One bag contains 5 white and 4 black balls. A | vedclass

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(D) Let $W_1$ be the event that a white ball is transferred from the first bag to the second,and $B_1$ be the event that a black ball is transferred.P

Vedclass · Accepted Answer

$\frac{4}{9}$

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(D) Let $W_1$ be the event that a white ball is transferred from the first bag to the second,and $B_1$ be the event that a black ball is transferred.Probability of transferring a white ball: $P(W_1) = \frac{5}{9}$.Probability of transferring a black ball: $P(B_1) = \frac{4}{9}$.If a white ball is transferred,the second bag now contains $8$ white and $9$ black balls (total $17$). The probability of drawing a white ball from the second bag is $P(W_2|W_1) = \frac{8}{17}$.If a black ball is transferred,the second bag now contains $7$ white and $10$ black balls (total $17$). The probability of drawing a white ball from the second bag is $P(W_2|B_1) = \frac{7}{17}$.Using the law of total probability,the required probability is:$P(W_2) = P(W_1) 	imes P(W_2|W_1) + P(B_1) 	imes P(W_2|B_1)$$P(W_2) = (\frac{5}{9} 	imes \frac{8}{17}) + (\frac{4}{9} 	imes \frac{7}{17})$$P(W_2) = \frac{40}{153} + \frac{28}{153} = \frac{68}{153} = \frac{4}{9}$.

One bag contains $5$ white and $4$ black balls. Another bag contains $7$ white and $9$ black balls. $A$ ball is transferred from the first bag to the second and then a ball is drawn from the second. The probability that the ball drawn is white,is

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