Q76 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics

It is given that in a random experiment events A and B are such that $\mathrm{P}(\mathrm{A})=\frac{1}{4}, P(A \mid B)=\frac{1}{2}$ and $P(B \mid A)=\frac{2}{3}$ then $P(B)=$ $\frac{1}{6}$

  • Correct answer
    A
    $\frac{1}{3}$
  • B
    $\frac{2}{3}$
  • C
    $\frac{1}{2}$
  • D
    $\frac{1}{2}$

Solution

We know that, $P\left(\frac{B}{A}\right)=\frac{P(A \cap B)}{P(A)}$ $\Rightarrow P\left(\frac{B}{A}\right) \cdot P(A)=P(A \cap B) \Rightarrow \frac{2}{3} \times \frac{1}{4}=\frac{1}{6}=P(A \cap B)$ And $P\left(\frac{A}{B}\right)=\frac{P(A \cap B)}{P(B)} \Rightarrow P(B)=\frac{P(A \cap B)}{P(A / B)}$ $\Rightarrow P(B)=\frac{1 / 6}{1 / 2}=\frac{1}{6} \times \frac{2}{1}=\frac{1}{3} .$

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

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