1 bag contains 4 white and 2 black balls. Ano | vedclass

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(C) Let $B_1$ be the first bag and $B_2$ be the second bag.In $B_1$,total balls $= 4 + 2 = 6$. Probability of white $(W_1) = \frac{4}{6} = \frac{2}{3}

Vedclass · Accepted Answer

$\frac{13}{24}$

Vedclass · Answer

(C) Let $B_1$ be the first bag and $B_2$ be the second bag.In $B_1$,total balls $= 4 + 2 = 6$. Probability of white $(W_1) = \frac{4}{6} = \frac{2}{3}$,probability of black $(Bl_1) = \frac{2}{6} = \frac{1}{3}$.In $B_2$,total balls $= 3 + 5 = 8$. Probability of white $(W_2) = \frac{3}{8}$,probability of black $(Bl_2) = \frac{5}{8}$.The event 'one is white and one is black' occurs in two mutually exclusive cases: $(W_1 	ext{ and } Bl_2)$ or $(Bl_1 	ext{ and } W_2)$.Probability $= P(W_1) 	imes P(Bl_2) + P(Bl_1) 	imes P(W_2)$$= (\frac{2}{3} 	imes \frac{5}{8}) + (\frac{1}{3} 	imes \frac{3}{8})$$= \frac{10}{24} + \frac{3}{24} = \frac{13}{24}$.

$1$ bag contains $4$ white and $2$ black balls. Another bag contains $3$ white and $5$ black balls. If $1$ ball is drawn from each bag,find the probability that $1$ ball is white and $1$ ball is black.

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