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

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==Solution==
 
==Solution==
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The product is zero if and only if zero is one of the numbers that is chosen. 2 numbers are chosen out of 6 total, so the chance that the product is 0 is <math>\frac{2}{6}=\textbf{(D)} \frac{1}{3}</math>.
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Revision as of 10:54, 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

The product is zero if and only if zero is one of the numbers that is chosen. 2 numbers are chosen out of 6 total, so the chance that the product is 0 is $\frac{2}{6}=\textbf{(D)} \frac{1}{3}$.

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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