{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-probability-01", "mko_id": "mko-probability", "title": { "zh-Hant": "有限樣本空間上的機率：驗證權重模型、計算事件機率與補事件", "en": "Probability on finite sample spaces: validating a weight model, computing event probabilities and complements" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-probability-01", "source": { "mko_id": "mko-probability", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "07f2b780895b31a69725da388129f9624510427b4a2937b15a61418f94e267ca", "english_sha256": "916cfce506d3d89365ae56463d792e878efa37eeafd9f269c15a87f152202f75" }, "mode": "worked_example", "audience_prerequisites": [ "finite_sum_rules", "set_operations_def", "subset_def", "order_basics", "frac_rules", "numeric_evaluation" ], "goal": { "zh-Hant": "S={1,2,3,4}，事件族為全冪集，權重 w_1=1/2、w_2=1/3、w_3=1/6、w_4=0。驗證這是合法的有限機率模型；由定義證明 P(∅)=0、P(S)=1、互斥可加與 P(S∖E)=1−P(E)；計算 P({1,3})，說明它不等於 |E|/|S|；說明 {4} 非空但機率為 0；並判定兩組候選權重 (1/2,1/3,1/3,0) 與 (1/2,2/3,−1/6,0) 都不合法。", "en": "Let S={1,2,3,4} with the full powerset as event family and weights w_1=1/2, w_2=1/3, w_3=1/6, w_4=0. Verify that this is a legal finite probability model; prove from the definition the identities P(∅)=0 and P(S)=1, additivity for disjoint events, and the complement identity P(S∖E)=1−P(E); compute P({1,3}) and explain why it differs from |E|/|S|; show that {4} is nonempty yet has probability 0; and decide that the candidate weights (1/2,1/3,1/3,0) and (1/2,2/3,−1/6,0) are both illegal." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "S={1,2,3,4}，四個互異的結果標籤；事件族 F=𝒫(S)；權重 w_1=1/2、w_2=1/3、w_3=1/6、w_4=0（MKO 假設 a1、a2）。", "en": "S={1,2,3,4}, four distinct outcome labels; event family F=𝒫(S); weights w_1=1/2, w_2=1/3, w_3=1/6, w_4=0 (MKO assumptions a1, a2)." } }, { "id": "G2", "statement": { "zh-Hant": "要計算的事件：E={1,3}、Z={4}，以及 A={1,2}、C={2,3}。", "en": "Events to compute: E={1,3}, Z={4}, and A={1,2}, C={2,3}." } }, { "id": "G3", "statement": { "zh-Hant": "比較用的候選權重：w′=(1/2,1/3,1/3,0) 與 w″=(1/2,2/3,−1/6,0)，依序對應結果 1、2、3、4。", "en": "Candidate weights for comparison: w′=(1/2,1/3,1/3,0) and w″=(1/2,2/3,−1/6,0), for outcomes 1, 2, 3, 4 in order." } } ], "steps": [ { "id": "S01", "uses": [ "G1" ], "before_tex": "(S,\\mathcal{P}(S),w)", "after_tex": "P(E)=\\sum_{s\\in E}w_s\\quad(E\\subseteq S),\\ \\text{provided}\\ w_s\\geq 0,\\ \\sum_{s\\in S}w_s=1", "relation": "definition", "rule": "mko_prob_def", "conditions": [ "S\\ \\text{finite, nonempty}" ], "explanation": { "zh-Hant": "依 MKO，有限機率模型由結果的權重決定：事件的機率是其成員權重之和。這個定義只有在權重合法（每個非負、總和恰為 1）時才給出機率律，所以必須先驗證權重（S03、S04），再計算任何事件。", "en": "By the MKO, a finite probability model is determined by the weights of the outcomes: the probability of an event is the sum of the weights of its members. This definition gives a probability law only when the weights are legal (each nonnegative, summing to exactly 1), so the weights must be validated first (S03, S04) before any event is computed." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "mko_prob_def" ], "explanation": { "zh-Hant": "此步只寫出定義與合法性條件，不推出新事實。", "en": "This step only states the definition and the legality conditions; no new fact is derived." } }, "leaf_basis": "mko_prob_def + def_intro" }, { "id": "S02", "uses": [ "G1", "G2", "subset_def" ], "before_tex": "E=\\{1,3\\},\\ Z=\\{4\\},\\ A=\\{1,2\\},\\ C=\\{2,3\\}", "after_tex": "E,Z,A,C\\subseteq S,\\ \\text{hence}\\ E,Z,A,C\\in\\mathcal{P}(S)", "relation": "implies", "rule": "subset_def", "conditions": [], "explanation": { "zh-Hant": "S 有限、非空，四個標籤互異。