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Suppose that
BOX – I contains 8 red, 3 blue and 5 green balls
BOX – II contains 24 red, 9 blue and 15 green balls
BOX – III contains 1 blue, 12 green and 3 yellow balls
BOX – IV contains 10 green, 16 orange and 6 white balls
A ball is chosen randomly for Box – I; call that ball b. If b is red then a ball is chosen randomly from Box – II, if b is blue then a ball is chosen randomly from Box – III, and if b is green then a ball is chosen randomly from Box – IV. The conditional probability of the event ‘one of the chosen balls is white’ given that the event ‘at least one of the chosen ball is green’ has happened, is equal to

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$\frac{1}{8}$

$\frac{3}{16}$

$\frac{5}{52}$

$\frac{15}{256}$

answer is C.

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Detailed Solution

$P (\frac{W_{1}}{G_{12}}) = \frac{P (\frac{W}{G})}{P (G)} = \frac{\frac{5}{16} . \frac{6}{12}}{\frac{5}{16} .1 + \frac{8}{16} . \frac{15}{48} + \frac{3}{16} . \frac{12}{16}} = \frac{15}{80 + 40 + 36} = \frac{5}{52}$

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