2007 AMC 10A Problem 16

Whether $ad - bc$ is even depends only on which numbers are odd and which are even. From $0$ to $2007$ there are $2008$ numbers, exactly half odd and half even. So each of $a, b, c, d$ is odd with chance $\frac12$.

1. When is $ad$ odd? Only when $a$ and $d$ are both odd. Chance: $\frac12 \times \frac12 = \frac14$. So $ad$ is even with chance $\frac34$.
2. Same for $bc$: odd with chance $\frac14$, even with chance $\frac34$. It uses different letters from $ad$, so the two products do not affect each other.
3. When is $ad - bc$ even? Look at the four cases:

So it is even exactly when the two products are both odd or both even.
4. Add those two chances.
$$\frac14 \times \frac14 + \frac34 \times \frac34 = \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac58.$$

The answer is $\boxed{\textbf{(E)}\ \frac 58}$.

For questions about "even" or "odd", replace every number by just its type, odd or even. A huge range of numbers then shrinks to a few cases. The surprise here is that products are even much more often than odd, which pushes the answer above one half.

Find the chance that $ad - bc$ is odd instead. That needs one product odd and the other even: $\frac14 \times \frac34 + \frac34 \times \frac14 = \frac{6}{16} = \frac38$. So it is even with chance $1 - \frac38 = \frac58$.

Only parity matters. The range $0$ to $2007$ contains $2008$ integers, half odd and half even, so each of $a, b, c, d$ is odd with probability $\frac12$, independently.

A product is odd only when both factors are odd, so
$$P(ad \text{ odd}) = \frac12 \cdot \frac12 = \frac14, \qquad P(ad \text{ even}) = \frac34,$$
and the same for $bc$. Since $ad$ and $bc$ involve disjoint sets of variables, they are independent.

$ad - bc$ is even exactly when $ad$ and $bc$ have the same parity:
$$P = P(\text{both odd}) + P(\text{both even}) = \left(\frac14\right)^2 + \left(\frac34\right)^2 = \frac{1}{16} + \frac{9}{16} = \frac{10}{16} = \frac58.$$

The answer is $\boxed{\textbf{(E)}\ \frac 58}$.