2007 AMC 10A Problem 16

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

Questions about evenness of an expression reduce to parity of the inputs, and independence lets us treat each product separately. The asymmetry (products are even $\frac34$ of the time) is what pushes the answer above $\frac12$. Reducing a huge sample space to a handful of parity classes is the standard first move for "is it even" probabilities.

Complement: $ad - bc$ is odd exactly when one product is odd and the other even, with probability $2 \cdot \frac14 \cdot \frac34 = \frac38$. So the desired probability is $1 - \frac38 = \frac58$.

• Assuming $ad$ and $bc$ are each odd or even with probability $\frac12$, which gives $\frac12$; a product of two random integers is odd only $\frac14$ of the time.
• Worrying about the exact range $0$ to $2007$; only the fact that it splits evenly between odd and even matters.