各事件的每個元素都在 S={1,2,3,4} 中，所以都是 S 的子集合；事件族是全冪集，因此它們都是事件（MKO 假設 a2：E 必須是 S 的子集合）。", "en": "S is finite and nonempty with four distinct labels. Every element of each of these sets lies in S={1,2,3,4}, so each is a subset of S; the event family is the full powerset, so they are all events (MKO assumption a2: E must be a subset of S)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "subset_def", "G1" ], "explanation": { "zh-Hant": "逐一檢查有限集合的元素，是完整檢查。", "en": "The elements of these finite sets are checked one by one; this is a complete check." } }, "leaf_basis": "subset_def + set_builder" }, { "id": "S03", "uses": [ "G1", "order_basics" ], "before_tex": "w_1=\\tfrac12,\\ w_2=\\tfrac13,\\ w_3=\\tfrac16,\\ w_4=0", "after_tex": "w_s\\geq 0\\ \\text{for all}\\ s\\in S", "relation": "implies", "rule": "order_basics", "conditions": [], "explanation": { "zh-Hant": "1/2、1/3、1/6 都是正有理數，w_4=0 也滿足 ≥0。四個結果逐一檢查，所以所有權重非負。", "en": "1/2, 1/3 and 1/6 are positive rationals and w_4=0 satisfies ≥0 as well. All four outcomes are checked, so every weight is nonnegative." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "order_basics", "numeric_evaluation" ], "explanation": { "zh-Hant": "非負是合法性的一半；總和條件另在 S04 檢查。", "en": "Nonnegativity is half of legality; the sum condition is checked separately in S04." } }, "leaf_basis": "order_basics + numeric_evaluation" }, { "id": "S04", "uses": [ "G1", "frac_rules" ], "before_tex": "\\sum_{s\\in S}w_s=\\tfrac12+\\tfrac13+\\tfrac16+0", "after_tex": "\\tfrac36+\\tfrac26+\\tfrac16+\\tfrac06=\\tfrac66=1", "relation": "equivalent", "rule": "frac_rules", "conditions": [], "explanation": { "zh-Hant": "化為共分母 6：1/2=3/6、1/3=2/6、1/6=1/6、0=0/6，相加得 (3+2+1+0)/6=6/6=1。總和恰為 1，連同 S03，權重合法，P 是 S 上的機率律。", "en": "Use the common denominator 6: 1/2=3/6, 1/3=2/6, 1/6=1/6, 0=0/6, and the sum is (3+2+1+0)/6=6/6=1. The sum is exactly 1, so together with S03 the weights are legal and P is a probability law on S." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "frac_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "通分與約分互為反向，精確有理運算可重算（N1、N2）；不使用近似小數。", "en": "Bringing to a common denominator and reducing are inverse moves, and the exact rational arithmetic is recomputable (N1, N2); no approximate decimals are used." } }, "leaf_basis": "frac_rules + numeric_evaluation" }, { "id": "S05", "uses": [ "S01", "sum_split" ], "before_tex": "P(\\varnothing)", "after_tex": "\\sum_{s\\in\\varnothing}w_s=0", "relation": "equivalent", "rule": "sum_split", "conditions": [], "explanation": { "zh-Hant": "依定義 P(∅) 是空集合上的權重和，空和為 0，所以 P(∅)=0。", "en": "By definition P(∅) is the sum of weights over the empty set, and an empty sum is 0, so P(∅)=0." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "sum_split", "finite_sum_rules" ], "explanation": { "zh-Hant": "空和的值是定義的一部分，兩式表示同一個數。", "en": "The value of an empty sum is part of the definition; both expressions name the same number." } }, "leaf_basis": "sum_split + finite_sum_rules" }, { "id": "S06", "uses": [ "S01", "S04" ], "before_tex": "P(S)", "after_tex": "\\sum_{s\\in S}w_s=1", "relation": "equivalent", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "依定義 P(S) 是全部權重之和，由 S04 等於 1。", "en": "By definition P(S) is the sum of all the weights, which equals 1 by S04." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "def_unfold", "S04" ], "explanation": { "zh-Hant": "展開定義並代入 S04，雙向成立。", "en": "Unfolding the definition and substituting S04 works both ways." } }, "leaf_basis": "def_unfold + eq_replace" }, { "id": "S07", "uses": [ "S01", "sum_split" ], "before_tex": "A',C'\\subseteq S,\\ A'\\cap C'=\\varnothing", "after_tex": "P(A'\\cup C')=P(A')+P(C')", "relation": "implies", "rule": "sum_split", "conditions": [ "A'\\cap C'=\\varnothing" ], "explanation": { "zh-Hant": "對任意互斥事件 A′、C′，A′∪C′ 上的權重和可拆成兩部分之和（sum_split），所以 P(A′∪C′)=P(A′)+P(C′)。這是有限情形的可加性；互斥足以保證這個等式，而不互斥時一般不能保證，S15 給出失敗的例子。