Bayes' Law

Suppose we have $\{A_1,\dots ,A_N\}$ such that

$$\begin{array}{rl}& 1)\bigcup _{n=1}^N{A}_n=S\\ & 2)A_n\cap A_m=\varnothing \forall n\ne m\end{array}$$

Then

[!Bayes’ Law]

Pr(A_{m}|B) = \frac{Pr(B|A_{m})Pr(A_{m})}{\sum_{n=1}^{N}Pr(B|A_{n})Pr(A_{n})}