AMC 8 · 2009 · #10

Grade 7 probability

Archetype Overlap / Complement Counting └ Sub-archetype At-Least-One Complement: 1 − P(None)

probability-basicarea-rectanglesfraction-arithmetic identify-subproblemseasier-related-problem ↑ Prerequisites: area-rectanglesfraction-arithmetic

📏 Short solution 💡 3 insights 📊 Diagram

Problem

An 8 × 8 checkerboard has 64 unit squares. One square is chosen at random. What is the probability that the chosen square does not touch the outer edge of the board?

Pick an answer.

(A)

$\frac{1}{16}$

(B)

$\frac{7}{16}$

(C)

$\frac{1}2$

(D)

$\frac{9}{16}$

(E)

$\frac{49}{64}$

AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The board is right there to look at, so Tool #1 (Draw a Diagram) is the natural way in: outline the squares that touch the outer edge — they form a one-square-thick "frame" around the board — and what's left in the middle is what we want. To make the framing rule crisp before counting on the 8 × 8 board, Tool #9 (Solve an Easier Problem) on a small 4 × 4 board shows the pattern: removing the border ring of a 4 × 4 leaves a 2 × 2 interior, i.e. an (n-2) × (n-2) inside an n × n. Then we apply that same picture to n = 8 and divide.

1STEP 1

Count the whole board first: 8 rows of 8 make 8 × 8 = 64 equally likely squares.

total = 8 × 8 = 64

2STEP 2

Test a smaller 4 × 4 board: peel its edge and a 2 × 2 center is left — so the interior is (n-2) × (n-2).

(4 - 2) × (4 - 2) = 2 × 2 = 4

3STEP 3

Same picture on the 8 × 8: the squares off the edge fill a (8-2) × (8-2) = 6 × 6 block, or 36 squares.

(8 - 2) × (8 - 2) = 6 × 6 = 36

4STEP 4

Make the probability favorable / total, $\frac{36}{64}$ , then divide both by 4 to reduce it to $\frac{9}{16}$ .

P =

\frac{36}{64}

= (36 ÷ 4)/(64 ÷ 4) =

\frac{9}{16}

→ (D)

Answer

\frac{9}{16}

Sanity-check by counting the border the other way: the outer ring of an 8 × 8 board has 4 × 8 - 4 = 28 squares (four sides of 8, minus the four corners counted twice). Interior = 64 - 28 = 36, matching the 6 × 6 count. The probability

\frac{36}{64}

\frac{9}{16}

is a bit more than half, which fits the picture — the interior is clearly larger than the border. Choices (A)

\frac{1}{16}

and (E)

\frac{49}{64}

are wildly off, and (C)

\frac{1}{2}

would mean border and interior were equal, which the diagram shows they aren't.

💡Key takeaway

This AMC 8 problem only needs Grade 7 probability — count favorable squares, divide by total, and reduce the fraction.

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