Let $0 \leq a, b, c, d \leq 2005$ be integers. What is the probability that $a b+c d$ is an even number?
Solution. The probability that an integer falling between 0 and 2005 is even is $\frac{1003}{2006}=\frac{1}{2}$. The probability that it is odd is the same. The product of two such numbers can only be odd if both are odd, which has a probability of $\frac{1}{2} \cdot \frac{1}{2}=\frac{1}{4}$. In all other cases, the product is even, which therefore has a probability of $1-\frac{1}{4}=\frac{3}{4}$.

The probability that $a b+c d$ is even can be satisfied in two ways:

1) Both $a b$ and $c d$ are even, which has a probability of $\frac{3}{4} \cdot \frac{3}{4}=\frac{9}{16}$;
2) Both $a b$ and $c d$ are odd, which has a probability of $\frac{1}{4} \cdot \frac{1}{4}=\frac{1}{16}$.

Thus, the desired probability is $\frac{9}{16}+\frac{1}{16}=\frac{10}{16}=0.625$.