Question

A bag has four pair of balls of four distinct colours. If four balls are picked at random (without replacement), the probability that there is atleast one pair among them have the same colour is

• A. $\frac{1}{7!}$
• B. $\frac{8}{35}$
• C. $\frac{19}{35}$
• D. $\frac{27}{35}$

Answer 1

The correct option is D

$\frac{27}{35}$

Explanation for the correct option:

Find the probability of getting same colour balls:

Total number of balls $=4$ pairs

$=8$ balls

The total number of ways of picking four balls at random $={}^{8}C_4$

The number of ways of picking 4 balls, in which none of the balls have the same colour $={}^{2}C_1\times {}^{2}C_1\times {}^{2}C_1\times {}^{2}C_1$

Thus, the probability of picking the same color balls $={}^{2}C_1\times {}^{2}C_1\times {}^{2}C_1\times \frac{{}^{2}C_1}{{}^{8}C_4}$

$=\frac{2\times 2\times 2\times 2}{{}^{8}C_4}$

$=\frac{16}{{}^{8}C_4}$ $\left[\mathbf{\because }{}^{\mathbf{n}}\mathit{C}_{\mathbf{r}}\mathbf{}\mathbf{=}\mathbf{}\frac{\mathbf{n}\mathbf{!}}{\mathbf{(}\mathbf{n}\mathbf{-}\mathbf{r}\mathbf{)}\mathbf{!}\mathbf{r}\mathbf{!}}\right]$

$=\frac{8}{35}$

$\therefore$The required probability $=1-\frac{8}{35}$

$=\frac{27}{35}$

Hence, option ‘D’ is correct.