def",
        "conditions": [],
        "explanation": {
          "zh-Hant": "Z 非空（6∈Z），但 P(Z)=w(6)=0。MKO 明示：P(B)=0 時本條目不定義條件機率，即使 B 非空。式子 P(A∩Z)/P(Z) 的分母為 0，而 0 沒有倒數，所以 P(A|Z) 沒有定義；答案不是 0，也不是 1。",
          "en": "Z is nonempty (6∈Z), yet P(Z)=w(6)=0. The MKO states that this entry leaves the conditional probability undefined when P(B)=0, even if B is nonempty. The expression P(A∩Z)/P(Z) has denominator 0, and 0 has no inverse, so P(A|Z) is undefined; the answer is neither 0 nor 1."
        },
        "reverse_audit": {
          "kind": "premise_audit",
          "required_ids": [
            "mko_cond_prob_def",
            "mul_inverse_nonzero"
          ],
          "explanation": {
            "zh-Hant": "判定只依 P(Z)=0；非空不足以條件化。不宣稱其他理論中零機率條件化的處理。",
            "en": "The verdict rests only on P(Z)=0; nonemptiness is not enough for conditioning. Nothing is claimed about how other theories handle conditioning on null events."
          }
        },
        "leaf_basis": "mko_cond_prob_def + mul_inverse_nonzero + weight_model"
      },
      {
        "id": "S15",
        "uses": [
          "S01",
          "S07",
          "S09",
          "S11",
          "S12",
          "S13",
          "S14"
        ],
        "before_tex": "\\text{S01 to S14}",
        "after_tex": "P(A\\mid B)=\\tfrac{1}{3},\\ P(B\\mid A)=\\tfrac{3}{5},\\ P(A\\mid Z)\\ \\text{undefined}",
        "relation": "implies",
        "rule": "and_intro_elim",
        "conditions": [],
        "explanation": {
          "zh-Hant": "彙整：模型合法（S01）；P(B)=3/4>0，P(A|B)=1/3（S07），且在 [0,1] 中（S09）；P(A)=5/12>0，P(B|A)=3/5≠1/3（S11）；計數比 1/4 與乘積 5/16 都不是 P(A|B)（S12、S13）；P(Z)=0，P(A|Z) 無定義（S14）。",
          "en": "Summary: the model is valid (S01); P(B)=3/4>0 and P(A|B)=1/3 (S07), which lies in [0,1] (S09); P(A)=5/12>0 and P(B|A)=3/5≠1/3 (S11); neither the count ratio 1/4 nor the product 5/16 is P(A|B) (S12, S13); P(Z)=0, so P(A|Z) is undefined (S14)."
        },
        "reverse_audit": {
          "kind": "sufficient_goal_reduction",
          "required_ids": [
            "S01",
            "S07",
            "S11",
            "S12",
            "S13",
            "S14",
            "and_intro_elim"
          ],
          "explanation": {
            "zh-Hant": "目標 G：題目各項。足夠條件 H：所列各步的結論；H⇒G 只是合取，H 由 G1、G2 正向證成，沒有把 G 當作前提。",
            "en": "Goal G: the items of the task. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction, H is proved forwards from G1 and G2, and G is never used as a premise."
          }
        },
        "leaf_basis": "and_intro_elim"
      }
    ],
    "result": {
      "zh-Hant": "權重 (1,2,3,2,4,0)/12 構成合法的有限模型。P(B)=3/4>0，P(A∩B)=1/4，所以 P(A|B)=1/3；P(A)=5/12>0，所以 P(B|A)=3/5，兩個方向不同。等可能計數比 1/4 與乘積 P(A)P(B)=5/16 都不是 P(A|B)。Z={6} 非空但 P(Z)=0，所以 P(A|Z) 在本條目中沒有定義。",
      "en": "The weights (1,2,3,2,4,0)/12 form a valid finite model. P(B)=3/4>0 and P(A∩B)=1/4, so P(A|B)=1/3; P(A)=5/12>0, so P(B|A)=3/5, and the two directions differ. Neither the equiprobable count ratio 1/4 nor the product P(A)P(B)=5/16 is P(A|B). Z={6} is nonempty but P(Z)=0, so P(A|Z) is undefined in this entry."
