AMC 10 · 2004 · #5

Grade 7 probability

Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position

probability-basiccombinations-basic systematic-enumerationidentify-subproblems ↑ Prerequisites: fraction-arithmetic

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

The figure is a 3×3 square grid of 9 dots. Three of the dots are chosen at random, and every set of three dots is equally likely. Find the probability that the three chosen dots all lie on one straight line.

Pick an answer.

(A)

$\frac{1}{21}$

(B)

$\frac{1}{14}$

(C)

$\frac{2}{21}$

(D)

$\frac{1}{7}$

(E)

$\frac{2}{7}$

AMC 10 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

Because every three-dot set is equally likely, the probability is just (collinear sets) ÷ (all sets), so the problem splits into two separate counts (Tool #7): a denominator (how many ways to pick 3 dots from 9) and a numerator (how many of those picks are straight lines). The denominator is a plain "choose 3 from 9" count. The numerator is where care is needed, so a systematic list (Tool #2) sweeps the grid by direction — rows, then columns, then diagonals — with a quick diagram (Tool #1) so no line is missed or double-counted.

1STEP 1

Set up probability as a fraction

Every three-dot set is equally likely, so P = (line sets) ÷ (all sets) — two separate counts: the bottom first, then the top.

P = (collinear sets)/(all sets)

2STEP 2

Count all three-dot sets

Order does not matter: 9×8×7=504 ordered picks, and each trio is counted 6 times, so 504÷6 gives 84 sets.

(9×8×7)/(3×2×1)=504/6=84

3STEP 3

List the straight lines through three dots

Sweep by direction: 3 rows, 3 columns, and only the 2 corner-to-corner diagonals hold 3 dots — 8 line sets in all.

3 rows+3 columns+2 diagonals=8

4STEP 4

Divide and simplify

The probability is 8/84; top and bottom share the factor 4, so it reduces to 2/21 — choice (C).

8/84=(8÷4)/(84÷4)=2/21 → (C)

Answer

2/21

The probability 2/21 is about 0.10, a small chance — which fits the picture, since three dots dropped randomly rarely land in a perfect line. The counting also explains the trap answers: forgetting the two diagonals leaves 6 lines and gives 6/84=1/14, which is the tempting choice (B). Only when the diagonals are included does the count reach 8 and the probability become 2/21, confirming (C).

💡Key takeaway

For a probability with equally likely picks, count the winners over the total — and count straight lines by sweeping rows, then columns, then diagonals so you never miss one.

Set up probability as a fraction
Count all three-dot sets
List the straight lines through three dots
Divide and simplify

Set up probability as a fraction

Count all three-dot sets

List the straight lines through three dots

Divide and simplify

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