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

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==Solution==
 
==Solution==
 
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>
 
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>36</math> possibilities. We want <math>0</math> so one of the multiples is <math>0</math>. There are <math>6</math> possibilities where <math>0</math> is chosen for the first number and there are <math>6</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\frac {6+6}{36}=\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 16:19, 23 November 2016

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
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. AMC logo.png

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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