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(C) This problem is solved using Bayes' theorem.Let $E_1$ be the event of selecting bag $A$ and $E_2$ be the event of selecting bag $B$.Since one bag

Vedclass · Accepted Answer

$\frac{3}{7}$

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(C) This problem is solved using Bayes' theorem.Let $E_1$ be the event of selecting bag $A$ and $E_2$ be the event of selecting bag $B$.Since one bag is selected at random,$P(E_1) = \frac{1}{2}$ and $P(E_2) = \frac{1}{2}$.Let $G$ be the event of drawing a green ball.Probability of drawing a green ball from bag $A$: $P(G|E_1) = \frac{4}{7}$.Probability of drawing a green ball from bag $B$: $P(G|E_2) = \frac{3}{7}$.We need to find the probability that the ball came from bag $B$ given that it is green,i.e.,$P(E_2|G)$.Using Bayes' theorem:$P(E_2|G) = \frac{P(E_2) \cdot P(G|E_2)}{P(E_1) \cdot P(G|E_1) + P(E_2) \cdot P(G|E_2)}$$P(E_2|G) = \frac{\frac{1}{2} 	imes \frac{3}{7}}{(\frac{1}{2} 	imes \frac{4}{7}) + (\frac{1}{2} 	imes \frac{3}{7})}$$P(E_2|G) = \frac{\frac{3}{14}}{\frac{4}{14} + \frac{3}{14}} = \frac{\frac{3}{14}}{\frac{7}{14}} = \frac{3}{7}$.

Bag $A$ contains $4$ green and $3$ red balls and bag $B$ contains $4$ red and $3$ green balls. One bag is selected at random and a ball is drawn,which is found to be green. What is the probability that it came from bag $B$?

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