Difference between revisions of "2011 AMC 12A Problems/Problem 10"
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== Solution == | == Solution == | ||
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> | 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> | ||
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+ | ==Video Solution== | ||
+ | |||
+ | https://www.youtube.com/watch?v=6tlqpAcmbz4 | ||
+ | ~Shreyas S | ||
== 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}} | ||
+ | {{AMC10 box|year=2011|num-b=13|num-a=15|ab=A}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Latest revision as of 14:51, 5 May 2021
Contents
Problem
A pair of standard -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?
Solution
For the circumference to be greater than the area, we must have , or . Now since is determined by a sum of two dice, the only possibilities for are thus and . In order for two dice to sum to , they most both show a value of . The probability of this happening is . In order for two dice to sum to , one must show a and the other must show a . Since this can happen in two ways, the probability of this event occurring is . The sum of these two probabilities now gives the final answer:
Video Solution
https://www.youtube.com/watch?v=6tlqpAcmbz4 ~Shreyas S
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 |
2011 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.