A bag contains 6 balls. Two balls are drawn from it at random and...

Let the bag contains black balls$\left(B\right)$ and non-black balls \left(\overset{¯}{B}\right)$, then we can have the following possibilities:</p><p>\left\{\left(6B, 0\overset{¯}{B}\right), \left(5B, 1\overset{¯}{B}\right), \left(4B, 2\overset{¯}{B}\right), \left(3B, 3\overset{¯}{B}\right), \left(2B, 4\overset{¯}{B}\right)\right\}$

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

Required probability

$= \dfrac{\dfrac{1}{5} \times \dfrac{C2 6}{C2 6} + \dfrac{1}{5} \times \dfrac{C2 5}{C2 6}}{\dfrac{1}{5} \times \dfrac{C2 6}{C2 6} + \dfrac{1}{5} \times \dfrac{C2 5}{C2 6} + \dfrac{1}{5} \times \dfrac{C2 4}{C2 6} + \dfrac{1}{5} \times \dfrac{C2 3}{C2 6} + \dfrac{1}{5} \times \dfrac{C2 2}{C2 6}}$

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