# It is given that the events A and B are such that $P(A) = \large\frac{1}{4}\: \: P \bigg( \large\frac{A}{B} \bigg) \large\frac{1}{2}\: and \: P \bigg( \large\frac{B}{A} \bigg) = \large\frac{2}{3}$ Then $P(B)$ is

$\begin {array} {1 1} (A)\;\large\frac{1}{6} & \quad (B)\;\large\frac{1}{3} \\ (C)\;\large\frac{2}{3} & \quad (D)\;\large\frac{1}{2} \end {array}$

Given that $P(A)=\large\frac{1}{4}$
$P \bigg( \large\frac{A}{B} \bigg) = \large\frac{1}{2}\: \: \: \: P \bigg( \large\frac{B}{A} \bigg) = \large\frac{2}{3}$
We know $P \bigg( \large\frac{A}{B} \bigg)= \large\frac{P(A \cap B)}{P(B)}$
and $P \bigg( \large\frac{B}{A} \bigg)= \large\frac{P(B \cap A)}{P(A)}$
$\therefore P(B) = \large\frac{P \bigg( \Large\frac{B}{A} \bigg).P(A)}{P\bigg( \large\frac{A}{B}\bigg)}$
$= \large\frac{ \bigg( \Large\frac{2}{3} \bigg) \bigg( \Large\frac{1}{4} \bigg)}{\bigg( \Large\frac{1}{2} \bigg) }$
$= \large\frac{1}{3}$
Ans : (B)