AMC 8 · 2009 · #10
Grade 7 probabilityProblem
On a checkerboard composed of 64 unit squares, what is the probability that a randomly chosen unit square does not touch the outer edge of the board?
Pick an answer.
AMC 8 2009 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: An $8 \times 8$ 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?
Givens: The board is an $8 \times 8$ grid of unit squares, $64$ in total; A square "touches the outer edge" if it sits on the top row, bottom row, leftmost column, or rightmost column; Every square is equally likely to be chosen; Answer choices: (A) $\tfrac{1}{16}$, (B) $\tfrac{7}{16}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{9}{16}$, (E) $\tfrac{49}{64}$
Unknowns: The probability that the chosen square does not touch the outer edge
Understand
Restated: An $8 \times 8$ 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?
Givens: The board is an $8 \times 8$ grid of unit squares, $64$ in total; A square "touches the outer edge" if it sits on the top row, bottom row, leftmost column, or rightmost column; Every square is equally likely to be chosen; Answer choices: (A) $\tfrac{1}{16}$, (B) $\tfrac{7}{16}$, (C) $\tfrac{1}{2}$, (D) $\tfrac{9}{16}$, (E) $\tfrac{49}{64}$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #9 Solve an Easier Problem
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 $8 \times 8$ board, Tool #9 (Solve an Easier Problem) on a small $4 \times 4$ board shows the pattern: removing the border ring of a $4 \times 4$ leaves a $2 \times 2$ interior, i.e. an $(n-2) \times (n-2)$ inside an $n \times n$. Then we apply that same picture to $n = 8$ and divide.
Execute — Answer: D
3.OA.A.1 Step 1 - Count the total number of equally likely squares.
- The board is $8$ rows of $8$ squares, so the sample space has $8 \times 8 = 64$ squares.
💡 An $8 \times 8$ array is $8$ equal groups of $8$, which is the Grade 3 meaning of multiplication.
3.OA.A.1 Step 2 - Try the same setup on a smaller board first to see the pattern.
- On a $4 \times 4$ board, mark every square that touches the outer edge.
- The unmarked squares form a $2 \times 2$ block in the center.
- So on a $4 \times 4$ board the interior is $(4-2) \times (4-2) = 2 \times 2$.
💡 Peeling off the border of an $n \times n$ board takes one square from each side, leaving an $(n - 2) \times (n - 2)$ interior — easier to see on a tiny board first.
3.OA.A.1 Step 3 - Apply the same picture to the $8 \times 8$ board.
- The squares that do not touch the outer edge form a $(8 - 2) \times (8 - 2) = 6 \times 6$ block in the middle, which has $36$ squares.
💡 Once the border ring is removed in the diagram, what remains is just another rectangle, so we can multiply rows by columns again.
7.SP.C.7 Step 4 - Form the probability as favorable $/$ total and simplify the fraction.
- Divide top and bottom by their greatest common factor, $4$.
💡 With every square equally likely, the probability of an event is just the fraction of squares that satisfy it — then reduce.
3.OA.A.1 Count the total number of equally likely squares. The board is $8$ rows of $8$ s 3.OA.A.1 Try the same setup on a smaller board first to see the pattern. On a $4 \times 4 3.OA.A.1 Apply the same picture to the $8 \times 8$ board. The squares that do not touch 7.SP.C.7 Form the probability as favorable $/$ total and simplify the fraction. Divide to Review
Reasonableness: Sanity-check by counting the border the other way: the outer ring of an $8 \times 8$ board has $4 \times 8 - 4 = 28$ squares (four sides of $8$, minus the four corners counted twice). Interior $= 64 - 28 = 36$, matching the $6 \times 6$ count. The probability $\tfrac{36}{64} = \tfrac{9}{16}$ is a bit more than half, which fits the picture — the interior is clearly larger than the border. Choices (A) $\tfrac{1}{16}$ and (E) $\tfrac{49}{64}$ are wildly off, and (C) $\tfrac{1}{2}$ would mean border and interior were equal, which the diagram shows they aren't.
Alternative: Tool #11 (Eliminate Possibilities) works on the choices directly. The interior must be a $k \times k$ block strictly smaller than $8 \times 8$, so the numerator $k^2$ must be a perfect square less than $64$: $1, 4, 9, 16, 25, 36, 49$. The probability is that perfect square over $64$. Only $\tfrac{36}{64} = \tfrac{9}{16}$ matches an answer choice from this list (choice D); $\tfrac{49}{64}$ (E) corresponds to a $7 \times 7$ interior, which would mean stripping only one row and one column — not a border.
CCSS standards used (min grade 7)
3.OA.A.1Interpret products of whole numbers (Counting the $8 \times 8 = 64$ squares on the full board and the $6 \times 6 = 36$ squares in the interior as rectangular arrays.)4.NF.A.1Explain why a fraction $a/b$ is equivalent to $(n \times a)/(n \times b)$ (Reducing $\tfrac{36}{64}$ to $\tfrac{9}{16}$ by dividing numerator and denominator by the common factor $4$.)7.SP.C.7Develop a probability model and use it to find probabilities of events (Treating the $64$ unit squares as equally likely outcomes and computing the probability of "not on the edge" as $\tfrac{36}{64}$.)
⭐ This AMC 8 problem only needs Grade 7 probability — count favorable squares, divide by total, and reduce the fraction.
⭐ This AMC 8 problem only needs Grade 7 probability — count favorable squares, divide by total, and reduce the fraction.
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