Question

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

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

Answer 1

The correct option is B

$\frac{1}{3}$

$\frac{P(A\cap B)}{P\left(B\right)}=\frac{1}{2},\frac{P(A\cap B)}{P\left(A\right)}=\frac{2}{3},$
Hence $\frac{P\left(A\right)}{P\left(B\right)}=\frac{3}{4},$ (But P(A) = 1/4)
$\Rightarrow P\left(B\right)=\frac{1}{3}.$