    },
    "negative_controls": [
      {
        "claim": {
          "zh-Hant": "Z={6} 非空，所以 P(A|Z) 有定義",
          "en": "Z={6} is nonempty, so P(A|Z) is defined"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P(Z)=w(6)=0；MKO 要求分母嚴格大於零。",
          "en": "P(Z)=w(6)=0; the MKO requires a strictly positive denominator."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(Z)=0，所以 P(A|Z)=0",
          "en": "P(Z)=0, so P(A|Z)=0"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "分母為 0 時式子沒有定義，不是 0。",
          "en": "With denominator 0 the expression is undefined, not 0."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(B|A)=P(A|B)=1/3",
          "en": "P(B|A)=P(A|B)=1/3"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "P(B|A)=(1/4)/(5/12)=3/5≠1/3。",
          "en": "P(B|A)=(1/4)/(5/12)=3/5≠1/3."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(A|B)=|A∩B|/|B|=1/4",
          "en": "P(A|B)=|A∩B|/|B|=1/4"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "模型不是等可能（w(5)=4/12、w(6)=0）；正確值 3/9=1/3。",
          "en": "The model is not equiprobable (w(5)=4/12, w(6)=0); the correct value is 3/9=1/3."
        }
      },
      {
        "claim": {
          "zh-Hant": "P(A|B)=P(A)P(B)=5/16",
          "en": "P(A|B)=P(A)P(B)=5/16"
        },
        "expected": "reject",
        "reason": {
          "zh-Hant": "條件機率是 P(A∩B)/P(B)=1/3，不是兩個機率的乘積。",
          "en": "The conditional probability is P(A∩B)/P(B)=1/3, not a product of two probabilities."
        }
      }
    ],
    "status": "candidate",
    "gaps": [],
    "verification": {
      "structure": "checked_local_script",
      "mathematics": "candidate_argument",
      "formalization": "not_formalized"
    },
    "rule_glossary": [
      {
        "id": "prob_axioms",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "機率公理：P(A)≥0；P(Ω)=1；兩兩互斥事件的（可數）聯集機率等於各機率之和。由此 P(∅)=0、P(Aᶜ)=1−P(A)。",
          "en": "Probability axioms: P(A)≥0; P(Ω)=1; the probability of a (countable) union of pairwise disjoint events is the sum of their probabilities. Hence P(∅)=0 and P(Aᶜ)=1−P(A)."
        }
      },
      {
        "id": "set_operations_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合運算：x∈A∪B ⇔ x∈A 或 x∈B；x∈A∩B ⇔ x∈A 且 x∈B；x∈A∖B ⇔ x∈A 且 x∉B；在全集 U 中 Aᶜ=U∖A。",
          "en": "Set operations: x∈A∪B ⇔ x∈A or x∈B; x∈A∩B ⇔ x∈A and x∈B; x∈A∖B ⇔ x∈A and x∉B; in a universe U, Aᶜ=U∖A."
        }
      },
      {
        "id": "subset_def",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "子集：A⊆B 表示 A 的每個元素都是 B 的元素，即 ∀x (x∈A⇒x∈B)。",
          "en": "Subset: A⊆B means every element of A is an element of B, i.e. ∀x (x∈A⇒x∈B)."
        }
      },
      {
        "id": "frac_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "分數運算（分母皆非零）：a/c+b/c=(a+b)/c；(a/b)(c/d)=(ac)/(bd)；(a/b)/(c/d)=(ad)/(bc)（另需 c≠0）；a/b=(ak)/(bk)（k≠0）。",
          "en": "Fraction rules (all denominators nonzero): a/c+b/c=(a+b)/c; (a/b)(c/d)=(ac)/(bd); (a/b)/(c/d)=(ad)/(bc) (also needing c≠0); a/b=(ak)/(bk) for k≠0."
