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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>3/36=1/12</math>
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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>.

Revision as of 23:13, 15 February 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?

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

Solution

We want the area, $\pi r^2$, to be less than the circumference, $2 \pi r$:

\begin{align*} \pi r^2 &< 2 \pi r \\ r &< 2 \end{align*}

If $r<2$ then the dice must show $(1,1),(1,2),(2,1)$ which are $3$ choices out of a total possible of $6 \times 6 =36$, so the probability is $\frac{3}{36}=\boxed{\frac{1}{12} \ \mathbf{(B)}}$.

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