Question

Events $E_1$ and $E_2$ from a partition of the sample space S. A is any event such that $P(E_1) = P(E_2) = \frac{1}{2}, P(E_2/A) = \frac{1}{2}$ and $P(A/E_2)=\frac{2}{3}$, then $P(E_1/A)$ is

• A. $\frac{1}{2}$
• B. $\frac{2}{3}$
• C. 1
• D. $\frac{1}{4}$

Answer 1

Answer: (A)

$\frac{1}{2}$

Answer 2

Let $P\left(A \mid E_{1}\right)=x$
By Bayes' theorem,
$P\left(E_{2} |A\right)=\frac{P\left(E_{2}\right) P\left(A |E_{2}\right)}{P\left(E_{1}\right) P\left(A |E_{1}\right)+P\left(E_{2}\right) P\left(A |E_{2}\right)}$
$\Rightarrow \frac{1}{2}=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{2}\right) x+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}$
$\Rightarrow x=\frac{2}{3}$
$\therefore P\left(E_{1} |A\right)=\frac{P\left(E_{1}\right) P\left(A |E_{1}\right)}{P\left(E_{1}\right) P\left(A | E_{1}\right)+P\left(E_{2}\right) P\left(A | E_{2}\right)}$
$=\frac{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}{\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)+\left(\frac{1}{2}\right)\left(\frac{2}{3}\right)}=\frac{1}{2}$