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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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The correct answer is:
$P(A \cup C)=\frac{7}{13}$ when $A, B, C$ are exhaustive and mutually exclusive events
Let $P(B)=x$
Then, $P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$
Now, if $A, B, C$ are exhaustive and mutually exclusive, then $P(A \cup B \cup C)=1$,
$\begin{aligned} & P(A \cap B)=P(B \cap C)=P(C \cap A)=P(A \cap B \cap C)=0 \\ & \therefore \quad P(A \cup B \cup C)=1\end{aligned}$
$\begin{array}{ll}\Rightarrow & P(A)+P(B)+P(C)=1 \\ \Rightarrow & \frac{2}{3} x+x+\frac{1}{2} x=1 \\ \Rightarrow & \frac{4 x+6 x+3 x}{6}=1 \\ \Rightarrow & P(A)=\frac{2}{3} \times \frac{6}{13}=\frac{12}{39} \\ \therefore & P(C)=\frac{1}{2} \times \frac{6}{13}=\frac{3}{13} \\ \text { and } & \\ \therefore & P(A \cup C)=P(A)+P(C)=\frac{12}{39}+\frac{3}{13}\end{array}$
$=\frac{12+9}{39}=\frac{21}{39}=\frac{7}{13}$
Then, $P(A)=\frac{2}{3} x$ and $P(C)=\frac{1}{2} x$
Now, if $A, B, C$ are exhaustive and mutually exclusive, then $P(A \cup B \cup C)=1$,
$\begin{aligned} & P(A \cap B)=P(B \cap C)=P(C \cap A)=P(A \cap B \cap C)=0 \\ & \therefore \quad P(A \cup B \cup C)=1\end{aligned}$
$\begin{array}{ll}\Rightarrow & P(A)+P(B)+P(C)=1 \\ \Rightarrow & \frac{2}{3} x+x+\frac{1}{2} x=1 \\ \Rightarrow & \frac{4 x+6 x+3 x}{6}=1 \\ \Rightarrow & P(A)=\frac{2}{3} \times \frac{6}{13}=\frac{12}{39} \\ \therefore & P(C)=\frac{1}{2} \times \frac{6}{13}=\frac{3}{13} \\ \text { and } & \\ \therefore & P(A \cup C)=P(A)+P(C)=\frac{12}{39}+\frac{3}{13}\end{array}$
$=\frac{12+9}{39}=\frac{21}{39}=\frac{7}{13}$
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