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Let $A$ and $B $ be two events such that $P\left( {\overline {A \cup B} } \right) = \frac{1}{6}\;,P\left( {A \cap B} \right) = \frac{1}{4}$ and $P\left( {\bar A} \right) = \frac{1}{4}$ where $\bar A$ stands for the complement of the event $A$. Then the events $A$ and$B$ are
independent but not equally likely..
independent but equally likely.
mutually exclusive and independent.
equally likely but not independent.
Solution
$P(\overline{A \cup B})=\frac{1}{6}$
$P(A \cap B)=\frac{1}{4}$
$P(\bar{A})=\frac{1}{4}$
$\text { clearly } \mathrm{P}(\mathrm{A})=\frac{3}{4}$
$\text { and } \mathrm{P}(\mathrm{A} \cup \mathrm{B})=\frac{5}{6}$
$P(A)+P(B)-P(A \cap B)=\frac{5}{6}$
$\frac{3}{4}+P(B)-\frac{1}{4}=\frac{5}{6}$
$P(B)=\frac{5}{6}-\frac{3}{4}+\frac{1}{4}=\frac{1}{3}$
$P(A \cap B)=\frac{1}{4}$
$P(A) \cdot P(B)=\frac{3}{4} \frac{1}{3}=\frac{1}{4} \text { so independent }$
$\mathrm{P}(\mathrm{A}) \neq \mathrm{P}(\mathrm{B}) \text { so not equally likely }$