重疊部分機率為 0 時，即使兩事件不互斥，等式仍可能成立。", "en": "For any disjoint events A′ and C′, the weight sum over A′∪C′ splits into the two partial sums (sum_split), so P(A′∪C′)=P(A′)+P(C′). This is additivity in the finite case: disjointness guarantees the equality, whereas without it the equality is not guaranteed, as S15 illustrates. The equality may still hold for overlapping events when their intersection has probability zero." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "sum_split", "set_operations_def" ], "explanation": { "zh-Hant": "需要 A′∩C′=∅；若有重疊，共同的權重會被加兩次。", "en": "It needs A′∩C′=∅; with an overlap the common weights would be added twice." } }, "leaf_basis": "sum_split + set_operations_def + forall_intro" }, { "id": "S08", "uses": [ "S06", "S07" ], "before_tex": "P(S\\setminus E)", "after_tex": "1-P(E)", "relation": "equivalent", "rule": "L_complement", "conditions": [ "E\\subseteq S" ], "explanation": { "zh-Hant": "依引理 L_complement，補事件的機率是 1 減去原事件的機率；補事件是相對同一個 S 取的。證明在 S08.1、S08.2。", "en": "By Lemma L_complement, the probability of the complement is 1 minus that of the event, with the complement taken relative to the same S. The proof is S08.1 and S08.2." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "L_complement" ], "explanation": { "zh-Hant": "兩邊是同一個數；引理對每個 E⊆S 成立。", "en": "Both sides are the same number; the lemma holds for every E⊆S." } }, "leaf_basis": "sum_split + set_operations_def + eq_congruence" }, { "id": "S08.1", "uses": [ "set_operations_def", "subset_def" ], "before_tex": "E\\subseteq S", "after_tex": "E\\cap(S\\setminus E)=\\varnothing,\\quad E\\cup(S\\setminus E)=S", "relation": "implies", "rule": "set_operations_def", "conditions": [ "E\\subseteq S" ], "explanation": { "zh-Hant": "S∖E 的元素不在 E 中，所以兩者互斥。S 的元素要嘛在 E、要嘛不在 E（因此在 S∖E），所以聯集包含 S；反過來 E⊆S、S∖E⊆S，所以聯集正好是 S。", "en": "Elements of S∖E are not in E, so the two sets are disjoint. Each element of S is either in E or not (and then in S∖E), so the union contains S; conversely E⊆S and S∖E⊆S, so the union is exactly S." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "set_operations_def", "subset_def", "set_extensionality" ], "explanation": { "zh-Hant": "需要 E⊆S；若 E 含 S 以外的元素，E∪(S∖E) 就不是 S。", "en": "It needs E⊆S; if E contained elements outside S, E∪(S∖E) would not be S." } }, "leaf_basis": "set_operations_def + subset_def + set_extensionality" }, { "id": "S08.2", "uses": [ "S08.1", "S07", "S06" ], "before_tex": "P(S)=P(E)+P(S\\setminus E)", "after_tex": "P(S\\setminus E)=1-P(E)", "relation": "equivalent", "rule": "eq_congruence", "conditions": [], "explanation": { "zh-Hant": "由 S08.1 與 S07，P(S)=P(E)+P(S∖E)；由 S06，左邊是 1。兩邊同減 P(E) 得 P(S∖E)=1−P(E)。S08.1 與 S08.2 合起來證得引理 L_complement。", "en": "By S08.1 and S07, P(S)=P(E)+P(S∖E); by S06 the left side is 1. Subtracting P(E) from both sides gives P(S∖E)=1−P(E). S08.1 and S08.2 together prove Lemma L_complement." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "eq_congruence", "add_inverse" ], "explanation": { "zh-Hant": "同減與同加 P(E) 互為反運算；所用的 P(S)=1 來自 S04 的權重驗證。", "en": "Subtracting and adding P(E) undo each other; the fact P(S)=1 used here comes from the weight check in S04." } }, "leaf_basis": "eq_congruence + add_inverse + sum_split" }, { "id": "S09", "uses": [ "S01", "S02", "G1" ], "before_tex": "P(\\{1,3\\})", "after_tex": "w_1+w_3=\\tfrac12+\\tfrac16=\\tfrac46=\\tfrac23", "relation": "equivalent", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "E={1,3}，依定義加總成員的權重：w_1+w_3=1/2+1/6=3/6+1/6=4/6=2/3。", "en": "E={1,3}; by definition add the weights of its members: w_1+w_3=1/2+1/6=3/6+1/6=4/6=2/3." