Conditional Probability Question

[GreenPrint]

Homework Statement

A population has n men and n women. If you where to take 4 people out of the population to form a group what's the probability that there are exactly the same number of men as women in the group .

Homework Equations

The Attempt at a Solution

Ok so I thought of this to some extent and am lost.

The population size is 2n

(2n)(2n-1)(2n-2)(2n-3) different possible groups if you consider each person of the population to be distinct people without regards to the condition of gender.

When you take gender into consideration there are

2*2*2*2 = 2^4 = 16 different combinations when take the gender condition into account and consider each individual to be distinct (that is GGBB is different from BBGG)

If I consider each man and women to be the same there are a total of 5 different combinations...
BBBB
BBBG
BBGG
BGGG
GGGG

I know the following formula

P=(A|B) = \frac{P(A \bigcap B)}{P(B)}

I'm assuming that I need to use this formula. The only problem is that I don't know how to come up with A and B.

Thanks for any help that nay one can provide.

 

[Dick]

Homework Statement

A population has n men and n women. If you where to take 4 people out of the population to form a group what's the probability that there are exactly the same number of men as women in the group .

Homework Equations

The Attempt at a Solution

Ok so I thought of this to some extent and am lost.

The population size is 2n

(2n)(2n-1)(2n-2)(2n-3) different possible groups if you consider each person of the population to be distinct people without regards to the condition of gender.

When you take gender into consideration there are

2*2*2*2 = 2^4 = 16 different combinations when take the gender condition into account and consider each individual to be distinct (that is GGBB is different from BBGG)

If I consider each man and women to be the same there are a total of 5 different combinations...
BBBB
BBBG
BBGG
BGGG
GGGG

I know the following formula

P=(A|B) = \frac{P(A \bigcap B)}{P(B)}

I'm assuming that I need to use this formula. The only problem is that I don't know how to come up with A and B.

Thanks for any help that nay one can provide.

It's just the number of ways to pick 2 women from n women times the number of ways to pick 2 men from n men divided by the number of ways to pick 4 people from 2n people regardless of gender, isn't it? You seem to be really over complicating this.

 

[GreenPrint]

It's just the number of ways to pick 2 women from n women times the number of ways to pick 2 men from n men divided by the number of ways to pick 4 people from 2n people regardless of gender, isn't it? You seem to be really over complicating this.

Ya I guess your right about this. I don't know if I can do this notation but can I do

(N_{K})^{2}

Square N choose K in this notation? I have never seen this before don't see why I can't use this notation.

I couldn't figure out how to do this in itex code

 

[Dick]

Ya I guess your right about this. I don't know if I can do this notation but can I do

(N_{K})^{2}

Square N choose K in this notation? I have never seen this before don't see why I can't use this notation.

I couldn't figure out how to do this in itex code

I'm not exactly sure what the jist of your question is, but without even using tex, if C(n,k) is the number of ways to choose k objects from n, then what's your answer? And then write it explicitly using a formula for C(n,k).

 

[GreenPrint]

my answer is

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

 

[Dick]

my answer is

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

I would write it as C(n,2)^2. But that's not terribly important. And the denominator should be C(2n,4), yes? You know C(n,k)=n!/(k!(n-k)!), right? I would guess they want you to use that to simplify [C(n,2)*C(n,2)]/C(2n,4).

 

[Ray Vickson]

my answer is

[C(n,2)*c(n,2)]/C(n,4)

I was wondering if you can write

C(n,2)*C(n,2) as C^2(n,2)?

If your denominator were C(2n,4) then, yes, your answer would be correct. This is just a so-called hypergeometric probability.