Difference between revisions of "2013 AMC 10B Problems/Problem 12"
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<math> \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 </math> | <math> \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 </math> | ||
− | ==Solution== | + | ==Solutions== |
− | In a regular pentagon, there are 5 sides with the same length | + | ===Solution 1=== |
− | the element picked, with 9 total elements remaining. Therefore, the probability is <math>\boxed{\textbf{(B) }\frac{4}{9}}</math> | + | 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=== | ||
+ | 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; | ||
+ | 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. | ||
+ | |||
+ | Summing these cases up gives us a probability of <math>\boxed{\textbf{(B) }\frac{4}{9}}</math>. | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | ===Solution 4=== | ||
+ | |||
+ | The problem is simply asking how many ways are there to choose two sides or two diagonals. Hence, the probability is <cmath>\dfrac{\binom{5}{2} + \binom{5}{2}}{\binom{10}{2}} = \dfrac{10+10}{45} = \dfrac{20}{45}=\boxed{\textbf{(B) }\dfrac49}</cmath>. | ||
+ | |||
+ | ~MrThinker | ||
+ | |||
== 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}} |
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 .
~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 |
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