State & prove Baye’s theorem

Bayes Theorem :
Statement : Suppose$A_1,A_2,....A_n$ are n mutually exclusive and exhaustive events of a random  experiment associated with the sample space such that $P\left(A\right) \ne 0$ and for any event A such that

$P\left(\frac{A_k}{A}\right)=\frac{P\left(A_k\right)P\left(A/A_k\right)}{\sum _{i=1}^n P\left(A_i\right)P\left(A/A_i\right)}\text{ for any }k=1,2,....n$

Proof : Given $P\left(A\right) \ne 0$ for $i = 1,2,3,....n$
Since $A_1,A_2,...A_n$ are mutually execlusive

$A_i\cap A_j=\varphi \mathrm{\&}1\le i,j\le n$

Since $A_1, A_2-----A_n$ are exhaustive events then 

$A_1\cup A_2\cup ----\cup A_n=S$

Here for any event A of S, we have

$\left(A\cap A_i\right),\left(A\cap A_j\right)\text{ are also exclusive for }i\ne j$

$P\left(A\right)=P(A\cap S)=P\left[A\cap \left(A_1\cup A_2\cup ----\cup A_n\right)\right]$

$=P\left[\left(A\cap A_1\right)\cup \left(A\cap A_2\right)+\dots .P\left(A\cap A_n\right)\right]$

$=P\left(A\cap A_1\right)+P\left(A\cap A_2\right)+---+P\left(A\cap A_n\right)$

(by Axiam of additivity)

$=P\left(A_1\right)P\left(A/A_1\right)+P\left(A_2\right)P\left(A/A_2\right)+---+P\left(A_n\right)P\left(A/A_n\right)$

$=\sum _{i=1}^n P\left(A_i\right)P\left(A/A_i\right)$

Consider for $1\le k\le n,P\left(A_K/A\right)=\frac{P\left(A_K\cap A\right)}{P\left(A\right)}$

(by defi. Of conditional probability)

$=\frac{P\left(\left(A_K\right)P\left(A/A_K\right)\right.}{P\left(A\right)}=\frac{P\left(A_K\right)P\left(A/A_K\right)}{\sum _{i=1}^n P\left(A_i\right)P\left(A/A_i\right)}$