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My problem is: There are 10 boy and 10 girl and we choose couples(boy-girl) randomly. We know 3 guys (Aladar,Bela,Csaba) and 3 girls (Aranka,Bori,Cecilia). What is the probability that Aladar is not in a pair with Aranka, Bori nor Cecilia and Bela in not in a pair with Bori and Cecilia and Csaba is not in a pair with Cecilia. So it is restricted to make pairs where any of the listed couples dance together.

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I think maybe I should calculate the probability that 'a' choose 'a' or 'b' or 'c', it's 3/10. And I should add the probability that 'b' choose 'b' or 'c' and 'c' choose 'c', these are 2/10 and 1/10. If I add them together I get 6/10. There I would subtract the probability of the cases I counted multiple times. These are P(a->b|b->b), P(a->c|b->c), P(a->c|c->c), P(b->c|c->c) and all these probabilities are 1/100 so I subtract all of them from 60/100 to get 56/100. This is the opposite of the probability I search for so the solution is 44/100. That's my idea, but it can be wrong.

I'm looking forward your answers!


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What are your own thoughts on this? –  Henning Makholm Dec 20 '13 at 20:03
I've just got an idea, but I don't know if it is a right one. –  Andrew Dec 20 '13 at 21:07
Sorry, Daniel McLaury - it was an accident:( –  Andrew Dec 20 '13 at 22:31

1 Answer 1

up vote 2 down vote accepted

I think your plan of attack can work if you're very careful to correct appropriately for double-counted cases, and then re-correct for cases you've corrected for twice, etc.

Your implementation of it seems to be wrong, though, because you're getting a different answer from the one I'm getting by a simpler method, namely just to count all of the valid matchings:

How many ways is there to choose a partner for Aladar? Once Aladar has a partner, how many ways to choose a partner for Bela? Once the the first two boys have partners, how many for Csaba to choose among?

Multiply, and then compare this to the number of ways there are to choose partners without the restrictions.

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Thank you very much. I think I know what was wrong with my answer, but yours is more simple. –  Andrew Dec 21 '13 at 21:05

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