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(D) Given, $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$.Since $A, B$ and $C$ are mutually exclusive and exhaustive events, their sum of pro

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$\frac{7}{13}$

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(D) Given, $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$.Since $A, B$ and $C$ are mutually exclusive and exhaustive events, their sum of probabilities is $1$:$P(A) + P(B) + P(C) = 1$Substituting the given relations in terms of $P(A)$:$P(A) + \frac{3}{2} P(A) + \frac{1}{2} \left( \frac{3}{2} P(A) ight) = 1$$P(A) \left( 1 + \frac{3}{2} + \frac{3}{4} ight) = 1$$P(A) \left( \frac{4 + 6 + 3}{4} ight) = 1$$\frac{13}{4} P(A) = 1 \implies P(A) = \frac{4}{13}$Now, calculate $P(C)$:$P(C) = \frac{3}{4} P(A) = \frac{3}{4} 	imes \frac{4}{13} = \frac{3}{13}$Since $A$ and $C$ are mutually exclusive, $P(A \cup C) = P(A) + P(C)$:$P(A \cup C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$

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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