# 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 (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 $1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.$

### Solution 3 (Casework)

There are two different cases in which the product is zero; either the first number we select is zero, or the second one is. We consider these cases separately.

Case 1: $0$ is the first number chosen

There is a $\frac{1}{6}$ chance of selecting zero as the first number. At this point, the product will be zero no matter the choice of the second number, so there is a $\frac{1}{6}$ of getting the desired product in this case.

Case 2: $0$ is the second number chosen

There is a $\frac{5}{6}$ chance of choosing a number that is NOT zero as the first number. From there, there is a $\frac{1}{5}$ chance of picking zero from the remaining 5 numbers. Thus, there is a $\frac{5}{6} \cdot \frac{1}{5} = \frac{1}{6}$ chance of getting a product of 0 in this case.

Adding the probabilities from the two distinct cases up, we find that there is a $\frac{1}{6} + \frac{1}{6} = \boxed{\textbf{(D)} \ \frac{1}{3}}$ chance of getting a product of zero.

## Video Solution (CREATIVE THINKING!!!)

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## See Also

 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.