Question

It is given that the events A and $B$ are such that $P\left(A\right)=\frac{1}{4},P\left(A|B\right)=\frac{1}{2}$ and $P\left(B|A\right)=\frac{2}{3}$. Then $P\left(B\right)$ is

• A. $\frac{2}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{6}$
• D. $\frac{1}{3}$

Answer 1

Answer: (D)

$\frac{1}{3}$

$P\left(A\right)=\frac{1}{4}$
$P(A|B)=\frac{P\left(A\cap B\right)}{P\left(B\right)}=\frac{1}{2}$ ... (i)
$P(B|A)=\frac{P\left(A\cap B\right)}{P\left(A\right)}=\frac{2}{3}=\frac{P\left(A\cap B\right)}{1/4}=P\left(A\cap B\right)=\frac{1}{6}$.
Putting the value of $P\left(A\cap B\right)$ in (i) $,P\left(B\right)=2\times \frac{1}{6}=\frac{1}{3}$