Question

If two different numbers are taken from the set $0,1,2,3,......,10;$ then the probability that their sum, as well as absolute difference, are both multiples of$4$, is.

• A. $\frac{12}{55}$
• B. $\frac{14}{45}$
• C. $\frac{7}{55}$
• D. $\frac{6}{55}$

Answer 1

The correct option is D

$\frac{6}{55}$

Explanation for the correct answer:

Finding the required probability:

Given,

Total number of ways $={}^{11}C_2$

$$\begin{gathered} =\frac{11!}{2!\left(9\right)!}\left[\because {}^n{C}_r=\frac{n!}{r!\left(n-r\right)!}\right] \\ =55 \end{gathered}$$

Favorable ways that their sum, as well as absolute difference, are both multiples of$4$, are $(0,4),(0,8),(4,8),(2,6),(2,10),(6,10)$

Therefore, the required probability $=\frac{\mathrm{Favorable}\mathrm{ways}}{\mathrm{Total}\mathrm{number}\mathrm{of}\mathrm{ways}}$

$=\frac{6}{55}$

Hence, the correct option is D.