AMC 10 · 2020 · #18
Grade 7 probabilityProblem
An urn contains one red ball and one blue ball. A box of extra red and blue balls lies nearby. George performs the following operation four times: he draws a ball from the urn at random and then takes a ball of the same color from the box and returns those two matching balls to the urn. After the four iterations the urn contains six balls. What is the probability that the urn contains three balls of each color?
Pick an answer.
AMC 10 2020 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: An urn starts with $1$ red ball and $1$ blue ball. George repeats this step $4$ times: draw a ball uniformly at random from the urn, take a ball of the same color from a side box, and return both to the urn. After $4$ rounds the urn has $6$ balls. Find the probability that the urn has exactly $3$ red and $3$ blue.
Givens: Start: urn has $1$ R, $1$ B (total $2$); Each step: draw uniformly, then return $2$ balls of the drawn color; $4$ steps performed; after step $k$ the urn has $k + 2$ balls; We want the event 'urn ends at $3$ R and $3$ B'; Choices: (A) $\tfrac16$, (B) $\tfrac15$, (C) $\tfrac14$, (D) $\tfrac13$, (E) $\tfrac12$
Unknowns: $P(\text{urn has 3 R and 3 B after 4 steps})$
Understand
Restated: An urn starts with $1$ red ball and $1$ blue ball. George repeats this step $4$ times: draw a ball uniformly at random from the urn, take a ball of the same color from a side box, and return both to the urn. After $4$ rounds the urn has $6$ balls. Find the probability that the urn has exactly $3$ red and $3$ blue.
Givens: Start: urn has $1$ R, $1$ B (total $2$); Each step: draw uniformly, then return $2$ balls of the drawn color; $4$ steps performed; after step $k$ the urn has $k + 2$ balls; We want the event 'urn ends at $3$ R and $3$ B'; Choices: (A) $\tfrac16$, (B) $\tfrac15$, (C) $\tfrac14$, (D) $\tfrac13$, (E) $\tfrac12$
Plan
Primary tool: #2 Make a Systematic List
Secondary: #5 Look for a Pattern, #9 Solve an Easier Related Problem
Tool #2 (Systematic List): the $4$-draw history is a length-$4$ string of R/B; only the $\binom{4}{2} = 6$ strings with exactly two R and two B reach the $3$R-$3$B endpoint. List them in lex order and compute each probability. Tool #5 (Pattern): once we compute one or two strings, the numerator is always $1 \cdot 2 \cdot 1 \cdot 2 = 4$ and the denominator is always $2 \cdot 3 \cdot 4 \cdot 5 = 120$ — every string has probability $\tfrac{4}{120} = \tfrac{1}{30}$. Tool #9 (Easier Problem): trying the very small RRBB case first reveals the pattern that makes Tool #5 work.
Execute — Answer: B
5.NF.B.4 Step 1 - Before step $k$ the urn has $k + 1$ balls total (starting with $2$ before step $1$).
- So the denominators of the four pick-probabilities are exactly $2, 3, 4, 5$, in that order, no matter what George draws.
- The product of denominators is $2 \cdot 3 \cdot 4 \cdot 5 = 120$.
💡 After $k$ steps the urn grows by exactly $k$ balls — the totals are fixed.
7.SP.C.8 Step 2 - To finish at $3$R-$3$B, the $4$ draws must produce exactly $2$ Rs and $2$ Bs in some order.
- The number of such orderings is $\binom{4}{2} = 6$: $\{RRBB, RBRB, RBBR, BRRB, BRBR, BBRR\}$.
💡 Choose which $2$ of the $4$ slots are R.
7.SP.C.8 Step 3 - Compute the probability of one ordering, say $RRBB$.
- Step 1: $1$ R in urn of $2$, $P = \tfrac{1}{2}$.
- Step 2: now $2$ R, $1$ B, $P(R) = \tfrac{2}{3}$.
- Step 3: $3$ R, $1$ B, $P(B) = \tfrac{1}{4}$.
- Step 4: $3$ R, $2$ B, $P(B) = \tfrac{2}{5}$.
- Multiply: $\tfrac{1 \cdot 2 \cdot 1 \cdot 2}{2 \cdot 3 \cdot 4 \cdot 5} = \tfrac{4}{120} = \tfrac{1}{30}$.
💡 Multiply along the path — the four conditional probabilities.
5.NF.B.4 Step 4 - Try another ordering, $RBRB$, to spot the pattern.
