# 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

1. Identify the total number of ways to select two different numbers from the set:

The set has 6 elements. The number of ways to choose 2 different numbers from 6 is given by the combination formula: $\binom{6}{2} = \frac{6 \times 5}{2 \times 1} = 15$.

2. Identify the favorable outcomes:

For the product to be zero, one of the chosen numbers must be zero. The set contains one zero (0). To have a product of zero, we need to choose 0 and any other number from the remaining five numbers $-2, -1, 3, 4, 5$.

The number of ways to choose 0 and one other number from the remaining five is 5.

3. Calculate the probability:

The probability is the number of favorable outcomes divided by the total number of outcomes: $\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of outcomes}} = \frac{5}{15} = \frac{1}{3}$.

Thus, the probability that the product is $0$ is $\boxed{\textbf{(D)} \ \frac{1}{3}}.$

~GeometryMystery

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

## Video Solution (CREATIVE THINKING!!!)

~Education, the Study of Everything

~ pi_is_3.14

~savannahsolver