**Question** : Let A and B be two events such that P\left( \overline{A\cup B}\right) =\dfrac{1}{6}\cdot P\left( A\cap B\right) =\dfrac{1}{4}andP\left( \overline{A}\right) =\dfrac{1}{4} where \overline{A} stands for complement of event . Then events A and B are

(A) Equally likely but not independent

(B) Equally likely and mutually exclusive

(C) Mutually exclusive and independent

(D) Independent but not equally likely

**Answer **: option (D)

**Solution **:

We have, P\left( \overline{A\cup B}\right) =\dfrac{1}{6}\cdot P\left( A\cap B\right) =\dfrac{1}{4}and P\left( \overline{A}\right) =\dfrac{1}{4},

P\left( A\right) =1-P\left( \overline{A}\right) =1-\dfrac{1}{4}=\dfrac{3}{4}therefore, P\left( \overline{A\cup B}\right) =1-P\left( A\cup B\right) P\left( \overline{A\cup B}\right)=1-\left[ P\left( A\right) +P\left( B\right) -P\left( A\cap B\right) \right]

\Rightarrow \dfrac{1}{6}=1-\dfrac{3}{4}-P\left( B\right) +\dfrac{1}{4}on simplification we get, P\left( B\right) =1-\dfrac{1}{2}-\dfrac{1}{6}=\dfrac{1}{3}

Since P\left( A\cap B\right) =P\left( A\right) P\left( B\right) and P\left( A\right) \neq P\left( B\right) therefore A and B are independent but not equally likely.