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

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It is clear that only a diameter of <math>2</math> and <math>3</math> would result in the circumference being larger than the radius.  
 
It is clear that only a diameter of <math>2</math> and <math>3</math> would result in the circumference being larger than the radius.  
  
For <math>2,</math> the radius is <math>1</math> so <math>2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi </math> The area is <math> \pi r^2 \rightarrow \pi 1^2 \right arrow \pi </math> Thus, <math> 2\pi > \pi </math> so we need snake eyes or <math>2</math> <math>1's</math> and the probability is <math> \dfrac{1}{6} \cdot  \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>
+
For <math>2,</math> the radius is <math>1</math> so <math>2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi </math> The area is <math> \pi r^2 \rightarrow \pi 1^2 \rightarrow \pi </math> Thus, <math> 2\pi > \pi </math> so we need snake eyes or <math>2</math> <math>1's</math> and the probability is <math> \dfrac{1}{6} \cdot  \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>
  
 
By the same work using diameter of <math>3</math>, we find that the circumference is greater than the area. So <math>3</math> would be found by rolling a <math>1</math> and a <math>2</math> so the probability is  <math> \dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>.  
 
By the same work using diameter of <math>3</math>, we find that the circumference is greater than the area. So <math>3</math> would be found by rolling a <math>1</math> and a <math>2</math> so the probability is  <math> \dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36} </math>.  

Revision as of 22:43, 9 February 2011

Problem

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?

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

Solution

It is clear that only a diameter of $2$ and $3$ would result in the circumference being larger than the radius.

For $2,$ the radius is $1$ so $2\pi r \rightarrow 2\pi(1) \rightarrow 2\pi$ The area is $\pi r^2 \rightarrow \pi 1^2 \rightarrow \pi$ Thus, $2\pi > \pi$ so we need snake eyes or $2$ $1's$ and the probability is $\dfrac{1}{6} \cdot  \dfrac{1}{6} \rightarrow \dfrac{1}{36}$

By the same work using diameter of $3$, we find that the circumference is greater than the area. So $3$ would be found by rolling a $1$ and a $2$ so the probability is $\dfrac{1}{6} \cdot \dfrac{1}{6} \rightarrow \dfrac{1}{36}$.

Thus, those are the only two cases however there are $2$ ways to roll a $3$ with $2$ die so the probability is $\dfrac{1}{36} + \dfrac{1}{36} + \dfrac{1}{36} \rightarrow \dfrac{3}{36} \rightarrow \dfrac{1}{12} \rightarrow \boxed{B}$

See also

2011 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 9
Followed by
Problem 11
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 12 Problems and Solutions
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