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 use the linearity of expectation. The prize is 100 if the first draw is crimson and 200 if the fourth draw is crimson. The expected value ( E[X] ) is the sum of the expected values for each prize.
- Probability of drawing a crimson ball on the first draw: There are 4 crimson balls out of 9 total, so the probability is ( \frac{4}{9} ).
- Probability of drawing a crimson ball on the fourth draw: Due to symmetry in permutations without replacement, this probability is also ( \frac{4}{9} ).
Thus, the expected value is: [ E[X] = 100 \cdot \frac{4}{9} + 200 \cdot \frac{4}{9} = \frac{400 + 800}{9} = \frac{1200}{9} = \frac{400}{3} ] Here, ( p = 400 ) and ( q = 3 ), which are coprime. Therefore, ( p \cdot q^2 = 400 \cdot 3^2 = 3600 ).
[/THOUGHT] [PYTHON] from fractions import Fraction
Calculate expected value components
e1 = Fraction(4, 9) * 100 e4 = Fraction(4, 9) * 200 total = e1 + e4
Extract p and q
p = total.numerator q = total.denominator
Compute p * q^2
result = p * (q ** 2) print(result) [/PYTHON]
[THOUGHT] The Python output confirms that ( p \cdot q^2 = 3600 ). The expected value calculation aligns with the derived fraction ( \frac{400}{3} ), where ( p = 400 ) and ( q = 3 ). Thus, the final answer is: \boxed{3600} [/THOUGHT]