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] To find the expected value of the total prize money ( X ), we need to consider the probabilities of drawing a crimson ball on the first and fourth draws. The prize for drawing a crimson ball on the first draw is 100 units, and for the fourth draw is 200 units. The expected value ( E[X] ) is the sum of the expected prizes from these two events.
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First Draw: The probability of drawing a crimson ball on the first draw is ( \frac{4}{9} ), and the prize is 100 units. The expected prize from the first draw is ( 100 \times \frac{4}{9} ).
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Fourth Draw: The probability of drawing a crimson ball on the fourth draw is more complex because it depends on the outcomes of the first three draws. However, by symmetry, the probability of drawing a crimson ball on any specific draw (including the fourth) is the same as the first draw, which is ( \frac{4}{9} ). The expected prize from the fourth draw is ( 200 \times \frac{4}{9} ).
Thus, the total expected value ( E[X] ) is: [ E[X] = 100 \times \frac{4}{9} + 200 \times \frac{4}{9} = \frac{400}{9} + \frac{800}{9} = \frac{1200}{9} = \frac{400}{3}. ]
The fraction ( \frac{400}{3} ) is already in its simplest form, so ( p = 400 ) and ( q = 3 ). The value of ( p \cdot q^2 ) is ( 400 \times 3^2 = 400 \times 9 = 3600 ).
[/THOUGHT] [PYTHON]
Verification of the expected value calculation
from fractions import Fraction
Probability of crimson on first draw
prob_first = Fraction(4, 9) prize_first = 100
Probability of crimson on fourth draw (same as first due to symmetry)
prob_fourth = Fraction(4, 9) prize_fourth = 200
Expected value calculation
E_X = prize_first * prob_first + prize_fourth * prob_fourth p = E_X.numerator q = E_X.denominator result = p * q * q
print(result) [/PYTHON]
[THOUGHT] The Python output confirms that the expected value calculation is correct, yielding ( p \cdot q^2 = 3600 ).
[/THOUGHT] \boxed{3600}