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

Revision as of 08:53, 17 August 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

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

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 $\boxed{\textbf{(B)}\ \frac27}$

Video Solution

https://www.youtube.com/watch?v=ZVD_PZYjonQ

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 21: / Line 21: @@
 ==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>
+==Video Solution==
+https://www.youtube.com/watch?v=ZVD_PZYjonQ
 ==See Also==
 {{AMC8 box|year=2008|num-b=18|num-a=20}}
 {{MAA Notice}}

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