2016 AMC 8 Problems/Problem 13

Problem

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 $0$ ?

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

Solutions

Solution 1

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 $30$ possibilities, because the two numbers that being multiplied are being picked at the same time, so there are $5$ possibilities that zero is being chosen because another number is already being chosen. We want $0$ to be the product so one of the numbers is $0$ . There are $5$ possibilities where $0$ is chosen for the first number and there are $5$ ways for $0$ to be chosen as the second number. We seek $\boxed{\textbf{(D)} \, \frac{1}{3}}$ .

Solution 3 (Complementary Counting)

Because the only way the product of the two numbers is $0$ is if one of the numbers we choose is $0,$ we calculate the probability of NOT choosing a $0.$ We get $\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.$ Therefore our answer is $1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.$

2016 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 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.

Page

Toolbox

Search