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

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(A) Let $E_1, E_2, E_3$ be the events of choosing Bag $I$,Bag $II$,and Bag $III$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(

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$\frac{7}{15}$

Vedclass · Answer

(A) Let $E_1, E_2, E_3$ be the events of choosing Bag $I$,Bag $II$,and Bag $III$ respectively. Since the bag is chosen at random,$P(E_1) = P(E_2) = P(E_3) = \frac{1}{3}$.Let $A$ be the event that the drawn ball is black.The probabilities of drawing a black ball from each bag are:$P(A|E_1) = \frac{3}{2+3} = \frac{3}{5}$$P(A|E_2) = \frac{1}{4+1} = \frac{1}{5}$$P(A|E_3) = \frac{7}{3+7} = \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}{30}}{\frac{6}{30} + \frac{2}{30} + \frac{7}{30}} = \frac{7}{15}$.

Box	White	Green	Red	Probability
$A$	$1$	$2$	$3$	$\frac{1}{2}$
$B$	$2$	$3$	$1$	$\frac{1}{3}$
$C$	$3$	$1$	$2$	$\frac{1}{6}$

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