# Homework Help: Probability Question

1. Mar 22, 2010

### mistermath

1. The problem statement, all variables and given/known data
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.

2. Relevant 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.

3. 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.

2. Mar 23, 2010

### vela

Staff Emeritus
Perhaps you can solve for the probability of the complement instead more easily.

3. Mar 23, 2010

### mistermath

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. Mar 24, 2010

### vela

Staff Emeritus
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. Mar 28, 2010

### awkward

This is a problem in derangements. The answer is $$\frac{D(4)}{4!}$$ where D(4) is the number of derangments of 4 objects. See

http://en.wikipedia.org/wiki/Derangements