{ "schema_version": "ocme-reviewed-question-reader-v1", "question_id": "q-20261013-advance-20261024-r03-finite-markov-chain-0075", "pilot_number": "R0021-11400", "number": 11400, "family_id": "qf-mod24-markov-two-step-entry", "title": { "zh-Hant": "第 11400 題：有限狀態Markov鏈", "en": "Question 11400: Two-Step Probability by Summing Every Intermediate State" }, "original_question": { "schema_version": "ocme-question-v0.1", "id": "q-20261013-advance-20261024-r03-finite-markov-chain-0075", "batch_id": "batch-20261013-advance-20261024-r03-finite-markov-chain-foundations", "language": "zh-Hant", "source_mko_ids": [ "mko-finite-markov-chain" ], "title_zh": "有限狀態Markov鏈", "stem_zh": "rows=current state且columns=next state。操作=markov-two-step-entry；資料={\"transition\":[[[25,27],[2,27]],[[13,14],[1,14]]],\"from\":0,\"to\":1}。請依 [from,to,middle_state_terms,two_step_probability] 欄序回答。", "answer": { "kind": "expression", "value": "(0, 1, [50/729, 1/189], 377/5103)", "explanation_zh": "依賴結構核對=(0, 1, [50/729, 1/189], 377/5103)；依完整明示資料重算；精確有限切片不自動升格為一般無限結論；轉移列非負且精確和為1；不推出平穩遍歷或長期收斂。" }, "difficulty": { "band": "undergraduate_core", "level": 3, "skills": [ "有限狀態Markov鏈" ] }, "variant": { "family_id": "qf-mod24-markov-two-step-entry", "seed": 202610243075, "parameters": { "transition": [ [ [ 25, 27 ], [ 2, 27 ] ], [ [ 13, 14 ], [ 1, 14 ] ] ], "from": 0, "to": 1, "domain": "Exact_finite_time_homogeneous_Markov_kernel", "operation": "markov-two-step-entry", "semantic_template": "qf-mod24-markov-two-step-entry/{\"from\":0,\"to\":1,\"transition\":[[[25,27],[2,27]],[[13,14],[1,14]]]}" } }, "verification": { "status": "candidate", "checks": [ "producer_exact_finite_model", "producer_source_snapshot_check", "producer_substantive_template_deduplication", "producer_json_schema" ], "warning_zh": "產生端候選須由主端重算；不構成人工審定、一般無限定理、Markov長期性、PDE適定性、Lean執行或形式Evidence。" }, "provenance": { "generator_role": "question_producer", "generator_model": "model_unknown", "generated_at": "2026-10-13", "contract_version": "ocme-question-factory-v0.1" } }, "original_english_edition": { "schema_version": "ocme-question-localization-v0.1", "question_id": "q-20261013-advance-20261024-r03-finite-markov-chain-0075", "batch_id": "batch-20261013-advance-20261024-r03-finite-markov-chain-foundations", "locale": "en", "translation_state": "ai_candidate", "source": { "question_sha256": "582e76542f3aba69002222257b8759c67647194ca6b31dc2cec9c3f591dc781f", "family_id": "qf-mod24-markov-two-step-entry", "seed": 202610243075, "semantic_template": "qf-mod24-markov-two-step-entry/{\"from\":0,\"to\":1,\"transition\":[[[25,27],[2,27]],[[13,14],[1,14]]]}" }, "fields": { "title": "Two-Step Probability by Summing Every Intermediate State", "stem": "Use one fixed finite time-homogeneous transition table=[[25/27, 2/27], [13/14, 1/14]]. Rows denote the current state and columns the next state; row entries must be nonnegative and sum exactly to 1 before stepping. State order stays fixed. Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs. No stationary, irreducibility, ergodicity or long-term convergence claim follows. Require a qualified kernel. from=0, to=1. Give (from,to,every P[from][middle]*P[middle][to] term in middle-state order,sum). This conditional transition entry does not require an initial law.", "answer": { "kind": "expression", "value": "(0, 1, [50/729, 1/189], 377/5103)", "explanation": "Sum over all intermediate states using exact products, rather than multiplying two unrelated marginal probabilities. This exact finite slice is not automatically a general infinite conclusion. The exact result, retaining the requested order, is (0, 1, [50/729, 1/189], 377/5103)." }, "skills": [ "Two-Step Probability by Summing Every Intermediate State" ], "verification_warning": "This is a candidate exercise. Mechanical recomputation checks only the stated conditions and this instance; it is neither a formal proof nor human review." } }, "givens": [ { "id": "G001", "text": { "zh-Hant": "rows=current state且columns=next state。