Urn A contains 6 white and 2 black balls; urn | vedclass

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(B) Let $U_A, U_B,$ and $U_C$ be the events of choosing urn $A, B,$ and $C$ respectively. Since the urn is chosen at random, $P(U_A) = P(U_B) = P(U_C)

Vedclass · Accepted Answer

$\frac{5}{8}$

Vedclass · Answer

(B) Let $U_A, U_B,$ and $U_C$ be the events of choosing urn $A, B,$ and $C$ respectively. Since the urn is chosen at random, $P(U_A) = P(U_B) = P(U_C) = \frac{1}{3}$.Let $W$ be the event of drawing a white ball.The conditional probabilities are:$P(W|U_A) = \frac{6}{6+2} = \frac{6}{8} = \frac{3}{4}$$P(W|U_B) = \frac{5}{5+3} = \frac{5}{8}$$P(W|U_C) = \frac{4}{4+4} = \frac{4}{8} = \frac{1}{2}$Using the Law of Total Probability:$P(W) = P(U_A)P(W|U_A) + P(U_B)P(W|U_B) + P(U_C)P(W|U_C)$$P(W) = \frac{1}{3} 	imes \frac{3}{4} + \frac{1}{3} 	imes \frac{5}{8} + \frac{1}{3} 	imes \frac{1}{2}$$P(W) = \frac{1}{3} 	imes (\frac{6}{8} + \frac{5}{8} + \frac{4}{8}) = \frac{1}{3} 	imes \frac{15}{8} = \frac{5}{8}$.

Urn $A$ contains $6$ white and $2$ black balls; urn $B$ contains $5$ white and $3$ black balls and urn $C$ contains $4$ white and $4$ black balls. If an urn is chosen at random and a ball is drawn at random from it, then the probability that the ball drawn is white is

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