Competition · AMC preparation · step 4 of 4

AMC 8 · 2009 · #10

Grade 7 probability
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 all 64 squares

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

total = 8 × 8 = 64
2STEP 2

Try a smaller board

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

Apply it to the big board

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

Simplify the probability

Make the probability favorable / total, 3664\frac{36}{64}, then divide both by 4 to reduce it to 916\frac{9}{16}.

P = 36/64 = (36 ÷ 4)/(64 ÷ 4) = 9/16 → (D)
Answer
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 3664\frac{36}{64} = 916\frac{9}{16} is a bit more than half, which fits the picture — the interior is clearly larger than the border. Choices (A) 116\frac{1}{16} and (E) 4964\frac{49}{64} are wildly off, and (C) 12\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.

  • Count all 64 squares
  • Try a smaller board
  • Apply it to the big board
  • Simplify the probability

A parent dashboard for the family lives at sensimlab.com.