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If $A$ and $B$ are independent events of a random experiment such that $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$, then $P(A)$ is equal to (Here, $\overrightarrow{\mathbf{E}}$ is the complement of the event $E$ )
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1219 Upvotes
Verified Answer
The correct answer is:
$\frac{1}{3}$
Given that, $P(A \cap B)=\frac{1}{6}$ and $P(\bar{A} \cap \bar{B})=\frac{1}{3}$ Since, $A$ and $B$ are independent.
$$
\begin{aligned}
& \therefore \quad P(A) P(B)=\frac{1}{6} \text { and } P(\bar{A}) P(\bar{B})=\frac{1}{3} \\
& \Rightarrow \quad[1-P(A)][1-P(B)]=\frac{1}{3} \\
& \Rightarrow 1-[P(A)+P(B)]+P(A) P(B)=\frac{1}{3} \\
& \Rightarrow \quad 1+\frac{1}{6}-\frac{1}{3}=P(A)+P(B) \\
& \Rightarrow \quad P(A)+P(B)=\frac{5}{6} \\
& \Rightarrow \quad P(A)=\frac{1}{2}, P(B)=\frac{1}{3} \\
& \text { and } P(A)=\frac{1}{3}, P(B)=\frac{1}{2} \\
&
\end{aligned}
$$
Hence, options (2) and (3) are correct.
$$
\begin{aligned}
& \therefore \quad P(A) P(B)=\frac{1}{6} \text { and } P(\bar{A}) P(\bar{B})=\frac{1}{3} \\
& \Rightarrow \quad[1-P(A)][1-P(B)]=\frac{1}{3} \\
& \Rightarrow 1-[P(A)+P(B)]+P(A) P(B)=\frac{1}{3} \\
& \Rightarrow \quad 1+\frac{1}{6}-\frac{1}{3}=P(A)+P(B) \\
& \Rightarrow \quad P(A)+P(B)=\frac{5}{6} \\
& \Rightarrow \quad P(A)=\frac{1}{2}, P(B)=\frac{1}{3} \\
& \text { and } P(A)=\frac{1}{3}, P(B)=\frac{1}{2} \\
&
\end{aligned}
$$
Hence, options (2) and (3) are correct.
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