        }
      },
      {
        "id": "numeric_evaluation",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "明確整數或有理數算式的精確求值；可以獨立重算。",
          "en": "Exact evaluation of an explicit integer or rational expression; independently recomputable."
        }
      },
      {
        "id": "finite_sum_rules",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限和：Σ_{i=1}^{0} a_i=0、Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}；Σ(a_i+b_i)=Σa_i+Σb_i；Σ c·a_i=c·Σa_i。",
          "en": "Finite sums: Σ_{i=1}^{0} a_i=0 and Σ_{i=1}^{n+1} a_i=Σ_{i=1}^{n} a_i + a_{n+1}; Σ(a_i+b_i)=Σa_i+Σb_i; Σ c·a_i=c·Σa_i."
        }
      },
      {
        "id": "order_basics",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "實數的序：任兩數恰有 u<v、u=v、u>v 之一；u<v 且 v<w 推出 u<w；u<v 推出 u+w<v+w；u<v 且 w>0 推出 uw<vw；u<v 且 w<0 推出 uw>vw。",
          "en": "Order of the reals: exactly one of u<v, u=v, u>v holds; u<v and v<w give u<w; u<v gives u+w<v+w; u<v and w>0 give uw<vw; u<v and w<0 give uw>vw."
        }
      },
      {
        "id": "set_extensionality",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "集合相等：A=B 當且僅當 A⊆B 且 B⊆A，即兩者有相同元素。",
          "en": "Set equality: A=B if and only if A⊆B and B⊆A, i.e. they have the same elements."
        }
      },
      {
        "id": "mul_inverse_nonzero",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "非零數的倒數：在 ℚ、ℝ、ℂ 中，若 u≠0，存在 1/u 使 u·(1/u)=1；對 u≠0 定義 v/u 為 v·(1/u)。u=0 時沒有倒數，v/0 不定義。",
          "en": "Inverse of a nonzero number: in ℚ, ℝ or ℂ, if u≠0 there is 1/u with u·(1/u)=1, and v/u means v·(1/u). 0 has no inverse, so v/0 is undefined."
        }
      },
      {
        "id": "and_intro_elim",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "合取：由 P 與 Q 得 P∧Q；由 P∧Q 可分別得 P 與 Q。",
          "en": "Conjunction: from P and Q infer P∧Q; from P∧Q infer P and infer Q."
        }
      },
      {
        "id": "mko_cond_prob_def",
        "kind": "cited",
        "source": {
          "mko_id": "mko-conditional-probability",
          "sha256": "7b0f4824aa1f3e4c64ea306c5bed04843566b8e4bbdf96d94dbc1367dbb68419",
          "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f"
        },
        "statement": {
          "zh-Hant": "在非空有限樣本空間 S、全冪集事件族 F=𝒫(S) 與已歸一化非負有理權重下，對 A、B⊆S 且 P(B)>0，定義 C(A,B)=P(A|B)=P(A∩B)/P(B)。若 P(B)=0，本條目不定義這個條件機率，即使 B 非空。",
          "en": "In a nonempty finite sample space S with event family F=𝒫(S) and normalized nonnegative rational weights, let A,B⊆S with P(B)>0. Define C(A,B)=P(A|B)=P(A∩B)/P(B). If P(B)=0, this entry leaves that conditional probability undefined, even when B is nonempty."
        }
      },
      {
        "id": "weight_model",
        "kind": "prerequisite",
        "statement": {
          "zh-Hant": "有限權重模型：w:S→ℚ，對每個 s 有 w(s)≥0，且 Σ_{s∈S} w(s)=1；事件 E⊆S 的機率是 P(E)=Σ_{s∈E} w(s)（空和為 0）。",
          "en": "Finite weighted model: w:S→ℚ with w(s)≥0 for every s and Σ_{s∈S} w(s)=1; the probability of an event E⊆S is P(E)=Σ_{s∈E} w(s) (the empty sum is 0)."
