2016 AMC 8 Problems/Problem 13: Difference between revisions
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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> | 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}=\textbf{(D)}</math> | ||
{{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 11:40, 23 November 2016
Two different numbers are randomly selected from the set 
and multiplied together. What is the probability that the product is 
?
Solution
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, 
| 2016 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
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