A bag contains 8 balls, whose colours are either white or black. 4...

Total Possible cases of balls in the bag could be:

$\left(2W, 6B\right), \left(3W, 5B\right), \left(4W, 4B\right), \left(5W, 3B\right), \left(6W, 2B\right)$

The favourable case is$to choose2W and 2B from$the bag having \left(4W, 4B\right)

So,$the required probability is given by</p><p>$P\left(E\right) = \dfrac{\dfrac{C2 4 \times C2 4}{C2 8}}${\dfrac{C2 2 \times C2 6}{C2 8} + \dfrac{C2 3 \times C2 5}{C2 8} + \dfrac{C2 4 \times C2 4}{C2 8} + \dfrac{C2 5 \times C2 3}{C2 8} + \dfrac{C2 6 \times C2 2}{C2 8}}$

$⇒ P \left(E\right) = \dfrac{6 \times 6}{15 + 3 \times 10 + 36 + 30 + 15}</p><p>$⇒ P \left(E\right) = \dfrac{36}{126}

$⇒ P \left(E\right) = \dfrac{2}{7}</p>$