Bag I contains 3 red and 4 black balls. Bag II contains 5 red and 6 black balls. If one ball is drawn at random from one of the bags and it is found to be red , then the probability that it was drawn from bag II is

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$\frac{37}{68}$

$\frac{31}{68}$

$\frac{41}{68}$

$\frac{35}{68}$

answer is B.

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Detailed Solution

$P (B_{1}) = P (B_{2}) = \frac{1}{2}$
$Now, P (\frac{R}{B_{1}}) = \frac{3}{7}, P (\frac{R}{B_{2}}) = \frac{5}{11}$
Required probability
$\begin{aligned} P (\frac{B_{2}}{R}) & = \frac{P (B_{2}) P (\frac{R}{B_{2}})}{P (B_{1}) P (\frac{R}{B_{1}}) + P (B_{2}) P (\frac{R}{B_{2}})} \\ = \frac{\frac{1}{2} (\frac{5}{11})}{\frac{1}{2} (\frac{3}{7} +) \frac{1}{2} (\frac{5}{11})} \\ = \frac{35}{68} \end{aligned}$