# Statistics questions

1. Sep 28, 2007

### bethanyd

My friend came up with some solutions but I have doubts about them and was wondering if someone could help a bit.

1. The problem statement, all variables and given/known data
A committee of 4 people is randomly selected from 5 married couples. What is the probability that a husband and wife do not belong to that committee?

3. The attempt at a solution
I have All possible combinations = 4C10
then, divided into groups / sets:
P("all men"), P("3 men, 1 woman"), P("2 men, 2 women"), P("1 man, 3 women"), P("all women").

Then P(final answer) = [P("all men") + P("3 men, 1 woman") + P("2 men, 2 women") + P("1 man, 3 women") + P("all women")] / All possible combinations

My question, or more specifically, the part I am confused about is this: what stops a couple from being on the committee? Please explain this answer, I don't quite get it.

Thank you!

2. Sep 28, 2007

### EnumaElish

With a single-sex committee you cannot have H&W team. But, with the other combinations, there may be a situation where a couple is included, or not. For example, if the committee has 3 men & 1 woman, there is a probability that one of the 3 men will be the woman's husband. Your formula does not reflect that probability.

3. Sep 29, 2007

### bethanyd

Hey

Okay, so here is the solution to this problem I have come up with:

P("all men") = (5, 4)
P("3 men, 1 woman") = (5, 3)(2, 1)
P("2 men, 2 women") = (5, 2)(3, 2)
P("1 man, 3 women") = (5, 1)(4, 3)
P("all women") = (5, 4)

And total possibilities: (10, 4)

Then, as I said above, P(final answer) = [P("all men") + P("3 men, 1 woman") + P("2 men, 2 women") + P("1 man, 3 women") + P("all women")] / All possible combinations

Please let me know if this makes sense / looks correct.

Thank you!
Beth

4. Sep 29, 2007

### Staff: Mentor

Hi Beth, you have split your people up wrong. You have 2 groups of 5 (5 men, 5 women), the right way to split them up is into 5 groups of 2 (the Smith's, the Wilson's, the Doe's, ...). For the purpose of this problem, what matters is which couple a given individual belongs to, not which gender.

-Dale

5. Sep 30, 2007

### bethanyd

Hey

Would it be possible to give a bit more guidance? I don't quite understand this. I have 5 groups of two couples each. 1 person from the 2 in the group can be selected... so would it be something like:

(2, 1)(2, 1)(2, 1)(2, 1)(2, 1) / (10, 4)

Beth

6. Sep 30, 2007

### bethanyd

Actually, I have another possibility I thought of.

Since P("husband and wife not being on the committee") = 1 - P("husband and wife on same committee").

There are 5 couples, ie, 5 sets of husband and wife, and (10, 4) total possibilities.
Could P("husband and wife on same committee") = 5 / (10, 4) ?

Thanks!
Beth