2016 AMC 8 Problems/Problem 13: Difference between revisions

Revision as of 09:47, 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 $0$ ?

$\textbf{(A) }\dfrac{1}{6}\qquad\textbf{(B) }\dfrac{1}{5}\qquad\textbf{(C) }\dfrac{1}{4}\qquad\textbf{(D) }\dfrac{1}{3}\qquad \textbf{(E) }\dfrac{1}{2}$

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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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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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>(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>
+<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==
+{{solution}}
+{{AMC8 box|year=2016|num-b=12|num-a=14}}
+{{MAA Notice}}

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2016 AMC 8 Problems/Problem 13: Difference between revisions

Revision as of 09:47, 23 November 2016

Solution