Difference between revisions of "2011 AMC 8 Problems/Problem 18"
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==Solution== | ==Solution== | ||
| + | There are <math>6\cdot6=36</math> ways to roll the two dice, and <math>6</math> of them result in two of the same number. Out of the remaining <math>36-6=30</math> ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are <math>30/2=15</math> ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is | ||
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| + | <cmath>\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}</cmath> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2011|num-b=17|num-a=19}} | {{AMC8 box|year=2011|num-b=17|num-a=19}} | ||
Revision as of 20:45, 25 November 2011
A fair 6-sided die is rolled twice. What is the probability that the first number that comes up is greater than or equal to the second number?
Solution
There are 
ways to roll the two dice, and 
of them result in two of the same number. Out of the remaining 
ways, the number of rolls where the first dice is greater than the second should be the same as the number of rolls where the second dice is greater than the first. In other words, there are 
ways the first roll can be greater than the second. The probability the first number is greater than or equal to the second number is
![\[\frac{15+6}{36}=\frac{21}{36}=\boxed{\textbf{(D)}\ \frac{7}{12}}\]](http://latex.artofproblemsolving.com/7/7/c/77cb6a30fa7c22cae1bb082935df6ba861b53d64.png)
See Also
| 2011 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 19 | |
| 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 | ||
