Bag I contains 3 red, 4 black and 3 white balls and Bag II contains 2 red, 5 black and 2white balls. One ball is transferred from Bag I to Bag II and then a ball is draw from Bag II.The ball so drawn is found to be black in colour. Then the probability, that the transferred ballis red, is :

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$\frac{1}{6}$

$\frac{5}{18}$

$\frac{3}{10}$

$\frac{4}{9}$

answer is B.

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Detailed Solution

$3 R 4 B 3 W 2 R 5 B 2 W$
A : Drown ball from boy II is black
B : Red ball transferred
$\begin{array}{l} P (\frac{B}{A}) = \frac{P (A \cap B)}{P (A)} \\ = \frac{\frac{3}{9} \times \frac{5}{10}}{\frac{3}{9} \times \frac{5}{10} + \frac{4}{9} \times \frac{6}{10} + \frac{3}{9} \times \frac{5}{10}} \\ = \frac{15}{15 + 24 + 15} = \frac{15}{54} = \frac{5}{18} \end{array}$

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