Problem Analysis #25

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Problem Statement

Suppose the following conditions hold. A prize-drawing event uses a box containing 4 crimson balls and 5 blue balls. You are given four opportunities to draw a ball, without replacement.Under these conditions, you receive a prize of 100(10,000-won units) if you draw a crimson ball on the first draw, and 200(10,000-won units) if you draw a crimson ball on the fourth draw. Let XX be the random variable representing the total prize money obtained after drawing four balls from the box. If the expected value of XX is written aspq,\frac{p}{q},where pp and qq are relatively prime positive integers, find the value ofpq2.p \cdot q^2.oindent

Ground Truth
3600
Execution Trace
Reasoning Process

[THOUGHT] The random variable X = 100 * I_first_crimson + 200 * I_fourth_crimson, where I are indicators. By linearity, E[X] = 100 * P(first crimson) + 200 * P(fourth crimson). Due to symmetry in sampling without replacement, P(first crimson) = 4/9 and P(fourth crimson) = 4/9. Thus E[X] = 300 * (4/9) = 1200/9 = 400/3. Here p=400, q=3, and p * q^2 = 400 * 9 = 3600. Verify with Python using fractions. [/THOUGHT] [PYTHON] from fractions import Fraction

p1 = Fraction(4, 9) p4 = Fraction(4, 9) ex = 100 * p1 + 200 * p4 p = ex.numerator q = ex.denominator result = p * q**2 print(result) [/PYTHON]

Tool Output
[PYTHON OUTPUT] 3600 [/PYTHON OUTPUT]
Reasoning Process

[THOUGHT] The Python code confirms the result is 3600. [/THOUGHT] \boxed{3600}

Explanation

The reasoning begins by defining the random variable X as the total prize money: X = 100 * I_first + 200 * I_fourth, where I_first is the indicator for drawing a crimson ball on the first draw, and I_fourth is the indicator for the fourth draw. The expected value is computed using linearity of expectation: E[X] = 100 * P(I_first = 1) + 200 * P(I_fourth = 1).

Next, the probabilities are determined. With 4 crimson balls out of 9 total balls and draws without replacement, the probability of drawing a crimson ball on the first draw is 4/9. Due to the symmetry of sampling without replacement, the probability for the fourth draw is also 4/9, as each ball is equally likely to be in any position in the sequence.

Thus, E[X] = 100 * (4/9) + 200 * (4/9) = 300 * (4/9) = 1200/9, which simplifies to 400/3. Here, 400 and 3 are relatively prime, so p = 400 and q = 3. The required value is p * q^2 = 400 * 3^2 = 400 * 9 = 3600.

To verify, Python code using the fractions module computes these values exactly: p1 = Fraction(4, 9), p4 = Fraction(4, 9), ex = 100 * p1 + 200 * p4, then extracts p and q from ex and calculates p * q**2, yielding 3600, confirming the result.

Final Model Output
3600

Comparative Analysis (P25) (Average across runs)

Avg Time Taken

Gemini-3-Pro-Preview
35.8
Grok-4.1-fast
31.5
HCX-007(high)
28.2
Deepseek-V3.2
27.9
Llama-VARCO-8B-Instruct
27.7
Solar-Pro-2 (31B)(high)
24.3
Claude-Opus-4.5
14.3
A.X-4.0 (72B)
13.5
GPT-5.1 (high)
13.4
EXAONE-4.0.1-32B (high)
3.9

Avg Token Usage

Gemini-3-Pro-Preview
5595.0
Grok-4.1-fast
5445.0
Solar-Pro-2 (31B)(high)
5220.0
HCX-007(high)
4007.0
Claude-Opus-4.5
3853.0
Deepseek-V3.2
3682.0
EXAONE-4.0.1-32B (high)
3480.0
GPT-5.1 (high)
2933.0
Llama-VARCO-8B-Instruct
1847.0
A.X-4.0 (72B)
1715.0