Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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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> | <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 1== |
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> | ||
Revision as of 09:25, 27 November 2016
Two different numbers are randomly selected from the set 
and multiplied together. What is the probability that the product is 
?
Solution 1
The product can only be if one of the numbers is 0. Once we chose , there are 
ways we can chose the second number, or 
. There are 
ways we can chose 
numbers randomly, and that is 
. So, 
so the answer is 
Solution 2
There are a total of 
possibilities, because the numbers are different. We want to be the product so one of the numbers is . There are possibilities where is chosen for the first number and there are ways for to be chosen as the second number. We seek 
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
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| All AJHSME/AMC 8 Problems and Solutions | ||
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