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Let $A$ and $B$ be two events such that $P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4}$ and $P(\bar{A})=\frac{1}{4}$, where $\bar{A}$ stands for complement of event $A$. Then events $A$ and $B$ are
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The correct answer is:
independent but not equally likely
independent but not equally likely
$$
\begin{aligned}
& P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} \text { and } P(\bar{A})=\frac{1}{4} \\
& \Rightarrow P(A \cup B)=5 / 6 P(A)=3 / 4 \\
& \text { Also } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow P(B)=5 / 6-3 / 4+1 / 4=1 / 3 \\
& P(A) P(B)=3 / 4-1 / 3=1 / 4=P(A \cap B)
\end{aligned}
$$
Hence $A$ and $B$ are independent but not equally likely.
\begin{aligned}
& P(\overline{A \cup B})=\frac{1}{6}, P(A \cap B)=\frac{1}{4} \text { and } P(\bar{A})=\frac{1}{4} \\
& \Rightarrow P(A \cup B)=5 / 6 P(A)=3 / 4 \\
& \text { Also } P(A \cup B)=P(A)+P(B)-P(A \cap B) \\
& \Rightarrow P(B)=5 / 6-3 / 4+1 / 4=1 / 3 \\
& P(A) P(B)=3 / 4-1 / 3=1 / 4=P(A \cap B)
\end{aligned}
$$
Hence $A$ and $B$ are independent but not equally likely.
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