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There are three urns, labeled, Urn 1, Urn 2 and Urn 3. Urn 1 contains 2 white balls and 2 black balls, Urn 2 contains 1 white ball and 3 black balls and Urn 3 contains 3 white balls and 1 black ball. Consider two coins with probability of obtaining head in their single trails as 0.2 and 0.3. The two coins are tossed independently once and an urn is selected according to the following scheme: Urn 1 is selected if 2 heads are obtained; Urn 3 is selected if 2 tails are obtained, otherwise Urn 2 is selected. A ball is then drawn at random from the selected urn. Then P(Urn 1 is selected | the ball drawn is white) is equal to

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$\frac{1}{18}$

$\frac{1}{9}$

$\frac{6}{109}$

$\frac{12}{109}$

answer is A.

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Detailed Solution

$P (W) = (0.2) (0.3) (0.5) + (0.7) (0.8) (0.75) + ((0.2) (0.7) + (0.8) (0.3)) (0.25)$

$= 0.03 + 0.42 + 0.095 = 0.545$

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