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Difference between revisions of "2011 AMC 12A Problems/Problem 10"
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== Solution ==
 
== Solution ==
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For the circumference to be greater than the area, we must have <math>\pi d > \pi (\frac{d}{2})^2</math>, or <math>d<4</math>. Now since <math>d</math> is determined by a sum of two dice, the only possibilities for <math>d</math> are thus <math>2</math> and <math>3</math>. In order for two dice to sum to <math>2</math>, they most both show a value of <math>1</math>. The probability of this happening is <math>\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}</math>. In order for two dice to sum to <math>3</math>, one must show a <math>1</math> and the other must show a <math>2</math>. Since this can happen in two ways, the probability of this event occurring is <math>2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}</math>. The sum of these two probabilities now gives the final answer: <math>\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{textbf{B}}</math>
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For the circumference to be greater than the area, we must have <math>\pi d > \pi \left( \frac{d}{2} \right) ^2</math>, or <math>d<4</math>. Now since <math>d</math> is determined by a sum of two dice, the only possibilities for <math>d</math> are thus <math>2</math> and <math>3</math>. In order for two dice to sum to <math>2</math>, they most both show a value of <math>1</math>. The probability of this happening is <math>\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}</math>. In order for two dice to sum to <math>3</math>, one must show a <math>1</math> and the other must show a <math>2</math>. Since this can happen in two ways, the probability of this event occurring is <math>2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}</math>. The sum of these two probabilities now gives the final answer: <math>\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{\textbf{B}}</math>
  
 
== See also ==
 
== See also ==
 
{{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}
 
{{AMC12 box|year=2011|num-b=9|num-a=11|ab=A}}

Revision as of 20:24, 10 February 2011

Problem

A pair of standard $6$-sided 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

For the circumference to be greater than the area, we must have $\pi d > \pi \left( \frac{d}{2} \right) ^2$, or $d<4$. Now since $d$ is determined by a sum of two dice, the only possibilities for $d$ are thus $2$ and $3$. In order for two dice to sum to $2$, they most both show a value of $1$. The probability of this happening is $\frac{1}{6} \times \frac{1}{6} = \frac{1}{36}$. In order for two dice to sum to $3$, one must show a $1$ and the other must show a $2$. Since this can happen in two ways, the probability of this event occurring is $2 \times \frac{1}{6} \times \frac{1}{6} = \frac{2}{36}$. The sum of these two probabilities now gives the final answer: $\frac{1}{36} + \frac{2}{36} = \frac{3}{36} = \frac{1}{12} \rightarrow \boxed{\textbf{B}}$

See also

2011 AMC 12A (Problems • Answer Key • Resources)
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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