two integers are chosen at random between $0$ and $10$ what is the probability that they differ by no more than $5$?
There are $\overbrace{6+7+\dots+10+11}^{\frac{17}2\cdot6}+\overbrace{10+\dots+7+6}^{\frac{16}2\cdot5}=91$ ordered pairs that differ by at most $5$ out of $11^2$ ordered pairs. That gives a probability of $$ \frac{91}{121} $$
Clarification: I read "differ by no more than $5$" to mean $\left|x-y\right|\le5$. Then for $0\le x\le5$, there are $x+6$ choices for $y$ and for $6\le x\le10$, there are $16-x$ choices for $y$.
Why The Discrete Probability Might be Greater than the Continuous
In the image above, the red squares hang down and to the right of the associated points. Thus, the red area, $91$, divided by $121$ represents the discrete probability. The area inside the black hexagon, $75$, divided by $100$ represents the continuous probability.
The red area is definitely greater than the area of the hexagon, but it is being divided by a larger number, so it is hard to tell which will be greater.