{"schema_version":"ocme-site-results-v1","locale":"zh-Hant","page_type":"derivations","facts":{"questions":36000,"reviewed":36000,"display_candidates":0,"knowledge":302,"guides":299,"paths":59,"batches":360,"families":1441,"atlas":302,"core":80,"evidence":9,"formal_evidence":5,"formal_subjects":4,"lean_files":6,"lean_toolchain":"leanprover/lean4:v4.30.0","question_state":"candidate","english_state":"ai_candidate","question_Lean":"NOT_RUN","canonical_truth_promotion":false},"filters":{"q":"mko-sample-space"},"kind":"guides","total":1,"pages":1,"page":1,"entries":[{"id":"derivation-mko-sample-space-01","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"},"summary":{"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)}."},"mko_ids":["mko-sample-space"]}],"title":"概念推導導讀"}