[THOUGHT] The problem asks for the expected value of the total prize money $X$. Let $C_i$ be the event that the ball drawn on the $i$-th draw is crimson. The total prize $X$ is given by $X = 100 \cdot \mathbb{I}(C_1) + 200 \cdot \mathbb{I}(C_4)$, where $\mathbb{I}(\cdot)$ is the indicator function. By linearity of expectation, $E[X] = 100 \cdot P(C_1) + 200 \cdot P(C_4)$.

The box contains 4 crimson balls and 5 blue balls, for a total of 9 balls. The draws are without replacement.

I need to calculate $P(C_1)$ and $P(C_4)$. $P(C_1)$ is the probability that the first ball drawn is crimson. $P(C_4)$ is the probability that the fourth ball drawn is crimson.

Since the draws are without replacement and the balls are indistinguishable apart from color, the probability of drawing a crimson ball at any specific position $k$ (where $1 \le k \le 9$) is the same. So, $P(C_1) = P(C_4) = \frac{\text{Number of crimson balls}}{\text{Total number of balls}}$.

Let's verify this using Python to be absolutely sure, although it's a standard probability result. I will simulate the drawing process or calculate the probabilities using combinations.

Total balls $N = 9$. Crimson balls $K = 4$. $P(C_1) = \frac{4}{9}$. $P(C_4) = \frac{4}{9}$.

Then $E[X] = 100 \cdot \frac{4}{9} + 200 \cdot \frac{4}{9} = 300 \cdot \frac{4}{9} = \frac{1200}{9} = \frac{400}{3}$.

So $p = 400$, $q = 3$. They are relatively prime since 400 is not divisible by 3 (sum of digits is 4). The value to find is $p \cdot q^2 = 400 \cdot 3^2 = 400 \cdot 9 = 3600$.

I will write a Python script to calculate the exact probabilities by iterating through all permutations or using hypergeometric logic to confirm $P(C_4)$.

[/THOUGHT] [PYTHON] from itertools import permutations from fractions import Fraction

Setup

crimson_count = 4 blue_count = 5 total_balls = crimson_count + blue_count

Balls represented as 1 for Crimson, 0 for Blue

balls = [1] * crimson_count + [0] * blue_count

We draw 4 balls without replacement.

However, the probability of the k-th ball being Crimson is just Total_Crimson / Total_Balls due to symmetry.

Let's verify P(C_1) and P(C_4) by simulation or calculation.

Calculation approach:

P(C_1) is obviously 4/9.

P(C_4): Consider the sequence of 4 draws.

Total permutations of 4 balls from 9 is P(9, 4).

Number of permutations where 4th is Crimson:

Pick a Crimson for 4th spot: 4 ways.

Pick remaining 3 balls for spots 1, 2, 3 from remaining 8 balls: P(8, 3) ways.

Probability = (4 * P(8, 3)) / P(9, 4)

def nPr(n, r): import math return math.factorial(n) // math.factorial(n - r)

total_perms = nPr(total_balls, 4)

For C1 (1st is Crimson)

Pick C for 1st: 4 ways

Pick 3 from remaining 8: P(8, 3)

count_c1 = crimson_count * nPr(total_balls - 1, 3) prob_c1 = Fraction(count_c1, total_perms)

For C4 (4th is Crimson)

Pick C for 4th: 4 ways

Pick 3 from remaining 8 for spots 1, 2, 3: P(8, 3)

count_c4 = crimson_count * nPr(total_balls - 1, 3) prob_c4 = Fraction(count_c4, total_perms)

print(f"{prob_c1=}") print(f"{prob_c4=}")

expected_value = 100 * prob_c1 + 200 * prob_c4 print(f"{expected_value=}")

p = expected_value.numerator q = expected_value.denominator print(f"{p=}, {q=}")

result = p * (q**2) print(f"{result=}") [/PYTHON] [THOUGHT] The code output confirms the theoretical reasoning. $P(C_1) = 4/9$ $P(C_4) = 4/9$ $E[X] = 100(4/9) + 200(4/9) = 300(4/9) = 1200/9 = 400/3$. $p = 400, q = 3$. $p$ and $q$ are relatively prime (gcd(400, 3) = 1). We need to find $p \cdot q^2$. $400 \cdot 3^2 = 400 \cdot 9 = 3600$.

The logic holds. The "without replacement" condition doesn't change the marginal probability of the $k$-th draw being a specific color, provided we don't condition on previous draws. The expectation is linear, so we just sum the marginal expectations.

Final check: Prize = 100 if 1st is Crimson. Prize = 200 if 4th is Crimson. Total Prize $X$. $E[X] = E[100 I(C_1) + 200 I(C_4)] = 100 P(C_1) + 200 P(C_4)$. $P(C_1) = 4/9$. $P(C_4) = 4/9$. Sum = $100(4/9) + 200(4/9) = 300(4/9) = 400/3$. $p=400, q=3$. $p \cdot q^2 = 400 \cdot 9 = 3600$.

The result seems solid. [/THOUGHT] \boxed{3600}

[THOUGHT] The Python code confirmed the probabilities $P(C_1) = 4/9$ and $P(C_4) = 4/9$. The expected value is $E[X] = 100(4/9) + 200(4/9) = 300(4/9) = 400/3$. Thus $p=400$ and $q=3$. These are relatively prime. The value to compute is $p \cdot q^2 = 400 \cdot 3^2 = 400 \cdot 9 = 3600$. [/THOUGHT] \boxed{3600}