Box A contains 2 black and 3 red balls, while box B contains 3 black and 4 red balls. Out of these two boxes one is selected at random, and the probability of choosing box A is double that of box B. If a red ball is drawn from the selected box then the probability that it has come from box B is

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$\frac{13}{41}$

$\frac{21}{41}$

$\frac{12}{31}$

$\frac{10}{31}$

answer is B.

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Detailed Solution

$P (A) = 2. P (B) \Rightarrow P (A) = \frac{2}{3}$

$P (B) = \frac{1}{3}$

$P (R / A) = \frac{3}{5}, P (R / B) = \frac{4}{7}$

$P (B / R) = \frac{\frac{1}{3} \times \frac{4}{7}}{\frac{2}{3} \times \frac{3}{5} + \frac{1}{3} \times \frac{4}{7}}$

$= \frac{10}{31}$

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