# It is given that the events A and B are such that P(A) = 1/ 4 , P(A/B) = 1/ 2 and P(B/A) = 2 /3 . Then P(B) is

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

(a) $\frac{1}{2}$

(b) $\frac{2}{3}$

(c) $\frac{1}{6}$

(d) $\frac{1}{3}$

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Answer : (d) $\frac{1}{3}$

Given,

P(B/A) =$\frac{P(A\,\cap\,B)}{P(A)}$

$P(A\,\cap\,B)= P(B/A)\times P(A)=\frac{2}{3}\times \frac{1}{4} = \frac{1}{6}$

Now P(A/B) = $\frac{P(A\,\cap\,B)}{P(B)}$

$\frac{1}{2} = \frac{1}{6}\times \frac{1}{P(B)} \implies P(B)=\frac{2}{6}= \frac{1}{3}$