Problem Analysis #25

[THOUGHT] To solve this problem, we need to calculate the expected value of the random variable $X$ which represents the total prize money obtained after drawing four balls. The prize money rules are as follows:

• 100 (10,000-won units) if a crimson ball is drawn first.
• 200 (10,000-won units) if a crimson ball is drawn fourth.

We have 4 crimson balls and 5 blue balls, for a total of 9 balls. We will draw 4 balls without replacement. Let's break down the calculation:

1. Expected Prize from Drawing a Crimson Ball First: The probability of drawing a crimson ball first is the number of crimson balls divided by the total number of balls:

If this happens, the prize is 100 (10,000-won units). Thus, the expected prize from drawing a crimson ball first is:

2. Expected Prize from Drawing a Crimson Ball Fourth: To find the probability of drawing a crimson ball fourth, we need to consider all sequences of draws where the fourth ball is crimson. This is more complex because the fourth draw's probability depends on the outcomes of the first three draws. However, by symmetry in the drawing process without replacement, the probability of drawing a crimson ball in any specific draw (including the fourth) is the same as the initial probability of drawing a crimson ball:

If this happens, the prize is 200 (10,000-won units). Thus, the expected prize from drawing a crimson ball fourth is:

3. Total Expected Prize $E(X)$: The total expected prize $E(X)$ is the sum of the expected prizes from the two events:

The expected value of $X$ is $\frac{400}{3}$, where 400 and 3 are relatively prime. We need to find the value of $p \cdot q^2$, where $\frac{p}{q} = \frac{400}{3}$, so $p = 400$ and $q = 3$.

Thus, we calculate:

[/THOUGHT] $\boxed{3600}$