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 December 17th, 2008, 10:32 AM #1 Newbie   Joined: Sep 2008 Posts: 9 Thanks: 0 Probability Question - 12 girls and 10 boys dividing into... Need help desperately: In the class there are 10 boys and 12 girls The students randomly divide into pairs, what's the probability they'll divide into 10 mixed pairs (1 boy and 1 girl) with 1 pair of 2 girls? I've been going crazy over this...I've tried to do 11!x2!x2!x2!x2!x2!x2!x2!x2!x2!x2!x2! but I don't think it's the right answer. What am I missing? Help, please?
 December 17th, 2008, 12:47 PM #2 Senior Member   Joined: Nov 2007 Posts: 258 Thanks: 0 Re: Probability Question - 12 girls and 10 boys dividing into... You probably know that $P(A\cap B)= P(B)P(A|B) = P(A)P(B|A)$ which generalizes easily to unions of more than two sets. Here if $B_1,...,B_{10}$ are the events "Boy j pairs up with a girl" we have $P(\bigcap_{j=1}^{10} B_j) = P(B_1)*P(B_2|B_1)*P(B_3|B_1 \cap B_2)*\: \: ...\: \: *P(B_{10} | B_1 \cap ... \cap B_9)$ $= (12/24)*(11/22)*(10/20)*\: \: ...\: \: *(3/6) = (1/2)^{10} = \frac{1}{1024}$ (because after k boys have hoooked up, there are 2k people out and 12-k girls remaining) Hope that helps! Nice problem, it's not obvious that the probability is 1/2 at every pick. Although I'm sure in real life it's a bit bigger
 December 21st, 2008, 10:17 AM #3 Newbie   Joined: Sep 2008 Posts: 9 Thanks: 0 Re: Probability Question - 12 girls and 10 boys dividing into... Thank you, helped me understand
 December 21st, 2008, 12:46 PM #4 Newbie   Joined: Sep 2008 Posts: 9 Thanks: 0 Re: Probability Question - 12 girls and 10 boys dividing into... Okay, that's what I came up with: 12/22*10/21*11/20*9/19*10/18*8/17*9/16*7/15*8/14*6/13*7/12*5/11*6/10*4/9*5/8*3/7*4/6*2/5*3/4*1/3*2/2*1/1 = 1/646646 2!*2!*2!*2!*2!*2!*2!*2!*2!*2!*2!=2048 1/646646*2048= 2048/323323*100 = 0.6334% Hope I'm right

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