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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 need up with two segments of the same length are if you choose two sides, or two diagonals. Thus, there are <math>2 * \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. Of these, the ways in which you need up with two 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 04:40, 30 October 2020

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. Of these, the ways in which you need up with two 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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