Suppose the following conditions hold. A prize-drawing event uses a box containing 4 crimson balls and 5 blue balls. You are given four opportunities to draw a ball, without replacement.Under these conditions, you receive a prize of 100(10,000-won units) if you draw a crimson ball on the first draw, and 200(10,000-won units) if you draw a crimson ball on the fourth draw. Let be the random variable representing the total prize money obtained after drawing four balls from the box. If the expected value of is written aswhere and are relatively prime positive integers, find the value ofoindent
[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}