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Difference between revisions of "2013 AMC 10B Problems/Problem 12"

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==Solution==
 
==Solution==
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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>
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===Solution 1===
 +
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>.
 +
===Solution 2===
 +
Alternatively, we can divide this problem into two cases.
 +
Case 1: Side
 +
In this case, there is a <math>\frac{5}{10}</math> chance of picking a side, and a <math>\frac{4}{9}</math> chance of picking another side.
 +
Case 2: Diagonal
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This case is similar to the first, for again, there is a <math>\frac{5}{10}</math> chance of picking a diagonal, and a <math>\frac{4}{9}</math> chance of picking another diagonal.
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 +
Summing these cases up gives us a probability of <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 15:49, 31 December 2017

Problem

Let $S$ be the set of sides and diagonals of a regular pentagon. A pair of elements of $S$ are selected at random without replacement. What is the probability that the two chosen segments have the same length?

$\textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5$

Solution

Solution 1

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 $\boxed{\textbf{(B) }\frac{4}{9}}$.

Solution 2

Alternatively, we can divide this problem into two cases. Case 1: Side In this case, there is a $\frac{5}{10}$ chance of picking a side, and a $\frac{4}{9}$ chance of picking another side. Case 2: Diagonal This case is similar to the first, for again, there is a $\frac{5}{10}$ chance of picking a diagonal, and a $\frac{4}{9}$ chance of picking another diagonal.

Summing these cases up gives us a probability of $\boxed{\textbf{(B) }\frac{4}{9}}$.

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

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