2011 AMC 10B Problem 13

The product is positive when both numbers have the same sign

The interval has $20$ negative units and $10$ positive units

A selected number is negative with probability $\frac{20}{30}=\frac{2}{3}$ and positive with probability $\frac{10}{30}=\frac{1}{3}.$ (Selecting exactly $0$ has probability $0.$) The product is positive exactly when both numbers have the same sign, so the probability is $$\left(\dfrac23\right)^2+\left(\dfrac13\right)^2=\dfrac49+\dfrac19=\dfrac59.$$

Thus, the correct answer is D .