Question
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 $$P\left( {\overline A } \right) = \frac{1}{4},$$ where $$\overline A $$ stand for complement of event $$A.$$ Then events $$A$$ and $$B$$ are
A.
equally likely and mutually exclusive
B.
equally likely but not independent
C.
independent but not equally likely
D.
mutually exclusive and independent
Answer :
independent but not equally likely
Solution :
$$\eqalign{
& P\left( {\overline {A \cup B} } \right) = \frac{1}{6},P\left( {A \cap B} \right) = \frac{1}{4}\,{\text{and }}P\left( {\overline A } \right) = \frac{1}{4} \cr
& \Rightarrow \,\,P\left( {A \cup B} \right) = \frac{5}{6}P\left( A \right) = \frac{3}{4} \cr
& {\text{Also }} \Rightarrow \,P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& \Rightarrow \,\,P\left( B \right) = \frac{5}{6} - \frac{3}{4} + \frac{1}{4} = \frac{1}{3} \cr
& \Rightarrow \,\,P\left( A \right)P\left( B \right) = \frac{3}{4} - \frac{1}{3} = \frac{1}{4} \cr
& = P\left( {A \cap B} \right) \cr} $$
Hence $$A$$ and $$B$$ are independent but not equally likely.