2011 AMC 10B Problem 13

Only the signs of the two numbers matter for whether the product is positive, so first find how likely each sign is.

1. The interval from $-20$ to $10$ is $30$ long.
2. The negative part, from $-20$ to $0$, is $20$ long. So a number is negative with probability $\frac{20}{30} = \frac{2}{3}$.
3. The positive part, from $0$ to $10$, is $10$ long. So a number is positive with probability $\frac{10}{30} = \frac{1}{3}$. (Landing exactly on $0$ has probability $0$, so we can ignore it.)
4. A product is positive when the two numbers have the same sign.
5. Both negative: $\frac{2}{3} \times \frac{2}{3} = \frac{4}{9}$.
6. Both positive: $\frac{1}{3} \times \frac{1}{3} = \frac{1}{9}$.
7. These two cases cannot happen at once, so add them: $\frac{4}{9} + \frac{1}{9} = \frac{5}{9}$.

The answer is $\boxed{\textbf{(D)}\ \frac{5}{9}}$.

For a number picked evenly from an interval, the chance it lands in a piece is the piece's length divided by the whole length. The two picks do not affect each other, so you multiply their chances. Asking "which signs make the product positive?" turns a question about infinitely many numbers into a simple two-outcome question.

Find the chance the product is negative and subtract from $1$. A negative product needs one negative and one positive number, in either order: $\frac{2}{3} \times \frac{1}{3} + \frac{1}{3} \times \frac{2}{3} = \frac{4}{9}$. So the answer is $1 - \frac{4}{9} = \frac{5}{9}$.

The interval $[-20, 10]$ has length $30$. A uniformly chosen number is negative with probability $\frac{20}{30} = \frac{2}{3}$ and positive with probability $\frac{10}{30} = \frac{1}{3}$ (being exactly $0$ has probability $0$).

The product is positive exactly when the two numbers have the same sign. Since the choices are independent,
$$P(\text{both negative}) + P(\text{both positive}) = \left(\frac{2}{3}\right)^2 + \left(\frac{1}{3}\right)^2 = \frac{4}{9} + \frac{1}{9} = \frac{5}{9}.$$

The answer is $\boxed{\textbf{(D)}\ \frac{5}{9}}$.