Question

Two different number are taken from the $0$ to $10$. The probability that their sum and positive differences are both multiple of $4$ is

• A. $\frac{6}{55}$
• B. $\frac{7}{55}$
• C. $\frac{1}{11}$
• D. $\frac{4}{55}$

Answer 1

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

Selecting two different numbers from above set
$={}^{11}{C}_2=55$
For sum and positive difference of both the selected numbers to be divisible of $4$, both the number should be even.
So total number of favorable events
$=(0,4),(2,6),(4,8),(6,10),(0,8),(2,10)$
So the required probability =$\frac{6}{55}$