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There are n applicants for the director of computing. The

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There are n applicants for the director of computing. The [#permalink] New post 05 Nov 2003, 13:56
There are n applicants for the director of computing. The applicants are interviewed independently by each member of the three-person search committee and ranked from 1 to n. A candidate will be hired if he or she is ranked first by at least two of the three interviewers. Find the probability that a candidate will be accepted if the members of the committee really have no ability at all to judge the candidates and just rank the candidates randomly. In particular, compare this probability for the case of three candidates and the case of ten candidates.
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Re: Bhai's Probability 11 [#permalink] New post 06 Nov 2003, 09:05
Bhai wrote:
There are n applicants for the director of computing. The applicants are interviewed independently by each member of the three-person search committee and ranked from 1 to n. A candidate will be hired if he or she is ranked first by at least two of the three interviewers. Find the probability that a candidate will be accepted if the members of the committee really have no ability at all to judge the candidates and just rank the candidates randomly. In particular, compare this probability for the case of three candidates and the case of ten candidates.



Probability of getting first rank = 1/n

The candidate will be accepted if two or three of the judges rank him as first

= Prob (two judges rank him first) + Prob (three judges rank him first)

= 3C2*(1/n)^2 + 3C3*(1/n)^3

= (3n+1)/n^3
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Re: Bhai's Probability 11 [#permalink] New post 08 Nov 2003, 16:54
prashant wrote:
Bhai wrote:
There are n applicants for the director of computing. The applicants are interviewed independently by each member of the three-person search committee and ranked from 1 to n. A candidate will be hired if he or she is ranked first by at least two of the three interviewers. Find the probability that a candidate will be accepted if the members of the committee really have no ability at all to judge the candidates and just rank the candidates randomly. In particular, compare this probability for the case of three candidates and the case of ten candidates.



Probability of getting first rank = 1/n

The candidate will be accepted if two or three of the judges rank him as first

= Prob (two judges rank him first) + Prob (three judges rank him first)

= 3C2*(1/n)^2 + 3C3*(1/n)^3

= (3n+1)/n^3


Prashant,

A slight disagreement with you.

In the question stem word atleast TWO is used.

So the required probability = (probability of any two member will give the candidate 1st rank) * (probability that the third member does not give 1st rank to the candidate) + probability that all the members give 1st rank to the candidate
= 3C2*(1/n)^2* (1-1/n) + 3C3*(1/n)^3 = (3n-2)/n^3

Do you agree?
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Re: Bhai's Probability 11 [#permalink] New post 19 Nov 2003, 16:27
I will go with am1974's answer
am1974 wrote:
Prashant,

A slight disagreement with you.

In the question stem word atleast TWO is used.

So the required probability = (probability of any two member will give the candidate 1st rank) * (probability that the third member does not give 1st rank to the candidate) + probability that all the members give 1st rank to the candidate
= 3C2*(1/n)^2* (1-1/n) + 3C3*(1/n)^3 = (3n-2)/n^3

Do you agree?
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 [#permalink] New post 19 Dec 2003, 18:43
Three people are interviewing and can give ratings from 1 to n. I used the following logic. if 3 bits represent ( bit can take value 0 or 1 ) then we have total combinations to be 2^3
Hence in this case we have total combinations = n^3
Let us call the interviewers as I1, I2 and I3

Desired combinations
I1 I2 I3
1 2 x -> x can range from 2 to n ( n-1 values )
1 x 1 -> x can take n-1 values
x 1 1 -> x can take n-1 values
1 1 1 -> only one value

total desired events = n-1 + n-1 + n-1 + 1 = 3n-2
So probability = (3n-2)/n^3
Am I doing anything wrong here ?
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 [#permalink] New post 19 Dec 2003, 23:43
(3n-2)/n^3 seems to be the right answer. At least that is what i got...

We can check it working the other way round :-

p(<=2) i.e probability where rank 1 is given by any three is less than 2.

Probability <2= 1-{3*(1-1/n)^2(1/n)+(1-1/n)^3} which gives the above result.
  [#permalink] 19 Dec 2003, 23:43
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There are n applicants for the director of computing. The

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