AMC 8 · 2014 · #12

Grade 7 probability

Archetype Small-Domain Constrained Search └ Sub-archetype Sum-Target Search

probability-basicpermutations-basicsystematic-enumeration systematic-enumeration ↑ Prerequisites: permutations-basicprobability-basic

📏 Short solution 💡 2 insights

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Problem

A magazine shows 3 celebrities and 3 unlabeled baby photos. A reader pairs each celebrity with one baby photo by guessing. What is the probability that all three pairings are correct?

Pick an answer.

(A)

$frac{1}{9}$

(B)

$frac{1}{6}$

(C)

$frac{1}{4}$

(D)

$frac{1}{3}$

(E)

$frac{1}{2}$

AMC 8 2014 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

There are only a handful of possible matchings, so Tool #2 (Make a Systematic List) lets us write out every possible guess in order and just count. To make sure the counting rule is right, we first try Tool #9 (Solve an Easier Problem) on the 2-celebrity version — small enough to list by hand — and notice that the count is 2 × 1 = 2. The same idea extends to 3 × 2 × 1 = 6 for the real problem. Once we have the 6 orderings, exactly one is the fully-correct match, so the probability is $\frac{1}{6}$ .

1STEP 1

Warm up on the easier 2-celebrity case: listing every pairing gives exactly 2 × 1 = 2 ways.

2 × 1 = 2 pairings

2STEP 2

List every matching for 3 celebrities — ABC, ACB, BAC, BCA, CAB, CBA — exactly 3 × 2 × 1 = 6 pairings.

3 × 2 × 1 = 6 pairings

3STEP 3

Of the 6 orderings, only ABC pairs everyone with their own photo, so there is only 1 fully-correct pairing.

favorable outcomes = 1

4STEP 4

All 6 pairings are equally likely, so probability = favorable/total = $\frac{1}{6}$ .

P(all correct) =

\frac{1}{6}

→ (B)

Answer

\frac{1}{6}

A probability of

\frac{1}{6}

≈ 16.7% feels right: the reader has to nail three guesses in a row with no information, and there are 6 ways to scramble three items. Compared to

\frac{1}{2}

(a coin flip) it is much smaller, and compared to

\frac{1}{9}

it is bigger — which makes sense because

\frac{1}{9}

would assume the three guesses are independent (

\frac{1}{3}

\frac{1}{3}

\frac{1}{3}

), but they are not: once celebrity 1 is matched, only 2 photos are left for celebrity 2, and then only 1 for celebrity 3.

💡Key takeaway

List all 6 ways to match 3 celebrities to 3 baby photos — only 1 is fully right, so the probability is $\frac{1}{6}$ .

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