What is the Probability of a Group of 4 Getting a Job Out of 12 Applicants?

[noreturn2]

Homework Statement

Assume a job has 12 applicants, and 4 job openings. I want me and my 3 friends to all get the job. What is this probability.

Homework Equations

! Factorial and Permuation & combinatoins

The Attempt at a Solution

Number of possible solutions: 12C4 = 495 Possible ways
Number of Possible Solutions of friends: 8 C 4 = 70 Ways

Possibility: 70/495 = 14.14%

Which just seemed to low to be right where did I go wrong?/

 

[magoo]

If there is only 4 job openings, how do you expect to have you and your 4 friends all get the job when there is only 4 jobs available?

 

[verty]

My advice is to think in terms of permutations rather than combinations. I won't say more at this time.

 

[noreturn2]

My advice is to think in terms of permutations rather than combinations. I won't say more at this time.

I do not think this would be a permutation because we could careless about the order right? We could have my friend would be app 3 and friend 2 be app 5. But they could hire app 2 5 3 9 etc.

12 nPr 4 = 11880
4! = 24

24/11880 = .002%This would be: 12!/(12-4)! = .002% which could be right but seems low still

Side note this is right!

If there is only 4 job openings, how do you expect to have you and your 4 friends all get the job when there is only 4 jobs available?

Well I would need to find how how many different ways we could hire an application which would be 12!, and the number of ways my friends could be hired would be 4!, but that number was way to low for a % possibility.

 

[Ray Vickson]

Homework Statement

Assume a job has 12 applicants, and 4 job openings. I want me and my 4 friends to all get the job. What is this probability.

Homework Equations

! Factorial and Permuation & combinatoins

The Attempt at a Solution

Number of possible solutions: 12C4 = 495 Possible ways
Number of Possible Solutions of friends: 8 C 4 = 70 Ways

Possibility: 70/495 = 14.14%

Which just seemed to low to be right where did I go wrong?/

The probability you want is zero, because you want 5 people to get 4 jobs.

If you had said that you want the probability that all 4 jobs are filled from a particular group of 5 people, that would make sense.

Your argument against using permutations does not hold; you can always divide by 4! to get rid of the"order" effect.

If I were doing it I would avoid both permutations and combinations, at least to start with. Assuming we fill the 4 jobs one after another, the probability the first job is filled from our group is 5/12. That leaves 4 in our group and 11 candidates remaining, so the probability the second job is also filled from our group is 4/11.

Continue like that.

 

[noreturn2]

The probability you want is zero, because you want 5 people to get 4 jobs.

If you had said that you want the probability that all 4 jobs are filled from a particular group of 5 people, that would make sense.

Your argument against using permutations does not hold; you can always divide by 4! to get rid of the"order" effect.

If I were doing it I would avoid both permutations and combinations, at least to start with. Assuming we fill the 4 jobs one after another, the probability the first job is filled from our group is 5/12. That leaves 4 in our group and 11 candidates remaining, so the probability the second job is also filled from our group is 4/11.

Continue like that.

So I worded it wrong, there is only 3 friends and myself then. Just summarized the problem instead of writing it all out and wrote it wrong.

 

[Ray Vickson]

So I worded it wrong, there is only 3 friends and myself then. Just summarized the problem instead of writing it all out and wrote it wrong.

So, if the group (you and three friends) is of size 4, you want to find the probability of a particular subset of size 4 taken from a set of size 12. That is $P = 1/_{12}C_4$, because $_{12}C_4$ is the number of subsets of size 4, and only one of those is the correct subset.