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] We are given a box with 4 crimson balls and 5 blue balls, total 9 balls. We draw 4 times without replacement. The prize is 100 (in 10,000 won units) for a crimson ball on the first draw and 200 for a crimson ball on the fourth draw. The total prize (X) is the sum of these prizes, so (X = 100 \cdot I_1 + 200 \cdot I_4), where (I_1) is the indicator for the first draw being crimson and (I_4) is the indicator for the fourth draw being crimson.
The expected value is: [ E[X] = E[100 \cdot I_1] + E[200 \cdot I_4] = 100 \cdot P(I_1 = 1) + 200 \cdot P(I_4 = 1) ]
Since the draws are without replacement, the probability that the first ball is crimson is: [ P(I_1 = 1) = \frac{4}{9} ]
By symmetry, the probability that the fourth ball is crimson is also: [ P(I_4 = 1) = \frac{4}{9} ]
Thus: [ E[X] = 100 \cdot \frac{4}{9} + 200 \cdot \frac{4}{9} = \frac{400}{9} + \frac{800}{9} = \frac{1200}{9} = \frac{400}{3} ]
The expected value is (\frac{400}{3}), where (p = 400) and (q = 3) are relatively prime. Therefore: [ p \cdot q^2 = 400 \cdot 3^2 = 400 \cdot 9 = 3600 ] \boxed{3600}
[THOUGHT] We must use the Python tool at least once. The expected value of X is 400/3, so p=400, q=3, and p*q^2=3600. We'll compute this in Python. [/THOUGHT] [PYTHON] from fractions import Fraction frac = Fraction(400, 3) p = frac.numerator q = frac.denominator result = p * q**2 print(result) [/PYTHON]
\boxed{3600}