{ "schema_version": "ocme-derivation-reader-v1", "guide_id": "derivation-mko-sample-space-01", "mko_id": "mko-sample-space", "title": { "zh-Hant": "樣本空間：擲兩次硬幣的兩種觀察尺度與不等可能的結果", "en": "Sample space: two observation scales for two coin tosses, and outcomes that are not equally likely" }, "contract": { "schema_version": "ocme-derivation-guide-v0.1", "guide_id": "derivation-mko-sample-space-01", "source": { "mko_id": "mko-sample-space", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "sha256": "c58b0d07252ece257a4143fd10f0f25b249023f4ed50bc6d393f4752a91267f7", "english_sha256": "ba452c266ef8c6ab4b549a3ccb6bc04a19fc7fafef1934d17f821c9ef4c04756" }, "mode": "worked_example", "audience_prerequisites": [ "cartesian_product_def", "finite_counting_basics", "prob_axioms", "image_preimage_def", "finite_prob_weights" ], "goal": { "zh-Hant": "隨機試驗：把一枚硬幣擲兩次。(1) 若依次記錄結果，寫出樣本空間 S，證明它包含所有可能基本結果且恰有 4 個元素；(2) 若只記錄正面次數，寫出樣本空間 S′={0,1,2}；(3) 在另外給定的機率模型 P（S 的每個有序結果機率 1/4）下，算出 S′ 各結果的機率 1/4、1/2、1/4，說明同一樣本空間中的結果不必等可能；(4) 以 S 上另一個合法的非均勻模型 Q 說明樣本空間本身不決定機率；(5) 分辨結果 (H,T) 與事件 {(H,T),(T,H)}。", "en": "Random experiment: toss a coin twice. (1) Recording the results in order, write the sample space S, and prove that it contains every possible elementary outcome and has exactly 4 elements; (2) recording only the number of heads, write the sample space S′={0,1,2}; (3) under a separately given probability model P (each ordered outcome in S has probability 1/4), compute the probabilities 1/4, 1/2, 1/4 of the outcomes in S′, showing that outcomes in one sample space need not be equally likely; (4) use another valid, non-uniform model Q on S to show that the sample space by itself does not fix probabilities; (5) distinguish the outcome (H,T) from the event {(H,T),(T,H)}." }, "givens": [ { "id": "G1", "statement": { "zh-Hant": "試驗：把同一枚硬幣擲兩次。每次擲的結果是 H（正面）或 T（反面），沒有其他結果（不考慮硬幣立起等情況）（假設 a1）。", "en": "Experiment: toss the same coin twice. Each toss results in H (heads) or T (tails), with no other result (landing on edge is excluded) (assumption a1)." } }, { "id": "G2", "statement": { "zh-Hant": "觀察尺度一：依次記錄兩次結果，成為有序對（第一次，第二次）。", "en": "Observation scale 1: record both results in order, as an ordered pair (first, second)." } }, { "id": "G3", "statement": { "zh-Hant": "觀察尺度二：只記錄兩次中出現 H 的次數。", "en": "Observation scale 2: record only the number of H among the two tosses." } }, { "id": "G4", "statement": { "zh-Hant": "機率模型 P（額外資料，不由樣本空間推出）：S 的每個有序結果機率都是 1/4，例如硬幣公正且兩次獨立時採用。", "en": "Probability model P (additional data, not derived from the sample space): each ordered outcome in S has probability 1/4, as used for example when the coin is fair and the tosses are independent." } }, { "id": "G5", "statement": { "zh-Hant": "另一個機率模型 Q（額外資料）：Q((H,H))=9/25，Q((H,T))=6/25，Q((T,H))=6/25，Q((T,T))=4/25，例如 H 的機率為 3/5 且兩次獨立時採用。", "en": "Another probability model Q (additional data): Q((H,H))=9/25, Q((H,T))=6/25, Q((T,H))=6/25, Q((T,T))=4/25, as used for example when H has probability 3/5 and the tosses are independent." } } ], "steps": [ { "id": "S01", "uses": [ "G1", "G2" ], "before_tex": "\\text{experiment: two tosses, recorded in order}", "after_tex": "\\text{outcome}=(a,b),\\ a,b\\in\\{H,T\\}", "relation": "definition", "rule": "mko_sample_space", "conditions": [], "explanation": { "zh-Hant": "依 MKO，樣本空間取決於試驗與觀察方式，所以先固定兩者：擲兩次，依次記錄。於是一個基本結果是有序對 (a,b)，a 是第一次結果，b 是第二次結果，各自是 H 或 T（G1）。", "en": "By the MKO, the sample space depends on the experiment and on how it is observed, so both are fixed first: two tosses, recorded in order. An elementary outcome is then an ordered pair (a,b), where a is the first result and b the second, each H or T (G1)." