Q269 / 270 · NDA 2017 (Phase 1) · Mathematics

Let A and B are two mutually exclusive events with $\mathrm{P}(\mathrm{A})$ $=\frac{1}{3}$ and $\mathrm{P}(\mathrm{B})=\frac{1}{4}$. What is the value of $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}}) ?$

  • A
    $\frac{1}{6}$
  • B
    $\frac{1}{4}$
  • C
    $\frac{1}{3}$
  • Correct answer
    D
    $\frac{1}{12}$

Solution

$\mathrm{A} \& \mathrm{~B}$ are mutualy exclusive events. i.e., $\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})=0$ We know, $\mathrm{P}(\overline{\mathrm{A}} \cap \overline{\mathrm{B}})=\mathrm{P}(\overline{\mathrm{A} \cup \mathrm{B}})$ $\begin{aligned} & =1-\mathrm{P}(\mathrm{~A} \cup \mathrm{~B}) \\ & =1-[\mathrm{P}(\mathrm{~A})+\mathrm{P}(\mathrm{~B})-\mathrm{P}(\mathrm{~A} \cap \mathrm{~B})] \\ & =1-\left[\frac{1}{3}+\frac{1}{4}-0\right] \\ & =1-\left[\frac{7}{12}\right]=\frac{5}{12} \end{aligned}$

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

Relevant Questions