AMC 8 · 2006 · #17

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Logic-Deduction / Elimination

probability-basicparityfraction-multiplicationsystematic-enumeration caseworkidentify-subproblems ↑ Prerequisites: probability-basicparity

📏 Medium solution 💡 3 insights 📊 Diagram

Problem

Jeff spins three spinners P, Q, R and adds the three numbers. Spinner P has {1, 2, 3}, spinner Q has {2, 4, 6, 8}, and spinner R has {1, 3, 5, 7, 9, 11}. Each region on a given spinner is equally likely. What is the probability that the sum P + Q + R is odd?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{1}{3}$

(C)

$\frac{1}{2}$

(D)

$\frac{2}{3}$

(E)

$\frac{3}{4}$

AMC 8 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find an Invariant

The actual values do not matter — only their parities. Looking at the three spinners, two of them have parities that never change: every number on Q is even, and every number on R is odd. Tool #11 (Find an Invariant) names that. With Q fixed even and R fixed odd, the parity of the sum depends on only one thing: the parity of P. Tool #7 (Identify Subproblems) reduces the three-spinner question to a single subproblem — the probability that spinner P lands on an even number.

1STEP 1

Tag each spinner's parity: every Q value is even and every R value is odd.

P: {1_odd, 2_even, 3_odd}, Q: {2, 4, 6, 8} all even, R: {1, 3, 5, 7, 9, 11} all odd

2STEP 2

Q adds even and R adds odd, so the sum is odd exactly when P is even.

P + Q + R ≡ P + even + odd ≡ P + odd (mod 2)

3STEP 3

Among P's three equal regions, exactly one — the 2 — is even.

P(P is even) =

\frac{1}{3}

4STEP 4

Multiplying the three independent probabilities (two of them 1) gives $\frac{1}{3}$ — choice (B).

P(sum is odd) =

\frac{1}{3}

× 1 × 1 =

\frac{1}{3}

→ (B)

Answer

\frac{1}{3}

Cross-check by trying every parity case for P. If P is odd (prob

\frac{2}{3}

), the sum is odd + even + odd = even. If P is even (prob

\frac{1}{3}

), the sum is even + even + odd = odd. The two cases cover all outcomes and their probabilities add to

\frac{2}{3}

\frac{1}{3}

= 1, so the probability of an odd sum is exactly

\frac{1}{3}

, matching answer (B). The number

\frac{1}{3}

also makes sense: Q and R are both "stuck" on a fixed parity, so the answer must be a probability that already lives on spinner P — and the only P-fractions in sight are

\frac{1}{3}

and

\frac{2}{3}

💡Key takeaway

Spinner Q is always even and spinner R is always odd — those two parities are locked in, so the only spinner that decides the sum's parity is P. The answer is just the probability that P lands even: $\frac{1}{3}$ .

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