Competition · AMC preparation · step 4 of 4

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 (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 (1/4)⁴ = 1/256 — then scale up while checking overlaps.

1STEP 1

Count all possible tilings

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

Count tilings for one spot

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 the four positions

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

Check for overlap

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

Add and form the probability

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

P = (4 · 4⁵)/4⁹ = 4⁶/4⁹ = 1/4³ = 1/64 → (C)

Answer

1/64

Probability check: a single fixed 2 × 2 diamond requires 4 specific tile choices, each with chance

\frac{1}{4}

, giving (

\frac{1}{4}

)⁴ =

\frac{1}{256}

. With 4 mutually exclusive sub-grid locations, the probability becomes 4 · (

\frac{1}{256}

) =

\frac{4}{256}

\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

\frac{1}{256}

(one location) and

\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!

Count all possible tilings
Count tilings for one spot
List the four positions
Check for overlap
Add and form the probability

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Count all possible tilings

Count tilings for one spot

List the four positions

Check for overlap

Add and form the probability

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