Difference between revisions of "2011 AMC 12A Problems/Problem 10"

Revision as of 20:51, 3 July 2013

Problem

A pair of standard $6$ -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?

$\textbf{(A)}\ \frac{1}{36} \qquad \textbf{(B)}\ \frac{1}{12} \qquad \textbf{(C)}\ \frac{1}{6} \qquad \textbf{(D)}\ \frac{1}{4} \qquad \textbf{(E)}\ \frac{5}{18}$

Solution

For the circumference to be greater than the area, we must have $\pi d > \pi \left( \frac{d}{2} \right) ^2$ , or $d<4$ . Now since $d$ is determined by a sum of two dice, the only possibilities for $d$ are thus $2$ and $3$ . In order for two dice to sum to $2$ , they most both show a value of $1$ . The probability of this happening is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$ . In order for two dice to sum to $3$ , one must show a $1$ and the other must show a $2$ . Since this can happen in two ways, the probability of this event occurring is $2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$ . The sum of these two probabilities now gives the final answer: $\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{\textbf{B}}$

2011 AMC 12A (Problems • Answer Key • Resources)
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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 == See also ==
 {{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}
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Difference between revisions of "2011 AMC 12A Problems/Problem 10"

Revision as of 20:51, 3 July 2013

Problem

Solution

See also