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "G2", "mko_sample_space" ], "explanation": { "zh-Hant": "此步只明示試驗與觀察尺度（假設 a1），不推出新事實；換一種記錄方式會得到不同的樣本空間（見 S06）。", "en": "This step only makes the experiment and observation scale explicit (assumption a1) and derives no new fact; a different way of recording gives a different sample space (see S06)." } }, "leaf_basis": "mko_sample_space + def_intro" }, { "id": "S02", "uses": [ "S01" ], "before_tex": "\\{(a,b)\\mid a\\in\\{H,T\\},\\ b\\in\\{H,T\\}\\}", "after_tex": "\\{H,T\\}\\times\\{H,T\\}", "relation": "equivalent", "rule": "cartesian_product_def", "conditions": [], "explanation": { "zh-Hant": "所有可能的有序記錄，恰好是 a、b 各自走遍 {H,T} 的有序對；依笛卡兒積定義，這就是 {H,T}×{H,T}。", "en": "The possible ordered records are exactly the ordered pairs with a and b each ranging over {H,T}; by definition this is the Cartesian product {H,T}×{H,T}." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "cartesian_product_def" ], "explanation": { "zh-Hant": "這是笛卡兒積定義的直接改寫，兩個方向都由定義保證。", "en": "This directly rewrites the definition of a Cartesian product; both directions are given by the definition." } }, "leaf_basis": "cartesian_product_def + set_builder" }, { "id": "S03", "uses": [ "S02" ], "before_tex": "\\{H,T\\}\\times\\{H,T\\}", "after_tex": "S=\\{(H,H),(H,T),(T,H),(T,T)\\}", "relation": "equivalent", "rule": "case_analysis", "conditions": [], "explanation": { "zh-Hant": "對 a 分兩種情況（H 或 T），每種情況再對 b 分兩種，共得四個有序對。把它們列出，並把這個集合命名為 S。", "en": "Split on a (H or T), and in each case on b (H or T), obtaining four ordered pairs. List them and call this set S." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "case_analysis", "cartesian_product_def", "set_extensionality" ], "explanation": { "zh-Hant": "分情況窮盡 {H,T} 的兩個值，所以列舉的集合與笛卡兒積元素相同：每個列出的對都在積中，積的每個元素也都被列出。", "en": "The case split exhausts the two values of {H,T}, so the listed set and the product have the same elements: each listed pair is in the product, and each element of the product is listed." } }, "leaf_basis": "case_analysis + cartesian_product_def + set_extensionality" }, { "id": "S04", "uses": [ "S01", "S03" ], "before_tex": "x\\ \\text{a possible elementary outcome}", "after_tex": "x\\in S", "relation": "implies", "rule": "L_S_complete", "conditions": [], "explanation": { "zh-Hant": "依引理 L_S_complete（在 S04.1、S04.2 證明），每個可能基本結果都屬於 S。這正是 MKO 對樣本空間的要求 x∈S，也滿足假設 a2：S 包含所有可能基本結果。", "en": "By Lemma L_S_complete (proved in S04.1 and S04.2), every possible elementary outcome belongs to S. This is exactly the MKO's requirement x∈S, and it meets assumption a2: S contains all possible elementary outcomes." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "L_S_complete", "G1" ], "explanation": { "zh-Hant": "完整性依賴 G1 的「每次只有 H 或 T」；若試驗允許硬幣立起，S 就不完整。", "en": "Completeness rests on G1's 'each toss is H or T'; if the experiment allowed the coin to land on edge, S would not be complete." } }, "leaf_basis": "forall_intro + case_analysis + cartesian_product_def" }, { "id": "S04.1", "uses": [ "S01", "G1" ], "before_tex": "x\\ \\text{possible}", "after_tex": "x=(a,b),\\ a\\in\\{H,T\\},\\ b\\in\\{H,T\\}", "relation": "implies", "rule": "def_unfold", "conditions": [], "explanation": { "zh-Hant": "設 x 是任意一個可能基本結果。依 S01 的觀察尺度，x 是一個有序對 (a,b)；依 G1，每次擲的結果只能是 H 或 T，所以 a、b∈{H,T}。", "en": "Let x be an arbitrary possible elementary outcome. By the observation scale of S01, x is an ordered pair (a,b); by G1 each toss can only give H or T, so a, b∈{H,T}." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G1", "S01" ], "explanation": { "zh-Hant": "只用到 G1、G2 對試驗與記錄方式的描述；x 是任意的。", "en": "Only the descriptions of the experiment and of the record in G1 and G2 are used; x is arbitrary." } }, "leaf_basis": "def_unfold + forall_elim" }, { "id": "S04.2", "uses": [ "S04.1", "S02", "S03" ], "before_tex": "x=(a,b),\\ a,b\\in\\{H,T\\}", "after_tex": "x\\in S", "relation": "implies", "rule": "forall_intro", "conditions": [], "explanation": { "zh-Hant": "由 S02，(a,b)∈{H,T}×{H,T}；由 S03，這個積等於 S，所以 x∈S。