Three urns respectively contain 2 white and 3 | vedclass

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(B) Let $W_i$ and $B_i$ be the events of drawing a white and a black ball from urn $i$ respectively,for $i = 1, 2, 3$.Urn $1$: $P(W_1) = \frac{2}{5}$,

Vedclass · Accepted Answer

$\frac{37}{125}$

Vedclass · Answer

(B) Let $W_i$ and $B_i$ be the events of drawing a white and a black ball from urn $i$ respectively,for $i = 1, 2, 3$.Urn $1$: $P(W_1) = \frac{2}{5}$,$P(B_1) = \frac{3}{5}$.Urn $2$: $P(W_2) = \frac{3}{5}$,$P(B_2) = \frac{2}{5}$.Urn $3$: $P(W_3) = \frac{1}{5}$,$P(B_3) = \frac{4}{5}$.We need the probability of selecting $1$ black and $2$ white balls. This can happen in three mutually exclusive ways:$1$. $(B_1, W_2, W_3)$: $P(B_1) 	imes P(W_2) 	imes P(W_3) = \frac{3}{5} 	imes \frac{3}{5} 	imes \frac{1}{5} = \frac{9}{125}$.$2$. $(W_1, B_2, W_3)$: $P(W_1) 	imes P(B_2) 	imes P(W_3) = \frac{2}{5} 	imes \frac{2}{5} 	imes \frac{1}{5} = \frac{4}{125}$.$3$. $(W_1, W_2, B_3)$: $P(W_1) 	imes P(W_2) 	imes P(B_3) = \frac{2}{5} 	imes \frac{3}{5} 	imes \frac{4}{5} = \frac{24}{125}$.The total probability is the sum of these probabilities: $\frac{9}{125} + \frac{4}{125} + \frac{24}{125} = \frac{37}{125}$.

Three urns respectively contain $2$ white and $3$ black,$3$ white and $2$ black,and $1$ white and $4$ black balls. If one ball is drawn from each urn,then the probability that the selection contains $1$ black and $2$ white balls is

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