The product can only be $0$ if one of the numbers is $0$. Once we choose $0$, there are $5$ ways of choosing the second number, and there are 15 ways of choosing $2$ numbers randomly. Thus $\frac{5}{15} = \frac{1}{3}$. The answer is $D$. 
Two different numbers are randomly selected from the set $-2, -1, 0, 3, 4, 5$~and multiplied together. What is the probability that the product is $0$? ($2016$ AMC $8$ Problem, Question \#$13$) 
a. $\frac{1}{6}$ 
b. $\frac{1}{5}$ 
c. $\frac{1}{4}$ 
d. $\frac{1}{3}$ 
e. $\frac{1}{2}$ 