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