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(B) Let $W_1$ be the event of drawing a white ball from the first bag and $B_1$ be the event of drawing a black ball from the first bag.Total balls in

Vedclass · Accepted Answer

$\frac{5}{8}$

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(B) Let $W_1$ be the event of drawing a white ball from the first bag and $B_1$ be the event of drawing a black ball from the first bag.Total balls in the first bag $= 5 + 3 = 8$.$P(W_1) = \frac{5}{8}$ and $P(B_1) = \frac{3}{8}$.After transferring one ball,the second bag contains $2 + 4 + 1 = 7$ balls.Case $1$: If a white ball is transferred,the second bag now contains $2 + 1 = 3$ white and $4$ black balls.Probability of drawing a black ball from the second bag given $W_1$ is $P(B_2|W_1) = \frac{4}{7}$.Case $2$: If a black ball is transferred,the second bag now contains $2$ white and $4 + 1 = 5$ black balls.Probability of drawing a black ball from the second bag given $B_1$ is $P(B_2|B_1) = \frac{5}{7}$.Using the Law of Total Probability:$P(B_2) = P(W_1) 	imes P(B_2|W_1) + P(B_1) 	imes P(B_2|B_1)$$P(B_2) = (\frac{5}{8} 	imes \frac{4}{7}) + (\frac{3}{8} 	imes \frac{5}{7})$$P(B_2) = \frac{20}{56} + \frac{15}{56} = \frac{35}{56} = \frac{5}{8}$.

One bag contains $5$ white and $3$ black balls and a second bag contains $2$ white and $4$ black balls. One ball is drawn from the first bag and placed unseen in the second bag. What is the probability that a ball now drawn from the second bag is black?

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