操作=markov-two-step-entry；資料={\"transition\":[[[25,27],[2,27]],[[13,14],[1,14]]],\"from\":0,\"to\":1}。請依 [from,to,middle_state_terms,two_step_probability] 欄序回答。", "en": "Use one fixed finite time-homogeneous transition table=[[25/27, 2/27], [13/14, 1/14]]. Rows denote the current state and columns the next state; row entries must be nonnegative and sum exactly to 1 before stepping. State order stays fixed. Use exact rational arithmetic, reduce fractions to positive denominators and omit /1 for integer outputs. No stationary, irreducibility, ergodicity or long-term convergence claim follows. Require a qualified kernel. from=0, to=1. Give (from,to,every P[from][middle]*P[middle][to] term in middle-state order,sum). This conditional transition entry does not require an initial law." } } ], "steps": [ { "id": "S001", "uses": [ "G001" ], "before": "transition row0:raw rational pairs=[[25, 27], [2, 27]]", "after": "transition row0:normalized=[25/27, 2/27]", "relation": "implies", "rule_id": "exact_weight_normalization", "conditions": [], "explanation": { "zh-Hant": "每個原分母須非零；分數約分並令分母正，不以小數近似。", "en": "Every original denominator must be nonzero.Reduce fractions to positive denominators without decimal approximation." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S002", "uses": [ "G001", "S001" ], "before": "transition row0 total := sum(25/27, 2/27)", "after": "transition row0 total = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步已指定的角色逐項精確運算；保留符號、分母與原始配對。", "en": "Compute exactly with this step's specified roles,retaining signs,denominators and original pairings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 transition row0 total，不以小數近似替代。", "en": "Recompute transition row0 total with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "25/27", "2/27" ], "output": "1", "label": "transition row0 total" } }, { "id": "S003", "uses": [ "G001", "S002" ], "before": "transition row0:weights=[25/27, 2/27];sum=1", "after": "nonnegative=是;total_one=是;qualified=是", "relation": "implies", "rule_id": "separate_weight_guards", "conditions": [], "explanation": { "zh-Hant": "非負與總和恰為1分開核；零權重合法，負數或接近1不自動修補。", "en": "Check nonnegativity and exact total one separately.Zero weights are allowed;negative weights or near-one totals are not repaired." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S004", "uses": [ "G001", "S003" ], "before": "transition row1:raw rational pairs=[[13, 14], [1, 14]]", "after": "transition row1:normalized=[13/14, 1/14]", "relation": "implies", "rule_id": "exact_weight_normalization", "conditions": [], "explanation": { "zh-Hant": "每個原分母須非零；分數約分並令分母正，不以小數近似。", "en": "Every original denominator must be nonzero.Reduce fractions to positive denominators without decimal approximation." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S005", "uses": [ "G001", "S004" ], "before": "transition row1 total := sum(13/14, 1/14)", "after": "transition row1 total = 1", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步已指定的角色逐項精確運算；保留符號、分母與原始配對。", "en": "Compute exactly with this step's specified roles,retaining signs,denominators and original pairings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 transition row1 total，不以小數近似替代。", "en": "Recompute transition row1 total with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "13/14", "1/14" ], "output": "1", "label": "transition row1 total" } }, { "id": "S006", "uses": [ "G001", "S005" ], "before": "transition row1:weights=[13/14, 1/14];sum=1", "after": "nonnegative=是;total_one=是;qualified=是", "relation": "implies", "rule_id": "separate_weight_guards", "conditions": [], "explanation": { "zh-Hant": "非負與總和恰為1分開核；零權重合法，負數或接近1不自動修補。", "en": "Check nonnegativity and exact total one separately.Zero weights are allowed;negative weights or near-one totals are not repaired." