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HMMT Spring Guts Round, 2021

Algebra Difficulty 5.3 AIME, harder Prove it United States

Problem:

Almondine has a bag with $N$ 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 $\left\lfloor\frac{N}{3}\right\rfloor$ .

Solution

Solution:

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

Going from $3k \rightarrow 3k+1$ replaces $\frac{k}{3k-2} \rightarrow \frac{k+1}{3k+1}$ , which is smaller, and going from $3k+1 \rightarrow 3k+2$ replaces $\frac{k}{3k-1} \rightarrow \frac{k+1}{3k+2}$ , which is again smaller.

For this to be larger than $\frac{23}{100}$ , we find we need $0 > 7k^{2} - 207k + 46$ , and so $k = 29$ 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.