[THOUGHT] To find the expected value of the total prize money $X$, we use the linearity of expectation. The prize is 100 if the first draw is crimson and 200 if the fourth draw is crimson. The expected value $E[X]$ is the sum of the expected values for each prize.

1. Probability of drawing a crimson ball on the first draw: There are 4 crimson balls out of 9 total, so the probability is $\frac{4}{9}$.
2. Probability of drawing a crimson ball on the fourth draw: Due to symmetry in permutations without replacement, this probability is also $\frac{4}{9}$.

Thus, the expected value is:

Here, $p = 400$ and $q = 3$, which are coprime. Therefore, $p \cdot q^2 = 400 \cdot 3^2 = 3600$.

[/THOUGHT] [PYTHON] from fractions import Fraction

Calculate expected value components

e1 = Fraction(4, 9) * 100 e4 = Fraction(4, 9) * 200 total = e1 + e4

Extract p and q

p = total.numerator q = total.denominator

Compute p * q^2

result = p * (q ** 2) print(result) [/PYTHON]

[THOUGHT] The Python output confirms that $p \cdot q^2 = 3600$. The expected value calculation aligns with the derived fraction $\frac{400}{3}$, where $p = 400$ and $q = 3$. Thus, the final answer is: $\boxed{3600}$ [/THOUGHT]