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(A) Since $A$,$B$,and $C$ are mutually exclusive and exhaustive events,their sum of probabilities is $1$: $P(A) + P(B) + P(C) = 1$ Given $P(B) = \frac

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$\frac{4}{13}$

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(A) Since $A$,$B$,and $C$ are mutually exclusive and exhaustive events,their sum of probabilities is $1$: $P(A) + P(B) + P(C) = 1$ Given $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B) = \frac{1}{2} 	imes \frac{3}{2} P(A) = \frac{3}{4} P(A)$ Substituting these into the equation: $P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1$ Taking the common denominator as $4$: $\frac{4 P(A) + 6 P(A) + 3 P(A)}{4} = 1$ $\frac{13 P(A)}{4} = 1$ $P(A) = \frac{4}{13}$

If $A$,$B$,and $C$ are mutually exclusive and exhaustive events of a sample space $S$ such that $P(B) = \frac{3}{2} P(A)$ and $P(C) = \frac{1}{2} P(B)$,then $P(A) = $

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