AMC 10 · 2020 · #16

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 one red ball and one blue ball. This step is repeated four times: draw a ball uniformly at random from the urn, take a ball of the same colour from a side box, and return both to the urn. After four rounds the urn has six balls. Find the probability that it holds exactly three red and three blue.

Pick an answer.

(A)

$\frac16$

(B)

$\frac15$

(C)

$\frac14$

(D)

$\frac13$

(E)

$\frac12$

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 4/120 = 1/30. Tool #9 (Easier Problem): trying the very small RRBB case first reveals the pattern that makes Tool #5 work.

1STEP 1

Check the denominators

The urn grows by one, fixing the denominators.

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

2STEP 2

Count the orderings

Count orderings with two of each.

C(4, 2) = 6 orderings

3STEP 3

Compute one ordering

Compute one ordering's probability.

P(RRBB) = 1/2 · 2/3 · 1/4 · 2/5 = 4/120 = 1/30

4STEP 4

Compute another ordering

Try a different ordering too.

P(RBRB) = 1/2 · 1/3 · 2/4 · 2/5 = 4/120 = 1/30

5STEP 5

See they all agree

The numerator is order-independent.

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

6STEP 6

Add them up

Multiplying and adding gives one fifth.

6 · 1/30 = 6/30 = 1/5 → (B)

Answer

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 1/5 — and that is exactly the answer.

💡Key takeaway

This AMC 12 problem only needs Grade 7 list-the-cases probability you already know — every ordering of 2 R picks and 2 B picks has probability (1 · 2 · 1 · 2)/(2 · 3 · 4 · 5) = 1/30. There are 6 orderings, so the answer is 6 · 1/30 = 1/5, choice (B).