x 是任意的，依 ∀引入，每個可能基本結果都屬於 S。S04.1 與 S04.2 合起來證得引理 L_S_complete。", "en": "By S02, (a,b)∈{H,T}×{H,T}; by S03 this product equals S, so x∈S. Since x was arbitrary, ∀-introduction shows that every possible elementary outcome belongs to S. S04.1 and S04.2 together prove Lemma L_S_complete." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S02", "S03", "forall_intro" ], "explanation": { "zh-Hant": "需要 S03 的集合相等；不宣稱 S 的元素出現機率相同。", "en": "It needs the set equality of S03; it does not claim that the elements of S are equally likely." } }, "leaf_basis": "cartesian_product_def + eq_replace + forall_intro" }, { "id": "S05", "uses": [ "S02", "S03" ], "before_tex": "|\\{H,T\\}\\times\\{H,T\\}|", "after_tex": "|S|=2\\cdot 2=4", "relation": "implies", "rule": "finite_counting_basics", "conditions": [ "H\\neq T" ], "explanation": { "zh-Hant": "依有限計數 |A×B|=|A|·|B|，|S|=2·2=4。列舉中的四個對兩兩不同：例如 (H,T) 與 (T,H) 的第一坐標不同，所以是不同的結果；次序是記錄的一部分。", "en": "By finite counting |A×B|=|A|·|B|, |S|=2·2=4. The four listed pairs are pairwise distinct: for example (H,T) and (T,H) differ in the first coordinate, so they are different outcomes; the order is part of the record." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "finite_counting_basics", "cartesian_product_def" ], "explanation": { "zh-Hant": "需要 {H,T} 有兩個不同元素（H≠T）；結論只是個數，不涉及機率。", "en": "It needs {H,T} to have two distinct elements (H≠T); the conclusion is a count and says nothing about probabilities." } }, "leaf_basis": "finite_counting_basics + cartesian_product_def + numeric_evaluation" }, { "id": "S06", "uses": [ "G3", "S03" ], "before_tex": "\\text{ordered outcome }(a,b)\\in S", "after_tex": "h:S\\to S',\\ h(a,b)=\\text{number of }H\\text{ among }a,b,\\ S'=\\{0,1,2\\}", "relation": "definition", "rule": "def_intro", "conditions": [], "explanation": { "zh-Hant": "換成觀察尺度二（G3）：只記錄 H 的次數。兩次擲出的 H 次數只能是 0、1 或 2，所以第二個樣本空間是 S′={0,1,2}。兩個尺度由函數 h:S→S′ 聯繫，h 把有序結果送到它含有的 H 個數。", "en": "Switch to observation scale 2 (G3): record only the number of H. Two tosses can show 0, 1 or 2 heads, so the second sample space is S′={0,1,2}. The two scales are linked by the function h:S→S′ sending an ordered outcome to the number of H it contains." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G3", "def_intro" ], "explanation": { "zh-Hant": "此步只定義；S′ 的完整性在 S07 以 h 的值核對。", "en": "This step only defines; completeness of S′ is checked in S07 through the values of h." } }, "leaf_basis": "def_intro + function_def" }, { "id": "S07", "uses": [ "S03", "S04", "S06" ], "before_tex": "h(H,H),\\ h(H,T),\\ h(T,H),\\ h(T,T)", "after_tex": "2,\\ 1,\\ 1,\\ 0;\\quad h(S)=\\{0,1,2\\}=S'", "relation": "implies", "rule": "image_preimage_def", "conditions": [], "explanation": { "zh-Hant": "逐一計算：h(H,H)=2、h(H,T)=1、h(T,H)=1、h(T,T)=0。所以 h 的像 h(S)={0,1,2}=S′：每個可能的 H 次數都在 S′ 中（完整），S′ 的每個值也都可能出現。", "en": "Compute each value: h(H,H)=2, h(H,T)=1, h(T,H)=1, h(T,T)=0. So the image h(S)={0,1,2}=S′: every possible number of heads is in S′ (complete), and every value in S′ can occur." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S04", "image_preimage_def", "L_S_complete" ], "explanation": { "zh-Hant": "因 S 已含所有可能結果（S04），h(S) 就是所有可能的 H 次數；這是對 S 四個元素的完整計算，不是抽樣。", "en": "Since S contains every possible outcome (S04), h(S) is the set of all possible numbers of heads; this is a complete computation over the four elements of S, not a sample." } }, "leaf_basis": "image_preimage_def + numeric_evaluation" }, { "id": "S08", "uses": [ "S05", "S07" ], "before_tex": "|S|=4,\\ |S'|=3", "after_tex": "S\\neq S'", "relation": "implies", "rule": "set_extensionality", "conditions": [], "explanation": { "zh-Hant": "S 的元素是有序對，S′ 的元素是數；兩者的個數也不同（4 與 3）。