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

\frac{1}{8}

, equal to

\frac{1}{8}

, between

\frac{1}{8}

and

\frac{1}{3}

, equal to

\frac{1}{3}

, or above

\frac{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 $\frac{1}{8}$ , 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 $\frac{1}{2}$ , giving ( $\frac{1}{2}$ )³= $\frac{1}{8}$ 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 $\frac{1}{2}$ , so three odds gives $\frac{1}{8}$ — the clean benchmark the choices target.

\frac{1008}{2016}

\frac{1}{2}

, (

\frac{1}{2}

)³=

\frac{1}{8}

3STEP 3

Write the real without-replacement chance

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

\frac{1008}{2016}

\frac{1007}{2015}

\frac{1006}{2014}

4STEP 4

Compare the factors to one half

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

\frac{1007}{2015}

＜

\frac{1}{2}

\frac{1006}{2014}

＜

\frac{1}{2}

→ p ＜ (

\frac{1}{2}

)³=

\frac{1}{8}

→ (A)

Answer

p ＜

\frac{1}{8}

The benchmark

\frac{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

\frac{1}{2}

. A product where one factor equals

\frac{1}{2}

and two are below

\frac{1}{2}

must land under

\frac{1}{8}

, never at or above it. So p ＜

\frac{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 $\frac{1}{8}$ , and forbidding repeats makes the later odds rarer, so the real chance is just under $\frac{1}{8}$ .

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

Turn odd product into all odd

Build an easier benchmark

Write the real without-replacement chance

Compare the factors to one half

More like this