AMC 8 · 2023 · #23

Grade 7 geometry-2d
Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)
probability-basiccombinations-basicspatial-visualization complementary-countingsystematic-enumerationeasier-related-problem ↑ Prerequisites: probability-basicfraction-arithmetic
📏 Medium solution 💡 4 insights 📊 Diagram
Problem
A 3 × 3 grid is tiled by independently placing one of 4 gray-and-white triangular tiles in each of the 9 squares (each tile equally likely). We need the probability that somewhere inside the grid — in one of the four 2 × 2 sub-grids (top-left, top-right, bottom-left, bottom-right) — the four gray triangles meet at the shared corner to form a large gray diamond.

Pick an answer.

(A)
$\frac{1}{1024}$
(B)
$\frac{1}{256}$
(C)
$\frac{1}{64}$
(D)
$\frac{1}{16}$
(E)
$\frac{1}{4}$

AMC 8 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Make a Systematic List

We compute probability as favorableoutcomestotaloutcomes\frac{favorable outcomes}{total outcomes}. Tool #2 (Systematic List) gives us a clean accounting: list the four 2 × 2 sub-grid positions and count favorable tilings for each. Tool #7 (Identify Subproblems) splits the question into three independent pieces — total tilings, favorable tilings at one fixed position, and whether the four "diamond events" overlap. Tool #9 (Easier Related Problem) is the key insight: first solve the easier question "what is the probability of a diamond in ONE specific 2 × 2 sub-grid?" — it is (14\frac{1}{4})⁴ = 1256\frac{1}{256} — then scale up while checking overlaps.

1STEP 1

Count every equally likely tiling: 9 squares, each independently 1 of 4 tiles, so the sample space is 4⁹ by the multiplication principle.

4⁹ = 262,144
2STEP 2

Easier sub-problem first: a diamond in ONE fixed 2 × 2 forces its 4 tiles (1 way); the other 5 squares are free, giving 4⁵.

1 · 4⁵ = 1024 favorable tilings per fixed 2 × 2 position
3STEP 3

List all four diamond spots — top-left, top-right, bottom-left, bottom-right; by symmetry each has 4⁵ tilings placing a diamond there.

4 positions × 4⁵ tilings each (before checking overlaps)
4STEP 4

Any two 2 × 2 sub-grids share a square whose diamond orientation would conflict, so the four events are mutually exclusive.

Top-left diamond ∩ Top-right diamond = ∅, and similarly for every pair
5STEP 5

Disjoint events just add: 4·4⁵ = 4⁶ favorable, so P = 4649\frac{4⁶}{4⁹} = 164\frac{1}{64} → choice (C).

P = 4⋅4549\frac{4 · 4⁵}{4⁹} = 4649\frac{4⁶}{4⁹} = 143\frac{1}{4³} = 164\frac{1}{64} → (C)
Answer
164\frac{1}{64}
Probability check: a single fixed 2 × 2 diamond requires 4 specific tile choices, each with chance 14\frac{1}{4}, giving (14\frac{1}{4})⁴ = 1256\frac{1}{256}. With 4 mutually exclusive sub-grid locations, the probability becomes 4 · (1256\frac{1}{256}) = 4256\frac{4}{256} = 164\frac{1}{64}. This matches choice (C). The answer is small but not tiny — sensible for a 4-out-of-256 type event, and it sits between 1256\frac{1}{256} (one location) and 116\frac{1}{16} (which would have been the answer if all 4 locations were independent, which they are not).
💡Key takeaway

This AMC 8 problem only needs Grade 7 compound-event probability — count favorable tile arrangements, divide by the total — that you already know!

More like this

Same archetype — closest grade level first.