Probability — Class 12 Maths Solution

.: As $P(A \cap B) = \cfrac{1}{8} = \cfrac{1}{4} \times \cfrac{1}{2} = P(A) \times P(B)$

$\Rightarrow$ $A$ and $B$ are independent.

$\Rightarrow$ ${A^c}$ and ${B^c}$ are also independent.

$\Rightarrow$ $P\left( {{A^c} \cap {B^c}} \right) = P\left( {{A^c}} \right)P\left( {{B^c}} \right)$

$\Rightarrow$ $P\left( {{A^c} \cap {B^c}} \right) = (1 - P(A))(1 - P(B))$

$= \left( {1 - \cfrac{1}{4}} \right)\left( {1 - \cfrac{1}{2}} \right) = \cfrac{3}{4} \times \cfrac{1}{2} = \cfrac{3}{8}$