Difference between revisions of "2011 AMC 10A Problems/Problem 14"
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If <math>r<2</math> then the dice must show <math>(1,1),(1,2),(2,1)</math> which are <math>3</math> choices out of a total possible of <math>6 \times 6 =36</math>, so the probability is <math>\frac{3}{36}=\boxed{\frac{1}{12} \ \mathbf{(B)}}</math>. | If <math>r<2</math> then the dice must show <math>(1,1),(1,2),(2,1)</math> which are <math>3</math> choices out of a total possible of <math>6 \times 6 =36</math>, so the probability is <math>\frac{3}{36}=\boxed{\frac{1}{12} \ \mathbf{(B)}}</math>. | ||
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| + | == See Also == | ||
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| + | {{AMC10 box|year=2011|ab=A|num-b=13|num-a=15}} | ||
Revision as of 10:49, 8 May 2011
Problem 14
A pair of standard 6-sided fair dice is rolled once. The sum of the numbers rolled determines the diameter of a circle. What is the probability that the numerical value of the area of the circle is less than the numerical value of the circle's circumference?
Solution
We want the area, 
, to be less than the circumference, 
:
If 
then the dice must show 
which are 
choices out of a total possible of 
, so the probability is 
.
See Also
| 2011 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 13 |
Followed by Problem 15 | |
| 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 AMC 10 Problems and Solutions | ||
