A6 - bayes' law

bayes' law

assuming $P (E) > 0 :$

P (F | E) = \frac{P (E | F) P (F)}{P (E)}

proof:

P (F | E) = \frac{P (E \cap F)}{P (E)} = \frac{P (E | F) P (F)}{P (E)}

the law of total probability can be combined with bayes' law

extended bayes' law

let $F_{1}, F_{2}, \dots F_{n}$ partition $Ω$ , and $P (E) > 0 :$

P (F | E) = \frac{P (E | F) P (F)}{\sum_{j = 1}^{n} P (E | F_{j}) P (F_{j})}

eg: there is a rare cancel that occurs in one in a million people with a test for it is $99.9 %$ effective, if one tests positive:

\begin{aligned} P (C | +) & = \frac{P (+ | C) P (C)}{P (+ | C) P (C) + P (+ | \bar{C}) P (\bar{C})} \\ = \frac{0.999 \times 10^{- 6}}{0.999 \times 10^{- 6} + 10^{- 3} \times (1 - 10^{6})} \\ = \frac{1}{1001} \end{aligned}