Let the bag contains black balls$\left(B\right)$ and non-black balls $\left(\overline{B}\right)$, then we can have the following possibilities:

$$\begin{gathered} \left\{\left(6B,0\overline{B}\right), \left(5B,1\overline{B}\right), \left(4B,2\overline{B}\right), \left(3B,3\overline{B}\right), \\ \left(2B,4\overline{B}\right)\right\} \end{gathered}$$

All the events are equally likely, i.e., $\frac{1}{5}$.

Required probability

$=\frac{\frac{1}{5}\times \frac{{}^{6}C_2}{{}^{6}C_2}+\frac{1}{5}\times \frac{{}^{5}C_2}{{}^{6}C_2}}{\frac{1}{5}\times \frac{{}^{6}C_2}{{}^{6}C_2}+\frac{1}{5}\times \frac{{}^{5}C_2}{{}^{6}C_2}+\frac{1}{5}\times \frac{{}^{4}C_2}{{}^{6}C_2}+\frac{1}{5}\times \frac{{}^{3}C_2}{{}^{6}C_2}+\frac{1}{5}\times \frac{{}^{2}C_2}{{}^{6}C_2}}$

$=\frac{25}{35}=\frac{5}{7}$