2004 AMC 10A Problem 5

Below is the professionally curated solution for Problem 5 of the 2004 AMC 10A, from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2004 AMC 10A solutions, or check the answer key.

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Concepts:basic probability combinations lattice point

Difficulty rating: 1240

A set of three points is chosen randomly 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?

$\dfrac{1}{21}$

$\dfrac{1}{14}$

$\dfrac{2}{21}$

$\dfrac{1}{7}$

$\dfrac{2}{7}$

Solution:

The number of three-point sets is $\binom{9}{3} = 84.$

The collinear triples are the $3$ rows, the $3$ columns, and the $2$ main diagonals, for a total of $8.$

The probability is therefore $\dfrac{8}{84} = \dfrac{2}{21}.$

Thus, the correct answer is C.

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