Difference between revisions of "2013 AMC 10B Problems/Problem 12"
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===Solution 2=== | ===Solution 2=== | ||
Alternatively, we can divide this problem into two cases. | Alternatively, 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. | ||
− | Case 2: Diagonal | + | 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. | 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. | ||
Revision as of 16:50, 31 December 2017
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?
Solution
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
Alternatively, 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 .
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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