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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
$\frac{{13}}{{32}}$
$\frac{1}{4}$
$\frac{1}{{32}}$
$\frac{3}{{16}}$
Solution
(a) Let $P({W_i})$ and $P({B_i})$ be the probabilities of drawing one white and one black ball from the $i$ th box where $i = 1,\,\,2,\,\,3$ respectively. Hence
$P({W_1}) = \frac{3}{4},\,\,\,P({B_1}) = \frac{1}{4}$
$P({W_2}) = \frac{2}{4} = \frac{1}{2},\,\,\,P({B_2}) = \frac{2}{4} = \frac{1}{2}$
$P({W_3}) = \frac{1}{4},\,\,\,P({B_3}) = \frac{3}{4}$
Two white and one black ball may be drawn from $3$ boxes in the following three ways –
$\begin{array}{*{20}{c}}{}&{{\rm{Box}}\,1}&{{\rm{Box}}\,2}&{{\rm{Box}}\,3}\\{{\rm{Way}}\,1}&W&W&B\\{{\rm{Way}}\,2}&W&B&W\\{{\rm{Way}}\,3}&B&W&W\end{array}$
$\therefore $ Required probability
$ = 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{{18 + 6 + 2}}{{64}} = \frac{{13}}{{32}}$.
