AMC 8 · 2023 · #23

Grade 7 geometry-2d

probability-basiccombinations-basicspatial-visualization complementary-countingsystematic-enumerationeasier-related-problem ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Medium solution 💡 4 insights 📊 Diagram

Problem

Each square in a $3 \times 3$ grid is randomly filled with one of the $4$ gray and white tiles shown below on the right.

What is the probability that the tiling will contain a large gray diamond in one of the smaller $2 \times 2$ grids? Below is an example of such tiling.

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

Pick an answer.

(A)

$\frac{1}{1024}$

(B)

$\frac{1}{256}$

(C)

$\frac{1}{64}$

(D)

$\frac{1}{16}$

(E)

$\frac{1}{4}$

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Toolkit + CCSS Solution

Understand

Restated: A $3 \times 3$ 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 $2 \times 2$ 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.

Givens: Grid size: $3 \times 3$ (so $9$ individual squares); Each square gets one of $4$ tile orientations, chosen independently and uniformly at random; A "large gray diamond" appears inside a $2 \times 2$ sub-grid when each of its $4$ tiles has its gray triangle pointing toward the shared center corner; The $3 \times 3$ grid contains exactly $4$ possible $2 \times 2$ sub-grid positions; Answer choices: (A) $\tfrac{1}{1024}$, (B) $\tfrac{1}{256}$, (C) $\tfrac{1}{64}$, (D) $\tfrac{1}{16}$, (E) $\tfrac{1}{4}$

Unknowns: The probability that the random tiling contains a large gray diamond in at least one of the four $2 \times 2$ sub-grids

Understand

Plan

Primary tool: #2 Make a Systematic List

Secondary: #7 Identify Subproblems, #9 Solve an Easier Related Problem

We compute probability as (favorable outcomes) $/$ (total outcomes). Tool #2 (Systematic List) gives us a clean accounting: list the four $2 \times 2$ 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 $2 \times 2$ sub-grid?" — it is $(1/4)^4 = 1/256$ — then scale up while checking overlaps.

Execute — Answer: C

#7 Identify Subproblems 6.EE.A.1 Step 1

Count the total number of equally likely tilings.
Each of the $9$ squares is filled independently with $1$ of $4$ tiles, so by the multiplication principle the sample space has $4^9$ tilings.

$$4^9 = 262{,}144$$

💡 Independent choices multiply — $4$ options nine times means $4^9$ total tilings (a Grade 6 exponent expression).

#9 Solve an Easier Related Problem 7.SP.C.8 Step 2

Solve the easier related problem first: how many tilings put a diamond in ONE specific $2 \times 2$ sub-grid (say the top-left)?
The $4$ tiles inside that sub-grid are forced — each must point its gray triangle at the shared center — so there is exactly $1$ way for those $4$ squares.
The other $9 - 4 = 5$ squares are free, giving $4^5$ ways.

$1 \cdot 4^5 = 1024$ favorable tilings per fixed $2 \times 2$ position

💡 Pinning down a specific event in one location and letting the rest vary is the Grade 7 "compound events using organized counting" idea.

#2 Make a Systematic List 7.SP.C.8 Step 3

List the four possible diamond positions systematically — top-left, top-right, bottom-left, bottom-right $2 \times 2$ sub-grids.
By the symmetry argument from Step 2, each one has $4^5 = 1024$ tilings that place a diamond there.

$$4 \text{ positions} \times 4^5 \text{ tilings each (before checking overlaps)}$$

💡 Making the list of all four diamond locations is exactly Tool #2 (Systematic List).

#7 Identify Subproblems 7.SP.C.8 Step 4

Check whether any tiling can put a diamond in two different $2 \times 2$ sub-grids at once.
Take the top-left and top-right grids — they share the middle-top square.
For a top-left diamond, that shared square's gray triangle must point to the bottom-left (toward the center of the top-left $2 \times 2$).
For a top-right diamond, the SAME square's triangle must point to the bottom-right.
One square can hold only one tile, so the two events cannot both happen.
The same conflict appears for every pair of $2 \times 2$ sub-grids (any two of them share at least one square, and the diamond requirement on that shared square disagrees).
Therefore the four diamond events are mutually exclusive.

$$\text{Top-left diamond} \cap \text{Top-right diamond} = \varnothing,\; \text{and similarly for every pair}$$

💡 Spotting that the shared cell demands two different orientations is the Grade 7 "events can be incompatible" check before adding counts.

#7 Identify Subproblems 7.SP.C.8 Step 5

Because the four diamond events are mutually exclusive, the favorable count is just the sum of the four individual counts (no inclusion-exclusion correction is needed).
Then divide by the total to get the probability.

$$P = \dfrac{4 \cdot 4^5}{4^9} = \dfrac{4^6}{4^9} = \dfrac{1}{4^3} = \dfrac{1}{64} \;\Rightarrow\; \textbf{(C)}$$

💡 Adding disjoint cases and dividing by the sample-space size is the Grade 7 compound-event probability formula.

[1] #7 6.EE.A.1 Count the total number of equally likely tilings. Each of the $9$ squares is fil

[2] #9 7.SP.C.8 Solve the easier related problem first: how many tilings put a diamond in ONE sp

[3] #2 7.SP.C.8 List the four possible diamond positions systematically — top-left, top-right, b

[4] #7 7.SP.C.8 Check whether any tiling can put a diamond in two different $2 \times 2$ sub-gri

[5] #7 7.SP.C.8 Because the four diamond events are mutually exclusive, the favorable count is j

Review

Reasonableness: Probability check: a single fixed $2 \times 2$ diamond requires $4$ specific tile choices, each with chance $\tfrac{1}{4}$, giving $(1/4)^4 = 1/256$. With $4$ mutually exclusive sub-grid locations, the probability becomes $4 \cdot (1/256) = 4/256 = 1/64$. This matches choice (C). The answer is small but not tiny — sensible for a 4-out-of-256 type event, and it sits between $1/256$ (one location) and $1/16$ (which would have been the answer if all $4$ locations were independent, which they are not).

Alternative: Tool #16 (Change Focus / Complement) is not useful here because the complement "no diamond anywhere" is much harder to count directly. A cleaner alternative is Tool #5 (Look for a Pattern) combined with the per-cell view: for any specific $2 \times 2$ sub-grid, each of its $4$ cells has probability $\tfrac{1}{4}$ of being the right orientation, and these $4$ cells are independent, so $P = (1/4)^4 = 1/256$. Sum over the $4$ disjoint locations to get $4/256 = 1/64$.

CCSS standards used (min grade 7)

6.EE.A.1 Write and evaluate numerical expressions involving whole-number exponents (Computing the size of the sample space as $4^9$ and the favorable counts as $4^5$ and $4^6$ using whole-number exponents.)
7.SP.C.8 Find probabilities of compound events using organized lists, tables, and simulation (Counting favorable tilings for one fixed $2 \times 2$ diamond location, listing all $4$ locations, checking mutual exclusivity by inspecting shared cells, and combining into the compound-event probability $4 \cdot 4^5 / 4^9 = 1/64$.)

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

Problem

Toolkit + CCSS Solution

Understand

Plan

Execute — Answer: C

Review

CCSS standards used (min grade 7)

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