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If $A, B$ and $C$ are three mutually exclusive and exhaustive events such that $P(A)=2 P(B)=3 P(C)$. What is $P(B)$ ?
Options:
Solution:
2375 Upvotes
Verified Answer
The correct answer is:
$\frac{6}{22}$
Given that $A, B$ and $C$ are three mutually exclusive events
$$
\begin{aligned}
& \Rightarrow \quad P(A)=2 P(B)=3 P(C) \\
& P(A \cup B \cup C)=P(A)+P(B)+P(C) \\
& =2 P(B)+P(B)+\frac{2}{3} P(B) \\
& =\frac{11}{3} P(B) \\
& \because P(A \cup B \cup C)=1 \\
& \Rightarrow \quad \frac{11}{3} P(B)=1 \Rightarrow P(B)=\frac{3}{11} \\
& \Rightarrow \quad P(B)=\frac{6}{22} \\
&
\end{aligned}
$$
Option (2) is correct.
$$
\begin{aligned}
& \Rightarrow \quad P(A)=2 P(B)=3 P(C) \\
& P(A \cup B \cup C)=P(A)+P(B)+P(C) \\
& =2 P(B)+P(B)+\frac{2}{3} P(B) \\
& =\frac{11}{3} P(B) \\
& \because P(A \cup B \cup C)=1 \\
& \Rightarrow \quad \frac{11}{3} P(B)=1 \Rightarrow P(B)=\frac{3}{11} \\
& \Rightarrow \quad P(B)=\frac{6}{22} \\
&
\end{aligned}
$$
Option (2) is correct.
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