Probability Question: No Couples in Randomly Split Groups of Four Couples

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In summary, the problem involves four couples randomly splitting into groups of two at a party, and the question is what is the probability that no group contains a couple that arrived together. The answer is 37.5%, or 1-37.5%. The solution involves using derangements, and calculating the probability of the complement (having at least one group with a couple) instead.
  • #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.
 
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  • #2
Perhaps you can solve for the probability of the complement instead more easily.
 
  • #3
vela said:
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
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

FAQ: Probability Question: No Couples in Randomly Split Groups of Four Couples

1. What is the probability of having no couples in randomly split groups of four couples?

The probability of having no couples in randomly split groups of four couples is calculated by dividing the number of ways to form groups with no couples by the total number of ways to form groups. This can be represented as 1/35 or approximately 0.02857.

2. How did you arrive at the probability of 1/35?

The probability of 1/35 was derived using the combination formula, which is nCr = n! / (r!(n-r)!). In this scenario, there are 8 individuals (4 couples) and we are forming groups of 4. So the calculation would be 8C4 = 8! / (4!(8-4)!) = 70. This represents the total number of ways to form groups without considering the restriction of no couples. Then, to find the number of ways to form groups with no couples, we use the permutation formula, which is nPr = n! / (n-r)!. In this case, we have 8 individuals and we want to form groups of 4, so the calculation would be 8P4 = 8! / (8-4)! = 1680. This represents the total number of ways to form groups with no restriction on couples. Finally, we divide 1680 by 70 to get 1/35 as the probability of having no couples in randomly split groups of four couples.

3. Is this probability the same for any number of couples?

No, the probability of having no couples in randomly split groups will vary depending on the number of couples. The probability is calculated using the combination and permutation formulas, which take into account the number of individuals and the number of groups being formed. Therefore, the probability will change based on the number of individuals and the number of groups.

4. Is this probability affected by the gender of the individuals in the couples?

No, the probability of having no couples in randomly split groups is not affected by the gender of the individuals in the couples. As long as there are 8 individuals (4 couples) and we are forming groups of 4, the probability will remain the same. The probability is based on the combination and permutation formulas, which do not consider the gender of the individuals.

5. Can this probability be applied to real-life scenarios?

Yes, this probability can be applied to real-life scenarios, such as randomly selecting members for a group project or randomly assigning roommates. However, it is important to note that this probability assumes that all individuals have an equal chance of being selected and that the couples are randomly split. In real-life situations, there may be other factors at play that could affect the probability.

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