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 :

$\begin{aligned} & \mathrm{P}(4 \mathrm{~W} 4 \mathrm{~B} / 2 \mathrm{~W} 2 \mathrm{~B})= \\\\ & \frac{P(4 W 4 B) \times P(2 W 2 B / 4 W 4 B)}{P(2 W 6 B) \times P(2 W 2 B / 2 W 6 B)+P(3 W 5 B) \times P(2 W 2 B / 3 W 5 B)} \\ & +\ldots \ldots \ldots \ldots+P(6 W 2 B) \times P(2 W 2 B / 6 W 2 B)\end{aligned}$ <br/><br/>$\begin{aligned} & =\frac{\frac{1}{5} \times \frac{{ }^4 \mathrm{C}_2 \times{ }^4 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}}{\frac{1}{5} \times \frac{{ }^2 \mathrm{C}_2 \times{ }^6 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\frac{1}{5} \times \frac{{ }^3 \mathrm{C}_2 \times{ }^5 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}+\ldots+\frac{1}{5} \times \frac{{ }^6 \mathrm{C}_2 \times{ }^2 \mathrm{C}_2}{{ }^8 \mathrm{C}_4}} \\\\ & =\frac{2}{7}\end{aligned}$