2011 AMC 10A Problems/Problem 14: Difference between revisions
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<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> | <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>3/36=1/12</math> | |||
Revision as of 14:45, 14 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 
