U_1, U_2, U_3 are three urns. U_1 contains 5 | vedclass

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(B) Let $E_1, E_2, E_3$ be the events of choosing urns $U_1, U_2, U_3$ respectively. Since the urn is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \fr

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$\frac{23}{30}$

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(B) Let $E_1, E_2, E_3$ be the events of choosing urns $U_1, U_2, U_3$ respectively. Since the urn 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. We want to find the probability of not getting a black ball,which is $P(B^c) = 1 - P(B)$.The probability of drawing a black ball from each urn is:$P(B|E_1) = \frac{2}{5+3+2} = \frac{2}{10} = \frac{1}{5}$$P(B|E_2) = \frac{2}{4+4+2} = \frac{2}{10} = \frac{1}{5}$$P(B|E_3) = \frac{3}{3+4+3} = \frac{3}{10}$Using the law of total probability,$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) = \frac{1}{3} 	imes \frac{1}{5} + \frac{1}{3} 	imes \frac{1}{5} + \frac{1}{3} 	imes \frac{3}{10} = \frac{1}{15} + \frac{1}{15} + \frac{1}{10} = \frac{2}{15} + \frac{1}{10} = \frac{4+3}{30} = \frac{7}{30}$.Therefore,the probability of not getting a black ball is $P(B^c) = 1 - \frac{7}{30} = \frac{23}{30}$.

$U_1, U_2, U_3$ are three urns. $U_1$ contains $5$ red,$3$ white,$2$ black balls; $U_2$ contains $4$ red,$4$ white,$2$ black balls and $U_3$ contains $3$ red,$4$ white,$3$ black balls. If a ball is chosen at random from an urn chosen at random,then the probability of not getting a black ball is

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