A box 'A' contains 2 white, 3 red and 2 black balls. Another box 'B' contains 4 white, 2 red and 3 block balls. If two balls are drawn at random, without replacement, from a randomly selected box and one ball turns out to be white while the other ball turns out to be red, then the probability that both balls are drawn from box 'B' is

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$\frac{7}{16}$

$\frac{9}{16}$

$\frac{9}{32}$

$\frac{7}{8}$

answer is A.

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Detailed Solution

	BoxA	BoxB
White	2	4
Red	3	2
Black	2	3

$E_{1} =$ Selected Box I $P (E_{1}) = \frac{1}{2}$

$E_{2} =$ Selected Box II $P (E_{2}) = \frac{1}{2}$

A= two balls drawn from selected box one is white another red without replaced

$P (\frac{A}{E_{1}}) = \frac{2}{7} \times \frac{3}{6} \times 2 = \frac{2}{7} P (\frac{A}{E_{2}}) = \frac{4}{9} \times \frac{2}{8} \times 2 = \frac{2}{9} P (\frac{E_{2}}{A}) = \frac{\frac{1}{2} \times \frac{2}{9}}{\frac{1}{2} \times \frac{2}{9} + \frac{1}{2} \times \frac{2}{7}} = \frac{7}{16}$