Question

Two integers x and y are chosen with replacement out of the set $\left\{0,1,2,3,...........10\right\}$. Then the probability that $|x-y|>5$ is

• A. $\frac{81}{121}$
• B. $\frac{30}{121}$
• C. $\frac{25}{121}$
• D. $\frac{20}{121}$

Answer 1

Answer: (B)

Total number of selections $=11\times 11$
Now $|x-y|>5$ clearly $x\ne 5$
If $x=0$, then $y>5\Rightarrow 5$ favourable cases
If $x=1$, then $y>6\Rightarrow 4$ favourable cases
If $x=2$, then $y>7\Rightarrow 3$ favourable cases
If $x=3$, then $y>8\Rightarrow 2$ favourable cases
If $x=2$, then $y>9\Rightarrow 1$ favourable cases
Symmetical cases will be obtained for $x=6,7,8,9,10$
$\therefore$ Favourable cases $=30$
$\therefore$ Desired probability $\frac{30}{121}$