Question 104: Prior and Likelihood Must Both Enter a Bayes Reversal

Original question and givens

Use the same complete finite atomic probability law with aligned nonnegative exact masses=[1/11, 1/11, 1/11, 8/11], summing to 1, and canonical event evidence=[0, 2]. Event entries are increasing distinct outcome indices from 0 through 3. Every actual conditional probability requires a strictly positive denominator; a dedicated zero-denominator audit records undefined instead of dividing. Priors, likelihoods, intersections and evidence mass all come from this one law. Keep reduced exact fractions with positive denominators, omitting /1 for integers. Bayes conditioning does not imply causation. Hypothesis=[0, 1] must have positive prior, and evidence must have positive mass. Give (prior,likelihood,evidence_mass,joint,posterior,not_causal).

• G001 — Use the same complete finite atomic probability law with aligned nonnegative exact masses=[1/11, 1/11, 1/11, 8/11], summing to 1, and canonical event evidence=[0, 2]. Event entries are increasing distinct outcome indices from 0 through 3. Every actual conditional probability requires a strictly positive denominator; a dedicated zero-denominator audit records undefined instead of dividing. Priors, likelihoods, intersections and evidence mass all come from this one law. Keep reduced exact fractions with positive denominators, omitting /1 for integers. Bayes conditioning does not imply causation. Hypothesis=[0, 1] must have positive prior, and evidence must have positive mass. Give (prior,likelihood,evidence_mass,joint,posterior,not_causal).

Definition and scope

Use one complete finite probability law; both conditioning events have strictly positive mass.

• mko-bayes-theorem

Step-by-step derivation

P(Ω) := sum(1/11, 1/11, 1/11, 8/11)

P(Ω) = 1

H=[0, 1]; E=[0, 2]

H∩E=[0]

P(H) := sum(1/11, 1/11)

P(H) = 2/11

P(E) := sum(1/11, 1/11)

P(E) = 2/11

P(H∩E) := sum(1/11)

P(H∩E) = 1/11

P(E|H) := div(1/11, 2/11)

P(E|H) = 1/2

P(E|H)·P(H) := mul(1/2, 2/11)

P(E|H)·P(H) = 1/11

P(H|E) := div(1/11, 2/11)

P(H|E) = 1/2

P(H|E)=1/2; P(H∩E)=P(E|H)P(H)

P(H|E)=P(E|H)P(H)/P(E)=1/2

P(H|E)·P(E) := mul(1/2, 2/11)

P(H|E)·P(E) = 1/11

probability identity

no causal direction has been established

Reverse checks

• S011 · premise_and_result_audit — With zero conditioning mass the quotient is undefined, not a numerical 0/0 answer.
• S010 · premise_and_result_audit — Recompute P(H|E)·P(E) with exact rational arithmetic, not rounded decimals.
• S009 · premise_and_result_audit — Reversing a conditional does not mean swapping labels while keeping the old denominator.
• S008 · premise_and_result_audit — Recompute P(H|E) with exact rational arithmetic, not rounded decimals.
• S007 · premise_and_result_audit — Recompute P(E|H)·P(H) with exact rational arithmetic, not rounded decimals.
• S006 · premise_and_result_audit — Recompute P(E|H) with exact rational arithmetic, not rounded decimals.
• S005 · premise_and_result_audit — Recompute P(H∩E) with exact rational arithmetic, not rounded decimals.
• S004 · premise_and_result_audit — Recompute P(E) with exact rational arithmetic, not rounded decimals.
• S003 · premise_and_result_audit — Recompute P(H) with exact rational arithmetic, not rounded decimals.
• S002 · premise_and_result_audit — Do not multiply the event probabilities: independence has not been assumed.
• S001 · premise_and_result_audit — Recompute P(Ω) with exact rational arithmetic, not rounded decimals.

Result

(2/11, 1/2, 2/11, 1/11, 1/2, not_causal)

P(H|E)=1/2; the complete answer also retains the prior, likelihood, evidence and joint.

This question review and source binding

Review date: 2026-10-24

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