Q249 / 270 · NDA 2016 (Phase 1) · Mathematics
For two mutually exclusive events A and $\mathrm{B}, \mathrm{P}(\mathrm{A})=0.2$ and $\mathrm{P}(\overline{\mathrm{A}} \mathrm{C} \mathrm{~B})=0.3$. What is $\mathrm{P}(\mathrm{A} \mid(\mathrm{A} \cup \mathrm{B}))$ equal to?
- A$\frac{1}{2}$
- Correct answerB$\frac{2}{5}$
- C$\frac{2}{7}$
- D$\frac{2}{3}$
Solution
Events $A$ and $B$ are mutually exclusive.
$\begin{aligned}
& \text { Hence } P(A \cap B)=\phi=0 \\
& \begin{aligned}
\therefore P(A \cup B)=P(A)+P(B)...(1) \\
P(A)=0.2...[given]
\end{aligned} \\
& \begin{array}{l}
P(B)=P(\bar{A} \cap B)+P(A \cap B) \\
P(B)=P(\bar{A} \cap B) \quad[\because P(A \cap B)=0] \\
\quad=0.3 \\
P(A \cup B)=0.2+0.3=0.5 \\
P(A \mid(A \cup B))=\frac{P(A)}{P(A \cup B)}=\frac{0.2}{0.5}=\frac{2}{5} \\
P(A \mid(A \cup B))=\frac{2}{5}
\end{array}
\end{aligned}$
Hence, the correct answer is option: 'B'.