8. If $x \geq 0, y \geq 0$ are integers, randomly chosen with the constraint $x+y \leq 10$, what is the probability that $x+y$ is even?
Solution: $6 / 11$
For each $p \leq 10$, if $x+y=p, x$ can range from 0 to $p$, yielding $p+1$ ordered pairs $(x, y)$. Thus there are a total of $1+2+3+\cdots+11$ allowable ordered pairs $(x, y)$, but $1+3+5+\cdots+11$ of these pairs have an even sum. So the desired probability is
$$
\frac{1+3+5+\cdots+11}{1+2+3+\cdots+11}=\frac{6^{2}}{11 \cdot 6}=\frac{6}{11} .
$$