Question
If $E_1$ and $E_2$ are equally likely, mutually exclusive and exhaustive events and $P\left(A / E_1\right)=0.2$,
$P\left(A / E_2\right)=0.3$. Find $P\left(E_1 / A\right)$.
$P\left(A / E_2\right)=0.3$. Find $P\left(E_1 / A\right)$.
$\therefore P\left(E_1\right)=P\left(E_2\right)=\frac{1}{2}$
$ \begin{aligned} & P\left(A / E_1\right)=\frac{P\left(A \cap E_1\right)}{P\left(E_1\right)}=0.2=\frac{2}{10} \\ & P\left(A / E_2\right)=\frac{P\left(A \cap E_2\right)}{P\left(E_2\right)}=0.3=\frac{3}{10} \end{aligned} $ By Bayes' theorem,
$\begin{aligned} P\left(E_1 / A\right) & =\frac{P\left(E_1\right) \cdot P\left(A / E_1\right)}{P\left(E_1\right) \cdot P\left(A / E_1\right)+P\left(E_2\right) \cdot P\left(A / E_2\right)} \\ & =\frac{\frac{1}{2} \cdot\left(\frac{2}{10}\right)}{\frac{1}{2}\left(\frac{2}{10}\right)+\frac{1}{2}\left(\frac{3}{10}\right)} \cdots\left[\because P\left(E_1\right)=P\left(E_2\right)\right]\end{aligned}$
$\begin{aligned} & =\frac{2}{2+3} M \\ & =\frac{2}{5}=0.4\end{aligned}$
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