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 choose $2W$ and $2B$ from the bag having $\left(4W,4B\right)$

So, the required probability is given by

$P\left(E\right)=\frac{{\displaystyle \frac{{}^{4}C_2\times {}^{4}C_2}{{}^{8}C_2}}}{{\displaystyle \frac{{}^{2}C_2\times {}^{6}C_2}{{}^{8}C_2}}+{\displaystyle \frac{{}^{3}C_2\times {}^{5}C_2}{{}^{8}C_2}}+{\displaystyle \frac{{}^{4}C_2\times {}^{4}C_2}{{}^{8}C_2}}+{\displaystyle \frac{{}^{5}C_2\times {}^{3}C_2}{{}^{8}C_2}}+{\displaystyle \frac{{}^{6}C_2\times {}^{2}C_2}{{}^{8}C_2}}}$

$\Rightarrow P\left(E\right)=\frac{{\displaystyle 6\times 6}}{{\displaystyle 15+3\times 10+36+30+15}}$

$\Rightarrow P\left(E\right)=\frac{{\displaystyle 36}}{{\displaystyle 126}}$

$\Rightarrow P\left(E\right)=\frac{{\displaystyle 2}}{{\displaystyle 7}}$