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Question 3193: Conditional Probability with Explicit Equiprobability

On this page
  1. Original question and givens
  2. Definition and scope
  3. Step-by-step derivation
  4. Reverse checks
  5. Result
  6. This question review and source binding

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R0005-03193 · q-20261003-advance-20261004-r07-conditional-probability-0043

Original question and givens

The nonempty finite sample space is S={5, 6, 7, 8, 9, 10}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={5, 7, 9} and B={9, 10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

  • G001 — The nonempty finite sample space is S={5, 6, 7, 8, 9, 10}, with event family F=𝒫(S). All elementary outcomes are explicitly equiprobable. A={5, 7, 9} and B={9, 10} is nonempty. Compute P(A|B) as a reduced fraction with positive denominator.

Definition and scope

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.

Step-by-step derivation

S001 · implies

Before

S={5, 6, 7, 8, 9, 10};A={5, 7, 9};B={9, 10};F=𝒫(S)

After

events are declared subsets;conditioning needs P(B)>0

Conditions: none

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.

Step identifiers and dependencies

Rule: conditional_model_domain · Uses: G001

S002 · implies

Before

A={5, 7, 9};B={9, 10}

After

A∩B={9}

Conditions: none

The intersection keeps outcomes in both events;neither a union nor independence is assumed.

Step identifiers and dependencies

Rule: explicit_event_intersection · Uses: G001, S001

S003 · implies

Before

all elementary outcomes explicitly equiprobable

After

P(A∩B)=|A∩B|/6;P(B)=|B|/6

Conditions: none

Only explicit equiprobability permits the cardinality ratio;nonempty B then has positive probability.

Step identifiers and dependencies

Rule: equiprobable_counting_premise · Uses: G001, S002

S004 · calculation

Before

uniform conditional probability := div(1, 2)

After

uniform conditional probability = 1/2

Conditions: none

Cancel the common positive denominator |S|,leaving intersection count divided by conditioning count.

Step identifiers and dependencies

Rule: exact_rational_arithmetic · Uses: G001, S003

Reverse checks

  • S004 · premise_and_result_audit — Recompute uniform conditional probability with exact rational arithmetic, not rounded decimals.
  • S003 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S002 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.
  • S001 · premise_and_result_audit — Check against the original data and this step's premises; a one-way implication is not an equivalence.

Result

1/2

The full substitutions and premise checks give the original answer; source identity,format and status are unchanged.

This question review and source binding

Review date: 2026-10-26

Original questions, answers, step expressions, numeric ASTs and historical records are retained. Source candidate labels remain historical and are not overwritten as formal proof.

Source hashes and review receipts
  • input commit: 0f7be61e2c22317a4075561dc822c87d881a50db
  • original question SHA-256: d2114d6958e6b46e0d69fd2723cc560c524e216059ae8f0c0d67b035212d042c
  • candidate record SHA-256: fafd9bc08dfd1bcf27df921d7c66fa1fb04b29704b414039e0988fc2cd387c89
  • semantic review receipt SHA-256: 5d74de11610ebc6037a4fc9946d77154d4ed18389f8f0feeb823148b6320c4c2
  • question-first witness SHA-256: 1fba4ab0f13c2cfcc4d9fdebe492c442d43dd45700394a443daa28f333355ed9
  • Lean: NOT_RUN · human review: NOT_PERFORMED
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