Q78 / 125 · BITSAT 2009 · Mathematics

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

  • A
    \(\frac{1}{6}\)
  • Correct answer
    B
    \(\frac{1}{3}\)
  • C
    \(\frac{2}{3}\)
  • D
    \(\frac{1}{2}\)

Solution

\(P(A)=1 / 4, P(A / B)=\frac{1}{2}, P(B / A)=2 / 3\) By conditional probability, \(\begin{aligned} & P(A \cap B)=P(A) P(B / A)=P(B) P(A / B) \\ & \Rightarrow \frac{1}{4} \times \frac{2}{3}=P(B) \times \frac{1}{2} \Rightarrow P(B)=\frac{1}{3} \end{aligned}\)

Hence, the correct answer is option: 'B'.

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