Competition · AMC preparation · step 4 of 4

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

Record each spinner's parity

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

Reduce using parity

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)

Adding an odd number flips parity, adding an even number keeps it. So adding one even and one odd flips the parity of P exactly once: P even → sum odd; P odd → sum even.

The three spun numbers total an odd number exactly when spinner P lands on an even number.

▸ Why?

However the spinners fall, Q always adds an even number and R always adds an odd number, so the whole sum apart from P is locked as complete pairs plus a single leftover unit — an odd amount.

▸ Why?

Every face of spinner Q — 2, 4, 6, 8 — splits into two equal whole groups with nothing left over, which is exactly what makes a number even.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

No face of spinner R — 1, 3, 5, 7, 9, 11 — can split into two equal whole groups; one unit is always left over, which is exactly what makes a number odd.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

▸ Why?

Because the three amounts can be added in any grouping, we may first fold Q and R together into that fixed odd part and then read the whole sum's parity from what P adds on top of it.

▸ Why?

The grouping in which three piles are pushed together does not change their total, so settling Q and R first is allowed.

Associativity of additionAdding three piles, it does not matter which two you push together first.

▸ Why?

An even P splits into complete pairs with no leftover of its own, so it leaves the standing odd part untouched and the total stays odd — while an odd P brings its own leftover unit that pairs off with the fixed one, cancelling the leftover and making the total even.

▸ Why?

An even value on P splits into two equal whole groups with no single unit left over, so it can only add complete pairs and never disturbs the leftover already there.

Even and odd (parity)A whole number is even when it splits into two equal whole groups (a factor of 2), else odd; odd×odd stays odd and even×anything is even.

4.OA.B.4Identify Subproblems

3STEP 3

Find the one probability

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

P(P is even) = 1/3

4STEP 4

Check with the full product

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

P(sum is odd) = 1/3 × 1 × 1 = 1/3 → (B)

Answer

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: $\frac{1}{3}$ .

Record each spinner's parity
Reduce using parity
Find the one probability
Check with the full product

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Record each spinner's parity

Reduce using parity

Find the one probability

Check with the full product

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