If A, B and C are mutually exclusive and exha | vedclass

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(D) Given that $A, B$ and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. Given $P(B) = \frac{3}{2} P(A)$ and $P(C)

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$\frac{7}{13}$

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(D) Given that $A, B$ and $C$ are mutually exclusive and exhaustive events,we have $P(A) + P(B) + P(C) = 1$. 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 sum equation: $P(A) + \frac{3}{2} P(A) + \frac{3}{4} P(A) = 1$ $P(A) \left( 1 + \frac{6}{4} + \frac{3}{4} ight) = 1$ $P(A) \left( \frac{4+6+3}{4} ight) = 1$ $\frac{13}{4} P(A) = 1 \implies P(A) = \frac{4}{13}$. Now,$P(C) = \frac{3}{4} P(A) = \frac{3}{4} 	imes \frac{4}{13} = \frac{3}{13}$. 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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