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(B) Let $E$ be the event that the ball is white.Let $B_1$ be the event of choosing bag $x$ and $B_2$ be the event of choosing bag $y$.Since a bag is p

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$7/15$

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(B) Let $E$ be the event that the ball is white.Let $B_1$ be the event of choosing bag $x$ and $B_2$ be the event of choosing bag $y$.Since a bag is picked at random,$P(B_1) = P(B_2) = \frac{1}{2}$.Probability of picking a white ball from bag $x$ is $P(E|B_1) = \frac{3}{3+2} = \frac{3}{5}$.Probability of picking a white ball from bag $y$ is $P(E|B_2) = \frac{2}{2+4} = \frac{2}{6} = \frac{1}{3}$.Using the law of total probability,$P(E) = P(B_1)P(E|B_1) + P(B_2)P(E|B_2)$.$P(E) = \frac{1}{2} 	imes \frac{3}{5} + \frac{1}{2} 	imes \frac{1}{3} = \frac{3}{10} + \frac{1}{6} = \frac{9+5}{30} = \frac{14}{30} = \frac{7}{15}$.

$A$ bag $x$ contains $3$ white balls and $2$ black balls and another bag $y$ contains $2$ white balls and $4$ black balls. $A$ bag and a ball out of it are picked at random. The probability that the ball is white,is

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