2026 AIME II Problem 5

Question

An urn contains

n

marbles. Each marble is either red or blue, and there are at least

7

marbles of each color. When

7

marbles are drawn randomly from the urn without replacement, the probability that exactly

4

of them are red equals the probability that exactly

5

of them are red. Find the sum of the five least values of

n

for which this is possible.

Answer

Write both probabilities with binomial coefficients,

\binom{r}{4}\binom{b}{3}

and

\binom{r}{5}\binom{b}{2},

and cancel the common factors.

The equation reduces to

5(b - 2) = 3(r - 4),

so

r \equiv 4 \pmod 5;

remember both colors need at least

7

marbles.

Say there are

r

red and

b

blue marbles,

r, b \ge 7.

The condition is

\binom{r}{4}\binom{b}{3} = \binom{r}{5}\binom{b}{2}.

Since

\binom{r}{5} = \binom{r}{4}\frac{r-4}{5}

and

\binom{b}{3} = \binom{b}{2}\frac{b-2}{3},

cancelling gives

\frac{b - 2}{3} = \frac{r - 4}{5},

that is,

b = \frac{3r - 2}{5}.

So

r \equiv 4 \pmod 5,

and

b \ge 7

requires

3r - 2 \ge 35,

so

r \ge 14.

The five smallest choices are

r = 14, 19, 24, 29, 34

with

b = 8, 11, 14, 17, 20,

giving

n = 22, 30, 38, 46, 54.

The sum is

22 + 30 + 38 + 46 + 54 = 190.

Problem 5 in other years