2023 AMC 12B Problem 19

Question

Each of

2023

balls is placed in 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}

Each ball lands independently; use a parity filter, writing bin

i

odd as

\tfrac{1-(-1)^{k_i}}{2}

Summing over all assignments, the all-odd fraction is

\dfrac{3^{\,n}-3}{4\cdot 3^{\,n}};

see what it approaches for large

n

Counting assignments where all three bins are odd with the parity filter gives

\scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4}

for odd

n.

Dividing by the

3^n

total assignments, the probability is

\dfrac{3^n-3}{4\cdot 3^n},

which for

n=2023

is extremely close to

\tfrac14.

Thus, the correct answer is E.

Problem 19 in other years