Bag B_1 contains 4 white and 2 black balls. B | vedclass

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(D) Let $P(B_1)$ and $P(B_2)$ be the probabilities of selecting bag $B_1$ and bag $B_2$ respectively. Since a bag is chosen at random,$P(B_1) = P(B_2)

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$\frac{23}{42}$

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(D) Let $P(B_1)$ and $P(B_2)$ be the probabilities of selecting bag $B_1$ and bag $B_2$ respectively. Since a bag is chosen at random,$P(B_1) = P(B_2) = \frac{1}{2}$.Let $W$ be the event of drawing a white ball.The probability of drawing a white ball from bag $B_1$ is $P(W|B_1) = \frac{4}{4+2} = \frac{4}{6} = \frac{2}{3}$.The probability of drawing a white ball from bag $B_2$ is $P(W|B_2) = \frac{3}{3+4} = \frac{3}{7}$.Using the law of total probability:$P(W) = P(B_1) 	imes P(W|B_1) + P(B_2) 	imes P(W|B_2)$$P(W) = \frac{1}{2} 	imes \frac{2}{3} + \frac{1}{2} 	imes \frac{3}{7}$$P(W) = \frac{1}{3} + \frac{3}{14} = \frac{14+9}{42} = \frac{23}{42}$.

Bag $B_1$ contains $4$ white and $2$ black balls. Bag $B_2$ contains $3$ white and $4$ black balls. $A$ bag is chosen at random and a ball is drawn from it at random,then the probability that the ball drawn is white,is

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