Suppose there are 3 urns containing 2 white, 3 black balls, 4 white 1 black and 3 white, 2 black balls respectively. There is equal chance for selecting an urn. One ball is drawn from an urn chosen at random and it is found to be white. The prob. that it was drawn from the first urn is

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$\frac{2}{5}$

$\frac{3}{5}$

$\frac{2}{9}$

$\frac{4}{9}$

answer is C.

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Detailed Solution

Required probability = $\frac{\frac{1}{3} (\frac{2}{5})}{\frac{1}{3} (\frac{2}{5}) + \frac{1}{3} (\frac{4}{5}) + \frac{1}{3} (\frac{3}{5})} = \frac{2}{9} (b y B a y e s' t h e o r e m)$

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