Mathematics - Probability Question with Solution | TestHub

Given set is $\left\{0,1,2,3,...,10\right\}$.

So, total number of outcomes are $\left(11\times 11\right)=121$

Favourable outcomes are:

$\left(0,6\right),\left(0,7\right),\left(0,8\right),\left(0,9\right),\left(0,10\right),\left(1,7\right),\left(1,8\right),\left(1,9\right),\left(1,10\right),\left(2,8\right),\left(2,9\right),\left(2,10\right),\left(3,9\right),\left(3,10\right),\left(4,10\right)$ 

And the same number of elements when $x \text{and} y$ are interchanged.

So, the number of favourable outcomes are $30$.

So, the required probability is given by $P\left(E\right)=\frac{30}{121}$.