所以同一試驗在不同觀察尺度下有不同的樣本空間，正如 MKO 所說。", "en": "The elements of S are ordered pairs and those of S′ are numbers; even their sizes differ (4 and 3). So one experiment has different sample spaces under different observation scales, as the MKO says." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S05", "S07", "set_extensionality" ], "explanation": { "zh-Hant": "個數不同就足以說明兩集合不等。兩者都是合法的樣本空間，各自相對於自己的觀察尺度。", "en": "Different sizes are enough to show that the sets differ. Both are legitimate sample spaces, each relative to its own observation scale." } }, "leaf_basis": "set_extensionality + finite_counting_basics" }, { "id": "S09", "uses": [ "G4", "S03" ], "before_tex": "p(\\omega)=\\tfrac14\\ \\ (\\omega\\in S)", "after_tex": "P\\ \\text{is a probability on }S", "relation": "implies", "rule": "finite_prob_weights", "conditions": [ "p(\\omega)\\ge 0", "\\sum_{\\omega\\in S}p(\\omega)=1" ], "explanation": { "zh-Hant": "G4 是另外給定的模型資料，不是從 S 推出的。四個權重都 ≥0，總和 4·(1/4)=1，依 finite_prob_weights，P(E)=Σ_{ω∈E} p(ω) 是 S 上的機率。", "en": "G4 is model data given separately; it is not derived from S. The four weights are ≥0 and sum to 4·(1/4)=1, so by finite_prob_weights, P(E)=Σ_{ω∈E} p(ω) is a probability on S." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G4", "finite_prob_weights", "prob_axioms" ], "explanation": { "zh-Hant": "需要非負與總和為 1 兩個條件，N2 核對後者。此步不宣稱 G4 是唯一可能的模型（S13 給出另一個）。", "en": "It needs the two conditions nonnegativity and total 1; N2 checks the latter. It does not claim that G4 is the only possible model (S13 gives another)." } }, "leaf_basis": "finite_prob_weights + numeric_evaluation" }, { "id": "S10", "uses": [ "S07" ], "before_tex": "h^{-1}(\\{0\\}),\\ h^{-1}(\\{1\\}),\\ h^{-1}(\\{2\\})", "after_tex": "\\{(T,T)\\},\\ \\{(H,T),(T,H)\\},\\ \\{(H,H)\\}", "relation": "equivalent", "rule": "image_preimage_def", "conditions": [], "explanation": { "zh-Hant": "依原像定義，h⁻¹({k}) 是 S 中 H 個數為 k 的結果。讀 S07 的值：0 次只有 (T,T)；1 次有 (H,T) 與 (T,H)；2 次只有 (H,H)。三個原像兩兩不相交，聯集是 S。「觀察到 k 次正面」恰好是 S 中的事件 h⁻¹({k}) 發生。", "en": "By the definition of preimage, h⁻¹({k}) consists of the outcomes in S with k heads. Reading the values of S07: 0 heads only for (T,T); 1 head for (H,T) and (T,H); 2 heads only for (H,H). The three preimages are pairwise disjoint and their union is S. 'Observing k heads' happens exactly when the event h⁻¹({k}) in S occurs." }, "reverse_audit": { "kind": "inverse_equivalence", "required_ids": [ "image_preimage_def", "set_extensionality" ], "explanation": { "zh-Hant": "每個原像都由 S 的四個元素逐一判定，兩邊元素相同，雙向成立。", "en": "Each preimage is decided element by element over the four elements of S, so both sides have the same elements and the step holds both ways." } }, "leaf_basis": "image_preimage_def + set_extensionality + numeric_evaluation" }, { "id": "S11", "uses": [ "S09", "S10" ], "before_tex": "P'(k)=P(h^{-1}(\\{k\\}))", "after_tex": "P'(0)=\\tfrac14,\\ P'(1)=\\tfrac12,\\ P'(2)=\\tfrac14", "relation": "implies", "rule": "prob_axioms", "conditions": [ "\\{(H,T)\\}\\cap\\{(T,H)\\}=\\varnothing" ], "explanation": { "zh-Hant": "定義 S′ 上的權重 P′(k)=P(h⁻¹({k}))。依 S09 的 P 與有限可加性：P′(0)=P({(T,T)})=1/4；P′(1)=P({(H,T)})+P({(T,H)})=1/4+1/4=1/2；P′(2)=P({(H,H)})=1/4。三者都 ≥0，總和 1/4+1/2+1/4=1，依 finite_prob_weights，P′ 是 S′ 上的機率。", "en": "Define weights on S′ by P′(k)=P(h⁻¹({k})). By P from S09 and finite additivity: P′(0)=P({(T,T)})=1/4; P′(1)=P({(H,T)})+P({(T,H)})=1/4+1/4=1/2; P′(2)=P({(H,H)})=1/4. All three are ≥0 and sum to 1/4+1/2+1/4=1, so by finite_prob_weights, P′ is a probability on S′." