aime26-19
0.000
Proposed solution
To solve this problem, we start by defining the variables and setting up the equations based on the given conditions. Let \( R \) be the number of red marbles and \( B \) be the number of blue marbles in the urn, with \( R + B = n \). We know that \( R \geq 7 \) and \( B \geq 7 \).
We are given that the probability of drawing exactly 4 red marbles out of 7 is equal to the probability of drawing exactly 5 red marbles out of 7. We can express these probabilities using combinations:
\[
P(\text{exactly 4 red}) = \frac{\binom{R}{4} \binom{B}{3}}{\binom{n}{7}}
\]
\[
P(\text{exactly 5 red}) = \frac{\binom{R}{5} \binom{B}{2}}{\binom{n}{7}}
\]
Since these probabilities are equal, we have:
\[
\frac{\binom{R}{4} \binom{B}{3}}{\binom{n}{7}} = \frac{\binom{R}{5} \binom{B}{2}}
\]
By simplifying, we get:
\[
\binom{R}{4} \binom{B}{3} = \binom{R}{5} \binom{B}{2}
\]
Using the properties of combinations, we can rewrite the binomial coefficients:
\[
\frac{R(R-1)(R-2)(R-3)}{4!} \cdot \frac{B(B-1)(B-2)}{3!} = \frac{R(R-1)(R-2)(R-3)(R-4)}{5!} \cdot \frac{B(B-1)}{2!}
\]
Simplifying further, we get:
\[
\frac{(R-3)(R-4)}{4 \cdot 5} = \frac{B(B-1)}{2}
\]
\[
\frac{(R-3)(R-4)}{120} = \frac{B(B-1)}{2}
\]
\[
(R-3)(R-4) = 60 B(B-1)
\]
Next, we need to find integer solutions for \( R \) and \( B \) such that \( R + B = n \) and \( R, B \geq 7 \). We will test values for \( R \) starting from 7 upwards to find the smallest \( n \) that satisfies this equation.
**Case 1: \( R = 7 \)**
\[
(7-3)(7-4) = 60 B(B-1)
\]
\[
4 \cdot 3 = 60 B(B-1)
\]
\[
12 = 60 B(B-1)
\]
\[
B(B-1) = \frac{12}{60} = \frac{1}{5}
\]
This does not yield an integer solution for \( B \).
**Case 2: \( R = 8 \)**
\[
(8-3)(8-4) = 60 B(B-1)
\]
\[
5 \cdot 4 = 60 B(B-1)
\]
\[
20 = 60 B(B-1)
\]
\[
B(B-1) = \frac{20}{60} = \frac{1}{3}
\]
This does not yield an integer solution for \( B \).
**Case 3: \( R = 9 \)**
\[
(9-3)(9-4) = 60 B(B-1)
\]
\[
6 \cdot 5 = 60 B(B-1)
\]
\[
30 = 60 B(B-1)
\]
\[
B(B-1) = \frac{30}{60} = \frac{1}{2}
\]
This does not yield an integer solution for \( B \).
**Case 4: \( R = 10 \)**
\[
(10-3)(10-4) = 60 B(B-1)
\]
\[
7 \cdot 6 = 60 B(B-1)
\]
\[
42 = 60 B(B-1)
\]
\[
B(B-1) = \frac{42}{60} = \frac{7}{10}
\]
This does not yield an integer solution for \( B \).
**Case 5: \( R = 11 \)**
\[
(11-3)(11-4) = 60 B(B-1)
\]
\[
8 \cdot 7 = 60 B(B-1)
\]
\[
56 = 60 B(B-1)
\]
\[
B(B-1) = \frac{56}{60} = \frac{14}{15}
\]
This does not yield an integer solution for \( B \).
**Case 6: \( R = 12 \)**
\[
(12-3)(12-4) = 60 B(B-1)
\]
\[
9 \cdot 8 = 60 B(B-1)
\]
\[
72 = 60 B(B-1)
\]
\[
B(B-1) = \frac{72}{60} = \frac{6}{5}
\]
This does not yield an integer solution for \( B \).
**Case 7: \( R = 13 \)**
\[
(13-3)(13-4) = 60 B(B-1)
\]
\[
10 \cdot 9 = 60 B(B-1)
\]
\[
90 = 60 B(B-1)
\]
\[
B(B-1) = \frac{90}{60} = \frac{3}{2}
\]
This does not yield an integer solution for \( B \).
**Case 8: \( R = 14 \)**
\[
(14-3)(14-4) = 60 B(B-1)
\]
\[
11 \cdot 10 = 60 B(B-1)
\]
\[
110 = 60 B(B-1)
\]
\[
B(B-1) = \frac{110}{60} = \frac{11}{6}
\]
This does not yield an integer solution for \( B \).
**Case 9: \( R = 15 \)**
\[
(15-3)(15-4) = 60 B(B-1)
\]
\[
12 \cdot 11 = 60 B(B-1)
\]
\[
132 = 60 B(B-1)
\]
\[
B(B-1) = \frac{132}{60} = \frac{11}{5}
\]
This does not yield an integer solution for \( B \).
**Case 10: \( R = 16 \)**
\[
(16-3)(16-4) = 60 B(B-1)
\]
\[
13 \cdot 12 = 60 B(B-1)
\]
\[
156 = 60 B(B-1)
\]
\[
B(B-1) = \frac{156}{60} = \frac{13}{5}
\]
This does not yield an integer solution for \( B \).
**Case 11: \( R = 17 \)**
\[
(17-3)(17-4) = 60 B(B-1)
\]
\[
14 \cdot 13 = 60 B(B-1)
\]
\[
182 = 60 B(B-1)
\]
\[
B(B-1) = \frac{182}{60} = \frac{91}{30}
\]
This does not yield an integer solution for \( B \).
**Case 12: \( R = 18 \)**
\[
(18-3)(18-4) = 60 B(B-1)
\]
\[
15 \cdot 14 = 60 B(B-1)
\]
\[
210 = 60 B(B-1)
\]
\[
B(B-1) = \frac{210}{60} = \frac{7}{2}
\]
This does not yield an integer solution for \( B \).
**Case 13: \( R = 19 \)**