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "def_unfold", "frac_rules", "numeric_evaluation" ], "explanation": { "zh-Hant": "精確有理運算，可重算（N3）。", "en": "Exact rational arithmetic, recomputable (N3)." } }, "leaf_basis": "def_unfold + frac_rules + numeric_evaluation" }, { "id": "S10", "uses": [ "S09", "G1" ], "before_tex": "\\tfrac{|E|}{|S|}=\\tfrac24=\\tfrac12", "after_tex": "\\tfrac12\\neq\\tfrac23=P(E)", "relation": "implies", "rule": "mko_prob_def", "conditions": [ "w_1\\neq w_2" ], "explanation": { "zh-Hant": "若誤用有利結果數除以總結果數，會得 |E|/|S|=2/4=1/2，與正確值 2/3 不同。原因是本模型沒有宣告等可能：w_1=1/2≠1/3=w_2。MKO 明示只有所有結果等可能時才可用 |E|/|S|（等可能情形見 S13）。", "en": "Misusing 'favourable over total' gives |E|/|S|=2/4=1/2, which differs from the correct 2/3. The reason is that this model does not declare equiprobability: w_1=1/2≠1/3=w_2. The MKO states that |E|/|S| applies only when all outcomes are equiprobable (the equiprobable case is S13)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "mko_prob_def", "numeric_evaluation" ], "explanation": { "zh-Hant": "一組不等的權重（N5）就足以說明等可能不成立；兩個數值不同（N4）說明公式在此不適用。", "en": "One pair of unequal weights (N5) suffices to show equiprobability fails; the two different values (N4) show the formula does not apply here." } }, "leaf_basis": "mko_prob_def + numeric_evaluation" }, { "id": "S11", "uses": [ "S08", "S09", "G1" ], "before_tex": "P(S\\setminus E)=P(\\{2,4\\})", "after_tex": "w_2+w_4=\\tfrac13+0=\\tfrac13=1-\\tfrac23", "relation": "implies", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "S∖E={2,4}，直接加總得 1/3+0=1/3；由 L_complement 得 1−2/3=1/3。兩種算法一致，這是對 S09 的交叉核對。", "en": "S∖E={2,4}, and adding directly gives 1/3+0=1/3; by L_complement, 1−2/3=1/3. The two computations agree, which cross-checks S09." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S08", "S09", "def_unfold" ], "explanation": { "zh-Hant": "補事件相對同一個 S={1,2,3,4} 取（N6）；換成別的樣本空間就不能用 1−P(E)。", "en": "The complement is taken relative to the same S={1,2,3,4} (N6); in another sample space 1−P(E) could not be used." } }, "leaf_basis": "def_unfold + L_complement + numeric_evaluation" }, { "id": "S12", "uses": [ "S01", "S02", "G1" ], "before_tex": "Z=\\{4\\}", "after_tex": "Z\\neq\\varnothing,\\quad P(Z)=w_4=0", "relation": "implies", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "Z={4} 含結果 4，所以不是空集合；但 P(Z)=w_4=0。因此 P(E)=0 不能推出 E=∅：非空事件的成員權重若全為 0，機率也是 0（MKO para1）。反方向 E=∅ ⇒ P(E)=0 則由 S05 成立。", "en": "Z={4} contains the outcome 4, so it is not empty; yet P(Z)=w_4=0. Hence P(E)=0 does not imply E=∅: a nonempty event whose members all have weight 0 has probability 0 (MKO para1). The other direction, E=∅ ⇒ P(E)=0, does hold by S05." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "def_unfold", "S05" ], "explanation": { "zh-Hant": "一個反例即否定「P(E)=0 ⇒ E=∅」這個一般命題。", "en": "One counterexample refutes the general claim 'P(E)=0 ⇒ E=∅'." } }, "leaf_basis": "def_unfold + contradiction" }, { "id": "S13", "uses": [ "S01", "finite_sum_rules" ], "before_tex": "w_s=\\tfrac{1}{|S|}\\ \\text{for all}\\ s\\in S", "after_tex": "P(E)=\\tfrac{|E|}{|S|}", "relation": "implies", "rule": "L_equiprob", "conditions": [ "\\text{equiprobable}" ], "explanation": { "zh-Hant": "依引理 L_equiprob，若每個權重都等於 1/|S|，則對所有事件 E⊆S 都有 P(E)=|E|/|S|；證明在 S13.1。非均勻模型的個別事件仍可能恰好符合這個等式，不能從單一事件的等式反推權重均勻。例如若改用等可能權重 1/4，則 P({1,3})=2/4=1/2——但那是另一個模型，不是 G1 的模型。", "en": "By Lemma L_equiprob, if every weight equals 1/|S|, then P(E)=|E|/|S| holds for every event E⊆S; the proof is S13.1. An individual event in a nonuniform model may still happen to satisfy this equality, so one event's equality does not imply uniform weights. For instance, with the equiprobable weights 1/4 one would get P({1,3})=2/4=1/2, but that is a different model, not the model of G1." