{
  "schema_version": "ocme-reviewed-question-reader-v1",
  "question_id": "q-20261003-advance-20261004-r07-conditional-probability-0027",
  "pilot_number": "R0005-03177",
  "number": 3177,
  "family_id": "qf-conditional-uniform-probability",
  "title": {
    "zh-Hant": "第 3177 題：明示等可能的條件機率",
    "en": "Question 3177: Conditional Probability with Explicit Equiprobability"
  },
  "original_question": {
    "schema_version": "ocme-question-v0.1",
    "id": "q-20261003-advance-20261004-r07-conditional-probability-0027",
    "batch_id": "batch-20261003-advance-20261004-r07-conditional-probability-foundations",
    "language": "zh-Hant",
    "source_mko_ids": [
      "mko-conditional-probability"
    ],
    "title_zh": "明示等可能的條件機率",
    "stem_zh": "明示非空 S={-11, -10, -9, -8, -7}、事件族 F=𝒫(S)，所有基本結果明示等可能。A={-11, -10, -9, -8, -7}，B={-11, -10} 且 B 非空。求 P(A|B)。",
    "answer": {
      "kind": "expression",
      "value": "1/1",
      "explanation_zh": "交集數=2；條件數=2。因明示等可能且 B 非空，P(A|B)=|A∩B|/|B|=1/1；𝒫 是全冪集，P 是機率律。"
    },
    "difficulty": {
      "band": "upper_secondary",
      "level": 1,
      "skills": [
        "明示等可能的條件機率"
      ]
    },
    "variant": {
      "family_id": "qf-conditional-uniform-probability",
      "seed": 202610047027,
      "parameters": {
        "S": [
          -11,
          -10,
          -9,
          -8,
          -7
        ],
        "F": "powerset(S)",
        "A": [
          -11,
          -10,
          -9,
          -8,
          -7
        ],
        "B": [
          -11,
          -10
        ],
        "equiprobable": true,
        "operation": "uniform_conditional",
        "semantic_template": "qf-conditional-uniform-probability/{\"A\":[-11,-10,-9,-8,-7],\"B\":[-11,-10],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8,-7],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"
      }
    },
    "verification": {
      "status": "candidate",
      "checks": [
        "producer_exact_integer_rational_arithmetic",
        "producer_finite_enumeration",
        "producer_source_snapshot_check",
        "producer_substantive_template_deduplication"
      ],
      "warning_zh": "僅為產生端候選及自檢；主架構者須另做獨立驗收，不構成人工審定、普遍證明或形式 Evidence。"
    },
    "provenance": {
      "generator_role": "question_producer",
      "generator_model": "Codex independent peer (inherited model; exact runtime model not exposed)",
      "generated_at": "2026-10-03",
      "contract_version": "ocme-question-factory-v0.1"
    }
  },
  "original_english_edition": {
    "schema_version": "ocme-question-localization-v0.1",
    "question_id": "q-20261003-advance-20261004-r07-conditional-probability-0027",
    "batch_id": "batch-20261003-advance-20261004-r07-conditional-probability-foundations",
    "locale": "en",
    "translation_state": "ai_candidate",
    "source": {
      "question_sha256": "51cfd6470b419cdfbebff78054969548ba01aa8fde7596909d5729dba261ea43",
      "family_id": "qf-conditional-uniform-probability",
      "seed": 202610047027,
      "semantic_template": "qf-conditional-uniform-probability/{\"A\":[-11,-10,-9,-8,-7],\"B\":[-11,-10],\"F\":\"powerset(S)\",\"S\":[-11,-10,-9,-8,-7],\"equiprobable\":true,\"operation\":\"uniform_conditional\"}"
    },
    "fields": {
      "title": "Conditional Probability with Explicit Equiprobability",
      "stem": "The nonempty finite sample space is S={-11, -10, -9, -8, -7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-11, -10, -9, -8, -7} and B={-11, -10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.",
      "answer": {
        "kind": "expression",
        "value": "1/1",
        "explanation": "|A∩B|=2 and |B|=2. Explicit equiprobability and nonempty B give P(A|B)=|A∩B|/|B|=1/1. Here 𝒫 denotes the full powerset and P denotes the probability law."
      },
      "skills": [
        "Conditional Probability with Explicit Equiprobability"
      ],
      "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": "明示非空 S={-11, -10, -9, -8, -7}、事件族 F=𝒫(S)，所有基本結果明示等可能。A={-11, -10, -9, -8, -7}，B={-11, -10} 且 B 非空。求 P(A|B)。",
        "en": "The nonempty finite sample space is S={-11, -10, -9, -8, -7}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={-11, -10, -9, -8, -7} and B={-11, -10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator."
      }
    }
  ],
  "steps": [
    {
      "id": "S001",
      "uses": [
        "G001"
      ],
      "before": "S={-11, -10, -9, -8, -7};A={-11, -10, -9, -8, -7};B={-11, -10};F=𝒫(S)",
      "after": "events are declared subsets;conditioning needs P(B)>0",
      "relation": "implies",
      "rule_id": "conditional_model_domain",
      "conditions": [],
      "explanation": {
        "zh-Hant": "𝒫(S)是事件的全冪集，不是機率值；非空事件也可能機率0。只問B的條件化資格時，不另假設未給定的A。",
        "en": "𝒫(S) is the full event powerset,not a probability value.A nonempty event may have probability0.When checking B's eligibility,do not invent an unspecified A."
      },
      "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": "A={-11, -10, -9, -8, -7};B={-11, -10}",
      "after": "A∩B={-11, -10}",
      "relation": "implies",
      "rule_id": "explicit_event_intersection",
      "conditions": [],
      "explanation": {
        "zh-Hant": "交集只保留同時在A和B中的結果，並集或兩事件的獨立性都不是本步前提。",
        "en": "The intersection keeps outcomes in both events;neither a union nor independence is assumed."
      },
      "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": "S003",
      "uses": [
        "G001",
        "S002"
      ],
      "before": "all elementary outcomes explicitly equiprobable",
      "after": "P(A∩B)=|A∩B|/5;P(B)=|B|/5",
      "relation": "implies",
      "rule_id": "equiprobable_counting_premise",
      "conditions": [],
      "explanation": {
        "zh-Hant": "只有明示等可能時才能把機率比改成個數比；B非空保證其機率正。",
        "en": "Only explicit equiprobability permits the cardinality ratio;nonempty B then has positive probability."
      },
      "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": "uniform conditional probability := div(2, 2)",
      "after": "uniform conditional probability = 1",
      "relation": "calculation",
      "rule_id": "exact_rational_arithmetic",
      "conditions": [],
      "explanation": {
        "zh-Hant": "兩個共同正分母|S|相消，剩交集數除以條件數。",
        "en": "Cancel the common positive denominator |S|,leaving intersection count divided by conditioning count."
      },
      "reverse_audit": {
        "kind": "premise_and_result_audit",
        "explanation": {
          "zh-Hant": "以精確有理數重算 uniform conditional probability，不以小數近似替代。",
          "en": "Recompute uniform conditional probability with exact rational arithmetic, not rounded decimals."
        }
      },
      "arithmetic": {
        "operation": "div",
        "inputs": [
          "2",
          "2"
        ],
        "output": "1",
        "label": "uniform conditional probability"
      }
    }
  ],
  "math_model": {
    "result": "1",
    "intersection_count": 2,
    "conditioning_count": 2
  },
  "conclusion": {
    "canonical_value": "1/1",
    "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."
  },
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      "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."
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  "revision": {
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    "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.",
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}