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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
Independent but not equally likely
Mutually exclusive and independent
Equally likely and mutually exclusive
Equally likely but not independent
Solution
(a) $P\left( {\overline {A \cup B} } \right) = \frac{1}{6};\,\,P\left( {A \cap B} \right) = \frac{1}{4}$,
$P\left( {\bar A} \right) = \frac{1}{4} \Rightarrow P\left( A \right) = \frac{3}{4}$,
$P\left( {\overline {A \cup B} } \right) = 1 – P\left( {A \cup B} \right) = 1 – P\left( A \right) – P\left( B \right) + P\left( {A \cap B} \right)$
$⇒$ $\frac{1}{6} = \frac{1}{4} – P\left( B \right) + \frac{1}{4}$ $⇒$ $P\left( B \right) = \frac{1}{3}$.
Since $P\left( {A \cap B} \right) = P\left( A \right)P\left( B \right)$ and $P\left( A \right) \ne P\left( B \right)$
$\therefore$ $ A$ and $B$ are independent but not equally likely.
