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If $A, B$ and $C$ are mutually exclusive and exhaustive events of a random experiment such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then $P(A \cup C)$ equals to
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Verified Answer
The correct answer is:
$\frac{7}{13}$
Given, $\quad P(B)=\frac{3}{2} P(A)$ and
$$
P(C)=\frac{1}{2} P(B)
$$
Since, $A, B$ and $C$ are exclusive events.
$$
\begin{aligned}
& \therefore & P(A)+P(B)+P(C) & =1 \\
& \therefore & P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A) & =1 \\
& \Rightarrow & P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right) & =1 \\
& & \frac{13}{4} P(A) & =1 \\
& & P(A) & =\frac{4}{13} \\
& \therefore & P(C)=\frac{1}{2} \times \frac{3}{2} P(A) & =\frac{3}{4} \times \frac{4}{13}=\frac{3}{13}
\end{aligned}
$$
Also, $A, B$ and $C$ are mutually exclusive.
$$
\begin{aligned}
& \therefore P(A \cap B)=P(B \cap C)=P(C \cap A)=0 \\
& \therefore P(A \cup C)=P(A)+P(B)-0 \\
& =\frac{4}{13}+\frac{3}{13}=\frac{7}{13}
\end{aligned}
$$
$$
P(C)=\frac{1}{2} P(B)
$$
Since, $A, B$ and $C$ are exclusive events.
$$
\begin{aligned}
& \therefore & P(A)+P(B)+P(C) & =1 \\
& \therefore & P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A) & =1 \\
& \Rightarrow & P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right) & =1 \\
& & \frac{13}{4} P(A) & =1 \\
& & P(A) & =\frac{4}{13} \\
& \therefore & P(C)=\frac{1}{2} \times \frac{3}{2} P(A) & =\frac{3}{4} \times \frac{4}{13}=\frac{3}{13}
\end{aligned}
$$
Also, $A, B$ and $C$ are mutually exclusive.
$$
\begin{aligned}
& \therefore P(A \cap B)=P(B \cap C)=P(C \cap A)=0 \\
& \therefore P(A \cup C)=P(A)+P(B)-0 \\
& =\frac{4}{13}+\frac{3}{13}=\frac{7}{13}
\end{aligned}
$$
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