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May 18th, 2010, 01:25 PM   #1
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Expectation and Variance Question

Hi,

Can anyone help answering this question??

A college which has 120 male and 90 female students organizes an end-of-year ball. Each male student has, independently, probability 0.7 of going to the ball. The corresponding probability for females is 0.9. Males and females at the ball pair off until there is no-one of the opposite gender available. What is the expectation and variance for the number of males at the ball without a female partner? What extra assumption do you need to make in answering this question?

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May 18th, 2010, 02:13 PM   #2
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Re: Expectation and Variance Question

I get
13317914063183385152860742783713884725505010780166 72941997421755502667043210319640026719898824512636 38117452486544321470918809794832999512051913359073 09200707328415063814731244325594255095795476112477 72869669129075091300415910023598148580585456068528 0418074276117710255341/17632428509737537813642539535622820786902183233659 51365531331978947023821789340949272397231859649884 32310482214413697945577339443852406950188546640658 35006250000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 000000000000
with
Code:
sum(m=0,120,binomial(120,m)*(7/190)^m*(3/10)^(120-m)*sum(f=0,120,max(m-f,0)*binomial(90,f)*(9/10)^f*1/10^(90-f)))
although I suppose
Code:
sum(m=0,120,binomial(120,m)*(7/190)^m*(3/10)^(120-m)*sum(f=0,m-1,(m-f)*binomial(90,f)*(9/10)^f*1/10^(90-f)))
would be more efficient. (Code in in Pari/GP; the floating-point version is much faster than the agonizing 90 milliseconds these take.)
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