Probability — Class 12 Maths Solution

.: Total no. of balls $= 18$

(i) Let $E$ : When both balls are red with replacement, then

$P(E) = \cfrac{8}{{18}} \times \cfrac{8}{{18}} = \cfrac{{16}}{{81}}$

(ii) Let $E$ : When first ball is black \& second is red, then

$P(E) = \cfrac{{10}}{{18}} \times \cfrac{8}{{18}} = \cfrac{{20}}{{81}}$

(iii) Let $E$ : When one of them is black and other is red, then

$P(E) = \cfrac{{10}}{{18}} \times \cfrac{8}{{18}} + \cfrac{8}{{18}} \times \cfrac{{10}}{{18}} = 2 \times \left( {\cfrac{{20}}{{81}}} \right) = \cfrac{{40}}{{81}}$