How is Bayes' theorem derived from conditional probability for events of positive probability?

Starting from the equality $P(A)P$(B|A) = $P(B)P$(A|B), you can solve for either conditional probability by dividing by the corresponding marginal probability. Dividing both sides by $P(B)$ isolates P(A|B), giving P(A|B) = $P(A)P$(B|A)/P(B).

Conditions: Both A and B have positive probability for the ordinary conditionals used here.; The algebraic rearrangement requires the denominators $P(B)$ and $P(A)$ to be nonzero.

The denominator $P(B)$ in Bayes' theorem represents the total probability of the evidence (event B). In the context of a screening test, a positive result can arise from two mutually exclusive scenarios: a true positive (person has the disease and tests positive) or a false positive (person does not have the disease but tests positive).

Conditions: The evidence event B can occur via multiple mutually exclusive paths.; The law of total probability applies.

The positive predictive value, or the probability of having the disease given a positive test result ($P(D \mid +)$), is calculated by dividing the probability of a true positive by the total probability of a positive result. The total probability of a positive result includes both true positives and false positives.

Conditions: Prevalence $P(D) = 0.12$.; Sensitivity $P(+ \mid D) = 0.95$.; False positive rate $P(+ \mid D^c) = 0.06$.; The events are mutually exclusive and exhaustive for the disease status.

The probability is $1/3$. This is calculated using Bayes' theorem where event $A$ is 'two girls' and event $B$ is 'at least one girl'.

Conditions: Four equally likely ordered outcomes: GG, GB, BG, BB.; Event A: Two girls.; Event B: At least one girl.

The denominator $P(B)$ ensures that the posterior probability $P(A \mid B)$ sums to 1 over all possible mutually exclusive events $A$. It scales the product of the likelihood and prior to a valid probability distribution.

Conditions: Bayes' theorem is written in probability form: $P(A \mid B) = \frac{P(B \mid A)P(A)}{P(B)}$.