There are 3 bags which are known to contain 2 | vedclass

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(A) Let the events be defined as follows:$E_1$: Bag $I$ is chosen (containing $2$ white,$3$ black balls).$E_2$: Bag $II$ is chosen (containing $4$ whi

Vedclass · Accepted Answer

$\frac{7}{15}$

Vedclass · Answer

(A) Let the events be defined as follows:$E_1$: Bag $I$ is chosen (containing $2$ white,$3$ black balls).$E_2$: Bag $II$ is chosen (containing $4$ white,$1$ black ball).$E_3$: Bag $III$ is chosen (containing $3$ white,$7$ black balls).$A$: The ball drawn is black.Since the bags are chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.The conditional probabilities of drawing a black ball are:$P(A|E_1) = \frac{3}{5}$,$P(A|E_2) = \frac{1}{5}$,$P(A|E_3) = \frac{7}{10}$.We need to find the probability that the ball was drawn from the bag with the most black balls,which is Bag $III$. Using Bayes' Theorem:$P(E_3|A) = \frac{P(E_3)P(A|E_3)}{P(E_1)P(A|E_1) + P(E_2)P(A|E_2) + P(E_3)P(A|E_3)}$$P(E_3|A) = \frac{\frac{1}{3} 	imes \frac{7}{10}}{\frac{1}{3} 	imes \frac{3}{5} + \frac{1}{3} 	imes \frac{1}{5} + \frac{1}{3} 	imes \frac{7}{10}}$$P(E_3|A) = \frac{\frac{7}{10}}{\frac{3}{5} + \frac{1}{5} + \frac{7}{10}} = \frac{\frac{7}{10}}{\frac{6+2+7}{10}} = \frac{7}{15}$.

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