Question

If two different numbers are taken from the set $\{0,1,2,3,\cdots ,10\}$; then the probability that their sum as well as absolute difference are both multiple of 4, is:

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

Answer 1

The correct option is A $\frac{6}{55}$

From the given conditions,
$x_1+x_2=4\alpha \text{and}{x}_1-x_2=\pm 4\beta$
Solving the two equations, we get
$x_1=2(\alpha \pm \beta )$
i.e, $x_1$ is multiple of $2$.
$$\begin{array}{cc}{x}_1& x_2\\ 0& 4,8\\ 2& 6,10\\ 4& 0,8\\ 6& 2,10\\ 8& 0,4\\ 10& 2,6\end{array}$$
​​​​There are total 12 possibilities. Since $(x_1,x_2)\text{and}(x_2,x_1)$ are same in this case, there is only 6 possibilities.
Hence, required probability $=\frac{6}{{}^{11}{C}_2}=\frac{6}{55}$