2004 AMC 10A Problem 5

Attempt Problem 5 of the 2004 AMC 10A below, then check your answer against the professionally curated solution 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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5.

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?

121\dfrac{1}{21}

114\dfrac{1}{14}

221\dfrac{2}{21}

17\dfrac{1}{7}

27\dfrac{2}{7}

Answer: C
Concepts:basic probabilitycombinationslattice point
Difficulty rating: 1240
Solution:

The number of three-point sets is (93)=84. \binom{9}{3} = 84.

The collinear triples are the 33 rows, the 33 columns, and the 22 main diagonals, for a total of 8.8.

The probability is therefore 884=221. \dfrac{8}{84} = \dfrac{2}{21}.

Thus, the correct answer is C.

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