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If $A, B, C$ are three mutually exclusive and exhaustive events of an experiment such that $P(A)=2 P(B)=3 P(C)$, then $P(B)$ is equal to
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Verified Answer
The correct answer is:
$\frac{3}{11}$
$A, B, C$ are mutually exclusive and exhaustive events.
$\begin{aligned}
\therefore \quad P(A \cap B)=P(B \cap C) \\
=P(A \cap C)=0 &=P(A \cap B \cap C) \\
\text { and } P(A)+P(B)+P(C) &=1 \\
2 P(B)+P(B)+\frac{2}{3} P(B) &=1 \\
{[} &\because P(A)=2 P(B), 2 P(B)=3 P(C)] \\
&=\frac{11 P(B)}{3}=1 \\
P(B) &=\frac{3}{11}
\end{aligned}$
$\begin{aligned}
\therefore \quad P(A \cap B)=P(B \cap C) \\
=P(A \cap C)=0 &=P(A \cap B \cap C) \\
\text { and } P(A)+P(B)+P(C) &=1 \\
2 P(B)+P(B)+\frac{2}{3} P(B) &=1 \\
{[} &\because P(A)=2 P(B), 2 P(B)=3 P(C)] \\
&=\frac{11 P(B)}{3}=1 \\
P(B) &=\frac{3}{11}
\end{aligned}$
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