If A, B, C are three events of a sample space | vedclass

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Question

(A) Let $P(B)=x$. Then, $P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$. If $A, B, C$ are exhaustive and mutually exclusive, then $P(A \cup B \cup C)=1$

Vedclass · Accepted Answer

$P(A \cup C)=\frac{7}{13}$ when $A, B, C$ are exhaustive and mutually exclusive events

Vedclass · Answer

(A) Let $P(B)=x$. Then, $P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$. If $A, B, C$ are exhaustive and mutually exclusive, then $P(A \cup B \cup C)=1$. Since they are mutually exclusive, $P(A \cup B \cup C) = P(A) + P(B) + P(C) = 1$. Substituting the values: $\frac{2}{3} x + x + \frac{1}{2} x = 1$. $\frac{4x + 6x + 3x}{6} = 1 \Rightarrow \frac{13x}{6} = 1 \Rightarrow x = \frac{6}{13}$. Thus, $P(A) = \frac{2}{3} 	imes \frac{6}{13} = \frac{4}{13}$, $P(B) = \frac{6}{13}$, and $P(C) = \frac{1}{2} 	imes \frac{6}{13} = \frac{3}{13}$. Now, $P(A \cup C) = P(A) + P(C) = \frac{4}{13} + \frac{3}{13} = \frac{7}{13}$. Therefore, option $A$ is correct.

If $A, B, C$ are three events of a sample space such that $P(B)=\frac{3}{2} P(A)$ and $P(C)=\frac{1}{2} P(B)$, then which of the following is correct?

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