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If $\mathrm{A}$ and $\mathrm{B}$ are two events in a random experiment such that $\mathrm{P}(\mathrm{A})+\mathrm{P}(\mathrm{B})=2 \mathrm{P}(\mathrm{A} \cap \mathrm{B})$ then
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Verified Answer
The correct answer is:
$P(A)=P(B)$
$P(A)+P(B)=2 P(A \cap B)$
$\therefore \quad P(A)+P(B)-P(A \cap B)=P(A \cap B)$
But we know that
$\begin{aligned}
& P(A)+P(B)-P(A \cap B)=P(A \cup B) \\
& \therefore \quad P(A \cup B)=P(A \cap B) \\
& \Rightarrow \quad P(A)=P(B) .
\end{aligned}$
$\therefore \quad P(A)+P(B)-P(A \cap B)=P(A \cap B)$
But we know that
$\begin{aligned}
& P(A)+P(B)-P(A \cap B)=P(A \cup B) \\
& \therefore \quad P(A \cup B)=P(A \cap B) \\
& \Rightarrow \quad P(A)=P(B) .
\end{aligned}$
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