2008 AMC 8 Problems/Problem 19: Difference between revisions
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<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=18|num-a=20}} | {{AMC8 box|year=2008|num-b=18|num-a=20}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 17:26, 12 August 2022
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?
![[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]](http://latex.artofproblemsolving.com/7/b/4/7b4e4d5204c427f5ab2b242c550558317658b12d.png)
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
![\[\frac{8}{28} = \boxed{\textbf{(B)}\ \frac27}\]](http://latex.artofproblemsolving.com/2/d/8/2d8bb743056c3f8a324def77999d5c702eb2113a.png)
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 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 | ||
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
