Two integers xandy are chosen with replacement from the set 0, 1,...

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

So, total number of outcomes are \left(11 \times 11\right) = 121$</p><p>Favourable outcomes are:$

\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 $P\left(E\right)=\dfrac{30}{121}$.