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2023 AMC 12B Problem 19

Problem 19 of 25HarderAlgebraNumber TheoryCounting & Probability

Each of 20232023 balls is placed in one of 33 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

Counting assignments where all three bins are odd with the parity filter gives 18S{1,2,3}(1)S(32S)n=3n34 \scriptsize\frac{1}{8}\sum_{S\subseteq\{1,2,3\}}(-1)^{|S|}(3-2|S|)^n=\frac{3^n-3}{4} for odd n.n. Dividing by the 3n3^n total assignments, the probability is 3n343n,\dfrac{3^n-3}{4\cdot 3^n}, which for n=2023n=2023 is extremely close to 14.\tfrac14. Thus, the correct answer is E.

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Concepts: roots of unity · parity · basic probability

Problem text and solution from the LIVE past-contest archive. See also the AoPS wiki page for community solutions.