Question
Let $$A$$ and $$B$$ be two events such that $$P\left( {A \cap B} \right) = \frac{1}{3},\,P\left( {A \cup B} \right) = \frac{5}{6}$$ and $$P\left( {\overline A } \right) = \frac{1}{2}.$$ Then :
A.
$$A,\,B$$ are independent
B.
$$A,\,B$$ are mutually exclusive
C.
$$P\left( A \right) = P\left( B \right)$$
D.
$$P\left( B \right) \leqslant P\left( A \right)$$
Answer :
$$A,\,B$$ are independent
Solution :
$$\eqalign{
& P\left( {A \cup B} \right) = P\left( A \right) + P\left( B \right) - P\left( {A \cap B} \right) \cr
& \Rightarrow \,\frac{5}{6} = \left( {1 - \frac{1}{2}} \right) + P\left( B \right) - \frac{1}{3} \cr
& \Rightarrow \,P\left( B \right) = \frac{2}{3} \cr
& P\left( A \right).P\left( B \right) = \frac{1}{2}.\frac{2}{3} = \frac{1}{3} = P\left( {A \cap B} \right) \cr} $$
So, $$A$$ and $$B$$ are independent and therefore, not mutually exclusive