Question

A bag contains $8$ red and $7$ black balls. Two balls are drawn at random. The probability that both the balls are of the same colour is

• A. $\frac{14}{15}$
• B. $\frac{11}{15}$
• C. $\frac{7}{15}$
• D. $\frac{4}{15}$

Answer 1

The correct option is C

$\frac{7}{15}$

Explanation for the correct option:

Step 1. Total ways and favourable ways of getting same colour balls:

Number of ways of drawing $2$ balls out of $\left(8+7\right)$ balls $={}^{15}C_2$

$=\frac{15\times 14}{2!}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{=}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=105$

Number of ways of drawing $2$ red balls $={}^{8}C_2$

$=\frac{8\times 7}{2!}$

$=28$

Number of ways of drawing $2$ black balls $={}^{7}C_2$

$=\frac{7\times 6}{2!}$

$=21$

Step 2. Probability of getting balls of same colour

$\mathbf{Probability}\mathbf{=}\mathbf{}\frac{\mathbf{favourable}\mathbf{}\mathbf{ways}}{\mathbf{total}\mathbf{}\mathbf{ways}}$

$=\frac{{}^{8}C_2+{}^{7}C_2}{{}^{15}C_2}$

$=\frac{28+21}{105}$

$=\frac{49}{105}$

$=\frac{7}{15}$

Hence, Option (C) is correct.