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S09", "S10", "prob_axioms", "finite_prob_weights" ], "explanation": { "zh-Hant": "可加性只用在不相交的單點事件 {(H,T)} 與 {(T,H)}；P′ 的值完全依賴 G4 的模型，不依賴 S′ 本身。", "en": "Additivity is used only for the disjoint single-point events {(H,T)} and {(T,H)}; the values of P′ depend entirely on the model G4, not on S′ itself." } }, "leaf_basis": "prob_axioms + finite_prob_weights + def_intro + numeric_evaluation" }, { "id": "S12", "uses": [ "S11" ], "before_tex": "P'(1)=\\tfrac12,\\ P'(0)=\\tfrac14", "after_tex": "P'(1)\\neq P'(0)", "relation": "implies", "rule": "numeric_evaluation", "conditions": [], "explanation": { "zh-Hant": "1/2≠1/4，所以在這個模型中 S′ 的三個結果不是等可能的，儘管它們都是 S′ 的元素。屬於同一樣本空間並不推出等可能。若誤把 S′ 當作等可能，每個結果會得到 1/3，與 P′(1)=1/2 不符。", "en": "1/2≠1/4, so in this model the three outcomes of S′ are not equally likely, even though all three are elements of S′. Belonging to one sample space does not imply being equally likely. Wrongly treating S′ as equally likely would give 1/3 to each outcome, which disagrees with P′(1)=1/2." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S11", "numeric_evaluation" ], "explanation": { "zh-Hant": "只需比較兩個明確的有理數（N5）。結論只針對 G4 的模型；不宣稱 S′ 在任何模型下都不等可能。", "en": "Only two explicit rationals are compared (N5). The conclusion concerns the model G4; it is not claimed that S′ is non-uniform under every model." } }, "leaf_basis": "numeric_evaluation" }, { "id": "S13", "uses": [ "G5", "S03", "S09" ], "before_tex": "Q:\\ \\tfrac{9}{25},\\ \\tfrac{6}{25},\\ \\tfrac{6}{25},\\ \\tfrac{4}{25}", "after_tex": "Q\\ \\text{is a probability on }S,\\ Q\\neq P", "relation": "implies", "rule": "finite_prob_weights", "conditions": [ "\\text{weights}\\ge 0", "\\text{total}=1" ], "explanation": { "zh-Hant": "G5 的四個權重都 ≥0，總和 (9+6+6+4)/25=1，依 finite_prob_weights，Q 也是 S 上的機率。但 Q((H,H))=9/25≠4/25=Q((T,T))，所以 Q 不均勻，也不同於 P。同一個樣本空間 S 承載兩個不同的機率模型，所以 S 本身不附帶等可能的假設。這些權重對應 H 機率 3/5、兩次獨立的情形：(3/5)²=9/25，(3/5)(2/5)=6/25，(2/5)²=4/25。", "en": "The four weights of G5 are ≥0 and sum to (9+6+6+4)/25=1, so by finite_prob_weights Q is also a probability on S. But Q((H,H))=9/25≠4/25=Q((T,T)), so Q is not uniform and differs from P. One sample space S carries two different probability models, so S by itself carries no assumption of equal likelihood. These weights correspond to H having probability 3/5 with independent tosses: (3/5)²=9/25, (3/5)(2/5)=6/25, (2/5)²=4/25." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "G5", "finite_prob_weights", "numeric_evaluation" ], "explanation": { "zh-Hant": "Q 的合法性只需非負與總和為 1（N6）；與獨立性的對應只是說明權重的來源，論證不依賴它（N7 到 N9 只核對算術）。", "en": "Validity of Q needs only nonnegativity and total 1 (N6); the link to independence only explains where the weights come from, and the argument does not depend on it (N7 to N9 only check the arithmetic)." } }, "leaf_basis": "finite_prob_weights + numeric_evaluation" }, { "id": "S14", "uses": [ "S03", "S10" ], "before_tex": "(H,T)\\in S,\\ E=\\{(H,T),(T,H)\\}", "after_tex": "(H,T)\\in E\\subseteq S,\\ (H,T)\\neq E", "relation": "implies", "rule": "subset_def", "conditions": [], "explanation": { "zh-Hant": "(H,T) 是 S 的一個元素，即一個基本結果。「恰好一次正面」對應 E=h⁻¹({1})={(H,T),(T,H)}，它是 S 的子集合，也就是一個事件。(H,T)∈E，但 (H,T)≠E：一個是有序對，一個是含兩個有序對的集合。試驗結果為 (H,T) 時，事件 E 發生。", "en": "(H,T) is an element of S, that is an elementary outcome. 'Exactly one head' corresponds to E=h⁻¹({1})={(H,T),(T,H)}, a subset of S, that is an event. (H,T)∈E, but (H,T)≠E: one is an ordered pair, the other a set containing two ordered pairs. When the outcome is (H,T), the event E occurs." }, "reverse_audit": { "kind": "premise_audit", "required_ids": [ "S03", "S10", "subset_def" ], "explanation": { "zh-Hant": "E⊆S 由 E 的兩個元素都在 S 中得到（完整檢查）。