Boxes B_1, B_2,and B_3 contain balls as given | vedclass

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(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:For $B_1$: $P(W_

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$\frac{5}{12}$

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(A) Let $W_i$ and $B_i$ denote the events of drawing a white ball and a black ball from box $B_i$ respectively. The probabilities are:For $B_1$: $P(W_1) = \frac{1}{3}, P(B_1) = \frac{2}{3}$For $B_2$: $P(W_2) = \frac{3}{4}, P(B_2) = \frac{1}{4}$For $B_3$: $P(W_3) = \frac{2}{5}, P(B_3) = \frac{3}{5}$We want the probability of drawing two black balls and one white ball. This can happen in three mutually exclusive ways:$1$. $(W_1, B_2, B_3)$: $P(W_1) 	imes P(B_2) 	imes P(B_3) = \frac{1}{3} 	imes \frac{1}{4} 	imes \frac{3}{5} = \frac{3}{60}$$2$. $(B_1, W_2, B_3)$: $P(B_1) 	imes P(W_2) 	imes P(B_3) = \frac{2}{3} 	imes \frac{3}{4} 	imes \frac{3}{5} = \frac{18}{60}$$3$. $(B_1, B_2, W_3)$: $P(B_1) 	imes P(B_2) 	imes P(W_3) = \frac{2}{3} 	imes \frac{1}{4} 	imes \frac{2}{5} = \frac{4}{60}$The total probability is the sum of these probabilities:$P = \frac{3}{60} + \frac{18}{60} + \frac{4}{60} = \frac{25}{60} = \frac{5}{12}$

Boxes $B_1, B_2$,and $B_3$ contain balls as given below:
Box White Black
$B_1$ $1$ $2$
$B_2$ $3$ $1$
$B_3$ $2$ $3$

One ball is drawn at random from each box. Then,among the balls drawn,the probability that two are black and one is white,is

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Boxes $B_1, B_2$,and $B_3$ contain balls as given below:BoxWhiteBlack$B_1$$1$$2$$B_2$$3$$1$$B_3$$2$$3$One ball is drawn at random from each box. Then,among the balls drawn,the probability that two are black and one is white,is