Difference between revisions of "2011 AMC 12A Problems/Problem 10"
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== Problem == | == Problem == | ||
+ | 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? | ||
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+ | <math> \textbf{(A)}\ \frac{1}{36}\qquad\textbf{(B)}\ \frac{1}{12}\qquad\textbf{(C)}\ \frac{1}{6}\qquad\textbf{(D)}\ \frac{1}{4}\qquad\textbf{(E)}\ \frac{5}{18} </math> | ||
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== Solution == | == Solution == | ||
Revision as of 22:42, 9 February 2011
Problem
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
It is clear that only a diameter of and would result in the circumference being larger than the radius.
For the radius is so The area is $\pi r^2 \rightarrow \pi 1^2 \right arrow \pi$ (Error compiling LaTeX. ! Missing delimiter (. inserted).) Thus, so we need snake eyes or and the probability is
By the same work using diameter of , we find that the circumference is greater than the area. So would be found by rolling a and a so the probability is .
Thus, those are the only two cases however there are ways to roll a with die so the probability is
See also
2011 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 9 |
Followed by Problem 11 |
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 12 Problems and Solutions |