Probability exercise

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ceci77777

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Hi All,
I had a lot of difficulty trying to solve this exercise. Can someone help me please?

There are 10 candidates for a job. Interviews with each candidate are conducted independently by the the members of a three-person examining committee: each member of the Committee shall draw up a ranking from 1 to 10 of the candidates. A candidate is hired if he or she is ranked first by at least two of the three members of the examining committee. Find the probability that a specific candidate (for example Dr. Talbot) will be hired, assuming that the members of the committee are unable to evaluate the candidates and that they make their rankings completely at random.
 
If rankings are entirely at random then every candidate is equally likely to be chosen. Since there are 10 candidates the chance any one candidate is chosen is 1/10.
 
But the candidate has to be ranked first by at least two of the three members. I think I have to calculate the probability of when the candidate is first in two rankings, and when he's first in three rankings, but I don't know how to do that
 
ceci77777 said:
Hi All,
I had a lot of difficulty trying to solve this exercise. Can someone help me please?

There are 10 candidates for a job. Interviews with each candidate are conducted independently by the the members of a three-person examining committee: each member of the Committee shall draw up a ranking from 1 to 10 of the candidates. A candidate is hired if he or she is ranked first by at least two of the three members of the examining committee. Find the probability that a specific candidate (for example Dr. Talbot) will be hired, assuming that the members of the committee are unable to evaluate the candidates and that they make their rankings completely at random.
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Since it is clear that each candidate is equally likely to be hired, you need to find the probability that someone is hired at all! Or, taking the complement, the probability that no one is hired, because each examiner picks a different person to be first.
 
I thought of this calculation:

In the first part the candidate is hired thanks to two rankings (so he's first in two rankings), and in the second part the candidate is hired thanks to 3 rankings (he's first in three rankings). It is correct?
 

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ceci77777 said:
I thought of this calculation:

In the first part the candidate is hired thanks to two rankings (so he's first in two rankings), and in the second part the candidate is hired thanks to 3 rankings (he's first in three rankings). It is correct?
Click to expand...
Are you sure about the 3!'s?
 
irspow said:
You don’t have 3! (6) arrangements only three. The probability can be found as:

3(1/10)(1/10)(9/10)+3(1/10)(1/10)(1/10)

3*(9/1000+1/1000)

3/100

It is simpler though if you realize that the third pick can be anything:

3(1/10)(1/10)(10/10)= 3/100
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Are you sure of both of the 3's?
 
irspow said:
Ouch, I will delete. Thanks for catching that.
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Don't just give up. I didn't say you weren't close. And don't delete either; errors are part of the process.

On the other hand, we're trying to help @ceci77777 work this out, so giving a complete answer isn't our goal yet. And the answer in #5, like yours, is not far from a valid answer.

(I'm holding off on showing how to carry out my suggestion in #4 until we see some progress there; I wasn't certain of my answer until I got both methods to agree, which took some reworking of both. I made my share of mistakes, mostly unwritten.)
 
In the first part (in the formula before the “+”) I’ve put the 3! because let’s assume that there are 3 empty spaces, the first 1/10 can be put in three different spaces, the second 1/10 can be put in two different spaces, and for 9/10 remain only one spaces. What’s wrong with this reasoning? :(

Maybe in the second part I’ haven’t to put the 3! Because I have three identical 1/10?
 
Or maybe, like that? Beacause we have two 1/10, in the first part, that we can put in three different ways, and in the second part we have three 1/10 so I don't have to multiply at all
 

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ceci77777 said:
Or maybe, like that? Beacause we have two 1/10, in the first part, that we can put in three different ways, and in the second part we have three 1/10 so I don't have to multiply at all
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That's right. The first way, there are 3 ways to choose which one doesn't rate him #1; the second way, there is only one possibility (all rate him #1).

Now here's my way (which for some problems might be shorter, but isn't for this one):

The probability that no one is hired is the probability that each examiner picks a different person: 10P3/10^3 = 720/1000 = 18/25.

The probability that someone is hired is the complement, 1 - 18/25 = 7/25.

The probability that a given person is hired is 1/10 of that (since symmetry implies that each is equally likely; 7/25 * 1/10 = 7/250.

Your answer is (9*3+1)/1000 = 28/1000 = 7/250. When two methods give the same result, I start to trust my answer.
 
All right, thank you very much and thank you also for the second reasoning which is very interesting!❤
 
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