2023 AMC 10B Problem 21

Each of

2023

balls is randomly placed into one of

3

bins. Which of the following is closest to the probability that each of the bins will contain an odd number of balls?

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Solution

All

3^{2023}

assignments are equally likely. A sign filter counts the ones with every bin odd:

\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)

\cdot (s_1 + s_2 + s_3)^{2023}.

For

s = (1,1,1)

and

s = (-1,-1,-1),

both terms equal

3^{2023}.

For each other sign choice, the sum in parentheses is

1

-1,

and the full term equals

-1,

so these six terms total

-6.

Thus the count is

\frac{2 \cdot 3^{2023} - 6}{8} = \frac{3^{2023} - 3}{4}.

Dividing, the probability is

\frac{3^{2023} - 3}{4 \cdot 3^{2023}} = \frac{1}{4} - \frac{1}{4 \cdot 3^{2022}},

a hair under

\frac{1}{4}.

Thus, E is the correct answer.