2008 AMC 8 Problems/Problem 19

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 $_8 C _2 = 28$ 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 $\boxed{\textbf{(B)}\ \frac27}$

2008 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 21	Followed by Problem 20
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All AJHSME/AMC 8 Problems and Solutions

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2008 AMC 8 Problems/Problem 19

Contents

Problem

Solution 1

Solution 2

See Also