Difference between revisions of "2008 AMC 8 Problems/Problem 19"

Line 13: Line 13:
 
</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 03:49, 25 December 2012

Problem

Eight points are spaced around at intervals of one unit around a $2 \times 2$ square, as shown. Two of the $8$ points are chosen at random. What is the probability that the two points are one unit apart? [asy] size((50)); dot((5,0)); dot((5,5)); dot((0,5)); dot((-5,5)); dot((-5,0)); dot((-5,-5)); dot((0,-5)); dot((5,-5)); [/asy] $\textbf{(A)}\ \frac{1}{4}\qquad\textbf{(B)}\ \frac{2}{7}\qquad\textbf{(C)}\ \frac{4}{11}\qquad\textbf{(D)}\ \frac{1}{2}\qquad\textbf{(E)}\ \frac{4}{7}$

Solution

The two points are one unit apart at $8$ places around the edge of the square. There are $_8 C _2 = 28$ ways to choose two points. The probability is

\[\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}\]

See Also

2008 AMC 8 (ProblemsAnswer KeyResources)
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