Conditional probability

If you know B occurred, discard outcomes outside B and renormalize the remaining weights. For $P(B)>0$, $P(A|B)=P(A\cap B)/P(B)$. The denominator is the event after the vertical bar. Swapping A and B usually changes the question.

Both $P(A|B)P(B)$ and $P(B|A)P(A)$ describe the same intersection. Equating them gives $P(A|B)=P(B|A)P(A)/P(B)$. Think 'prior weight × compatibility with evidence, then normalize'. The denominator must include signals from every possible cause, including false positives.

If H and its complement exhaust the possibilities, $P(B)=P(B|H)P(H)+P(B|H^c)P(H^c)$. This is a weighted average, not an unweighted average of the two signal rates. A frequency table over an imagined population often makes the same calculation easier to audit.

Odds in favor of H compare H with its complement. Odds a:b therefore mean $P(H)=a/(a+b)$, not a/b. For example, 2:3 odds mean a probability of 2/5. When both likelihoods are positive, Bayes also says posterior odds equal prior odds multiplied by the likelihood ratio $P(B|H)/P(B|H^c)$. Keep odds and probability as different quantities when interpreting a signal.