# Difference between revisions of "2016 AMC 8 Problems/Problem 13"

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 $0$? $\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$

## Solution

The product can only be $0$ if one of the numbers is 0. Once we chose $0$, there are $5$ ways we can chose the second number, or $6-1$. There are $\dbinom{6}{2}$ ways we can chose $2$ numbers randomly, and that is $15$. So, $\frac{5}{15}=\frac{1}{3}$ so the answer is $\boxed{\textbf{(D)} \, \frac{1}{3}}$

## Solution 2

There are a total of $36$ possibilities. We want $0$ so one of the multiples is $0$. There are $6$ possibilities where $0$ is chosen for the first number and there are $6$ ways for $0$ to be chosen as the second number. We seek $\frac {6+6}{36}=\frac {1}{3}$

 2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 12 Followed byProblem 14 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. ## Solution 2

There are a total of $36$ possibilities. We want $0$ so one of the multiples is $0$. There are $6$ possibilities where $0$ is chosen for the first number and there are $6$ ways for $0$ to be chosen as the second number. We seek $\frac {6+6}{36}=\frac {1}{3}$