A bag contains 8 balls, whose colours are either white or black. 4 balls are drawn at random without replacement and it was found that 2 balls are white and other 2 balls are black. The probability that the bag contains equal number of white and black balls is:

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$\frac{1}{5}$

$\frac{1}{7}$

$\frac{2}{7}$

$\frac{2}{5}$

answer is B.

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Detailed Solution

$P (4 W 4 B / 2 W 2 B) = \frac{P (4 W 4 B) \times P (2 W 2 B / 4 W 4 B)}{\begin{array}{l} P (2 W 6 B) \times P (2 W 2 B / 2 W 6 B) + P (3 W 5 B) \times P (2 W 2 B / 3 W 5 B) \\ + \dots \dots \dots \dots + P (6 W 2 B) \times P (2 W 2 B / 6 W 2 B \end{array}}$

$= \frac{\frac{1}{5} \times \frac{{}^{4}C_{2} \times {}^{4}C_{2}}{{}^{8}C_{4}}}{\frac{1}{5} \times \frac{{}^{2}C_{2} \times {}^{6}C_{2}}{{}^{8}C_{4}} + \frac{1}{5} \times \frac{{}^{3}C_{2} \times {}^{5}C_{2}}{{}^{8}C_{4}} + \dots + \frac{1}{5} \times \frac{{}^{6}C_{2} \times {}^{2}C_{2}}{{}^{8}C_{4}}}$

$= \frac{2}{7}$

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