Maths Olympiad Prep

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, 2021

Algebra Difficulty 5.3 AIME, harder Prove it United States

Problem:

Almondine has a bag with NN balls, each of which is red, white, or blue. If Almondine picks three balls from the bag without replacement, the probability that she picks one ball of each color is larger than 23 percent. Compute the largest possible value of N3\left\lfloor\frac{N}{3}\right\rfloor.

Solution

Solution:

If k=N3k = \left\lfloor \frac{N}{3} \right\rfloor, then the maximum possible probability is 6k3(3k)(3k1)(3k2)\frac{6k^{3}}{(3k)(3k-1)(3k-2)}, with equality when there are kk balls of each of the three colors.

Going from 3k3k+13k \rightarrow 3k+1 replaces k3k2k+13k+1\frac{k}{3k-2} \rightarrow \frac{k+1}{3k+1}, which is smaller, and going from 3k+13k+23k+1 \rightarrow 3k+2 replaces k3k1k+13k+2\frac{k}{3k-1} \rightarrow \frac{k+1}{3k+2}, which is again smaller.

For this to be larger than 23100\frac{23}{100}, we find we need 0>7k2207k+460 > 7k^{2} - 207k + 46, and so k=29k = 29 is the maximal value.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.