2013 AMC 10B Problems/Problem 12
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 | |
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All AMC 10 Problems and Solutions |
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