AMC 8 · 2004 · #21

Grade 7 probability

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

probability-basicparitycomplementary-countingfraction-multiplication complementary-countingidentify-subproblems ↑ Prerequisites: probability-basicfraction-arithmetic

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Spinner A has four equal regions labeled 1, 2, 3, 4 and spinner B has three equal regions labeled 1, 2, 3. Both spinners are spun once. What is the probability that the product of the two numbers shown is even?

Pick an answer.

(A)

$\frac14$

(B)

$\frac13$

(C)

$\frac12$

(D)

$\frac23$

(E)

$\frac34$

AMC 8 2004 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

The event "product is even" breaks into three favorable cases (even×even, even×odd, odd×even), but its complement "product is odd" is a single case: both spinners land on an odd number. Tool #16 (Change Focus / Count the Complement) swaps the messy event for the clean one, computes the probability of the complement, and subtracts from 1. Tool #7 (Identify Subproblems) splits that complement into two independent subproblems — P(A odd) and P(B odd) — which multiply because the spinners are independent.

1STEP 1

The product is odd only when both spins are odd, so P(even) = 1 - P(both odd) — one case, not three.

P(product even) = 1 - P(A odd and B odd)

2STEP 2

Spinner A: two of its four labels (1, 3) are odd, so P(A odd) = $\frac{2}{4}$ = $\frac{1}{2}$ .

P(A odd) =

\frac{2}{4}

\frac{1}{2}

3STEP 3

Spinner B: two of its three labels (1, 3) are odd, so P(B odd) = $\frac{2}{3}$ .

P(B odd) =

\frac{2}{3}

4STEP 4

Independent, so P(both odd) = $\frac{1}{2}$ × $\frac{2}{3}$ = $\frac{1}{3}$ , and the even product is the rest: 1 - $\frac{1}{3}$ = $\frac{2}{3}$ , choice (D).

P(both odd) =

\frac{1}{2}

\frac{2}{3}

\frac{1}{3}

→ P(product even) = 1 -

\frac{1}{3}

\frac{2}{3}

→ (D)

Answer

\frac{2}{3}

Sanity check with the full 4 × 3 = 12 outcome grid. The products row by row are: A=1 → 1,2,3; A=2 → 2,4,6; A=3 → 3,6,9; A=4 → 4,8,12. Even products per row: 1 + 3 + 1 + 3 = 8. Probability =

\frac{8}{12}

\frac{2}{3}

, matching answer (D). The

\frac{2}{3}

also feels right: spinner A alone lands even half the time, which already forces the product to be even, so the answer must be at least

\frac{1}{2}

— and

\frac{2}{3}

comfortably exceeds it.

💡Key takeaway

Even products have many sub-cases, but odd products only happen one way — both spinners must land on odd. Count that single easy case ( $\frac{1}{3}$ ), then take the complement: 1 - $\frac{1}{3}$ = $\frac{2}{3}$ .

More like this