2013 AMC 10B Problems/Problem 12

Revision as of 16:49, 31 December 2017 by Pulusona (talk | contribs) (Solution)

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?

$\textbf{(A) }\frac{2}5\qquad\textbf{(B) }\frac{4}9\qquad\textbf{(C) }\frac{1}2\qquad\textbf{(D) }\frac{5}9\qquad\textbf{(E) }\frac{4}5$

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 $\boxed{\textbf{(B) }\frac{4}{9}}$.

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 $\boxed{\textbf{(B) }\frac{4}{9}}$.

See also

2013 AMC 10B (ProblemsAnswer KeyResources)
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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