## Bayes’ Theorem

One of the important uses of conditional probability is to find out the inverse probability. If an event $B$ can happen with mutually exclusive and exhaustive events (partitions) $A_1, A_2, \cdots, A_n$ whose probabilities are known and also the conditional probabilities $P(B/A_1)$, $P(B/A_2)$, $\cdots, P(B/A_n)$ are known, then with the help of these probabilities the inverse probabilities $P(A_1/B)$, $P(A_1/B)$, $\cdots, P(A_n/B)$ can be determined. This particular rule was given by British Mathematician Bayes. It is known as Bayes’ Theorem.

## Bayes’ Theorem

Let $A_1, A_2, \cdots, A_n$ are mutually exclusive and exhaustive events of the sample space $S$ and if $P(A_i)\neq 0$, $i=1,2,\cdots,n$, then for any event $B$ of the sample space $S$

$$\begin{equation*} P(A_i/B) =\frac{P(A_i) A(B/A_i)}{\sum_{i=1}^n P(A_i) A(B/A_i)} \end{equation*}$$ for $i=1,2,\cdots , n$.

### Proof

As $A_1, A_2, \cdots, A_n$ are mutually exclusive and exhaustive events of the sample space $S$, we have $A_1\cup A_2\cup \cdots \cup A_n=S$ and $A_1\cap A_2\cap \cdots\cap A_n=\phi$. Let $B$ be any event of $S$. From the Venn diagram it is clear that $B\cap S = B$. $$\begin{eqnarray*} \therefore B & = & B\cap (A_1\cup A_2\cup \cdots \cup A_n)\\ & = & (B\cap A_1)\cup (B\cap A_2)\cup \cdots \cup(B\cap A_n)\\ \therefore P(B) & = & P\big[(B\cap A_1)\cup (B\cap A_2)\cup \cdots \cup(B\cap A_n)\big] \end{eqnarray*}$$

But as $A_1, A_2, \cdots, A_n$ are mutually exclusive events, $(B\cap A_1), (B\cap A_2),\cdots, (B\cap A_n)$ are also mutually exclusive. Hence by Axiom (3) of probability, we have

$$\begin{eqnarray} \label{eqc1} \therefore P(B) & = & P(B\cap A_1)+ P(B\cap A_2)+ \cdots +P(B\cap A_n)\\\nonumber & = &P(A_1)\cdot P(B/A_1) + P(A_2)\cdot P(B/A_2)+ \cdots \\\nonumber & & \quad + P(A_2)\cdot P(B/A_n)\\ & = & \sum_{i=1}^n P(A_i)\cdot P(B/A_i) \end{eqnarray}$$

But, $$$$\label{eqc2} P(A_i/B) = \dfrac{P(A_i\cap B)}{P(B)}$$$$

From \eqref{eqc1} and \eqref{eqc2}, we have $$\begin{equation*} P(A_i/B) =\frac{P(A_i) P(B/A_i)}{\sum_{i=1}^n P(A_i) P(B/A_i)} \end{equation*}$$

for $i=1,2,\cdots , n$.