不宣稱結果與事件可以互換。", "en": "E⊆S follows because both elements of E lie in S (a complete check). It is not claimed that outcomes and events are interchangeable." } }, "leaf_basis": "subset_def + set_extensionality" }, { "id": "S15", "uses": [ "S04", "S05", "S07", "S08", "S11", "S12", "S13", "S14" ], "before_tex": "\\text{two observation scales and two models}", "after_tex": "S=\\{H,T\\}^2,\\ |S|=4;\\ S'=\\{0,1,2\\};\\ P'=(\\tfrac14,\\tfrac12,\\tfrac14);\\ Q\\neq P", "relation": "implies", "rule": "and_intro_elim", "conditions": [], "explanation": { "zh-Hant": "彙整：依次記錄時 S 有 4 個元素並包含所有可能結果（S04、S05）；只記錄正面次數時 S′={0,1,2}（S07、S08）；在 G4 之下 S′ 的機率是 1/4、1/2、1/4，不等可能（S11、S12）；S 可承載不同模型（S13）；(H,T) 是結果，{(H,T),(T,H)} 是事件（S14）。", "en": "Collect the results: recorded in order, S has 4 elements and contains every possible outcome (S04, S05); recording only the number of heads gives S′={0,1,2} (S07, S08); under G4 the outcomes of S′ have probabilities 1/4, 1/2, 1/4 and are not equally likely (S11, S12); S can carry different models (S13); (H,T) is an outcome and {(H,T),(T,H)} is an event (S14)." }, "reverse_audit": { "kind": "sufficient_goal_reduction", "required_ids": [ "S04", "S05", "S07", "S08", "S11", "S12", "S13", "S14", "and_intro_elim" ], "explanation": { "zh-Hant": "目標 G 是題目五項結論的合取；每一項都由所列步驟從已知正向證成，此步只做合取，沒有把 G 當作前提。", "en": "Goal G is the conjunction of the five conclusions of the problem; each is proved forwards from the givens by the listed steps, and this step only forms the conjunction; G is never used as a premise." } }, "leaf_basis": "and_intro_elim" } ], "result": { "zh-Hant": "依次記錄兩次擲硬幣時，樣本空間 S={(H,H),(H,T),(T,H),(T,T)} 包含所有可能基本結果，恰有 4 個元素；只記錄正面次數時樣本空間是 S′={0,1,2}。在另外給定的模型 P 下，S′ 的結果機率為 1/4、1/2、1/4，並不等可能；S 上也可以有非均勻的模型 Q。所以樣本空間只說明哪些結果可能，不決定機率。(H,T) 是結果，{(H,T),(T,H)} 是事件。", "en": "Recording two coin tosses in order, the sample space S={(H,H),(H,T),(T,H),(T,T)} contains every possible elementary outcome and has exactly 4 elements; recording only the number of heads gives the sample space S′={0,1,2}. Under the separately given model P, the outcomes of S′ have probabilities 1/4, 1/2, 1/4 and are not equally likely; S also carries the non-uniform model Q. So a sample space only says which outcomes are possible and does not fix probabilities. (H,T) is an outcome; {(H,T),(T,H)} is an event." }, "negative_controls": [ { "claim": { "zh-Hant": "S′={0,1,2} 是樣本空間，所以每個結果的機率都是 1/3", "en": "S′={0,1,2} is a sample space, so each outcome has probability 1/3" }, "expected": "reject", "reason": { "zh-Hant": "在模型 P 下 P′(1)=1/2≠1/3（S11）；樣本空間不附帶等可能假設。", "en": "Under the model P, P′(1)=1/2≠1/3 (S11); a sample space carries no assumption of equal likelihood." } }, { "claim": { "zh-Hant": "S 有 4 個元素，所以 (H,H) 的機率必為 1/4", "en": "S has 4 elements, so (H,H) must have probability 1/4" }, "expected": "reject", "reason": { "zh-Hant": "在同一個 S 上，模型 Q 給 (H,H) 機率 9/25（S13）。", "en": "On the same S, the model Q gives (H,H) probability 9/25 (S13)." } }, { "claim": { "zh-Hant": "依次記錄時，{(H,H),(H,T),(T,T)} 是樣本空間", "en": "When results are recorded in order, {(H,H),(H,T),(T,T)} is the sample space" }, "expected": "reject", "reason": { "zh-Hant": "(T,H) 是可能結果卻不在其中，違反假設 a2。", "en": "(T,H) is a possible outcome but is missing, violating assumption a2." } }, { "claim": { "zh-Hant": "(H,T) 與 {(H,T),(T,H)} 是同一個對象", "en": "(H,T) and {(H,T),(T,H)} are the same object" }, "expected": "reject", "reason": { "zh-Hant": "前者是有序對（結果），後者是含兩個元素的集合（事件）。", "en": "The first is an ordered pair (an outcome), the second a two-element set (an event)." } }, { "claim": { "zh-Hant": "S 與 S′ 不同，所以其中至少一個不是這個試驗的樣本空間", "en": "S and S′ differ, so at least one of them is not a sample space of this experiment" }, "expected": "reject", "reason": { "zh-Hant": "兩者各自相對於不同的觀察尺度（G2、G3），都是合法的樣本空間（S04、S07）。", "en": "Each is relative to a different observation scale (G2, G3), and both are legitimate sample spaces (S04, S07)." } } ], "status": "candidate", "gaps": [], "verification": { "structure": "checked_local_script", "mathematics": "candidate_argument", "formalization": "not_formalized" }, "rule_glossary": [ { "id": "cartesian_product_def", "kind": "prerequisite", "statement": { "zh-Hant": "笛卡兒積：A×B 