$\mathrm{A}$ and $\mathrm{B}$ are independent events with $\mathrm{P}(\mathrm{A})=\frac{1}{4}$ and $P(A \cup…

$\mathrm{A}$ and $\mathrm{B}$ are independent events with $\mathrm{P}(\mathrm{A})=\frac{1}{4}$ and $P(A \cup B)=2 P(B)-P(A)$, then $P(B)$ is
  1. $\frac{1}{4}$
  2. $\frac{3}{5}$
  3. $\frac{2}{3}$
  4. $\frac{2}{5}$

Solution

$\begin{array}{ll} & \mathrm{P}(\mathrm{A} \cup \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \\ \therefore \quad & \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A} \cap \mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \\ \therefore \quad & \mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{B})-\mathrm{P}(\mathrm{A}) \end{array}$ $\ldots[\because A$ and $B$ are independent events $]$ $\begin{array}{ll} \therefore \quad & \mathrm{P}(\mathrm{B})+\mathrm{P}(\mathrm{A}) \cdot \mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A}) \\ \therefore \quad \mathrm{P}(\mathrm{B})=\frac{2 \mathrm{P}(\mathrm{A})}{(1+\mathrm{P}(\mathrm{A}))}=\frac{2 \times \frac{1}{4}}{\left(1+\frac{1}{4}\right)}=\frac{2}{5} \end{array}$

Asked in: MHT CET 2023 (12 May Shift 1)

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