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2004 AMC 10A Problem 5

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2004 AMC 10A Problem 5AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 2 of 5 · about 2 min · position P1-10

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?
figure

  1. A)

    121\frac{1}{21}

  2. B)

    114\frac{1}{14}

  3. C)

    221\frac{2}{21}

  4. D)

    17\frac{1}{7}

  5. E)

    27\frac{2}{7}

Answer

C

Key insight

Only 8 of the C(9,3) = 84 triples are collinear: three rows, three columns, and two diagonals.

Solution

The grid has 99 points, so the number of equally likely three-point sets is
(93)=9⋅8⋅76=84.\binom{9}{3} = \frac{9 \cdot 8 \cdot 7}{6} = 84.

A line through three of these points must be a full row, a full column, or a main diagonal, since any other line hits at most two grid points. That gives 3+3+2=83 + 3 + 2 = 8 collinear triples.

The probability is 884=221\dfrac{8}{84} = \dfrac{2}{21}, so the answer is (C) 221\boxed{\textbf{(C)}\ \frac{2}{21}}.

Why this works

With a small, symmetric sample space, the safest method is to count the total with a combination and list the favorable outcomes by type. The "lines of three" on a 3×33 \times 3 board are exactly the eight winning lines of tic-tac-toe, a picture worth remembering.

The trap

Forgetting the two diagonals and counting only 6 lines, which gives 1/14, choice (B).

Common mistakes

  • Forgetting the two diagonals and counting only 6 lines, which gives 1/14, choice (B).
  • Using ordered triples (9⋅8⋅7=5049 \cdot 8 \cdot 7 = 504) for the total while counting unordered lines for the numerator.

Techniques

Organized listing / direct enumeration