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 by Problem 12	Followed by Problem 14
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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 $\frac {6+6}{36}=\frac {1}{3}$

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2016 AMC 8 Problems/Problem 13

Solution

Solution 2

Solution 2