Bag A contains 4 white and 2 black balls, bag | vedclass

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(C) Let $E_1, E_2, E_3$ be the events of selecting bag $A$, bag $B$, and bag $C$ respectively. Since the bag is chosen at random, $P(E_1) = P(E_2) = P

Vedclass · Accepted Answer

$\frac{1}{2}$

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(C) Let $E_1, E_2, E_3$ be the events of selecting bag $A$, bag $B$, and bag $C$ respectively. Since the bag is chosen at random, $P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.Let $B$ be the event of drawing a black ball.The probability of drawing a black ball from bag $A$ is $P(B|E_1) = \frac{2}{6} = \frac{1}{3}$.The probability of drawing a black ball from bag $B$ is $P(B|E_2) = \frac{3}{6} = \frac{1}{2}$.The probability of drawing a black ball from bag $C$ is $P(B|E_3) = \frac{4}{6} = \frac{2}{3}$.Using the law of total probability:$P(B) = P(E_1)P(B|E_1) + P(E_2)P(B|E_2) + P(E_3)P(B|E_3)$$P(B) = \left(\frac{1}{3} 	imes \frac{1}{3}ight) + \left(\frac{1}{3} 	imes \frac{1}{2}ight) + \left(\frac{1}{3} 	imes \frac{2}{3}ight)$$P(B) = \frac{1}{9} + \frac{1}{6} + \frac{2}{9} = \frac{3}{9} + \frac{1}{6} = \frac{1}{3} + \frac{1}{6} = \frac{2+1}{6} = \frac{3}{6} = \frac{1}{2}$.

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

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