|
|
| Line 1: |
Line 1: |
| − | ==Problem 14==
| + | #redirect [[2011 AMC 12A Problems/Problem 10]] |
| − | 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?
| |
| − | | |
| − | <math>\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}</math>
| |
| − | | |
| − | == Solution ==
| |
| − | | |
| − | We want the area, <math>\pi r^2</math>, to be less than the circumference, <math>2 \pi r</math>:
| |
| − | | |
| − | <cmath>\begin{align*}
| |
| − | \pi r^2 &< 2 \pi r \\
| |
| − | r &< 2
| |
| − | \end{align*}</cmath>
| |
| − | | |
| − | 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>.
| |
| − | | |
| − | == See Also ==
| |
| − | | |
| − | | |
| − | {{AMC10 box|year=2011|ab=A|num-b=13|num-a=15}}
| |
| − | {{MAA Notice}}
| |
