Competition · AMC preparation · step 4 of 4

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
📘 View easy version →
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 - 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

Count all outcomes

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

State the parity rule

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

Count the odd products

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 to get even products

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

Write the probability

Probability = favorable ÷ total = 59\frac{5}{9}, which matches choice (E).

P(even product) = 5/9 → (E)
Answer
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 13\frac{1}{3}, from Box 2 is 13\frac{1}{3}, so P(at least one 2) = 1 - 23\frac{2}{3} · 23\frac{2}{3} = 1 - 49\frac{4}{9} = 59\frac{5}{9}. This independently confirms 59\frac{5}{9} and rules out the smaller options (A) 19\frac{1}{9}, (B) 29\frac{2}{9}, (C) 49\frac{4}{9}, and the symmetric-looking (D) 12\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.

  • Count all outcomes
  • State the parity rule
  • Count the odd products
  • Subtract to get even products
  • Write the probability

A parent dashboard for the family lives at sensimlab.com.

More like this

Same problem type. Tags show what each one shares with this problem.