There are 3 bags A, B and C. Bag A contains 2 | vedclass

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(B) Let $E_1, E_2, E_3$ be the events of choosing bags $A, B, C$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{

Vedclass · Accepted Answer

$\frac{4}{9}$

Vedclass · Answer

(B) Let $E_1, E_2, E_3$ be the events of choosing bags $A, B, C$ 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 $P(B|E_1) = \frac{3}{2+3} = \frac{3}{5}$.The probability of drawing a black ball from bag $B$ is $P(B|E_2) = \frac{2}{4+2} = \frac{2}{6} = \frac{1}{3}$.The probability of drawing a black ball from bag $C$ is $P(B|E_3) = \frac{2}{3+2} = \frac{2}{5}$.By the law of total probability,$P(B) = P(E_1)P(B|E_1) + P(E_2)P(B|E_2) + P(E_3)P(B|E_3)$.$P(B) = \frac{1}{3} 	imes \frac{3}{5} + \frac{1}{3} 	imes \frac{1}{3} + \frac{1}{3} 	imes \frac{2}{5}$.$P(B) = \frac{1}{5} + \frac{1}{9} + \frac{2}{15} = \frac{9 + 5 + 6}{45} = \frac{20}{45} = \frac{4}{9}$.

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

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