A bag P contains 5 white marbles and 3 black marbles. Four marbles are drawn at random from P and are put in an empty bag Q. If a marble drawn at random from Q is found to be black, then the probability that all the three black marbles in P are transferred to the bag Q, is

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$\frac{1}{8}$

$\frac{1}{7}$

$\frac{6}{8}$

$\frac{7}{8}$

answer is D.

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Detailed Solution

$E_{1} =$ Transfer 3w, 1B

$E_{2} =$ Transfer 2W, 2B

$E_{3} =$ Transfer 1W, 3B

A=one ball drawn from Q it is black

$\begin{array}{rcl} P (E_{1}) & = & \frac{{}^{5}{C_{3}} \times {}^{3}{C_{1}}}{{}^{8}{C_{4}}} P (E_{2}) = \frac{{}^{5}{C_{2}} \times {}^{3}{C_{2}}}{{}^{8}{C_{4}}} \\ P (E_{3}) & = & \frac{{}^{5}{C_{1}} \times {}^{3}{C_{3}}}{{}^{8}{C_{4}}} \\ P (\frac{A}{E_{1}}) & = & \frac{1}{4} P (\frac{A}{E_{2}}) = \frac{2}{4}, P (\frac{A}{E_{3}}) = \frac{3}{4} \\ P (\frac{E_{3}}{A}) & = & \frac{\frac{5}{{}^{8}{C_{4}}} \times \frac{3}{4}}{\frac{30}{{}^{8}{C_{4}}} \times \frac{1}{4} + \frac{30}{{}^{8}{C_{4}}} \times \frac{2}{4} + \frac{5}{{}^{8}{C_{4}}} \times \frac{3}{4}} \\ = & \frac{15}{30 + 60 + 15} = \frac{1}{7} \end{array}$

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