AMC 10 · 2020 · #18

Grade 7 probability

Archetype Branching State Enumeration └ Sub-archetype Markov State Probability

probability-basicconditional-probabilitysymmetry-argumentfraction-arithmetic easier-related-problemcaseworksymmetry-argument ↑ Prerequisites: probability-basicconditional-probability

📏 Medium solution 💡 2 insights

Problem

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.

Pick an answer.

(A)

$\frac{1}{6}$

(B)

$\frac{1}{5}$

(C)

$\frac{1}{4}$

(D)

$\frac{1}{3}$

(E)

$\frac{1}{2}$

AMC 10 2020 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Make a Systematic List

Tool #2 (Systematic List): the 4-draw history is a length-4 string of R/B; only the C(4, 2) = 6 strings with exactly two R and two B reach the 3R-3B 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 · 2 · 1 · 2 = 4 and the denominator is always 2 · 3 · 4 · 5 = 120 — every string has probability $\frac{4}{120}$ = $\frac{1}{30}$ . Tool #9 (Easier Problem): trying the very small RRBB case first reveals the pattern that makes Tool #5 work.

1STEP 1

The urn holds 2, 3, 4, 5 balls before the four draws, so every path shares denominators multiplying to 120.

denominators = 2, 3, 4, 5; product = 120

2STEP 2

Ending 3R-3B needs exactly two R and two B draws; the number of orderings is C(4, 2) = 6.

C(4, 2) = 6 orderings

3STEP 3

For RRBB: $\frac{1}{2}$ · $\frac{2}{3}$ · $\frac{1}{4}$ · $\frac{2}{5}$ = $\frac{4}{120}$ = $\frac{1}{30}$ , one concrete path's probability.

P(RRBB) =

\frac{1}{2}

\frac{2}{3}

\frac{1}{4}

\frac{2}{5}

\frac{4}{120}

\frac{1}{30}

4STEP 4

Another order RBRB gives $\frac{1}{2}$ · $\frac{1}{3}$ · $\frac{2}{4}$ · $\frac{2}{5}$ = $\frac{4}{120}$ = $\frac{1}{30}$ — identical probability.

P(RBRB) =

\frac{1}{2}

\frac{1}{3}

\frac{2}{4}

\frac{2}{5}

\frac{4}{120}

\frac{1}{30}

5STEP 5

Each color's first pick contributes 1, its second contributes 2, so every ordering shares numerator 1 · 2 · 1 · 2 = 4.

numerator = 1_R · 2_R · 1_B · 2_B = 4 for all 6 orderings

6STEP 6

The 6 mutually exclusive orderings each carry $\frac{1}{30}$ , so the total is 6 · $\frac{1}{30}$ = $\frac{1}{5}$ , choice (B).

6 ·

\frac{1}{30}

\frac{6}{30}

\frac{1}{5}

→ (B)

Answer

\frac{1}{5}

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

\frac{1}{5}

— and that is exactly the answer.

💡Key takeaway

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 $\frac{1 · 2 · 1 · 2}{2 · 3 · 4 · 5}$ = $\frac{1}{30}$ . There are 6 orderings, so the answer is 6 · $\frac{1}{30}$ = $\frac{1}{5}$ , choice (B).

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