Probability Question

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  • #1
mistermath
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Homework Statement


Four couples go to a party. They decide to split randomly into groups of 2. What is the probability that no group has a couple that came together.


Homework Equations


I actually have an MS in math; feel free to use mathematics at any level. The answer is supposed to be 37.5% or 1-37.5 I have forgotten.


The Attempt at a Solution


I've tried several different approaches.

First: Total ways possible is 8! / 2!2!2!2! OR you can do: 8C2 * 6C2 * 4C2 * 2C2 (which is the same thing); I believe this is correct.

Total ways of getting no couples (this part is incorrect). 7P6 * 5P4 * 3P2 * 1P1 = 7*5*3*1.

Second: There are 8 people, label them 1-8. 1,2 = couple, 3,4 = couple, 5,6 = couple, 7,8 = couple. This gives us 28 different combinations of couples that are possible. 7+6+5+4+3+2+1(because 1 can be paired with 2-8 = 7 ways, 2 can be paired with 3-8 = 6 etc).

But I can't figure out all the different cases; mostly because I don't want to do it this way. I feel like there should be an easier way to solve this that I'm over looking. The book solves this problem by setting up an experiment.
 

Answers and Replies

  • #2
vela
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Perhaps you can solve for the probability of the complement instead more easily.
 
  • #3
mistermath
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Perhaps you can solve for the probability of the complement instead more easily.

The problem with looking at the complement is that it makes it even harder. The complement of none is at least 1. In this case, you'd need 1-P(One couple)+P(Two Couples)+P(Three Couples) + P(4 Couples).

But ouch, P(3 Couples).. what's that mean? Is it possible to have 3 couples and the 4th set not being a couple? nope cannot. So how do we handle that.
 
  • #4
vela
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That just means P(3) and P(4) aren't independent, so you can't just add the individual probabilities to get P(3 or 4).
 
  • #5
awkward
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