AMC 10 · 2023 · #21

Grade 8 probability

probability-basicparitygenerating-functionscombinations-basicpattern-recognition easier-related-problempattern-recognitioncomplementary-counting ↑ Prerequisites: probability-basicparitycombinations-basic

📏 Long solution 💡 4 insights

Problem

Drop 2023 distinguishable balls into 3 bins, each ball uniformly at random and independently. Find the answer choice closest to the probability that all three bins end up with an odd number of balls.

Pick an answer.

(A)

$frac{2}{3}$

(B)

$frac{3}{10}$

(C)

$frac{1}{2}$

(D)

$frac{1}{3}$

(E)

$frac{1}{4}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Solve an Easier Related Problem

We cannot count 3²⁰²³ outcomes directly. Tool #9 (Easier Problem) — replace 2023 with a small odd number like n = 1, 3, 5 — lets us compute the probability by hand. Tool #5 (Pattern) reveals the answer approaches $\frac{1}{4}$ very quickly. Tool #16 (Complement) gives the cleanest algebraic confirmation. Finally tool #3 (Eliminate) matches our limit to the closest of the multiple-choice values.

1STEP 1

Shrink 2023 to the smallest odd n=1: one bin gets the ball, two get zero, so all-odd is impossible — P₁ = 0.

n = 1: P₁ = 0

2STEP 2

For n=3, all-odd forces (1,1,1): 3!=6 of the 27 placements, so P₃ = $\frac{2}{9}$ ≈ 0.222.

n=3: P₃ =

\frac{3!}{3³}

\frac{6}{27}

\frac{2}{9}

3STEP 3

For n=5 the odd partitions are permutations of (1,1,3): 60 of 243 placements, so P₅ = $\frac{20}{81}$ ≈ 0.2469.

n=5: P₅ =

\frac{60}{243}

\frac{20}{81}

≈ 0.2469

4STEP 4

The values 0, 0.222, 0.2469 climb toward but stay below $\frac{1}{4}$ — conjecture P_n → $\frac{1}{4}$ for large odd n.

P₁ = 0, P₃ =

\frac{2}{9}

≈ 0.222, P₅ =

\frac{20}{81}

≈ 0.2469 ↗

\frac{1}{4}

5STEP 5

A parity/symmetry argument gives the exact P_n = $\frac{1}{4}$ (1 - $\frac{1}{3ⁿ⁻¹}$ ); for n=2023 the correction is astronomically tiny, so P ≈ $\frac{1}{4}$ .

P_n =

\frac{1}{4}

(1 -

\frac{1}{(3^ n-1)}

) →

\frac{1}{4}

6STEP 6

P₂₀₂₃ ≈ 0.25; among $\frac{2}{3}$ , $\frac{3}{10}$ , $\frac{1}{2}$ , $\frac{1}{3}$ , $\frac{1}{4}$ the closest is $\frac{1}{4}$ , choice (E).

P₂023 ≈

\frac{1}{4}

→ (E)

Answer

\frac{1}{4}

Three quick checks pass. (1) The small-case probabilities 0,

\frac{2}{9}

\frac{20}{81}

all satisfy the formula

\frac{1}{4}

(1 -

\frac{1}{3ⁿ⁻¹}

): for n=3,

\frac{1}{4}

(1 -

\frac{1}{9}

) =

\frac{1}{4}

\frac{8}{9}

\frac{2}{9}

— matches. For n=5,

\frac{1}{4}

(1 -

\frac{1}{81}

) =

\frac{20}{81}

— matches. (2) The limit

\frac{1}{4}

also has a heuristic: of the 2³ = 8 parity patterns of the three bin counts, only one is (O,O,O), giving a baseline

\frac{1}{8}

... but conditioning on the sum being odd halves the space, doubling to

\frac{1}{4}

. (3) 0.25 is closer to all the offered choices than the runners-up

\frac{3}{10}

=0.3 (gap 0.05) and

\frac{1}{3}

≈ 0.333 (gap 0.083).

💡Key takeaway

This AMC 10 problem only needs Grade 8 exponent reasoning you already know — try n = 1, 3, 5 balls instead of 2023 to see the probability climb 0, $\frac{2}{9}$ , $\frac{20}{81}$ , … toward $\frac{1}{4}$ . The general formula $\frac{1}{4}$ (1 - $\frac{1}{3ⁿ⁻¹}$ ) has a correction so tiny for n=2023 that the closest choice is (E) $\frac{1}{4}$ .