AMC 10 · 2016 · #12

Grade 7 probability
Archetype Overlap / Complement Counting └ Sub-archetype Parity Partition: Complement via Odd-Even Split
probability-basicparitycombinations-basic casework ↑ Prerequisites: probability-basic
📏 Medium solution 💡 3 insights
Problem
Pick three different whole numbers at random from 1 to 2016. Let p be the probability that their product is odd. Decide which statement about p is correct: is it below 1/8, equal to 1/8, between 1/8 and 1/3, equal to 1/3, or above 1/3?

Pick an answer.

(A)
$p<\dfrac{1}{8}$
(B)
$p=\dfrac{1}{8}$
(C)
$\dfrac{1}{8}<p<\dfrac{1}{3}$
(D)
$p=\dfrac{1}{3}$
(E)
$p>\dfrac{1}{3}$

AMC 10 2016 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Solve an Easier Related Problem

The question only asks how p compares to 1/8, not for its exact value, so an estimate beats a messy calculation. Tool #16 (Change Focus) does the key reframe: the product is odd exactly when all three picks are odd, so p is really the chance of drawing three odds. Tool #9 (Solve an Easier Related Problem) then builds a benchmark: if the picks were independent (allowing repeats), each is odd with probability 1/2, giving (1/2)³=1/8 exactly. Tool #7 (Identify Subproblems) handles the real without-replacement draw as three shrinking fractions, and comparing them to the benchmark settles the answer without ever finishing the arithmetic.

1STEP 1

Turn odd product into all odd

A product is odd only if every factor is odd, so all three picks must be odd; from 1 to 2016 exactly 1008 of the numbers are odd.

#odd=1008, #even=1008, total=2016
2STEP 2

Build an easier benchmark

Pretend repeats are allowed: each pick is odd with probability 1/2, so three odds gives 1/8 — the clean benchmark the choices target.

1008/2016=1/2, (1/2)³=1/8
3STEP 3

Write the real without-replacement chance

Without repeats, multiply the odd chances in order: 1008/2016·1007/2015·1006/2014, each fraction over a pool one smaller than the last.

p=1008/2016·1007/2015·1006/2014
4STEP 4

Compare the factors to one half

Each later factor is just under 1/2 (2×1007＜2015, 2×1006＜2014), so the product dips below 1/8 — choice (A).

1007/2015 ＜ 1/2, 1006/2014 ＜ 1/2 → p ＜ (1/2)³=1/8 → (A)
Answer
p ＜ 1/8
The benchmark 1/8 comes from treating the picks as independent. Removing a chosen odd number leaves proportionally fewer odds for the next pick, so each later factor falls strictly under 1/2. A product where one factor equals 1/2 and two are below 1/2 must land under 1/8, never at or above it. So p ＜ 1/8 is consistent, ruling out (B) through (E) and confirming (A).
💡Key takeaway

An odd product needs all three picks odd; pretending repeats are allowed gives exactly 1/8, and forbidding repeats makes the later odds rarer, so the real chance is just under 1/8.

  • Turn odd product into all odd
  • Build an easier benchmark
  • Write the real without-replacement chance
  • Compare the factors to one half

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