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S007", "uses": [ "G001", "S006" ], "before": "complete transition matrix=[[25/27, 2/27], [13/14, 1/14]];row sums=[1, 1]", "after": "nonnegative=是;row_stochastic=是;rows=current state, columns=next state", "relation": "implies", "rule_id": "whole_row_stochastic_kernel_convention", "conditions": [], "explanation": { "zh-Hant": "完整轉移表逐列非負且和1才合格；不以行和代替列和。固定時間齊次Markov模型，不推出平穩、遍歷或長期收斂。", "en": "Every complete row must be nonnegative with exact sum1;column sums are not the convention.This fixed time-homogeneous Markov model gives no stationary,ergodic or long-term convergence theorem." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S008", "uses": [ "G001", "S007" ], "before": "from=0;to=1;middle states=[0, 1]", "after": "two-step conditional entry=sum_k P[from][k]*P[k][to];no initial law needed", "relation": "implies", "rule_id": "markov_all_middle_states_chapman_kolmogorov", "conditions": [], "explanation": { "zh-Hant": "給定起點的兩步轉移以所有中間狀態分割，時間齊次與Markov前提才允許兩轉移機率相乘；不是任意邊際獨立。", "en": "Partition the conditional two-step transition over every middle state.The declared homogeneous Markov premise permits these two transition factors;this is not independence of arbitrary marginals." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "回到原始資料與本步條件核對，不把單向推論誤稱等價。", "en": "Check against the original data and this step's premises; a one-way implication is not an equivalence." } }, "arithmetic": null }, { "id": "S009", "uses": [ "G001", "S008" ], "before": "two-step middle0 term := mul(25/27, 2/27)", "after": "two-step middle0 term = 50/729", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步已指定的角色逐項精確運算；保留符號、分母與原始配對。", "en": "Compute exactly with this step's specified roles,retaining signs,denominators and original pairings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 two-step middle0 term，不以小數近似替代。", "en": "Recompute two-step middle0 term with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "25/27", "2/27" ], "output": "50/729", "label": "two-step middle0 term" } }, { "id": "S010", "uses": [ "G001", "S009" ], "before": "two-step middle1 term := mul(2/27, 1/14)", "after": "two-step middle1 term = 1/189", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步已指定的角色逐項精確運算；保留符號、分母與原始配對。", "en": "Compute exactly with this step's specified roles,retaining signs,denominators and original pairings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 two-step middle1 term，不以小數近似替代。", "en": "Recompute two-step middle1 term with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "mul", "inputs": [ "2/27", "1/14" ], "output": "1/189", "label": "two-step middle1 term" } }, { "id": "S011", "uses": [ "G001", "S010" ], "before": "two-step probability := sum(50/729, 1/189)", "after": "two-step probability = 377/5103", "relation": "calculation", "rule_id": "exact_rational_arithmetic", "conditions": [], "explanation": { "zh-Hant": "按本步已指定的角色逐項精確運算；保留符號、分母與原始配對。", "en": "Compute exactly with this step's specified roles,retaining signs,denominators and original pairings." }, "reverse_audit": { "kind": "premise_and_result_audit", "explanation": { "zh-Hant": "以精確有理數重算 two-step probability，不以小數近似替代。", "en": "Recompute two-step probability with exact rational arithmetic, not rounded decimals." } }, "arithmetic": { "operation": "sum", "inputs": [ "50/729", "1/189" ], "output": "377/5103", "label": "two-step probability" } } ], "math_model": { "result": [ 0, 1, [ "50/729", "1/189" ], "377/5103" ] }, "conclusion": { "canonical_value": "(0, 1, [50/729, 1/189], 377/5103)", "explanation": { "zh-Hant": "依上列完整代值及條件核對得到原題答案；原題編號、格式與狀態保持不變。", "en": "The full substitutions and premise checks give the original answer; source identity,format and status are unchanged." } }, "definition_note": { "zh-Hant": "沿用原題明示的論域、假設與來源 MKO 定義。