Q258 / 270 · NDA 2017 (Phase 2) · Mathematics

Let A and B be two events with $\mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{6}$ and $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12}$. What is $\mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})$ equal to?

  • A
    $\frac{1}{5}$
  • B
    $\frac{1}{7}$
  • Correct answer
    C
    $\frac{1}{8}$
  • D
    $\frac{1}{10}$

Solution

$\begin{aligned} & \mathrm{P}(\mathrm{A})=\frac{1}{3}, \mathrm{P}(\mathrm{B})=\frac{1}{6}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{12} \\ & \mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})=\frac{\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}})}{\mathrm{P}(\overline{\mathrm{A}})}\end{aligned}$ Now, $\mathrm{P}(\mathrm{B} \cap \overline{\mathrm{A}})=\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{6}-\frac{1}{12}=\frac{1}{12}$ $\mathrm{P}(\overline{\mathrm{A}})=1-\frac{1}{3}=\frac{2}{3}$ $\therefore \mathrm{P}(\mathrm{B} \mid \overline{\mathrm{A}})=\frac{\frac{1}{12}}{\frac{2}{3}}=\frac{1}{12} \times \frac{3}{2}=\frac{1}{8}$.

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

Relevant Questions