# Math Help - combinatorices

1. ## combinatorices

Hello, I have a small question please...

In a building there are 8 floors, in each floor, there are 4 flats. In each flat lives a couple ( man & woman ). We want to choose committee of 4 people ( a family can send only 1 representative ). What is the probability that the committee will contain 4 women ?
( assume that the chances of choosing men and women are equal )

thankes!

2. Originally Posted by WeeG
In a building there are 8 floors, in each floor, there are 4 flats. In each flat lives a couple ( man & woman ). We want to choose committee of 4 people ( a family can send only 1 representative ). What is the probability that the committee will contain 4 women ?
There are 32 flats in that building.
Because no two from the same flat can be selected, this is picking four flats: $\binom{32}{4}$.
There two ways to select a person from each of those four flats: $\left(2^4\right)$.

Can you finish?

3. The number of flats is irrelevant.

You choose 4 flats to the committee, all the flats are equal so it doesnt matter how many there are.

Of those flats you choose a man or a woman in each.

you can make $2^4$ different groups and only 1 all women group so the chance is 1/16.

4. thank you for the quick answer, both of you, but your answeres are different from one another....I am confused

5. Originally Posted by WeeG
thank you for the quick answer, both of you, but your answeres are different from one another....I am confused
No they do not differ from one another. They are the same.
I just did finish the solution. I wanted you to find out for yourself that the number of flats makes no difference. Do you see why?

6. no

7. Originally Posted by WeeG
Hello, I have a small question please...

In a building there are 8 floors, in each floor, there are 4 flats. In each flat lives a couple ( man & woman ). We want to choose committee of 4 people ( a family can send only 1 representative ). What is the probability that the committee will contain 4 women ?
( assume that the chances of choosing men and women are equal )

thankes!
there are 32 families (64 persons), so we have $64.62.60.58=2^4.32.31.30.29$ ways to choose committee of 4 people.
And we have $C^4_{32}=\frac{32.31.30.29}{4.3.2.1}$ways to choose group 4 women.
result: $\frac{1}{4! 2^4}$

8. There are $2^4\binom{32}{4}$ ways the select the committee.

There are $\binom{32}{4}$ ways the select the committee of all women.
We pick the four flats. Take the lady from each of the flats.

That gives $\frac{\binom{32}{4}}{2^4\binom{32}{4}}=\frac{1}{2^ 4}$.

As was pointed out by Haytham, the number of flats does not matter (it does have to be at least four).
Once the four flats a chosen, we can pick either wife or husband. That is $2^4$ ways.

9. now I got it, thanks ! it does not matter the number of flats...

I have just found out an addition to the question, I hope you don't mind....

b. what is the probability of all people in the committe being from the same floor ?

c. it is knows, that there are 3 families in the building with the same name, "smith". what is the probability of having two "smith"s in the committee ?

d. now let's assume we have 32 people ( 1 in each flat ), and still, 3 "smith"s. if we define a random variable of being a smith, what kind of variable is it and what's it's expectation ?

if you can help me with some of them, I'll really appreciate it !!

10. b.
there are 8 floor so 8 groups
there are also 32*31*30*29 possible groups, so
$P=\frac{8}{32*31*30*29}$

c. (in this one ill consider '2 smiths'. not 'at least 2 smiths')
we need 1 group of 3*2*29*28 groups
same possible groups, so
$P=\frac{3*2*29*28}{32*31*30*29}$

dont know/understand d.

11. Originally Posted by Haytham
b.
there are 8 floor so 8 groups
there are also 32*31*30*29 possible groups, so (this is not correct) there are $\binom{32}{4}$ groups.
$P=\frac{8}{32*31*30*29}$

c. (in this one ill consider '2 smiths'. not 'at least 2 smiths')
we need 1 group of 3*2*29*28 groups
same possible groups, so
$P=\frac{3*2*29*28}{32*31*30*29}$(this is not correct)
For part c. $\frac{\binom{3}{2}\binom{29}{2}}{\binom{32}{4}}$

12. thanks, I understand...

as for the last question, it's not very clear, but can it be HyperGeometric variable ?

13. Haytham smiles at his silly mistake