AMC 8 · 2007 · #25

Grade 7 probabilitygeometry-2d

Archetype Geometric Probability by Area └ Sub-archetype Area-Proportional Region Probability

probability-basicarea-circlesfraction-arithmeticparity caseworkidentify-subproblems ↑ Prerequisites: area-circlesprobability-basicparity

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

A dart board has an outer circle of radius 6 and an inner circle of radius 3. Three radii cut each circle into three congruent regions, labeled with point values 1 or 2. The inner circle has scores 1, 2, 2 (one 1 and two 2s). The outer ring has scores 2, 1, 1 (one 2 and two 1s). The chance a dart lands in a region is proportional to that region's area. Two independent darts are thrown. What is the probability that the sum of the two scores is odd?

Pick an answer.

(A)

$\frac{17}{36}$

(B)

$\frac{35}{72}$

(C)

$\frac{1}{2}$

(D)

$\frac{37}{72}$

(E)

$\frac{19}{36}$

AMC 8 2007 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Identify Subproblems

The two-dart probability is intimidating, but it breaks cleanly into smaller pieces. Tool #7 (Identify Subproblems) splits the work into three stages: (1) find the area of each region; (2) collapse those areas into a single-dart probability P(odd) and P(even); (3) combine the two darts. Tool #1 (Draw a Diagram) keeps the six regions and their scores straight — the inner disk and outer ring carry different amounts of area, so the 1s and 2s are not equally likely. Tool #2 (Systematic List) handles step (3): the two darts have only four parity outcomes (OO, OE, EO, EE), and exactly the mixed ones give an odd sum.

1STEP 1

Sketch the board: each inner region has area 3π while each outer-ring region has area 9π — three times as large.

A_outer disk = 36π, A_inner disk = 9π, A_ring = 27π, A_inner region = 3π, A_ring region = 9π

2STEP 2

Add every region scoring 1 — one inner (3π) plus two outer (18π) = 21π out of 36π — so P(odd) = $\frac{7}{12}$ .

P(odd) = 21π/36π =

\frac{21}{36}

\frac{7}{12}

3STEP 3

Add every region scoring 2 — two inner (6π) plus one outer (9π) = 15π — so P(even) = $\frac{5}{12}$ , and P(odd) + P(even) = 1.

P(even) = (6π + 9π)/36π =

\frac{15}{36}

\frac{5}{12}

4STEP 4

List the four parity cases OO, OE, EO, EE; only the mixed ones, where exactly one dart is odd, sum to odd.

Odd sum ⇔ exactly one dart is odd

5STEP 5

Multiply within each case and add the two: 2 · ( $\frac{7}{12}$ )( $\frac{5}{12}$ ) = $\frac{35}{72}$ , choice (B).

P(odd sum) = P(odd) P(even) + P(even) P(odd) = 2 ·

\frac{7}{12}

\frac{5}{12}

\frac{70}{144}

\frac{35}{72}

→ (B)

Answer

\frac{35}{72}

The answer

\frac{35}{72}

≈ 0.486 is just under

\frac{1}{2}

, which fits the setup: a slightly tilted coin should land odd-sum slightly less often than even-sum. Quick check via the complement: P(even sum) = P(O)² + P(E)² =

\frac{49}{144}

\frac{25}{144}

\frac{74}{144}

\frac{37}{72}

, and

\frac{35}{72}

\frac{37}{72}

\frac{72}{72}

= 1, exactly as required. Choice (C)

\frac{1}{2}

would only hold if odd and even were equally likely on a single dart, which they aren't (

\frac{7}{12}

≠

\frac{5}{12}

💡Key takeaway

Split a scary two-dart probability into three small subproblems — find each region's area, get the single-dart P(odd) and P(even), then list the four parity cases. The mixed ones give the odd sum, and the answer drops out as $\frac{35}{72}$ .

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