Difference between revisions of "2016 AMC 8 Problems/Problem 13"
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| + | ==Solution 2== | ||
| + | There are a total of <math>36</math> possibilities. We want <math>0</math> so one of the multiples is <math>0</math>. There are <math>6</math> possibilities where <math>0</math> is chosen for the first number and there are <math>6</math> ways for <math>0</math> to be chosen as the second number. We seek <math>\frac {6+6}{36}=\frac {1}{3}</math> | ||
Revision as of 16:18, 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, 
so the answer is 
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
Solution 2
There are a total of 
possibilities. We want so one of the multiples 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 
