2 Bayes’ theorem

We again consider the conditional probability statement:

$\text{P}\left(A|B\right)=\frac{\text{P}\left(A\cap B\right)}{\text{P}\left(B\right)}=\frac{\text{P}\left(A\cap B\right)}{\text{P}\left(B|A_1\right)\text{P}\left(A_1\right)+\text{P}\left(B|A_2\right)\text{P}\left(A_2\right)+\cdots +\text{P}\left(B|A_n\right)\text{P}\left(A_n\right)}$

in which we have used the theorem of Total Probability to replace $\text{P}\left(B\right)$ . Now

$\text{P}\left(A\cap B\right)=\text{P}\left(B\cap A\right)=\text{P}\left(B|A\right)\times \text{P}\left(A\right)$

Substituting this in the expression for $\text{P}\left(A|B\right)$ we immediately obtain the result

$\text{P}\left(A|B\right)=\frac{\text{P}\left(B|A\right)\times \text{P}\left(A\right)}{\text{P}\left(B|A_1\right)\text{P}\left(A_1\right)+\text{P}\left(B|A_2\right)\text{P}\left(A_2\right)+\cdots +\text{P}\left(B|A_n\right)\text{P}\left(A_n\right)}$

This is true for any event $A$ and so, replacing $A$ by $A_i$ gives the result, known as Bayes’ theorem as

$\text{P}\left(A_i|B\right)=\frac{\text{P}\left(B|A_i\right)\times \text{P}\left(A_i\right)}{\text{P}\left(B|A_1\right)\text{P}\left(A_1\right)+\text{P}\left(B|A_2\right)\text{P}\left(A_2\right)+\cdots +\text{P}\left(B|A_n\right)\text{P}\left(A_n\right)}$