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

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===Solution 3===
 
===Solution 3===
 
Another way to do this is to use combinations.
 
Another way to do this is to use combinations.
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We know that there are <math>\binom{10}{2} = 45</math> ways to select two segments. Of these, the ways in which you get 2 segments of the same length are if you choose two sides, or two diagonals. Thus, there are <math>2 \times \binom{5}{2}</math> = 20 ways in which you end up with two segments of the same length. <math>\frac{20}{45}</math> is equivalent to <math>\boxed{\textbf{(B) }\frac{4}{9}}</math>.
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We know that there are <math>\binom{10}{2} = 45</math> ways to select two segments. The ways in which you get 2 segments of the same length are if you choose two sides, or two diagonals. Thus, there are <math>2 \times \binom{5}{2}</math> = 20 ways in which you end up with two segments of the same length. <math>\frac{20}{45}</math> is equivalent to <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 18:20, 11 August 2021

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?

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

Solutions

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

Solution 3

Another way to do this is to use combinations. We know that there are $\binom{10}{2} = 45$ ways to select two segments. The ways in which you get 2 segments of the same length are if you choose two sides, or two diagonals. Thus, there are $2 \times \binom{5}{2}$ = 20 ways in which you end up with two segments of the same length. $\frac{20}{45}$ is equivalent to $\boxed{\textbf{(B) }\frac{4}{9}}$.

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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