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If $A$ and $B$ are two events such that $P(A \cup B)+P(A \cap B)=\frac{7}{8}$ and $P(A)=2 P(B)$, then $P(A)=$
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The correct answer is:
$\frac{7}{12}$
$P(A \cup B)+P(A \cap B)=P(A)+P(B)$ $=P(A)+\frac{P(A)}{2}$
$\Rightarrow \frac{7}{8}=\frac{3 P(A)}{2} \Rightarrow P(A)=\frac{7}{12}$
$\Rightarrow \frac{7}{8}=\frac{3 P(A)}{2} \Rightarrow P(A)=\frac{7}{12}$
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