本題已完成第二對話數理與中英語義覆核，並由原始參數逐題獨立重算模型、答案與每個算術步驟；原題仍為 candidate。", "en": "Use the original question domain, assumptions and cited MKO definitions. This item has second-context mathematical and bilingual review, with the model, answer and every arithmetic step independently recomputed from the original parameters; the source question remains a candidate." }, "source": { "input_commit": "0f7be61e2c22317a4075561dc822c87d881a50db", "question_binding": { "batch_id": "batch-20261013-advance-20261024-r03-finite-markov-chain-foundations", "batch_path": "public/data/questions/batches/batch-20261013-advance-20261024-r03-finite-markov-chain-foundations/questions.jsonl", "batch_sha256": "16a6ed294654ed9a3835f127395ae71742842a568c7ce11052f824b7aacebee4", "original_line": 75, "original_line_sha256": "f0e2b1271844add3d6e2fcc6c2ecf65b52eb60637cdd48aa9cc43f803ed108a6", "question_compact_sha256": "582e76542f3aba69002222257b8759c67647194ca6b31dc2cec9c3f591dc781f", "english_row_compact_sha256": "9ca59981a55a429ca6a6780e9e152e4767dbfed116ce8322d9ae4f4cd2e8ceb1", "english_manifest_path": "public/data/i18n/editions/batch-20261013-advance-20261024-r03-finite-markov-chain-foundations.en.json", "english_rows_path": "public/data/i18n/questions/batch-20261013-advance-20261024-r03-finite-markov-chain-foundations.en.jsonl" }, "mko_bindings": { "mko-finite-markov-chain": { "sha256": "713cef8fb3c1efc7627452592ab30fb69b9ebb7992c47aad78b9ae322ee7fd55", "english_sha256": "531d840a993901f295df6db2e167b57a16a1d3820e22c5212389593049a3ef38", "original_path": "public/data/mko/finite-markov-chain.json", "english_path": "public/data/i18n/editions/mko-finite-markov-chain.en.json" } }, "selected_manifest_sha256": "5233edba9654a6cffd8b6b2d83c932f82ded15f9c0b7c733bd049943e672acff", "selected_part_sha256": "18bbb89127e14ce3466c1a8730f1c3c6e22a8f59a9148433b2abc7f6ea274ee5", "original_candidate_sha256": "4fc4ee52a982aa8f04b870a73443b48af25508e714d5e4af0a7117c45a380568", "original_candidate_line_sha256": "4fc4ee52a982aa8f04b870a73443b48af25508e714d5e4af0a7117c45a380568" }, "historical_source_state": { "review_state": { "derivation": "candidate_authored_and_locally_checked", "source_question_state": "candidate", "independent_AI_or_human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN" }, "scope_note": { "zh-Hant": "來源綁定教學推導候選；精確核對不等於獨立AI／人工審查或一般形式證明。原題、答案、MKO、公式、AST與歷史不變。", "en": "Source-bound pedagogical derivation candidate.Exact checking is not independent AI/human review or general formal proof.Original questions,answers,MKOs,formulas,ASTs and history remain unchanged." }, "label": "Frozen producer metadata; current per-question review is separate below." }, "revision": { "id": "review-20261026-006-p22", "review_date": "2026-10-26", "owner_authorized_reader_publication": true, "scope": "One hundred sixty family prototypes and their mathematical/bilingual reasoning variants, stems, boundary statements and every distinct explanation shape were reviewed in a context independent of the producer; parameter substitutions, source bindings, full math models, both answer strings and every arithmetic step were independently reconstructed per question from the original parameters only, with negative controls. This grouped method does not claim a separate literal AI reading of every repeated sentence or an extra AI voter.", "review_receipts": { "semantic_review_sha256": "47d23d3735987d68e27733b051bdfa80a3616d44fb223b82bf2e6c8e46d49258", "question_first_witness_sha256": "a857db558000724020d2d5e815cb6bc3e2c9e2933df295bf33eb478d3c695139" }, "status": "second_context_reviewed", "mathematical_semantics": true, "bilingual_semantics": true, "steps_reviewed": 11, "method": "family_template_semantic_review_plus_independent_per_instance_reconstruction", "exact_question_first_check": { "model_sha256": "d4d77a6d90799f6fd1178e4c5afc4c73b34443324eb9998af07ff981f7b9ca7e", "arithmetic_steps": 5 }, "uses_answer_as_premise": false, "human_review": "NOT_PERFORMED", "formalization": "not_formalized", "Lean": "NOT_RUN", "Canonical_truth_promotion": false, "authority": "Owner-authorized reviewed reader publication; original question/candidate states remain historical and unchanged." } }