Question

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

• A. $\frac{15}{121}$
• B. $\frac{30}{121}$
• C. $\frac{60}{121}$
• D. $\frac{40}{121}$

Answer 1

The correct option is B $\frac{30}{121}$

Since $x$ and $y$ each can take values from $0$ to $10,$ so the total number of ways of selecting $x$ and $y$ is $11\times 11=121$.
Now, $|x-y|>5$
$\Rightarrow x-y<-5$ or $x-y>5$
When $x-y>5,$ we have following cases:

$\begin{array}{ccc}\text{Value of}x& \text{Value of}y& \text{Number of cases}\\ 6& \text{0}& \text{1}\\ 7& \text{0, 1}& \text{2}\\ \text{8}& \text{0, 1, 2}& \text{3}\\ \text{9}& \text{0, 1, 2, 3}& \text{4}\\ \text{10}& \text{0, 1, 2, 3, 4}& \text{5}\\ & \text{Total number of cases}& \text{15}\end{array}$
Similarly, we have $15$ cases for $x-y<-5.$
There are $30$ pairs of values of $x$ and $y$ satisfying these two inequalities.
So, favourable number of ways is $30.$
Hence, required probability is $\frac{30}{121}.$