AMC 10 · 2024 · #25

Grade 11 probabilitycounting
independent-eventsprobability-basicpermutations-basiccomplementary-counting caseworksystematic-enumerationidentify-subproblems ↑ Prerequisites: probability-basicfundamental-counting-principlecombinations-basic
📏 Long solution 💡 5 insights
Problem
Pablo paints 6 balls. Each ball gets a pattern (striped or dotted) and a color (red or blue), and all 12 of those choices are made by separate fair coin flips. He drops the 6 painted balls into an urn and Frida draws one at random. Depending on how the flips came out, the events "the drawn ball is red" and "the drawn ball is striped" may or may not be independent. The probability that they turn out independent is m/n in lowest terms. Find m.

Pick an answer.

(A)
243
(B)
245
(C)
247
(D)
249
(E)
251
How to solve
Strategy Organize Information in More Ways

The phrase "the probability that two events are independent" sounds tangled, so the work is to make it concrete. Tool #12 (Draw a Venn Diagram) does that first: pattern and color are two overlapping properties, so the 6 balls sort into the four boxes striped-red, striped-blue, dotted-red, dotted-blue, and Tool #4 (Introduce a Variable) names those four counts a, b, c, d. Independence then becomes one equation in those four numbers. Tool #15 (Organize Information in More Ways) is the pivot and drives everything after it: the equation first appears as the lopsided 6a=(a+b)(a+c), but replacing the 6 by a+b+c+d collapses it to the symmetric ad=bc — equal diagonal products in the two-way table. That symmetric form is what makes the count short. Tool #7 (Identify Subproblems) then splits the count into two clean halves, tables with an empty row or column and tables with none, and Tool #2 (Make a Systematic List) finishes the second half, where only two shapes of positive four-part sums can reach 6 at all.

1STEP 1

Sort the urn into four boxes

Balls fall into just four kinds.

a+b+c+d=6, a,b,c,d ≥ 0
2STEP 2

Turn independence into an equation

It becomes an equation in the counts.

a/6=(a+b)/6·(a+c)/6 ⟺ 6a=(a+b)(a+c)
3STEP 3

Rewrite as equal cross-products

It reduces to equal cross-products.

(a+b+c+d) a=(a+b)(a+c) ⟹ a²+ab+ac+ad=a²+ac+ab+bc ⟹ ad=bc
4STEP 4

Split into flat tables and full tables

An empty box splits it into two families.

a=0 → bc=0 → b=0 or c=0; and a=b=0 or c=d=0 or a=c=0 or b=d=0 each give ad=0=bc
5STEP 5

Count the flat tables

After removing overlaps, 252.

4 · 2⁶-4 · 1=256-4=252
6STEP 6

Count the full tables

Four shapes together give 720.

6!/1! 1! 2! 2!=720/4=180, 4 · 180=720
7STEP 7

Add, reduce, read off m

Reducing leaves numerator 243.

(252+720)/4096=972/4096=(2² · 3⁵)/2¹²=243/1024, m=243
Answer
243
Start by testing the condition on two nearby urns. The table (a,b,c,d)=(1,1,2,2) was counted as independent: it has a+c=3 red balls and a+b=2 striped balls, so P(red)=3/6=1/2, P(striped)=2/6=1/3, and P(both)=1/6, which is exactly 1/2·1/3 — independent, as claimed. The neighboring table (2,1,1,2) has the same total and the same shape but was rejected: there P(red)=3/6=1/2 and P(striped)=3/6=1/2, so the product is 1/4, while P(both)=2/6=1/3 — not independent. So the ad=bc test really does separate cases and is not quietly accepting everything. Next, a size check on the total: sorting all 4096 paint jobs by their table, 28 different tables satisfy ad=bc, carrying 252 paint jobs from the flat ones and 720 from the four full ones, and 252+720=972 is comfortably under 4096, giving a probability of about 0.237. Finally the answer choices themselves check the shape of the result: they are all odd, so m is odd, which forces n to be a power of 2 dividing 4096. Read over the denominator 1024, the five choices run from 243/1024≈ 0.2373 to 251/1024≈ 0.2451, and 972/4096=0.2373 lands exactly on the first one, (A). A curiosity worth noticing but not leaning on: 243/1024 happens to equal (3/4)⁵.
💡Key takeaway

Independence of "red" and "striped" is just the statement that the two diagonals of the striped-dotted by red-blue table have equal products, ad=bc, so count the tables that pass that test, weight each by how many of the 4096 flip outcomes build it, and 972/4096=243/1024 falls out.

  • Sort the urn into four boxes
  • Turn independence into an equation
  • Rewrite 6a=(a+b)(a+c) as ad=bc
  • Split into flat tables and full tables
  • Count the flat tables: 252
  • Count the full tables: 720
  • Add, reduce, read off m