AMC 10 · 2007 · #19

Grade 7 geometry-2d

Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split

probability-basicmodular-arithmeticcasework systematic-enumeration ↑ Prerequisites: probability-basic

📏 Long solution 💡 3 insights 📊 Diagram

Problem

A wheel with the numbers 1, 2, 3, 6, 7, 9 is spun twice, each number equally likely. The first result is divided by 4 and its remainder picks a column (1, 2, or 3). The second result is divided by 5 and its remainder picks a row (1, 2, 3, or 4). The board is 3 columns by 4 rows, shaded like a checkerboard, with shaded squares at (column, row) = (1,1), (3,1), (2,2), (1,3), (3,3), (2,4). Find the probability that the chosen column-and-row square is shaded.

Pick an answer.

(A)

$\frac{1}{3}$

(B)

$\frac{4}{9}$

(C)

$\frac{1}{2}$

(D)

$\frac{5}{9}$

(E)

$\frac{2}{3}$

AMC 10 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Make a Systematic List) — the column and row are each a remainder, so first list what remainder every wheel number gives. This turns a vague spinner into a clean table of how likely each column and each row is. Tool #1 (Draw a Diagram) — read the six shaded squares straight off the board as (column, row) pairs so we know exactly which outcomes count as a win. Tool #7 (Identify Subproblems) — because the two spins are independent, the probability of any one square is (its column's probability) times (its row's probability); sum that over the six shaded squares.

1STEP 1

List column remainders (÷4)

Divide each wheel number by 4: the remainders are 1,1,2,2,3,3 — two per column, so every column has chance 1/3.

P(col 1)=P(col 2)=P(col 3)=2/6=1/3

2STEP 2

List row remainders (÷5)

Divide by 5 instead: rows 1 and 2 each come from two numbers, rows 3 and 4 from one — chances 1/3, 1/3, 1/6, 1/6.

P(row 1)=P(row 2)=2/6=1/3, P(row 3)=P(row 4)=1/6

3STEP 3

Read the shaded squares

Shaded: (1,1),(3,1),(2,2) sit in rows 1 and 2; (1,3),(3,3),(2,4) sit in rows 3 and 4 — three each.

shaded: (1,1),(3,1),(2,2) | (1,3),(3,3),(2,4)

4STEP 4

Probability from the likely rows

Independent spins multiply: each shaded square in rows 1 and 2 is 1/3·1/3=1/9, and three of them give 1/3.

3×(1/3·1/3)=3·1/9=1/3

5STEP 5

Probability from the unlikely rows, then total

The other three squares are 1/3·1/6=1/18 each, adding 1/6. Total: 1/3+1/6=1/2, choice (C).

1/3+3·1/18=1/3+1/6=2/6+1/6=1/2 → (C)

Answer

1/2

The answer 1/2 is believable: the board is a perfect checkerboard, so exactly half of the twelve squares are shaded, and a quick symmetry argument confirms the weighting does not break that balance. Every column is equally likely, so within any single row the shaded and unshaded squares are hit with matching total weight. Because each row's three squares split evenly between shaded and unshaded across the checkerboard, no row can tip the balance, and the whole board stays at one-half regardless of the uneven row probabilities. That rules out the lopsided choices like (A) 1/3 or (E) 2/3, and the exact sum lands on (C).

💡Key takeaway

Sort the spinner numbers by their remainders to see how likely each column and row is, then multiply column-chance by row-chance for every shaded square and add — the checkerboard's balance makes it exactly 1/2.

List column remainders (÷4)
List row remainders (÷5)
Read the shaded squares
Probability from the likely rows
Probability from the unlikely rows, then total

List column remainders (÷4)

List row remainders (÷5)

Read the shaded squares

Probability from the likely rows

Probability from the unlikely rows, then total

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