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_equiprob", "finite_sum_rules" ], "explanation": { "zh-Hant": "計數公式是等可能假設下的推論，不是一般定義；G1 的模型不滿足此假設。", "en": "The counting formula is a consequence of the equiprobability assumption, not the general definition; the model of G1 does not satisfy that assumption." } }, "leaf_basis": "L_equiprob + finite_sum_rules" }, { "id": "S13.1", "uses": [ "finite_sum_rules", "S01" ], "before_tex": "\\sum_{s\\in E}\\tfrac{1}{|S|}", "after_tex": "|E|\\cdot\\tfrac{1}{|S|}=\\tfrac{|E|}{|S|}", "relation": "equivalent", "rule": "finite_sum_rules", "conditions": [ "|S|\\geq 1" ], "explanation": { "zh-Hant": "若每個 w_s=1/|S|，P(E) 是 |E| 個相同項 1/|S| 之和；依 finite_sum_rules 的常數倍規則（Σ c·1=c·Σ 1），等於 |E|·(1/|S|)=|E|/|S|。S 非空，所以 |S|≥1，分母不為零。這證得引理 L_equiprob。", "en": "If every w_s=1/|S|, P(E) is a sum of |E| equal terms 1/|S|; by the constant-multiple rule of finite_sum_rules (Σ c·1=c·Σ 1), it equals |E|·(1/|S|)=|E|/|S|. S is nonempty, so |S|≥1 and the denominator is nonzero. This proves Lemma L_equiprob." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "finite_sum_rules", "frac_rules" ], "explanation": { "zh-Hant": "常數項的和等於項數乘常數，雙向成立；需要 S 非空（N7 以 |E|=2、|S|=4 為例）。", "en": "A sum of a constant equals the number of terms times the constant, valid both ways; it needs S nonempty (N7 illustrates with |E|=2 and |S|=4)." } }, "leaf_basis": "finite_sum_rules + frac_rules" }, { "id": "S14", "uses": [ "G3", "S01" ], "before_tex": "w'=(\\tfrac12,\\tfrac13,\\tfrac13,0),\\quad w''=(\\tfrac12,\\tfrac23,-\\tfrac16,0)", "after_tex": "\\sum w'=\\tfrac76\\neq 1;\\quad w''_3=-\\tfrac16<0", "relation": "implies", "rule": "mko_prob_def", "conditions": [], "explanation": { "zh-Hant": "w′ 的權重都非負，但總和 3/6+2/6+2/6+0=7/6≠1，用它會得到 P(S)=7/6，不是機率律。w″ 的總和 3/6+4/6−1/6+0=6/6=1，但 w″_3=−1/6<0，會給出 P({3})<0。兩者都違反 MKO 的合法性條件，所以都不能用來計算機率。合法性兩個條件缺一不可。", "en": "The weights of w′ are nonnegative, but they sum to 3/6+2/6+2/6+0=7/6≠1, which would give P(S)=7/6, so it is not a probability law. The weights of w″ sum to 3/6+4/6−1/6+0=6/6=1, but w″_3=−1/6<0 would give P({3})<0. Both violate the MKO's legality conditions and cannot be used to compute probabilities. Neither of the two conditions may be dropped." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "mko_prob_def", "numeric_evaluation", "order_basics" ], "explanation": { "zh-Hant": "每個候選只需一個條件不成立即被拒絕（N8 至 N10）。", "en": "Each candidate is rejected as soon as one condition fails (N8 to N10)." } }, "leaf_basis": "mko_prob_def + order_basics + numeric_evaluation" }, { "id": "S15", "uses": [ "G2", "S01", "S06" ], "before_tex": "A=\\{1,2\\},\\ C=\\{2,3\\},\\ A\\cap C=\\{2\\}", "after_tex": "P(A)+P(C)=\\tfrac56+\\tfrac12=\\tfrac43\\neq 1=P(A\\cup C)", "relation": "implies", "rule": "def_unfold", "conditions": [ "A\\cap C\\neq\\varnothing" ], "explanation": { "zh-Hant": "A、C 有共同結果 2，不互斥。P(A)=1/2+1/3=5/6，P(C)=1/3+1/6=1/2，和為 4/3；但 A∪C={1,2,3}，P(A∪C)=1/2+1/3+1/6=1。差 1/3 正是共同結果 2 的權重被加了兩次。所以 S07 的互斥條件不能省略。", "en": "A and C share the outcome 2, so they are not disjoint. P(A)=1/2+1/3=5/6 and P(C)=1/3+1/6=1/2, summing to 4/3; but A∪C={1,2,3} and P(A∪C)=1/2+1/3+1/6=1. The difference 1/3 is exactly the weight of the shared outcome 2, counted twice. So the disjointness condition of S07 cannot be dropped." