If from each of the three boxes containing 3 | vedclass

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Question

(A) Let the boxes be $A$,$B$,and $C$.Box $A$ contains $3$ white and $1$ black ball. Probability of drawing white $P(W_A) = \frac{3}{4}$,black $P(B_A)

Vedclass · Accepted Answer

$\frac{13}{32}$

Vedclass · Answer

(A) Let the boxes be $A$,$B$,and $C$.Box $A$ contains $3$ white and $1$ black ball. Probability of drawing white $P(W_A) = \frac{3}{4}$,black $P(B_A) = \frac{1}{4}$.Box $B$ contains $2$ white and $2$ black balls. Probability of drawing white $P(W_B) = \frac{2}{4}$,black $P(B_B) = \frac{2}{4}$.Box $C$ contains $1$ white and $3$ black balls. Probability of drawing white $P(W_C) = \frac{1}{4}$,black $P(B_C) = \frac{3}{4}$.To get $2$ white and $1$ black ball,the possible cases are $(W, W, B)$,$(W, B, W)$,and $(B, W, W)$.Probability $= P(W_A) 	imes P(W_B) 	imes P(B_C) + P(W_A) 	imes P(B_B) 	imes P(W_C) + P(B_A) 	imes P(W_B) 	imes P(W_C)$$= (\frac{3}{4} 	imes \frac{2}{4} 	imes \frac{3}{4}) + (\frac{3}{4} 	imes \frac{2}{4} 	imes \frac{1}{4}) + (\frac{1}{4} 	imes \frac{2}{4} 	imes \frac{1}{4})$$= \frac{18}{64} + \frac{6}{64} + \frac{2}{64} = \frac{26}{64} = \frac{13}{32}$.

If from each of the three boxes containing $3$ white and $1$ black,$2$ white and $2$ black,and $1$ 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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