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Bag A contains 4 red and 3 green balls and B contains 3 red and 4 green balls. One bag is taken at random and ball is drawn and noted to be green. The probability that it comes from bag B is ____

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$\frac{4}{21}$

$\frac{3}{7}$

$\frac{2}{7}$

$\frac{4}{7}$

answer is C.

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Detailed Solution

$P (A) = P (B) = \frac{1}{2} P (G / A) = \frac{3}{7}, P (G / B) = \frac{4}{7}$

From Baye’s theorem

$∴ P (B / G) = \frac{P (B) P (G / B)}{P (A) P (G / A) + P (B) P (G / B)} = \frac{\frac{1}{2} . \frac{4}{7}}{\frac{1}{2} . \frac{3}{7} + \frac{1}{2} . \frac{4}{7}} = \frac{4}{7}$

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