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "def_unfold", "numeric_evaluation", "S07" ], "explanation": { "zh-Hant": "一個反例（N11 至 N13）說明沒有互斥時加法不成立；不影響 S07 對互斥事件的結論。", "en": "One counterexample (N11 to N13) shows that additivity fails without disjointness; it does not affect S07's conclusion for disjoint events." } }, "leaf_basis": "def_unfold + numeric_evaluation" }, { "id": "S16", "uses": [ "S04", "S05", "S06", "S08", "S09", "S10", "S12", "S14", "S15" ], "before_tex": "\\text{model}\\ (S,\\mathcal{P}(S),w)", "after_tex": "\\text{legal};\\ P(\\varnothing)=0,\\ P(S)=1;\\ P(\\{1,3\\})=\\tfrac23\\neq\\tfrac12;\\ P(\\{4\\})=0;\\ w',w''\\ \\text{illegal}", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "彙整：G1 的權重合法；P(∅)=0、P(S)=1、互斥可加、補事件 P(S∖E)=1−P(E)；P({1,3})=2/3，而非計數公式的 1/2；{4} 非空但機率為 0；w′、w″ 不合法；不互斥時不能直接相加。", "en": "Summary: the weights of G1 are legal; P(∅)=0, P(S)=1, additivity for disjoint events and P(S∖E)=1−P(E) hold; P({1,3})=2/3, not the counting value 1/2; {4} is nonempty with probability 0; w′ and w″ are illegal; overlapping events may not simply be added." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S04", "S05", "S06", "S08", "S09", "S10", "S12", "S14", "S15", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G：各項結論。足夠條件 H：所列各步的結論；H⇒G 只是合取。H 由定義與精確計算正向證成，沒有把 G 當作前提。", "en": "Goal G: all the stated conclusions. Sufficient condition H: the conclusions of the listed steps; H⇒G is just conjunction. H is proved forwards from the definition and exact computation; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "權重 (1/2,1/3,1/6,0) 非負且總和為 1，構成 S={1,2,3,4} 上的合法機率律。由定義得 P(∅)=0、P(S)=1、互斥事件可加、P(S∖E)=1−P(E)。P({1,3})=2/3，不是 |E|/|S|=1/2（本模型不等可能）；P({2,4})=1/3=1−2/3；{4} 非空但 P({4})=0。(1/2,1/3,1/3,0) 總和 7/6、(1/2,2/3,−1/6,0) 含負權重，都不合法。", "en": "The weights (1/2,1/3,1/6,0) are nonnegative and sum to 1, so they form a legal probability law on S={1,2,3,4}. From the definition, P(∅)=0, P(S)=1, disjoint events add, and P(S∖E)=1−P(E). P({1,3})=2/3, not |E|/|S|=1/2 (this model is not equiprobable); P({2,4})=1/3=1−2/3; {4} is nonempty but P({4})=0. The weights (1/2,1/3,1/3,0), summing to 7/6, and (1/2,2/3,−1/6,0), with a negative weight, are both illegal." }, "negative_controls": [ { "claim": { "zh-Hant": "P({1,3})=|{1,3}|/|S|=2/4=1/2", "en": "P({1,3})=|{1,3}|/|S|=2/4=1/2" }, "expected": "reject", "reason": { "zh-Hant": "結果不等可能（w_1=1/2≠1/3=w_2）；正確值是 w_1+w_3=2/3。", "en": "The outcomes are not equiprobable (w_1=1/2≠1/3=w_2); the correct value is w_1+w_3=2/3." } }, { "claim": { "zh-Hant": "P({4})=0，所以 {4} 是空事件", "en": "P({4})=0, so {4} is the empty event" }, "expected": "reject", "reason": { "zh-Hant": "{4} 含結果 4，非空；它的機率為 0 只因 w_4=0。", "en": "{4} contains the outcome 4 and is nonempty; its probability is 0 only because w_4=0." } }, { "claim": { "zh-Hant": "w′=(1/2,1/3,1/3,0) 的權重都非負，所以是合法模型", "en": "w′=(1/2,1/3,1/3,0) has only nonnegative weights, so it is a legal model" }, "expected": "reject", "reason": { "zh-Hant": "總和是 7/6≠1，會給出 P(S)=7/6。", "en": "The weights sum to 7/6≠1, which would give P(S)=7/6." } }, { "claim": { "zh-Hant": "P({1,2}∪{2,3})=P({1,2})+P({2,3})=4/3", "en": "P({1,2}∪{2,3})=P({1,2})+P({2,3})=4/3" }, "expected": "reject", "reason": { "zh-Hant": "兩事件共有結果 2，不互斥；正確值 P({1,2,3})=1。", "en": "The events share the outcome 2 and are not disjoint; the correct value is P({1,2,3})=1." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "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": "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": "order_basics", "kind": "prerequisite", "statement": { "zh-Hant": "實數的序：任兩數恰有 uv 之一；u0 推出 uwvw。", "en": "Order of the reals: exactly one of uv holds; u0 give uwvw." } }, { "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": "def_intro", "kind": "prerequisite", "statement": { "zh-Hant": "引入定義：可為一個命題、式子或物件取名；之後名稱與被定義者可互相替換。", "en": "Introducing a definition: a statement, expression or object may be given a name, after which the name and what it stands for are interchangeable." } }, { "id": "set_builder", "kind": "prerequisite", "statement": { "zh-Hant": "集合建構式：x∈{y∈A ∣ P(y)} 當且僅當 x∈A 且 P(x)。", "en": "Set-builder notation: x∈{y∈A ∣ P(y)} if and only if x∈A and