Q263 / 270 · NDA 2012 (Phase 2) · Mathematics

If A and B are two events such that $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}$, $\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}, \mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3}$ where $\overline{\mathrm{A}}$ is the complement of $A$, then what is $\mathrm{P}(\mathrm{B})$ equal to?

  • A
    $\frac{1}{3}$
  • Correct answer
    B
    $\frac{2}{3}$
  • C
    $\frac{1}{9}$
  • D
    $\frac{2}{9}$

Solution

$\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{3}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}$ $\begin{aligned} & \mathrm{P}(\overline{\mathrm{A}})=\frac{2}{3} \Rightarrow \mathrm{P}(\overline{\mathrm{A}})=1-\mathrm{P}(\mathrm{A}) \\ & \Rightarrow \mathrm{P}(\mathrm{A})=1-\mathrm{P}(\overline{\mathrm{A}})=1-\frac{2}{3}=\frac{1}{3}\end{aligned}$ $\mathrm{P}(\mathrm{A} \cup \mathrm{B})=\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})$ $\begin{aligned} & \Rightarrow \frac{3}{4}=\frac{1}{3}+\mathrm{P}(\mathrm{B})-\frac{1}{4} \\ & \Rightarrow \frac{3}{4}+\frac{1}{4}-\frac{1}{3}=\mathrm{P}(\mathrm{B})\end{aligned}$ $\begin{aligned} & \Rightarrow \frac{4}{4}-\frac{1}{3}=\mathrm{P}(\mathrm{B}) \\ & \Rightarrow \mathrm{P}(\mathrm{B})=1-\frac{1}{3}=\frac{2}{3}\end{aligned}$

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

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