There are three bags B_1, B_2 and B_3. The ba | vedclass

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Question

(A) Let $G$ be the event of choosing a green ball. The probabilities of selecting bags are $P(B_1) = \frac{3}{10}, P(B_2) = \frac{3}{10}, P(B_3) = \fr

Vedclass · Accepted Answer

$1, 2$

Vedclass · Answer

(A) Let $G$ be the event of choosing a green ball. The probabilities of selecting bags are $P(B_1) = \frac{3}{10}, P(B_2) = \frac{3}{10}, P(B_3) = \frac{4}{10}$.The conditional probabilities of choosing a green ball from each bag are:$P(G|B_1) = \frac{5}{5+5} = \frac{5}{10} = \frac{1}{2}$$P(G|B_2) = \frac{5}{3+5} = \frac{5}{8}$$P(G|B_3) = \frac{3}{5+3} = \frac{3}{8}$$(1)$ $P(B_3 \cap G) = P(G|B_3) 	imes P(B_3) = \frac{3}{8} 	imes \frac{4}{10} = \frac{12}{80} = \frac{3}{20}$. (Statement $1$ is correct)$(2)$ $P(G) = P(G|B_1)P(B_1) + P(G|B_2)P(B_2) + P(G|B_3)P(B_3)$$P(G) = (\frac{1}{2} 	imes \frac{3}{10}) + (\frac{5}{8} 	imes \frac{3}{10}) + (\frac{3}{8} 	imes \frac{4}{10}) = \frac{3}{20} + \frac{15}{80} + \frac{12}{80} = \frac{12+15+12}{80} = \frac{39}{80}$. (Statement $2$ is correct)$(3)$ $P(G|B_3) = \frac{3}{8}$. (Statement $3$ is correct)$(4)$ $P(B_3|G) = \frac{P(B_3 \cap G)}{P(G)} = \frac{3/20}{39/80} = \frac{3}{20} 	imes \frac{80}{39} = \frac{4}{13}$. (Statement $4$ is correct)Since statements $1, 2, 3, 4$ are all correct,the question implies selecting the correct set of options. Given the options provided,$1, 2$ is a subset of the correct statements.

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Bag $A$ contains $3$ white and $7$ red balls,and bag $B$ contains $3$ white and $2$ red balls. One bag is selected at random and a ball is drawn from it. What is the probability that the ball was drawn from bag $A$,given that the ball drawn is white?

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