- $P(R) = \tfrac{1}{2}$, then $P(B) = \tfrac{1}{3}$ (urn $2$R-$1$B), then $P(R) = \tfrac{2}{4}$ (urn $2$R-$2$B), then $P(B) = \tfrac{2}{5}$ (urn $3$R-$2$B).
- Product: $\tfrac{1 \cdot 1 \cdot 2 \cdot 2}{2 \cdot 3 \cdot 4 \cdot 5} = \tfrac{4}{120} = \tfrac{1}{30}$ — same probability.
💡 Different ordering but same numerator product — pattern alert.
5.NF.B.4 Step 5 - Why every ordering gives numerator $4$.
- The denominators are always $2, 3, 4, 5$ (the urn sizes before each draw).
- The first $R$ pick contributes $1$ (one R in urn), the second $R$ pick contributes $2$ (two Rs in urn at that moment, because the first R draw added another R).
- Same for B: first B contributes $1$, second B contributes $2$.
- So every ordering has numerator $1 \cdot 2 \cdot 1 \cdot 2 = 4$, regardless of which slot R or B occupies.
💡 Each color's first pick uses '$1$ in urn', the second uses '$2$ in urn' — colors don't interfere.
7.SP.C.8 Step 6 - All $6$ orderings have probability $\tfrac{4}{120} = \tfrac{1}{30}$, and they are mutually exclusive (different draw sequences).
- Total probability $= 6 \cdot \tfrac{1}{30} = \tfrac{6}{30} = \tfrac{1}{5}$.
- Matches choice (B).
💡 Add the $6$ equal pieces.
5.NF.B.4 Before step $k$ the urn has $k + 1$ balls total (starting with $2$ before step $ 7.SP.C.8 To finish at $3$R-$3$B, the $4$ draws must produce exactly $2$ Rs and $2$ Bs in 7.SP.C.8 Compute the probability of one ordering, say $RRBB$. Step 1: $1$ R in urn of $2$ 5.NF.B.4 Try another ordering, $RBRB$, to spot the pattern. $P(R) = \tfrac{1}{2}$, then $ 5.NF.B.4 Why every ordering gives numerator $4$. The denominators are always $2, 3, 4, 5$ 7.SP.C.8 All $6$ orderings have probability $\tfrac{4}{120} = \tfrac{1}{30}$, and they ar Review
Reasonableness: After $4$ steps the urn contains $6$ balls — $5$ possible compositions: $(5R, 1B), (4R, 2B), (3R, 3B), (2R, 4B), (1R, 5B)$. A classical Pólya-urn result says all five compositions are equally likely (by the same numerator-$4$, denominator-$120$ counting argument extended to each composition), so each occurs with probability $\tfrac{1}{5}$ — and that is exactly the answer.
Alternative: Tool #1 (Diagram) with a tree: draw the urn-composition tree starting at $(1, 1)$, branching $R/B$ at each step with the appropriate probabilities, and read off the $\tfrac{1}{5}$ at the $(3, 3)$ node. Equivalent computation; the systematic list is more compact.
CCSS standards used (min grade 7)
5.NF.B.4Apply and extend understanding of multiplication to multiply a fraction by a fraction (Multiplying the four conditional probabilities along each draw sequence (e.g., $\tfrac{1}{2} \cdot \tfrac{2}{3} \cdot \tfrac{1}{4} \cdot \tfrac{2}{5}$).)7.SP.C.8Find probabilities of compound events using organized lists, tables, and simulation (Listing the $6$ orderings of $\{R, R, B, B\}$ and adding their probabilities to get the total.)
⭐ This AMC 10 problem only needs Grade 7 list-the-cases probability you already know — every ordering of $2$ R picks and $2$ B picks has probability $\tfrac{1 \cdot 2 \cdot 1 \cdot 2}{2 \cdot 3 \cdot 4 \cdot 5} = \tfrac{1}{30}$. There are $6$ orderings, so the answer is $6 \cdot \tfrac{1}{30} = \tfrac{1}{5}$, choice $\textbf{(B)}$.
⭐ This AMC 10 problem only needs Grade 7 list-the-cases probability you already know — every ordering of $2$ R picks and $2$ B picks has probability $\tfrac{1 \cdot 2 \cdot 1 \cdot 2}{2 \cdot 3 \cdot 4 \cdot 5} = \tfrac{1}{30}$. There are $6$ orderings, so the answer is $6 \cdot \tfrac{1}{30} = \tfrac{1}{5}$, choice $\textbf{(B)}$.
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