Difference between revisions of "2012 UNCO Math Contest II Problems/Problem 2"
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== Solution == | == Solution == | ||
| − | + | The total possibilities when we roll 4 fair 6-sided dice is <math>6\cdot 6\cdot 6\cdot 6=1296.</math> We see that the only possible way for these dice to have a sum of <math>5</math> is to have <math>3</math> dice with ones on it and one dice with a <math>2</math> on it. This can be done in <math>4</math> ways. Thus, the probability is <cmath>\frac{4}{1296}=\boxed{\frac{1}{324}}</cmath> | |
| + | ~SharpApricot123 | ||
== See Also == | == See Also == | ||
| − | {{ | + | {{UNCO Math Contest box|n=II|year=2012|num-b=1|num-a=3}} |
[[Category:Introductory Probability Problems]] | [[Category:Introductory Probability Problems]] | ||
Latest revision as of 16:25, 18 November 2020
Problem
Four ordinary, six-sided, fair dice are tossed. What is the probability that the sum of the
numbers on top is 
?
Solution
The total possibilities when we roll 4 fair 6-sided dice is 
We see that the only possible way for these dice to have a sum of is to have 
dice with ones on it and one dice with a 
on it. This can be done in 
ways. Thus, the probability is ![\[\frac{4}{1296}=\boxed{\frac{1}{324}}\]](http://latex.artofproblemsolving.com/e/d/f/edf39bd87ec30fbe73f06f503977a53727459ab2.png)
~SharpApricot123
See Also
| 2012 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
| All UNCO Math Contest Problems and Solutions | ||