是所有有序對 (a,b)（a∈A、b∈B）的集合；(a,b)=(c,d) ⇔ a=c 且 b=d。", "en": "Cartesian product: A×B is the set of ordered pairs (a,b) with a∈A and b∈B; (a,b)=(c,d) ⇔ a=c and b=d." } }, { "id": "finite_counting_basics", "kind": "prerequisite", "statement": { "zh-Hant": "有限計數：互斥有限集合 A、B 有 |A∪B|=|A|+|B|；|A×B|=|A|·|B|；有限集之間存在雙射 ⇔ 元素個數相同。", "en": "Finite counting: for disjoint finite sets A, B, |A∪B|=|A|+|B|; |A×B|=|A|·|B|; there is a bijection between finite sets ⇔ they have the same number of elements." } }, { "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": "image_preimage_def", "kind": "prerequisite", "statement": { "zh-Hant": "像與原像：f(S)={f(a) ∣ a∈S}；f⁻¹(T)={a∈A ∣ f(a)∈T}。原像記號不預設 f 有反函數。", "en": "Image and preimage: f(S)={f(a) ∣ a∈S}; f⁻¹(T)={a∈A ∣ f(a)∈T}. The preimage notation does not presuppose that f has an inverse." } }, { "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": "case_analysis", "kind": "prerequisite", "statement": { "zh-Hant": "分情況論證：若 P∨Q 成立，且 P⇒R、Q⇒R 都已證，則 R 成立。所有情況必須涵蓋全部可能。", "en": "Proof by cases: if P∨Q holds and both P⇒R and Q⇒R are proved, then R holds. The cases must cover every possibility." } }, { "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": "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": "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": "forall_elim", "kind": "prerequisite", "statement": { "zh-Hant": "∀消去（代入特例）：若 ∀x∈D, P(x) 成立且 a∈D，則 P(a) 成立。", "en": "∀-elimination (instantiation): if ∀x∈D, P(x) holds and a∈D, then P(a) holds." } }, { "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": "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": "function_def", "kind": "prerequisite", "statement": { "zh-Hant": "函數：f:A→B 對每個 a∈A 指定恰好一個 f(a)∈B。兩函數相等 ⇔ 定義域、對應域相同且對每個 a 都有相同值。", "en": "Function: f:A→B assigns to each a∈A exactly one f(a)∈B. Two functions are equal ⇔ they have the same domain and codomain and agree at every 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": "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_sample_space", "kind": "cited", "source": { "mko_id": "mko-sample-space", "sha256": "c58b0d07252ece257a4143fd10f0f25b249023f4ed50bc6d393f4752a91267f7", "commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f" }, "statement": { "zh-Hant": "若隨機試驗的樣本空間為 S，則每個可能基本結果 x 都滿足 x∈S；空間本身不附帶各結果等可能的假設。", "en": "If a random experiment has sample space S, every possible elementary outcome x satisfies x∈S. Membership in the space does not itself assume that outcomes are equally likely." } }, { "id": "finite_prob_weights", "kind": "prerequisite", "statement": { "zh-Hant": "有限樣本空間上的點權重：若對每個結果 ω 給定 p(ω)≥0，且所有 p(ω) 的總和為 1，則對 S 的每個子集合 E 令 P(E)=Σ_{ω∈E} p(ω)，所得 P 滿足機率公理。", "en": "Point weights on a finite sample space: if each outcome ω is given p(ω)≥0 and all the p(ω) sum to 1, then setting P(E)=Σ_{ω∈E} p(ω) for every subset E of S gives a P that satisfies the probability axioms." } }, { "id": "L_S_complete", "kind": "lemma", "proved_in": [ "S04.1", "S04.2" ], "statement": { "zh-Hant": "在 G1、G2 之下，每個可能基本結果都屬於 S={(H,H),(H,T),(T,H),(T,T)}。", "en": "Under G1 and G2, every possible elementary outcome belongs to S={(H,H),(H,T),(T,H),(T,T)}." } } ], "revision_log": [], "numeric_checks": [ { "id": "N1", "step": "S05", "lhs": "2*2", "rhs": "4" }, { "id": "N2", "step": "S09", "lhs": "4*(1/4)", "rhs": "1" }, { "id": "N3", "step": "S11", "lhs": "1/4+1/4", "rhs": "1/2" }, { "id": "N4", "step": "S11", "lhs": "1/4+1/2+1/4", "rhs": "1" }, { "id": "N5", "step": "S12", "lhs": "1/2", "relation": "!=", "rhs": "1/4" }, { "id": "N6", "step": "S13", "lhs": "9/25+6/25+6/25+4/25", "rhs": "1" }, { "id": "N7", "step": "S13", "lhs": "(3/5)**2", "rhs": "9/25" }, { "id": "N8", "step": "S13", "lhs": "(3/5)*(2/5)", "rhs": "6/25" }, { "id": "N9", "step": "S13", "lhs": "(2/5)**2", "rhs": "4/25" }, { "id": "N10", "step": "S13", "lhs": "9/25", "relation": "!