nm - delete

[badluck]

This is going for any math people....but I can't think of how to do it for the life of me. It's been 3 years since any math classes.

Here it is (I need to know the steps on how to get the answer, not just the answer):


The 3 special seats have to be assigned to 7 men and 5 women. What is the probability that all 3 seats are assigned to men?

 

[BigPoppa]

If its simple as I think it is:

7/12 people are men. That is the probability of a man receiving the first seat. Now for them to receive all 3, is it not (7/12)^3?
Or would be it (7/12)(6/12)(5/12)?

 

[Kyteland]

There are 12 people total and 3 seats that need to be assigned. That means that there are CHOOSE(12,3) total ways to assign those 3 seats to the 12 people. The equation or CHOOSE(x,y) is (x!)/((y!)((x-y)!)). So CHOOSE(12,3) is 220.

There are 7 men and 3 seats. CHOOSE(7,3) is 35.

35/220 = 15.909090... %

 

[badluck]

I should say the answer is:

0.1666


I just don't know how to get there....


Kyteland: I thought that was how to do it as well....guess not. heh.

 

[Kyteland]

Originally posted by: BigPoppa
If its simple as I think it is:

7/12 people are men. That is the probability of a man receiving the first seat. Now for them to receive all 3, is it not (7/12)^3?
Or would be it (7/12)(6/12)(5/12)?

Almost right. It would be (7/12)(6/11)(5/10). You have to take in to account that there is one less seat every time.

 

[badluck]

Kyteland, this is how I got the same answer you did...but I guess its not right...

N(S) = 12C3 = 12!/9!*3! = 220

N(E) =7C3 7!/4!*3! = 35


N(E)
------
N(S)

= .1590

 

[Kyteland]

Originally posted by: badluck
Kyteland, this is how I got the same answer you did...but I guess its not right...

N(S) = 12C3 = 12!/9!*3! = 220

N(E) =7C3 7!/4!*3! = 35


N(E)
------
N(S)

= .1590

Yes it is right, unless there is more information you are not telling us, like there are 13 chairs or something.

I did this in excel.

men 7
women 5
total 12

women men arrangements
3 0 10
2 1 70
1 2 105
0 3 35
Total 220

This covers 100% of all possibilities. The answer is 35/220.

How do you know that the answer is 3/18 (0.16666)?

 

[notfred]

How do you assign 3 seats to twelve people? Stack 4 people per seat? Make 9 people stand?

 

[Kyteland]

Originally posted by: notfred
How do you assign 3 seats to twelve people? Stack 4 people per seat? Make 9 people stand?

I think what it means is that there are 12 seats, one per person, but that 3 of those seats are "special". If everyone takes a seat and has no idea before hand which of the 3 are marked, what is the probability that the 3 marked seats have men in them.

The answer is 35/220

 

[oog]

Originally posted by: Kyteland

Originally posted by: BigPoppa
If its simple as I think it is:

7/12 people are men. That is the probability of a man receiving the first seat. Now for them to receive all 3, is it not (7/12)^3?
Or would be it (7/12)(6/12)(5/12)?

Almost right. It would be (7/12)(6/11)(5/10). You have to take in to account that there is one less seat every time.

Note that this is the same answer as using the Choose method.

 

[maziwanka]

Originally posted by: Kyteland
There are 12 people total and 3 seats that need to be assigned. That means that there are CHOOSE(12,3) total ways to assign those 3 seats to the 12 people. The equation or CHOOSE(x,y) is (x!)/((y!)((x-y)!)). So CHOOSE(12,3) is 220.

There are 7 men and 3 seats. CHOOSE(7,3) is 35.

35/220 = 15.909090... %

indeed

 

[FFactory0x]

LET ME GUESS. STAT 201

 

[badluck]

it is right, unless there is more information you are not telling us, like there are 13 chairs or something

I'm not sure what the deal is...I gave you the problem verbatim. I agree though...we should be right!

How do you know that the answer is 3/18 (0.16666)?

This answer is given from the instructor.....He could be wrong!!!

 

[nd]

Kyteland is right.

You can do it either way. CHOOSE(7,3) / CHOOSE(12,3) or (7/12)*(6/11)*(5/10) will give you the correct answer of 0.15909~

 

[Kyteland]

Originally posted by: badluck
This answer is given from the instructor.....He could be wrong!!!

I'm going to go out on a limb here and say that he is wrong. The only way I can figure that he got that answer is to take CHOOSE(12,3)/PERMUTE(12,3), but that would definitely be the wrong way to do it.

CHOOSE(12,3)/PERMUTE(12,3)
12!/(3!*9!)/(12!/9!)
1/3!
1/6
0.16666666

Just to reassure you that the other answer is right:

There are 3 chairs. There are 7 men. There are 7 possible ways to put a man in the first chair, 6 possible ways to put one in the second and 5 ways to put one in the third. There are 7*6*5 = 210 ways for there to be all men in the three chairs.

There are 3 chairs. There are 12 people. There are 12 possible ways to put a person in the first chair, 11 possible ways to put one in the second and 10 ways to put one in the third. There are 12*11*10 = 1320 ways for there to be a person in the three chairs.

210/1320 = .159090...

his answer: 220/1320 = .166666....

 

[badluck]

Actually, after looking at it over again through your explanation....I understand why it is 12P3 in the first half of the problem, and 12C3 in the 2nd half of the problem.....This first part of the problem is an ordered selection (as you indicated). This means that it is not a combination problem since we care about order. Basically, it appears that we are being asked to rearrange elements in an ordered list.....

The 2nd half of the problem, order does not matter to us....making that part of the equation a combination problem.

Right? And, thanks for your help!!!