# Homework Help: Probability question

1. Sep 2, 2007

### haoku

1. The problem statement, all variables and given/known data
There are 8 opposite-sex married couples in a party and 6 people are chosen to win a prize.
What is the probability that they are 3 married couple?

2. Relevant equations
P=6!/C(16,6)

3. The attempt at a solution
I try to do the question as
P=6!/C(16,6)
because there are 6! way which how many ways that 6 person can be choose from and choosing 6 people from 16 people C(16,6)
I know this equation is incorrect, but I have no idea how to do this question.
Could you please help me?

Thanks

2. Sep 2, 2007

### Hurkyl

Staff Emeritus
Instead of trying to jump straight to the answer... try finding the first step.

3. Sep 2, 2007

### Hurkyl

Staff Emeritus
(Don't read this post until you've thought my last one over)

One thing I often do first is to simply rewrite the problem in a different form. If I let

P = the probability that they are 3 married couples

Q = the number of ways to choose 3 married couple
R = the number of ways to choose 6 people

then

P = Q / R.

(Aside: I wanted to point out that an unstated assumption of the problem is that the choices are made uniformly randomly; each possible way to choose 6 people is exactly as likely as any other way)

4. Sep 2, 2007

### haoku

The first step should be?
Total possible outcome: C(16,6)
Favourable outcome?

5. Sep 2, 2007

### haoku

The Q is what I stuck into

6. Sep 2, 2007

### haoku

Is the Q =8*7*6 ?
That is number if ways which 3 couples chosen from 8 couples

7. Sep 2, 2007

### Hurkyl

Staff Emeritus
note: the text in red is incorrect

Almost: you wrote down the number of permutations, not the number of combinations.

You can (correctly) work the problem by computing
permutations of 3 couples from 8 / permutations of 6 people from 18
or by computing
combinations of 3 couples from 8 / combinations of 6 people from 18

but mixing the two is wrong.

Although either of these ways work, using combinations is a direct translation of the problem, and is probably the way you should think about it.

Last edited: Sep 3, 2007
8. Sep 2, 2007

### haoku

Is the probability is:

P= C(8,3)/C(16,6)

?

Last edited: Sep 2, 2007
9. Sep 2, 2007

### Hurkyl

Staff Emeritus
How did you show that Q = C(6,3)? How did you show that R = C(16,6)?

10. Sep 2, 2007

### haoku

On more question:
permutations of 3 couples from 8 / permutations of 6 people from 18
is not equal to combinations of 3 couples from 8 / combinations of 6 people from 18
?

11. Sep 2, 2007

### Hurkyl

Staff Emeritus
Those numbers are equal.

12. Sep 3, 2007

### haoku

R =C(16,6) because that is the number of ways that 6 people can be chosen from 16 people, regardless the order.

Q=C(8,3) is the number of ways to choose 3 couple from 8 couple.
(I have correct it, C(6,3) to C(8,3))

I think I am still missing something, because the lucky draw choose person one by one, not couple by couple, I need to multiply something to C(8,3)

13. Sep 3, 2007

### Hurkyl

Staff Emeritus
So if you are confident in these calculations, and you are confident that P=Q/R, then you should be confident your answer!

In the problem you stated, the draw was simply to choose six. The method of choosing 6 people doesn't effect the number of combinations.

But if you want to work out the problem as if the order of drawing the people matters, then you can still do it. It would look something like

P = Q/R
Q = Number of permutations of 6 people from 16, such that the 6 form 3 married couple
R = Number of permutations of 6 people from 16

and then you'd split Q up into two pieces, maybe

Q = S T
S = Number of ways to choose 3 married couples from 8
T = Number of ways to arrange those 6 people

There are other ways you might effect this calculation: e.g.

Q = AB + CD + EF
A = < Number of ways to choose 1 couple from 8 >
B = < Number of ways to place those 2 people into 6 slots >
C = < Number of ways to choose 1 couple from the remaining 7 >
D = < Number of ways to place those 2 people into the remaining 4 slots >
E = < Number of ways to choose 1 couple from the remaining 6 >
F = < Number of ways to place those 2 people into the remaining 2 slots >

(note: I think this Q overcounts by a factor of 6. I'm too tired to work it out)

Last edited: Sep 3, 2007
14. Sep 3, 2007

### haoku

When looking at definition, C(a,b)/C(c,d) =[ a!d!(c-d)!]/[c!b!(a-b)!]
P(a,b)/P(c,d)=[a!(c-d)!]/[c!(a-b)!]

So, the two number is not equal.

15. Sep 3, 2007

### Hurkyl

Staff Emeritus
You're right, silly mistake on my part to say that. It's not the permutations of the couples that goes on the numerator, but instead the permutations of the people in the couples.

(p.s. I've made an addition to my previous post)

16. Sep 3, 2007

### haoku

Oh yes, I have some logic error that confuse myself. Thanks

17. Sep 3, 2007

### Hurkyl

Staff Emeritus
Argh, I've overcounted by a factor of 6, I think. I shouldn't do probability this late at night!

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