2011 AMC 10A Problems/Problem 14: Difference between revisions
No edit summary |
|||
| Line 9: | Line 9: | ||
<cmath>\begin{align*} | <cmath>\begin{align*} | ||
\pi r^2 &< 2 \pi r | \pi r^2 &< 2 \pi r \\ | ||
r &< 2 | r &< 2 | ||
\end{align*}</cmath> | \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>3/36=1/12</math> | 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> | ||
Revision as of 23:12, 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?
Solution
We want the area,
, to be less than the circumference,
:
If
then the dice must show
which are
choices out of a total possible of
, so the probability is