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

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<cmath>\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}</cmath>
 
<cmath>\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}</cmath>
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==Solution 2==
 
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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=18|num-a=20}}
 
{{AMC8 box|year=2008|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 15:37, 25 July 2022

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 1

The two points are one unit apart at $8$ places around the edge of the square. There are $8 \choose 2$$= 28$ ways to choose two points. The probability is

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

See Also

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by
Problem 18 		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

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