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(A) Let $G$ be the event that the selected ball is green. Let $A, B, C$ be the events of selecting boxes $A, B, C$ respectively.The probabilities of s

Vedclass · Accepted Answer

$\frac{1}{13}$

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(A) Let $G$ be the event that the selected ball is green. Let $A, B, C$ be the events of selecting boxes $A, B, C$ respectively.The probabilities of selecting the boxes are:$P(A) = \frac{1}{2}, P(B) = \frac{1}{3}, P(C) = \frac{1}{6}$The conditional probabilities of selecting a green ball from each box are:$P(G|A) = \frac{2}{1+2+3} = \frac{2}{6} = \frac{1}{3}$$P(G|B) = \frac{3}{2+3+1} = \frac{3}{6} = \frac{1}{2}$$P(G|C) = \frac{1}{3+1+2} = \frac{1}{6}$Using the law of total probability,the probability of selecting a green ball $P(G)$ is:$P(G) = P(A)P(G|A) + P(B)P(G|B) + P(C)P(G|C)$$P(G) = (\frac{1}{2} 	imes \frac{1}{3}) + (\frac{1}{3} 	imes \frac{1}{2}) + (\frac{1}{6} 	imes \frac{1}{6})$$P(G) = \frac{1}{6} + \frac{1}{6} + \frac{1}{36} = \frac{6+6+1}{36} = \frac{13}{36}$Using Bayes' theorem,the probability that the green ball was selected from box $C$ is $P(C|G)$:$P(C|G) = \frac{P(C)P(G|C)}{P(G)}$$P(C|G) = \frac{\frac{1}{6} 	imes \frac{1}{6}}{\frac{13}{36}} = \frac{\frac{1}{36}}{\frac{13}{36}} = \frac{1}{13}$

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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