=", "rhs": "4/25" }, { "id": "N11", "step": "S12", "lhs": "1/3", "relation": "!=", "rhs": "1/2" } ] }, "presentation": { "md_scope": { "zh-Hant": "在 MKO 的範圍內工作：試驗與觀察尺度明示（假設 a1），樣本空間包含指定描述下的所有可能基本結果（假設 a2）。機率值只在明示給定的模型（G4、G5）中計算，作為後續機率模型的額外資料，不從樣本空間推出。不處理連續或無限樣本空間與測度論；不改動 canonical MKO 或英文 edition。", "en": "Works inside the MKO's scope: the experiment and observation scale are explicit (assumption a1), and the sample space contains every possible elementary outcome under that description (assumption a2). Probability values are computed only in explicitly given models (G4, G5), as additional data of a later probability model, never derived from the sample space. It does not treat continuous or infinite sample spaces or measure theory, and does not change the canonical MKO or its English edition." }, "md_symbols": [ [ "H,\\ T", "正面、反面", "heads, tails" ], [ "S", "依次記錄兩次結果時的樣本空間", "the sample space when both results are recorded in order" ], [ "S'", "只記錄正面次數時的樣本空間", "the sample space when only the number of heads is recorded" ], [ "h", "把有序結果對應到正面次數的函數", "the function sending an ordered outcome to its number of heads" ], [ "P,\\ Q", "S 上兩個不同的機率模型", "two different probability models on S" ], [ "P'", "由 P 經 h 得到的 S′ 上的機率", "the probability on S′ obtained from P through h" ] ], "md_error_controls": [ { "title": "同一樣本空間不代表等可能 / One sample space does not mean equally likely", "zh": "MKO 指出常見誤解是認為同一樣本空間中的結果必定等可能。S′={0,1,2} 是合法的樣本空間，但在模型 P 下三個結果的機率是 1/4、1/2、1/4（S11），不是各 1/3。就連 S 上的均勻權重也是 G4 另外給的資料；G5 的 Q 在同一個 S 上並不均勻（S13）。", "en": "The MKO names the mistake of believing that outcomes in one sample space must be equally likely. S′={0,1,2} is a legitimate sample space, yet under the model P its three outcomes have probabilities 1/4, 1/2, 1/4 (S11), not 1/3 each. Even the uniform weights on S are separate data from G4; the model Q of G5 is not uniform on the same S (S13)." }, { "title": "結果不是事件 / An outcome is not an event", "zh": "MKO 提醒不要把單一結果與事件集合混為一談。(H,T) 是 S 的元素；「恰好一次正面」在 S 中是事件 {(H,T),(T,H)}，含兩個結果（S14）。同一句描述在 S′ 中對應結果 1，在 S 中對應一個事件；判斷時要先說清楚用的是哪個樣本空間。", "en": "The MKO warns against confusing a single outcome with an event set. (H,T) is an element of S; 'exactly one head' is, in S, the event {(H,T),(T,H)} with two outcomes (S14). The same description corresponds to the outcome 1 in S′ but to an event in S; one must first say which sample space is in use." }, { "title": "漏掉可能結果 / Omitting a possible outcome", "zh": "依次記錄時，(H,T) 與 (T,H) 是不同的結果（S05）。若把 S 寫成 {(H,H),(H,T),(T,T)}，就漏掉了可能結果 (T,H)，違反假設 a2；之後事件與機率計算都會失去正確的論域。L_S_complete 的證明（S04.1、S04.2）正是用來排除這種遺漏。", "en": "When results are recorded in order, (H,T) and (T,H) are different outcomes (S05). Writing S as {(H,H),(H,T),(T,T)} omits the possible outcome (T,H) and violates assumption a2; later event and probability computations would lose their correct domain. The proof of L_S_complete (S04.1, S04.2) is what rules out such an omission." } ], "md_open_items": [ { "zh-Hant": "G4、G5 是模型假設；本稿不討論它們是否符合某枚實際硬幣，那是統計問題。", "en": "G4 and G5 are modelling assumptions; this guide does not discuss whether they fit an actual coin, which is a statistical question." }, { "zh-Hant": "finite_prob_weights 作為前置知識使用，未在本稿中由機率公理推出。", "en": "finite_prob_weights is used as a prerequisite and is not derived from the probability axioms in this guide." } ] }, "revision": { "id": "production-20261019-adopted-96", "source_commit": "b28a5edbd46279b1db2a3f3f48bb3f447e18899f", "adopted_source_sha256": "c0d1f9b0e3d3e27afde0183074318b955b8122887189e0f0232556ac536804a1", "original_source_sha256": "c0d1f9b0e3d3e27afde0183074318b955b8122887189e0f0232556ac536804a1", "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." } }