        }
      },
      {
        "id": "L_split",
        "kind": "lemma",
        "proved_in": [
          "S09.1"
        ],
        "statement": {
          "zh-Hant": "在有限權重模型中，若 E⊆E′⊆S，則 P(E′)=P(E)+P(E′∖E)，因此 P(E)≤P(E′)。",
          "en": "In a finite weighted model, if E⊆E′⊆S then P(E′)=P(E)+P(E′∖E), and hence P(E)≤P(E′)."
        }
      }
    ],
    "revision_log": [],
    "numeric_checks": [
      {
        "id": "N1",
        "step": "S01",
        "lhs": "(1+2+3+2+4+0)/12",
        "rhs": "1"
      },
      {
        "id": "N2",
        "step": "S03",
        "lhs": "(3+2+4+0)/12",
        "rhs": "3/4"
      },
      {
        "id": "N3",
        "step": "S06",
        "lhs": "3/12",
        "rhs": "1/4"
      },
      {
        "id": "N4",
        "step": "S07",
        "lhs": "(1/4)/(3/4)",
        "rhs": "1/3"
      },
      {
        "id": "N5",
        "step": "S07",
        "lhs": "(1/3)*(3/4)",
        "rhs": "1/4"
      },
      {
        "id": "N6",
        "step": "S08",
        "lhs": "(3/12)/(9/12)",
        "rhs": "3/9"
      },
      {
        "id": "N7",
        "step": "S08",
        "lhs": "3/9",
        "rhs": "1/3"
      },
      {
        "id": "N8",
        "step": "S10",
        "lhs": "(2+4+0)/12",
        "rhs": "1/2"
      },
      {
        "id": "N9",
        "step": "S10",
        "lhs": "1/4+1/2",
        "rhs": "3/4"
      },
      {
        "id": "N10",
        "step": "S10",
        "lhs": "(1/2)/(3/4)",
        "rhs": "2/3"
      },
      {
        "id": "N11",
        "step": "S10",
        "lhs": "1/3+2/3",
        "rhs": "1"
      },
      {
        "id": "N12",
        "step": "S11",
        "lhs": "(2+3)/12",
        "rhs": "5/12"
      },
      {
        "id": "N13",
        "step": "S11",
        "lhs": "(1/4)/(5/12)",
        "rhs": "3/5"
      },
      {
        "id": "N14",
        "step": "S11",
        "lhs": "3/5",
        "relation": "!=",
        "rhs": "1/3"
      },
      {
        "id": "N15",
        "step": "S12",
        "lhs": "1/4",
        "relation": "!=",
        "rhs": "1/3"
      },
      {
        "id": "N16",
        "step": "S13",
        "lhs": "(5/12)*(3/4)",
        "rhs": "5/16"
      },
      {
        "id": "N17",
        "step": "S13",
        "lhs": "5/16",
        "relation": "!=",
        "rhs": "1/3"
      },
      {
        "id": "N18",
        "step": "S09",
        "lhs": "1/4",
        "relation": "<=",
        "rhs": "3/4"
      }
    ]
  },
  "presentation": {
    "md_scope": {
      "zh-Hant": "在 MKO 的範圍內工作：非空有限樣本空間、全冪集事件族、已歸一化的非負有理權重。只在 P(B)>0 時定義 P(A|B)=P(A∩B)/P(B)，並證明一個加法小引理（E⊆E′ 時 P(E′)=P(E)+P(E′∖E)）以說明 0≤P(A|B)≤1。不定義零機率條件化；不討論事件獨立性、貝氏定理或無限樣本空間；不改動 canonical MKO 或英文 edition。",
      "en": "Works inside the MKO's scope: a nonempty finite sample space, the full power set as event family, normalized nonnegative rational weights. Defines P(A|B)=P(A∩B)/P(B) only when P(B)>0, and proves a small additivity lemma (P(E′)=P(E)+P(E′∖E) for E⊆E′) to show 0≤P(A|B)≤1. Leaves conditioning on null events undefined; does not discuss independence, Bayes' theorem or infinite sample spaces; does not change the canonical MKO or its English edition."