P(x)." } }, { "id": "def_unfold", "kind": "prerequisite", "statement": { "zh-Hant": "展開定義：把一個已定義的詞換成它的定義內容（或反向收回），命題的真假不變。", "en": "Unfolding a definition: replacing a defined term by its definition (or folding it back) does not change whether a statement is true." } }, { "id": "eq_replace", "kind": "prerequisite", "statement": { "zh-Hant": "等值代換：命題中的一個式子可換成與它相等的式子，命題的真假不變。", "en": "Replacement of equals: an expression inside a statement may be replaced by an equal expression without changing whether the statement is true." } }, { "id": "forall_intro", "kind": "prerequisite", "statement": { "zh-Hant": "∀引入（全稱推廣）：若某命題對一個「任意的」元素 x 證成，而證明沒有對 x 加任何額外假設，則它對論域中所有 x 成立。", "en": "∀-introduction (generalization): if a statement is proved for an arbitrary element x, with no extra assumption on x, it holds for every x in the domain." } }, { "id": "eq_congruence", "kind": "prerequisite", "statement": { "zh-Hant": "等式同用：若 u=v，對兩邊做同一個有定義的運算 f，得 f(u)=f(v)。只對一邊做運算不屬於此規則。", "en": "Congruence of equality: if u=v, applying the same defined operation f to both sides gives f(u)=f(v). Operating on one side only is not an instance of this rule." } }, { "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": "add_inverse", "kind": "prerequisite", "statement": { "zh-Hant": "加法反元素：每個 u 都有 −u，使 u+(−u)=0；u−v 定義為 u+(−v)。", "en": "Additive inverse: every u has −u with u+(−u)=0; u−v means u+(−v)." } }, { "id": "contradiction", "kind": "prerequisite", "statement": { "zh-Hant": "反證法：若假設 ¬P 推出矛盾（某命題與其否定同時成立），則 P 成立。", "en": "Proof by contradiction: if assuming ¬P leads to a contradiction (a statement and its negation both hold), then P holds." } }, { "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_prob_def", "kind": "cited", "source": { "mko_id": "mko-probability", "sha256": "07f2b780895b31a69725da388129f9624510427b4a2937b15a61418f94e267ca", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "令 S 為非空有限樣本空間，事件族 F=𝒫(S)。指定各 s∈S 的有理權重 w_s≥0 且 Σ_(s∈S)w_s=1，則事件 E⊆S 的機率 P(E)=Σ_(s∈E)w_s。只有另明示所有結果等可能，才可用 P(E)=|E|/|S|。", "en": "Let S be a nonempty finite sample space and F=𝒫(S) its event family. Assign rational weights w_s≥0 to all s∈S with Σ_(s∈S)w_s=1. For E⊆S, P(E)=Σ_(s∈E)w_s. The formula P(E)=|E|/|S| applies only when all outcomes are explicitly equiprobable." } }, { "id": "sum_split", "kind": "prerequisite", "statement": { "zh-Hant": "有限和的分割：對有限集合 S 上的數 w_s 與互斥子集合 A、C⊆S，Σ_{s∈A∪C} w_s=Σ_{s∈A} w_s+Σ_{s∈C} w_s；空集合上的和為 0。這是 finite_sum_rules 經重新編號所得。", "en": "Splitting a finite sum: for numbers w_s on a finite set S and disjoint subsets A, C⊆S, Σ_{s∈A∪C} w_s=Σ_{s∈A} w_s+Σ_{s∈C} w_s; the sum over the empty set is 0. This follows from finite_sum_rules by reindexing." } }, { "id": "L_complement", "kind": "lemma", "proved_in": [ "S08.1", "S08.2" ], "statement": { "zh-Hant": "在合法的有限模型中，對每個事件 E⊆S，P(S∖E)=1−P(E)。", "en": "In a legal finite model, P(S∖E)=1−P(E) for every event E⊆S." } }, { "id": "L_equiprob", "kind": "lemma", "proved_in": [ "S13.1" ], "statement": { "zh-Hant": "若每個 w_s=1/|S|，則對每個 E⊆S，P(E)=|E|/|S|。", "en": "If every w_s=1/|S|, then P(E)=|E|/|S| for every E⊆S." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S04", "lhs": "1/2+1/3+1/6+0", "rhs": "1" }, { "id": "N2", "step": "S04", "lhs": "3+2+1+0", "rhs": "6" }, { "id": "N3", "step": "S09", "lhs": "1/2+1/6", "rhs": "2/3" }, { "id": "N4", "step": "S10", "lhs": "2/4", "relation": "!=", "rhs": "2/3" }, { "id": "N5", "step": "S10", "lhs": "1/2", "relation": "!