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 as
where and are relatively prime positive integers, find the value of
oindent
[THOUGHT] Let me analyze this problem step by step.
We have a box with 4 crimson balls and 5 blue balls (9 total). We draw 4 balls without replacement.
Prize structure:
- 100 (in 10,000-won units) if we draw a crimson ball on the first draw
- 200 (in 10,000-won units) if we draw a crimson ball on the fourth draw
Let X be the total prize money.
X can be:
- 0 (if neither first nor fourth draw is crimson)
- 100 (if only first draw is crimson)
- 200 (if only fourth draw is crimson)
- 300 (if both first and fourth draws are crimson)
I need to calculate the probability of each case and find E[X].
Let me use indicator random variables:
- Let I₁ = 1 if first draw is crimson, 0 otherwise
- Let I₄ = 1 if fourth draw is crimson, 0 otherwise
Then X = 100·I₁ + 200·I₄
E[X] = 100·E[I₁] + 200·E[I₄] = 100·P(first is crimson) + 200·P(fourth is crimson)
P(first is crimson) = 4/9
P(fourth is crimson) = ?
By symmetry in drawing without replacement, the probability that any specific position has a crimson ball is the same as the proportion of crimson balls in the box.
So P(fourth is crimson) = 4/9
Therefore E[X] = 100·(4/9) + 200·(4/9) = 400/9 + 800/9 = 1200/9 = 400/3
Let me verify this with Python by computing all possibilities. [/THOUGHT] [PYTHON] from itertools import permutations from fractions import Fraction
4 crimson balls (C) and 5 blue balls (B)
Total 9 balls, we draw 4 without replacement
balls = ['C'] * 4 + ['B'] * 5
Count all possible 4-ball draws (ordered)
total_outcomes = 0 total_prize = 0
We need to consider all permutations of choosing 4 balls from 9
from itertools import permutations
Generate all possible sequences of 4 draws
all_draws = list(permutations(balls, 4))
for draw in all_draws: prize = 0 if draw[0] == 'C': # First draw is crimson prize += 100 if draw[3] == 'C': # Fourth draw is crimson prize += 200 total_prize += prize total_outcomes += 1
expected_value = Fraction(total_prize, total_outcomes) print(f"Total outcomes: {total_outcomes}") print(f"Total prize sum: {total_prize}") print(f"Expected value: {expected_value}") print(f"p = {expected_value.numerator}, q = {expected_value.denominator}") print(f"p * q^2 = {expected_value.numerator * expected_value.denominator**2}") [/PYTHON]
[THOUGHT] The Python calculation confirms that the expected value E[X] = 400/3, where p = 400 and q = 3 are relatively prime positive integers. Therefore, p · q² = 400 × 3² = 400 × 9 = 3600. [/THOUGHT]