AMC 8 · 2015 · #7

Grade 7 probability

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

probability-basicparitysystematic-enumeration complementary-countingsystematic-enumeration ↑ Prerequisites: fraction-arithmeticparity

📏 Short solution 💡 3 insights

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Problem

Two boxes each hold three chips numbered 1, 2, 3. Pull one chip from each box at random and multiply the two numbers. What is the probability that the product is even?

Pick an answer.

(A)

$frac{1}{9}$

(B)

$frac{2}{9}$

(C)

$frac{4}{9}$

(D)

$frac{1}{2}$

(E)

$frac{5}{9}$

AMC 8 2015 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only 3 × 3 = 9 outcomes, small enough to list every pair explicitly — Tool #2 (Systematic List) is the cleanest fit for this size. The wording "product is even" is a classic trigger for Tool #16 (Complement): "even product" splits into many cases, but "odd product" needs both factors odd — only 2 × 2 = 4 pairs. We compute 1 - $\frac{4}{9}$ instead of summing five cases. Tool #3 (Eliminate) is a check: only one of the five fractions can match a count out of 9.

1STEP 1

Multiplication principle: Box 1 has 3 choices, Box 2 has 3, so there are 3 × 3 = 9 equally likely ordered pairs (a, b).

3 × 3 = 9 outcomes

2STEP 2

Parity rule: a × b is even unless both are odd, so count the easier complement instead — pairs with both numbers odd.

a × b is odd ⇔ a is odd AND b is odd

3STEP 3

Odd chips {1, 3} in each box give 2 × 2 = 4 odd-product pairs: (1,1), (1,3), (3,1), (3,3).

2 × 2 = 4 odd-product outcomes

4STEP 4

Subtract from the total: 9 - 4 = 5 pairs have an even product — (1,2), (2,1), (2,2), (2,3), (3,2).

9 - 4 = 5 even-product outcomes

5STEP 5

Probability = favorable ÷ total = $\frac{5}{9}$ , which matches choice (E).

P(even product) =

\frac{5}{9}

→ (E)

Answer

frac{5}{9}

Sanity check the answer's size. The product is forced to be even whenever a 2 is drawn from either box — that already covers a lot of pairs. The chance of drawing a 2 from Box 1 is

\frac{1}{3}

, from Box 2 is

\frac{1}{3}

, so P(at least one 2) = 1 -

\frac{2}{3}

\frac{2}{3}

= 1 -

\frac{4}{9}

\frac{5}{9}

. This independently confirms

\frac{5}{9}

and rules out the smaller options (A)

\frac{1}{9}

, (B)

\frac{2}{9}

, (C)

\frac{4}{9}

, and the symmetric-looking (D)

\frac{1}{2}

💡Key takeaway

When a problem says "the product is even," flip it around and count when it is odd instead — both numbers odd is just one tiny case to check.

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