    },
    "md_symbols": [
      [
        "S",
        "樣本空間 {1,2,3,4,5,6}",
        "the sample space {1,2,3,4,5,6}"
      ],
      [
        "w(s)",
        "樣本點 s 的權重",
        "the weight of the sample point s"
      ],
      [
        "P(E)",
        "事件 E 的機率 Σ_{s∈E} w(s)",
        "the probability Σ_{s∈E} w(s) of the event E"
      ],
      [
        "P(A\\mid B)",
        "在 B 之下 A 的條件機率，即 C(A,B)",
        "the conditional probability of A given B, i.e. C(A,B)"
      ],
      [
        "A^{c}",
        "S∖A",
        "S∖A"
      ]
    ],
    "md_error_controls": [
      {
        "title": "非空不等於正機率 / Nonempty does not mean positive probability",
        "zh": "MKO 指出的第一個誤解是把非空當正機率。Z={6} 非空，但 w(6)=0，所以 P(Z)=0，不能在 Z 上條件化（S14）。條件化前要檢查的是 P(B)>0（S04），不是 B≠∅。",
        "en": "The first misconception named by the MKO is treating nonempty as positive probability. Z={6} is nonempty, but w(6)=0, so P(Z)=0 and one cannot condition on Z (S14). Before conditioning, what must be checked is P(B)>0 (S04), not B≠∅."
      },
      {
        "title": "零機率條件不給零答案 / Conditioning on a null event does not give zero",
        "zh": "MKO 的第二個誤解是對零機率條件給零答案。P(A|Z) 的式子分母為 0，0 沒有倒數，所以本條目中 P(A|Z) 沒有定義；寫成 0 或 1 都是錯的。",
        "en": "The second misconception identified in the MKO is answering zero for a null conditioning event. The formula for P(A|Z) has denominator 0, which has no inverse, so P(A|Z) is undefined in this entry; writing 0 or 1 is wrong."
      },
      {
        "title": "兩個方向不能混同 / The two directions must not be confused",
        "zh": "MKO 的第三個誤解是混同兩個方向。P(A|B)=1/3 以 P(B) 為分母，P(B|A)=3/5 以 P(A) 為分母（S07、S11），兩者不同，且各自都要先確認分母為正。同樣，條件機率不是 P(A)P(B)=5/16（S13），在非等可能模型中也不是計數比 1/4（S12）。",
        "en": "The third misconception identified in the MKO is confusing the two directions. P(A|B)=1/3 has denominator P(B), while P(B|A)=3/5 has denominator P(A) (S07, S11); they differ, and each needs its own positive-denominator check. Likewise, conditional probability is not P(A)P(B)=5/16 (S13), nor, in a non-equiprobable model, the count ratio 1/4 (S12)."
      }
    ],
    "md_open_items": [
      {
        "zh-Hant": "本稿不討論事件獨立性；S13 只比較數值，不據此宣稱 A、B 是否獨立。",
        "en": "This guide does not discuss independence; S13 only compares numbers and makes no claim about whether A and B are independent."
      },
      {
        "zh-Hant": "零機率條件化在其他理論（例如以極限或測度論的條件期望）中另有處理，不在本條目範圍。",
        "en": "Conditioning on null events is handled elsewhere in other theories (for example via limits or measure-theoretic conditional expectation), outside this entry's scope."
      }
    ]
  },
  "revision": {
    "id": "production-20261019-adopted-96",
    "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f",
    "adopted_source_sha256": "0c101f249b943132cf4ae7daef5fdcdd38e0b9d888feacc77fc0a88e0e9198d3",
    "original_source_sha256": "0c101f249b943132cf4ae7daef5fdcdd38e0b9d888feacc77fc0a88e0e9198d3",
    "owner_adopted": true,
    "bilingual_sync": "complete_source_bound",
    "formal_Lean": "NOT_RUN",
    "independent_AI_concurrence": "NOT_MEASURED",
    "issue_ids": [],
    "scope": "Owner-adopted natural-language teaching guide; not a formal proof, canonical MKO rewrite or evidence-status promotion."
  }
}