A bag contains $6$ white and $4$ black balls. Two balls are drawn at random. The probability that they are of the same colours, is

The correct option is D

$\frac{7}{15}$

Explanation for the correct option:

Step 1. Let assume $S$ be the sample space

Then, the number of ways of picking 2 balls out of $(6+4)$ balls $=n\left(S\right)$

$={}^{10}C_2$

$=\frac{10\times 9}{2\times 1}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=45$

Step 2. Let the event of getting both balls of same color $=E$

then, the number of ways of getting ($2$ balls out of $6$ white balls) or ($2$ balls out of $4$ black balls) $=n\left(E\right)$

$={}^{6}C_2+{}^{4}C_2$

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

$=15+6$

$=21$

$\therefore$Required probability, $\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{\left(}\mathbf{E}\mathbf{\right)}}{\mathbf{n}\mathbf{\left(}\mathbf{S}\mathbf{\right)}}$

$=\frac{21}{45}$

$=\frac{7}{15}$

Hence, option (D) is Correct.