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

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(C) Since $A, B, C$ are mutually exclusive and exhaustive events,$P(A) + P(B) + P(C) = 1$. Given $P(B) = 2 P(C)$,we can write $P(C) = \frac{1}{2} P(B)

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$\frac{7}{13}$

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(C) Since $A, B, C$ are mutually exclusive and exhaustive events,$P(A) + P(B) + P(C) = 1$. Given $P(B) = 2 P(C)$,we can write $P(C) = \frac{1}{2} P(B)$. Also given $P(A) = \frac{2}{3} P(B)$. Substituting these into the sum: $\frac{2}{3} P(B) + P(B) + \frac{1}{2} P(B) = 1$. Finding a common denominator $(6)$: $\frac{4}{6} P(B) + \frac{6}{6} P(B) + \frac{3}{6} P(B) = 1$. $\frac{13}{6} P(B) = 1 \implies P(B) = \frac{6}{13}$. Then $P(A) = \frac{2}{3} 	imes \frac{6}{13} = \frac{4}{13}$ and $P(C) = \frac{1}{2} 	imes \frac{6}{13} = \frac{3}{13}$. Since $A$ and $C$ are mutually exclusive,$P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$.

If $A, B$ and $C$ are mutually exclusive and exhaustive events of a random experiment such that $P(A) = \frac{2}{3} P(B)$ and $P(B) = 2 P(C)$,then $P(A \cup C) =$

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