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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 at random,$P(B_1) = P(

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 at random,$P(B_1) = P(B_2) = \frac{1}{2}$.Let $E$ be the event of picking a black ball.For the first bag,the probability of picking a black ball is $P(E|B_1) = \frac{2}{3+2} = \frac{2}{5}$.For the second bag,the probability of picking a black ball is $P(E|B_2) = \frac{4}{2+4} = \frac{4}{6} = \frac{2}{3}$.Using the law of total probability,$P(E) = P(B_1) 	imes P(E|B_1) + P(B_2) 	imes P(E|B_2)$.$P(E) = \frac{1}{2} 	imes \frac{2}{5} + \frac{1}{2} 	imes \frac{2}{3} = \frac{1}{5} + \frac{1}{3} = \frac{3+5}{15} = \frac{8}{15}$.

$A$ bag contains $3$ white and $2$ black balls and another bag contains $2$ white and $4$ black balls. $A$ bag is chosen at random and a ball is picked up from it. The probability of the ball being black is:

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