[THOUGHT] Compute the expected prize X. Since X = 100·I(first draw is crimson) + 200·I(fourth draw is crimson), by linearity of expectation we need P(first is crimson) and P(fourth is crimson). By symmetry in sampling without replacement, each draw position has the same probability of being crimson, namely 4/9. I will verify by enumerating all 4-draw sequences from the urn and computing the average payout. [/THOUGHT] [PYTHON] import itertools from fractions import Fraction

4 crimson (C), 5 blue (B); draw 4 without replacement (ordered)

balls = ['C']*4 + ['B']*5

total_prob = Fraction(0,1) EV = Fraction(0,1)

enumerate all ordered 4-tuples without replacement

for seq in itertools.permutations(range(9), 4): drawn = [balls[i] for i in seq] # probability of any specific ordered 4-draw sequence without replacement prob = Fraction(1, 987*6) total_prob += prob payout = 0 if drawn[0] == 'C': payout += 100 if drawn[3] == 'C': payout += 200 EV += prob * payout

total_prob, EV, float(EV) print(total_prob) print(EV) print(EV.numerator * EV.denominator**2) [/PYTHON][THOUGHT] The expected prize is $E[X]=100\Pr(\text{1st crimson})+200\Pr(\text{4th crimson})$. By symmetry of drawing without replacement, each draw position has probability $4/9$ of being crimson. Thus $E[X]=100\cdot\frac49+200\cdot\frac49=\frac{400}{3}$. With $p=400,q=3$, we need $p q^2=400\cdot 9=3600$. [/THOUGHT] $\boxed{3600}$

[THOUGHT] From the Python output, $E[X]=\frac{400}{3}$, so $p=400$ and $q=3$. Therefore $p\cdot q^2 = 400\cdot 3^2 = 3600$. $\boxed{3600}$