Q54 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
If $A$ and $B$ are mutually exclusive events and if $\mathrm{P}(\mathrm{B})=\frac{1}{3}, \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{13}{21}$, then $\mathrm{P}(\mathrm{A})$ is equal to
- A
1/7 - B
4/7 - Correct answerC
2/7 - D
5/7
Solution
For mutually exclusive events
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7}$
$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B}) \Rightarrow \mathrm{P}(\mathrm{A})=\frac{2}{7}$
Hence, the correct answer is option: 'C'.