If from each of the three boxes containing 3 white and 1 black, 2 white and 2 black, 1 white and 3 black balls, one ball is drawn at random, then the probability that 2 white and 1 black ball will be drawn is

$\frac{13}{32}$

The contents of three boxes are

Let $W_i$ (i =  1,2, 3) be the event of drawing a white ball from ith box and $B_i (i= 1, 2, 3)$ be the event of drawing a black ball from ith box. Then,

Required probability

$= P(W_1 ∩ W_2 ∩ B_3) ∪ (W_1 ∩ B_2 ∩ W_3) (B_1 ∩ W_2 ∩ W_3)$

$= P(W_1 ∩ W_2 ∩ B_3) +P (W_1 ∩ B_2 ∩ W_3) +P(B_1 ∩ W_2 ∩ W_3)$

$= P(W_1)P(W_2)P(B_3)+P(W_1)P(B_2)P(W_3) +P(B_1)P(W_2)P(W_3)$

$=\frac{3}{4}×\frac{2}{4}×\frac{3}{4}+\frac{3}{4}×\frac{2}{4}×\frac{1}{4}+\frac{1}{4}×\frac{2}{4}×\frac{1}{4}=\frac{26}{64}=\frac{13}{32}$