2004 AMC 10A Problems/Problem 5

Problem

A set of three points is randomly chosen from the grid shown. Each three point set has the same probability of being chosen. What is the probability that the points lie on the same straight line?

$\mathrm{(A) \ } \frac{1}{21} \qquad \mathrm{(B) \ } \frac{1}{14} \qquad \mathrm{(C) \ } \frac{2}{21} \qquad \mathrm{(D) \ } \frac{1}{7} \qquad \mathrm{(E) \ } \frac{2}{7}$

Solution

There are $\binom{9}{3}$ ways to choose three points out of the 9 there. There are 8 combinations of dots such that they lie in a straight line: three vertical, three horizontal, and the diagonals.

$\dfrac{8}{\binom{9}{3}}=\dfrac{8}{84}=\dfrac{2}{21} \Rightarrow\boxed{\mathrm{(C)}\ \frac{2}{21}}$

Video Solution

https://youtu.be/jWqX7ruQwr0

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2004 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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All AMC 10 Problems and Solutions

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2004 AMC 10A Problems/Problem 5

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Problem

Solution

Video Solution

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