If from each of the three boxes containing 3 | vedclass

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Question

(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 $i = 1,\,\,2,\,\,3$ respectiv

Vedclass · Accepted Answer

$\frac{{13}}{{32}}$

Vedclass · Answer

(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 -
$\begin{array}{*{20}{c}}{}&{{m{Box}}\,1}&{{m{Box}}\,2}&{{m{Box}}\,3}\{{m{Way}}\,1}&W&W&B\{{m{Way}}\,2}&W&B&W\{{m{Way}}\,3}&B&W&W\end{array}$
$	herefore $ 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}}$.

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

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