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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) \(\frac{1}{3}\)


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) = \(\frac{P(A\,\cap\,B)}{P(B)}\)

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

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