A bag contains 4 white balls and 2 black ball | vedclass

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(C) Given: In the first bag,the number of white balls $= 4$ and the number of black balls $= 2$. Total balls in the first bag $= 4 + 2 = 6$.In the sec

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$\frac{5}{24}$

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(C) Given: In the first bag,the number of white balls $= 4$ and the number of black balls $= 2$. Total balls in the first bag $= 4 + 2 = 6$.In the second bag,the number of white balls $= 3$ and the number of black balls $= 5$. Total balls in the second bag $= 3 + 5 = 8$.We know that the probability of drawing one black ball from the first bag $= \frac{	ext{Number of black balls}}{	ext{Total number of balls}} = \frac{2}{6} = \frac{1}{3}$.Similarly,the probability of drawing one black ball from the second bag $= \frac{5}{8}$.Since the events are independent,the probability that both balls are black $= P(	ext{Black from bag } 1) 	imes P(	ext{Black from bag } 2) = \frac{1}{3} 	imes \frac{5}{8} = \frac{5}{24}$.

$A$ bag contains $4$ white balls and $2$ black balls. Another bag contains $3$ white balls and $5$ black balls. If one ball is drawn from each bag,then the probability that both the balls are black,is

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