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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
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The correct answer is:
$\frac{7}{13}$
Given, $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$
Since, $A, B$ and $C$ are exclusive events.
$\therefore \quad P(A)+P(B)+P(C)=1$ $\Rightarrow \quad P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A)=1$ $\Rightarrow \quad P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right)=1$ $\Rightarrow \quad \quad \frac{13}{4} P(A)=1 \Rightarrow \quad P(A)=\frac{4}{13}$ $\Rightarrow \quad P(C)=\frac{1}{2} \times \frac{3}{2} P(A)=\frac{3}{4} \times \frac{4}{13}=\frac{3}{13}$ Also, $A, B$ and $C$ are mutually exclusive. $\therefore \quad P(A \cap B)=P(B \cap C)=P(C \cap A)=0$ $\Rightarrow \quad P(A \cup C)=P(A)+P(C)-0=\frac{4}{13}+\frac{3}{13}=\frac{7}{13}$
Since, $A, B$ and $C$ are exclusive events.
$\therefore \quad P(A)+P(B)+P(C)=1$ $\Rightarrow \quad P(A)+\frac{3}{2} P(A)+\frac{1}{2} \times \frac{3}{2} P(A)=1$ $\Rightarrow \quad P(A)\left(1+\frac{3}{2}+\frac{3}{4}\right)=1$ $\Rightarrow \quad \quad \frac{13}{4} P(A)=1 \Rightarrow \quad P(A)=\frac{4}{13}$ $\Rightarrow \quad P(C)=\frac{1}{2} \times \frac{3}{2} P(A)=\frac{3}{4} \times \frac{4}{13}=\frac{3}{13}$ Also, $A, B$ and $C$ are mutually exclusive. $\therefore \quad P(A \cap B)=P(B \cap C)=P(C \cap A)=0$ $\Rightarrow \quad P(A \cup C)=P(A)+P(C)-0=\frac{4}{13}+\frac{3}{13}=\frac{7}{13}$
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