AMC 10 · 2015 · #9

Grade 7 probability
Archetype Branching State Enumeration └ Sub-archetype Ordered Placement Count
probability-basiccombinations-basicfundamental-counting-principle symmetry-argument ↑ Prerequisites: probability-basiccombinations-basic
📏 Medium solution 💡 2 insights
Problem
Items are drawn in turn until the last person takes what remains. Find the chance the last pair matches.

Pick an answer.

(A)
$\frac{1}{10}$
(B)
$\frac{1}{6}$
(C)
$\frac{1}{5}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$
How to solve
Strategy Make a Systematic List

The process has three stages, but Tool #2 (Make a Systematic List) collapses it into one flat sample space: every way of dealing the six marbles into Carol's pair, Claudia's pair, and Cheryl's pair. Once all those deals carry the same probability, the question becomes pure counting, and Tool #7 (Identify Subproblems) splits the favorable count by which color Cheryl ends up with. There is a famous one-line shortcut for this problem — "order doesn't matter, so let Cheryl draw first" — and it gets the right answer, but as usually written it is an assertion with no argument behind it. Tool #15 (Organize Information in More Ways) is used to actually earn it: re-organize the same 90 deals by what Cheryl is left holding, and show that all 15 possible leftover pairs come out equally likely.

1STEP 1

Keep the six marbles apart

Keeping items distinct makes deals countable.

{R₁, R₂, G₁, G₂, Y₁, Y₂} — six distinct marbles, two of each color
2STEP 2

Count every possible deal

There are 90 equally likely deals.

15 · 6 · 1 = 90 deals, each of probability 1/15 · 1/6 = 1/90
3STEP 3

Count the deals Cheryl likes

Of these, 18 leave a matching pair.

3 × (4 · 3)/2 = 3 × 6 = 18 favorable deals
4STEP 4

Prove that drawing last costs nothing

Drawing last costs nothing.

P(Cheryl holds a specified pair) = 6/90 = 1/15 for each of the 15 pairs → P = 3/15
5STEP 5

Divide and compare the two counts

Dividing gives 1/5, choice (C).

P = 18/90 = 3/15 = 1/5 → (C)
Answer
1/5
The value clears several checks. It lies strictly between 0 and 1, and it is small, as it must be: of the 15 pairs Cheryl could be left with, only 3 match, so the answer sits well below 1/3 — that alone rules out (D) 1/3 and (E) 1/2. The complement agrees: 12 of the 15 pairs are mixed, giving 12/15 = 4/5 for a mismatch, and 1/5 + 4/5 = 1. A third check uses different arithmetic entirely — draw the marbles out one at a time and ask for the chance that the first four drawn are exactly the four non-red marbles: 4/6 · 3/5 · 2/4 · 1/3 = 1/15, matching the per-pair 1/15 from step 4, and 3 colors gives 1/5 once more. Finally, (E) 1/2 is the trap that step 1 was built to avoid: pretend the two marbles of a color are identical, so that the color handfuls RR, GG, YY, RG, RY, GY look like six equally likely outcomes, and you get 3/6 = 1/2. Those six outcomes are not equally likely, which is why the marbles had to be labeled.
💡Key takeaway

Every one of the 15 pairs Cheryl could be left with is equally likely, and only 3 of them are matching colors, so her chance is 3/15 = 1/5.

  • Keep the six marbles apart
  • Count every possible deal
  • Count the deals Cheryl likes
  • Prove that drawing last costs nothing
  • Divide and compare the two counts

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