Difference between revisions of "2013 AMC 10B Problems/Problem 12"
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===Solution 2=== | ===Solution 2=== | ||
| − | + | We can divide this problem into two cases. | |
Case 1: Side; | 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. | 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. | ||
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===Solution 3=== | ===Solution 3=== | ||
| − | + | One way to do this is to use combinations. | |
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>. | 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>. | ||
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===Solution 4=== | ===Solution 4=== | ||
Latest revision as of 10:53, 27 April 2024
Contents
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?
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 
.
Solution 2
We can divide this problem into two cases.
Case 1: Side;
In this case, there is a 
chance of picking a side, and a 
chance of picking another side.
Case 2: Diagonal;
This case is similar to the first, for again, there is a chance of picking a diagonal, and a chance of picking another diagonal.
Summing these cases up gives us a probability of .
Solution 3
One way to do this is to use combinations.
We know that there are 
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 
= 20 ways in which you end up with two segments of the same length. 
is equivalent to .
Solution 4
The problem is simply asking how many ways are there to choose two sides or two diagonals. Hence, the probability is ![\[\dfrac{\binom{5}{2} + \binom{5}{2}}{\binom{10}{2}} = \dfrac{10+10}{45} = \dfrac{20}{45}=\boxed{\textbf{(B) }\dfrac49}\]](http://latex.artofproblemsolving.com/1/a/6/1a6a025604e4c99a96081d5ba05bec723037fa3d.png)
.
~MrThinker
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. 
