AMC 10 · 2006 · #17
Grade 7 probabilityBob and Alice each have a bag that contains one ball of each of the colors blue, green, orange, red, and violet. Alice randomly selects one ball from her bag and puts it into Bob's bag. Bob then randomly selects one ball from his bag and puts it into Alice's bag. What is the probability that after this process the contents of the two bags are the same?
Pick an answer.
AMC 10 2006 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: 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.
Givens: Each bag starts with exactly one ball of each of $5$ colors, so $5$ balls per bag; Move 1: Alice picks $1$ of her $5$ balls at random and puts it in Bob's bag; Move 2: Bob then picks $1$ of his balls at random and puts it in Alice's bag; Answer choices: (A) $\frac{1}{10}$, (B) $\frac{1}{6}$, (C) $\frac{1}{5}$, (D) $\frac{1}{3}$, (E) $\frac{1}{2}$
Unknowns: The probability that the two bags end up with identical contents
Understand
Restated: 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.
Givens: Each bag starts with exactly one ball of each of $5$ colors, so $5$ balls per bag; Move 1: Alice picks $1$ of her $5$ balls at random and puts it in Bob's bag; Move 2: Bob then picks $1$ of his balls at random and puts it in Alice's bag; Answer choices: (A) $\frac{1}{10}$, (B) $\frac{1}{6}$, (C) $\frac{1}{5}$, (D) $\frac{1}{3}$, (E) $\frac{1}{2}$
Plan
Primary tool: #16 Change Focus / Count the Complement
Secondary: #9 Solve an Easier Related Problem, #7 Identify Subproblems
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.
Execute — Answer: D
7.SP.C.8 Step 1 Track the two moves
- Split the process into its two stages.
- In Move 1 Alice sends one ball to Bob, so Bob's bag grows to $6$ balls and Alice's bag drops to $4$ balls, missing the color she sent.
- Whatever color Alice moved, Bob's bag now holds two balls of that color (his original one plus hers) and one each of the other four.
- In Move 2 Bob draws one of his $6$ balls and gives it back.
💡 Following the balls one move at a time shows exactly which color each bag is short of.
7.SP.C.8 Step 2 When are the bags identical?
- For the two bags to hold the same collection, each must again have all five colors, one apiece.
- Alice's bag is missing only the color she gave away, so the only ball that can fix it is one of that same color coming back.
- If Bob returns any other color, Alice ends with two of that color and still none of the sent color, so the bags cannot match.
- So 'same contents' happens exactly when Bob returns a ball of the color Alice originally moved.
💡 Alice's bag has exactly one hole, so only the matching color can plug it and even things out.
7.SP.C.7 Step 3 The moved color does not matter
- The five colors play identical roles, so it makes no difference which one Alice moved.
- Assume she moved red.
- Then Bob's bag holds two red balls and one each of blue, green, orange, and violet, for $6$ balls total.
- The whole question is now just: what is the chance Bob draws a red ball to send back?
💡 By symmetry one fixed color stands in for all of them, so we only solve the problem once.
7.SP.C.7 Step 4 Compute Bob's draw
- Bob picks one of his $6$ balls at random, and each is equally likely.
- Two of the $6$ are red, the color that makes the bags match, so the probability is $\frac{2}{6}=\frac{1}{3}$.
- That is the chance the contents end up the same, which is choice (D).
💡 With two favorable balls out of six equally likely ones, the odds are simply two-sixths.
7.SP.C.8 Split the process into its two stages. In Move 1 Alice sends one ball to Bob, so 7.SP.C.8 For the two bags to hold the same collection, each must again have all five colo 7.SP.C.7 The five colors play identical roles, so it makes no difference which one Alice 7.SP.C.7 Bob picks one of his $6$ balls at random, and each is equally likely. Two of the Review
Reasonableness: The answer $\frac{1}{3}$ 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 $\frac{1}{5}$ (one specific ball out of five) because the doubled color makes a hit easier, yet well below $\frac{1}{2}$. This rules out (A), (B), (C), and (E) and leaves (D).
Alternative: Skip the symmetry shortcut and count outcomes directly. There are $5$ equally likely colors Alice can move and $6$ equally likely balls Bob can return, giving $5\times 6=30$ equally likely (move, return) pairs. For each of Alice's $5$ choices, exactly $2$ of Bob's $6$ balls share that color, so $5\times 2=10$ pairs make the bags match. The probability is $\frac{10}{30}=\frac{1}{3}$, confirming (D).
CCSS standards used (min grade 7)
7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Tracking the two-move process and identifying the single condition (Bob returns the moved color) under which the bags match.)7.SP.C.7Develop probability models and use them to find probabilities of events (Treating Bob's six balls as equally likely and computing the chance of drawing one of the two red balls as $\frac{2}{6}=\frac{1}{3}$.)
⭐ 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 $\frac{2}{6}=\frac{1}{3}$.
⭐ 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 $\frac{2}{6}=\frac{1}{3}$.
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