=", "rhs": "1/3" }, { "id": "N6", "step": "S11", "lhs": "1/3+0", "rhs": "1-2/3" }, { "id": "N7", "step": "S13.1", "lhs": "2*(1/4)", "rhs": "2/4" }, { "id": "N8", "step": "S14", "lhs": "1/2+1/3+1/3+0", "rhs": "7/6" }, { "id": "N9", "step": "S14", "lhs": "1/2+2/3+(-1/6)+0", "rhs": "1" }, { "id": "N10", "step": "S14", "lhs": "-1/6", "relation": "<", "rhs": "0" }, { "id": "N11", "step": "S15", "lhs": "(1/2+1/3)+(1/3+1/6)", "rhs": "4/3" }, { "id": "N12", "step": "S15", "lhs": "1/2+1/3+1/6", "rhs": "1" }, { "id": "N13", "step": "S15", "lhs": "4/3-1", "rhs": "1/3" }, { "id": "N14", "step": "S03", "lhs": "1/6", "relation": ">=", "rhs": "0" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內：非空有限樣本空間、全冪集事件族、精確有理權重（非負且總和恰為一），事件機率是成員權重之和。不推論真實實驗的機率，不處理獨立性、條件機率或一般測度論（MKO para2）；不改動 canonical MKO 或英文 edition。", "en": "Within the MKO's scope: a nonempty finite sample space, the full powerset as event family, exact rational weights (nonnegative, summing to exactly one), and event probability as the sum of member weights. It infers no probabilities of real experiments and does not treat independence, conditional probability or general measure theory (MKO para2); the canonical MKO and its English edition are not changed." }, "md_symbols": [ [ "S", "樣本空間 {1,2,3,4}", "the sample space {1,2,3,4}" ], [ "\\mathcal{P}(S)", "S 的全冪集（事件族）", "the full powerset of S (the event family)" ], [ "w_s", "結果 s 的有理權重", "the rational weight of outcome s" ], [ "P(E)=\\sum_{s\\in E}w_s", "事件 E 的機率", "the probability of the event E" ], [ "S\\setminus E", "相對於 S 的補事件", "the complement of E relative to S" ], [ "|E|", "事件 E 的元素個數", "the number of elements of E" ] ], "md_error_controls": [ { "title": "未宣告等可能不能用 |E|/|S| / Without declared equiprobability, |E|/|S| is not available", "zh": "MKO 指出的第一個誤解是未聲明等可能就套「有利結果數除以總結果數」。本模型中 P({1,3})=2/3，而 |E|/|S|=1/2（S09、S10）。計數公式只是等可能假設下的推論（S13）。", "en": "The first error named in the MKO is to use 'favourable outcomes over total outcomes' without declared equiprobability. In this model P({1,3})=2/3, while |E|/|S|=1/2 (S09, S10). The counting formula is only a consequence of the equiprobability assumption (S13)." }, { "title": "機率為零不代表事件為空 / Probability zero does not mean the event is empty", "zh": "MKO 指出的第二個誤解是由 P(E)=0 推出 E=∅。{4} 非空但 P({4})=w_4=0（S12）。只有反方向 E=∅ ⇒ P(E)=0 成立（S05）。", "en": "The second error named in the MKO is to infer E=∅ from P(E)=0. {4} is nonempty but P({4})=w_4=0 (S12). Only the other direction, E=∅ ⇒ P(E)=0, holds (S05)." }, { "title": "先驗證權重；相加要互斥 / Validate the weights first; add only disjoint events", "zh": "計算前要確認每個權重非負且總和恰為 1：w′ 總和 7/6、w″ 有負權重，都不能用（S14）。互斥時可直接相加，不互斥時則一般不能直接相加（若重疊部分機率為 0，等式仍可成立）。例如：{1,2} 與 {2,3} 的機率相加得 4/3，大於 1（S15）。補事件要相對同一個 S 取（S11）。", "en": "Before computing, confirm that every weight is nonnegative and the weights sum to exactly 1: w′ sums to 7/6 and w″ has a negative weight, so neither may be used (S14). Disjoint events may be added directly; overlapping events generally may not (though the equality still holds when their intersection has probability zero). For example, adding the probabilities of {1,2} and {2,3} gives 4/3, more than 1 (S15). Complements are taken relative to the same S (S11)." } ], "md_open_items": [ { "zh-Hant": "sum_split 作為前置規則使用，是 finite_sum_rules 經重新編號的結果，本稿沒有寫出重新編號的細節。", "en": "sum_split is used as a prerequisite; it follows from finite_sum_rules by reindexing, and the details of the reindexing are not written out here." }, { "zh-Hant": "本稿不推論真實實驗的機率，也不處理獨立性、條件機率或一般測度論的事件族（MKO para2）。", "en": "This guide infers no probabilities of real experiments and does not treat independence, conditional probability or event families of general measure theory (MKO para2)." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "1207c52eb66a78870472c59aac433a88d6d082ed767b689177fc213797904c1d", "original_source_sha256": "1207c52eb66a78870472c59aac433a88d6d082ed767b689177fc213797904c1d", "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." } }