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(B) Let $B_1$ be the event of choosing the first bag and $B_2$ be the event of choosing the second bag. Since the bag is chosen randomly,$P(B_1) = P(B

Vedclass · Accepted Answer

$\frac{8}{15}$

Vedclass · Answer

(B) Let $B_1$ be the event of choosing the first bag and $B_2$ be the event of choosing the second bag. Since the bag is chosen randomly,$P(B_1) = P(B_2) = \frac{1}{2}$.In the first bag,there are $3$ white and $2$ black balls,so the total number of balls is $5$. The probability of picking a black ball from the first bag is $P(Black|B_1) = \frac{2}{5}$.In the second bag,there are $2$ white and $4$ black balls,so the total number of balls is $6$. The probability of picking a black ball from the second bag is $P(Black|B_2) = \frac{4}{6} = \frac{2}{3}$.Using the law of total probability,the probability of picking a black ball is $P(Black) = P(B_1) 	imes P(Black|B_1) + P(B_2) 	imes P(Black|B_2)$.$P(Black) = \frac{1}{2} 	imes \frac{2}{5} + \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

$A$ bag contains $3$ white and $2$ black balls and another bag contains $2$ white and $4$ black balls. $A$ ball is picked up randomly. The probability of its being black is

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