AMC 10 · 2006 · #17

Grade 7 probability
Archetype Branching State Enumeration └ Sub-archetype Constrained Sequence or Path Count
probability-basicsystematic-enumeration symmetry-argument ↑ Prerequisites: probability-basic
📏 Medium solution 💡 2 insights
Problem
Bob and Alice each own a bag holding one ball of each of five colors: blue, green, orange, red, and violet. Alice takes one ball at random from her bag and drops it into Bob's bag. Bob then takes one ball at random from his (now larger) bag and drops it into Alice's bag. Find the probability that, once both moves are done, the two bags hold exactly the same collection of balls.

Pick an answer.

(A)
$\frac{1}{10}$
(B)
$\frac{1}{6}$
(C)
$\frac{1}{5}$
(D)
$\frac{1}{3}$
(E)
$\frac{1}{2}$

AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve
Strategy Change Focus / Count the Complement

Chasing all the ways two bags could match looks messy, so Tool #16 (Change Focus) rewrites the goal: the bags can only match if Bob hands back the exact color Alice gave him, because Alice's bag is missing precisely that color. That single condition is the whole problem. Tool #7 (Identify Subproblems) separates the two moves so we can track what each bag holds after Move 1 before worrying about Move 2. Tool #9 (Solve an Easier Related Problem) then removes the clutter: since the five colors behave identically, we may assume Alice moved one specific color and just count Bob's draw, turning the question into a single easy probability.

1STEP 1

Track the two moves

After Move 1 Bob holds 6 balls, two of them the color Alice sent, while Alice holds 4, missing exactly that color.

Bob: 6 balls, Alice: 4 balls (one color missing)
2STEP 2

When are the bags identical?

Alice's bag has just one hole, so the bags match only if Bob hands back a ball of the very color she gave away.

bags match⇔ Bob returns Alice's moved color
3STEP 3

The moved color does not matter

All five colors play the same role, so say Alice moved red. Bob's bag is then red, red, blue, green, orange, violet.

Bob's bag={red, red, blue, green, orange, violet}
4STEP 4

Compute Bob's draw

Each of Bob's 6 balls is equally likely and 2 of them are red, so the probability is 2/6=1/3, choice (D).

P=(red balls)/(total balls)=2/6=1/3 → (D)
Answer
1/3
The answer 1/3 is a plausible probability between 0 and 1. A quick sanity check: Bob must return the one color Alice removed, and his bag holds two of the six balls in that color, so a success rate near one-third feels right — clearly more than 1/5 (one specific ball out of five) because the doubled color makes a hit easier, yet well below 1/2. This rules out (A), (B), (C), and (E) and leaves (D).
💡Key takeaway

The bags can only match if Bob hands back the same color Alice gave him, and since his bag now holds two of that color out of six, the chance is 2/6=1/3.

  • Track the two moves
  • When are the bags identical?
  • The moved color does not matter
  • Compute Bob's draw

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