Difference between revisions of "2013 AMC 10B Problems/Problem 12"
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==Solution== | ==Solution== | ||
| − | In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as | + | In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is <math>\boxed{\textbf{(B) }\frac{4}{9}}</math> |
| − | the element picked, with 9 total elements remaining. Therefore, the probability is <math>\boxed{\textbf{(B) }\frac{4}{9}}</math> | + | |
== See also == | == See also == | ||
{{AMC10 box|year=2013|ab=B|num-b=11|num-a=13}} | {{AMC10 box|year=2013|ab=B|num-b=11|num-a=13}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 18:12, 30 December 2017
Problem
Let 
be the set of sides and diagonals of a regular pentagon. A pair of elements of are selected at random without replacement. What is the probability that the two chosen segments have the same length?
Solution
In a regular pentagon, there are 5 sides with the same length, and 5 diagonals with the same length. Picking an element at random will leave 4 elements with the same length as the element picked, with 9 total elements remaining. Therefore, the probability is 
See also
| 2013 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 11 |
Followed by Problem 13 | |
| 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. 
