Difference between revisions of "2008 AMC 8 Problems/Problem 19"
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<math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7} </math> | <math> \textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7} </math> | ||
− | ==Solution== | + | ==Solution 1== |
The two points are one unit apart at <math>8</math> places around the edge of the square. There are <math>_8 C _2 = 28</math> ways to choose two points. The probability is | The two points are one unit apart at <math>8</math> places around the edge of the square. There are <math>_8 C _2 = 28</math> ways to choose two points. The probability is | ||
<cmath>\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}</cmath> | <cmath>\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}</cmath> | ||
+ | |||
+ | ===Solution 2=== | ||
+ | Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is <math>\boxed{\textbf{(B)}\ \frac27}</math> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=21|num-a=20}} | {{AMC8 box|year=2008|num-b=21|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 19:01, 17 June 2017
Contents
Problem
Eight points are spaced around at intervals of one unit around a square, as shown. Two of the points are chosen at random. What is the probability that the two points are one unit apart?
Solution 1
The two points are one unit apart at places around the edge of the square. There are ways to choose two points. The probability is
Solution 2
Arbitrarily pick a point in the grid. Clearly, we see two options for the other point to be placed, so the answer is
See Also
2008 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 21 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.