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>?
 
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>?
  
<math>(A)</math>  <math>\frac{1}{6}</math>          <math>(B)</math> <math>\frac{1}{5}</math>          <math>(C)</math> <math>\frac{1}{4}</math>            <math>(D)</math> <math>\frac{1}{3}</math>          <math>(E)</math> <math>\frac{1}{2}</math>
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<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>
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==Solution==
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{{solution}}
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{{AMC8 box|year=2016|num-b=12|num-a=14}}
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{{MAA Notice}}

Revision as of 10:47, 23 November 2016

13. 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

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2016 AMC 8 (ProblemsAnswer KeyResources)
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

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