Wikipedia has an article on:
 P(A|B)\equiv \frac{P(B|A) P(A)}{P(B|A) P(A)+ P(B| not A)(1- P(A))}


 P(A|B)\equiv \frac{P(B|A) P(A)}{P(B) }

See also Bayesian Priors

Ad blocker interference detected!

Wikia is a free-to-use site that makes money from advertising. We have a modified experience for viewers using ad blockers

Wikia is not accessible if you’ve made further modifications. Remove the custom ad blocker rule(s) and the page will load as expected.