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

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13. Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?
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== Problem ==
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Two different numbers are randomly selected from the set <math>{ - 2, -1, 0, 3, 4, 5}</math> and multiplied together. What is the probability that the product is <math>0</math>?
  
 
<math>\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}</math>
 
<math>\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}</math>
  
==Solution==
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== Solutions==
{{solution}}
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===Solution 1===
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The product can only be <math>0</math> if one of the numbers is 0. Once we chose <math>0</math>, there are <math>5</math> ways we can chose the second number, or <math>6-1</math>. There are <math>\dbinom{6}{2}</math> ways we can chose <math>2</math> numbers randomly, and that is <math>15</math>. So, <math>\frac{5}{15}=\frac{1}{3}</math> so the answer is  <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.
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===Solution 2===
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There are a total of <math>30</math> possibilities, because the two numbers that being multiplied are being picked at the same time, so there are <math>5</math> possibilities that zero is being chosen because another number is already being chosen. We want <math>0</math> to be the product so one of the numbers is <math>0</math>. There are <math>5</math> possibilities where <math>0</math> is chosen for the first number and there are <math>5</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\boxed{\textbf{(D)} \, \frac{1}{3}}</math>.
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===Solution 3 (Complementary Counting)===
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Because the only way the product of the two numbers is <math>0</math> is if one of the numbers we choose is <math>0,</math> we calculate the probability of NOT choosing a <math>0.</math> We get <math>\frac{5}{6} \cdot \frac{4}{5} = \frac{2}{3}.</math> Therefore our answer is <math>1 - \frac{2}{3} = \boxed{\textbf{(D)} \ \frac{1}{3}}.</math>
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{{AMC8 box|year=2016|num-b=12|num-a=14}}
 
{{AMC8 box|year=2016|num-b=12|num-a=14}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 03:42, 16 January 2021

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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