To solve the problem, we are given a box containing 4 crimson balls and 5 blue balls (total of 9 balls). We draw four balls without replacement, and the random variable $X$ represents the total prize money obtained.

Step 1: Understand the Prize Structure

• If the first draw is crimson, you receive 100 (units).
• If the fourth draw is crimson, you receive 200 (units).

Let:

• $I_1 = 1$ if the first ball is crimson, else 0
• $I_4 = 1$ if the fourth ball is bolder, else 0

Then,

We are to compute:

Step 2: Compute the Expectations

In a random permutation of all 9 balls, each position is equally likely to be any specific ball. Therefore, the probability that the first ball is crimson is:

Similarly, by symmetry (since all positions are equally likely to be any ball), the probability that the fourth draw is crimson is also:

Thus:

Step 3: Final Computation

We are told that the expected value of $X$ is expressed as $\frac{p}{q}$ in lowest terms.

We are to compute:

✅ Final Answer

[THOUGHT] To solve the problem, we need to compute the expected value of the total prize money $X$, where:

• $X = 100 \cdot I_1 + 200 \cdot I_4$
• $I_1 = 1$ if the first draw is crimson (otherwise 0)
• $I_4 = 1$ if the fourth draw is crimson (otherwise 0)

The box contains 4 crimson and 5 blue balls (total 9). Since all positions are equally likely to be crimson in a random permutation of the 9 balls, the probability that any specific draw (e.g., first or fourth) is crimson is $\frac{4}{9}$.

Thus:

This fraction is in lowest terms since $\gcd(400, 3) = 1$. So $p = 400$, $q = 3$, and:

We verify this calculation using Python to ensure accuracy.

[PYTHON] from fractions import Fraction

Calculate expected value of X

E_X = Fraction(4, 9) * 100 + Fraction(4, 9) * 200 print(E_X) # Output: 400/3

Extract p and q

p = E_X.numerator q = E_X.denominator

Compute p * q^2

result = p * (q ** 2) print(result) # Output: 3600 [/PYTHON]

[THOUGHT] I need to compute the expected value of the total prize money $X$, where:

• Prize is 100 if first draw is crimson (4 crimson out of 9)
• Prize is 200 if fourth draw is crimson (also 4/9 probability due to symmetry)

So expectation $E[X] = 100 \cdot \frac{4}{9} + 200 \cdot \frac{4}{9} = \frac{400 + 800}{9} = \frac{1200}{9} = \frac{400}{3}$

Now I need to verify this calculation using Python to ensure no arithmetic errors. [PYTHON] from fractions import Fraction

Probabilities for first and fourth draws being crimson

p_first_crimson = Fraction(4, 9) p_fourth_crimson = Fraction(4, 9)

Expected value of X

E_X = 100 * p_first_crimson + 200 * p_fourth_crimson

Get p and q in reduced fraction form

p_val = E_X.numerator q_val = E_X.denominator

Calculate p * q^2

result = p_val * (q_val ** 2) print(f"E[X] = {E_X}") print(f"p = {p_val}, q = {q_val}") print(f"p * q^2 = {result}") [/PYTHON]

[THOUGHT]

[PYTHON] from fractions import Fraction

Define the expectation calculation

E_X = Fraction(4, 9) * 100 + Fraction(4, 9) * 200

Extract p and q in reduced form

p = E_X.numerator q = E_X.denominator

Compute p * q^2

result = p * (q ** 2)

print(f"E[X] = {E_X}") print(f"p = {p}, q = {q}") print(f"p * q^2 = {result}") [/PYTHON]

To solve the problem, we are given a scenario involving a box with 4 crimson and 5 blue balls (total of 9), and we draw 4 balls without replacement. The random variable $X$ represents the total prize money obtained, where:

• You receive 100 if the first draw is crimson
• You receive 200 if the fourth draw is crimson

Thus, the expected value of $X$ is:

Step 1: Compute Individual Probabilities

In a random permutation of the 9 balls, each position has an equal chance of being any of the 9 balls. Therefore, for any specific draw (e.g., first, second, third, or fourth), the probability that it is crimson is:

This is because the process of drawing without replacement preserves the symmetry across all positions in the permutation.

So:

• $\mathbb{P}(\text{first is crimson}) = \frac{4}{9}$
• $\mathbb{P}(\text{fourth is crimson}) = \frac{4}{9}$

Step 2: Compute the Expected Value

This fraction is in lowest terms since $\gcd(400, 3) = 1$, so $p = 400$, $q = 3$.

Step 3: Final Computation

We are asked to compute:

✅ Final Answer