Q254 / 270 · NDA 2021 (Phase 2) · Mathematics

If $A$ and $B$ are two events such that $P(\operatorname{not} A)=\frac{7}{10}, P($ not $B)=\frac{3}{10}$ and $P(A \mid B)=\frac{3}{14}$, then what is $P(B \mid A)$ equal to?

  • A
    $\frac{11}{14}$
  • B
    $\frac{9}{14}$
  • C
    $\frac{1}{4}$
  • Correct answer
    D
    $\frac{1}{2}$

Solution

$\begin{aligned} & P(\operatorname{not} A)=\frac{7}{10}, P(\operatorname{not} B)=\frac{3}{10}, P(A / B)=\frac{3}{14} \\ & \rightarrow \operatorname{Now}, P(A / B)=\frac{P(A \cap B)}{P(B)} \Rightarrow P(A \cap B)=P(B) P(A / B) \\ & \therefore P(A \cap B)=[1-P(\operatorname{not} B)] P(A / B)=\left(1-\frac{3}{10}\right)\left(\frac{3}{14}\right) \\ & =\frac{7}{10} \times \frac{3}{14}=\frac{3}{20} \\ & \rightarrow P(B / A)=\frac{P(A \cap B)}{P(A)}=\frac{P(A \cap B)}{1-P(\operatorname{not} A)}=\frac{\frac{3}{20}}{1-\frac{7}{10}} \\ & =\frac{3}{20} \times \frac{10}{3} \\ & \therefore P(B / A)=1 / 2\end{aligned}$

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

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