Difference between revisions of "2008 AMC 8 Problems/Problem 19"
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</asy> | </asy> | ||
<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== | ||
+ | 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> | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2008|num-b=21|num-a=20}} | {{AMC8 box|year=2008|num-b=21|num-a=20}} |
Revision as of 02:49, 25 December 2012
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
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
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 |