Difference between revisions of "2011 AMC 12A Problems/Problem 10"
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== Problem == | == Problem == | ||
− | A pair of standard 6-sided | + | A pair of standard <math>6</math>-sided 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? |
− | <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> | + | <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> | ||
== Solution == | == Solution == |
Revision as of 02:33, 10 February 2011
Problem
A pair of standard -sided 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 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 |