Question

Each of $2023$ balls is 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?

$\dfrac{2}{3}$

$\dfrac{3}{10}$

$\dfrac{1}{2}$

$\dfrac{1}{3}$

$\dfrac{1}{4}$

Solution:

All $3^{2023}$ assignments are equally likely. A sign filter counts the ones with every bin odd: $\frac{1}{8}\sum_{s \in \{\pm 1\}^3}(s_1 s_2 s_3)(s_1 + s_2 + s_3)^{2023}.$ Only $s = (1,1,1)$ and $s = (-1,-1,-1)$ carry weight, each $\pm 3^{2023};$ the other six terms total $-6.$ So 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.

2023 AMC 10B Problem 21

Solution:

Problem 21 in Other Years