Three urns A,B,and C contain 7 red,5 black; 5 | vedclass

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(B) Let $E_1, E_2, E_3$ be the events of selecting urns $A, B, C$ respectively. Since the urn is selected at random,$P(E_1) = P(E_2) = P(E_3) = \frac{

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$\frac{5}{18}$

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(B) Let $E_1, E_2, E_3$ be the events of selecting urns $A, B, C$ respectively. Since the urn is selected at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.Let $X$ be the event of drawing a black ball.The probabilities of drawing a black ball from each urn are:$P(X|E_1) = \frac{5}{12}$$P(X|E_2) = \frac{7}{12}$$P(X|E_3) = \frac{6}{12}$Using Bayes' theorem,the probability that the ball is drawn from urn $A$ given that it is black is:$P(E_1|X) = \frac{P(E_1)P(X|E_1)}{P(E_1)P(X|E_1) + P(E_2)P(X|E_2) + P(E_3)P(X|E_3)}$$P(E_1|X) = \frac{\frac{1}{3} \cdot \frac{5}{12}}{\frac{1}{3} \cdot \frac{5}{12} + \frac{1}{3} \cdot \frac{7}{12} + \frac{1}{3} \cdot \frac{6}{12}}$$P(E_1|X) = \frac{5}{5 + 7 + 6} = \frac{5}{18}$.

Three urns $A$,$B$,and $C$ contain $7$ red,$5$ black; $5$ red,$7$ black; and $6$ red,$6$ black balls,respectively. One of the urns is selected at random and a ball is drawn from it. If the ball drawn is black,then the probability that it is drawn from urn $A$ is:

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