Check GMAT Club Decision Tracker for the Latest School Decision Releases https://gmatclub.com/AppTrack

 It is currently 22 May 2017, 16:49

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Events & Promotions

###### Events & Promotions in June
Open Detailed Calendar

# There are n applicants for the director of computing. The

Author Message
SVP
Joined: 16 Oct 2003
Posts: 1805
Followers: 5

Kudos [?]: 154 [0], given: 0

There are n applicants for the director of computing. The [#permalink]

### Show Tags

05 Nov 2003, 14:56
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

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.
Manager
Joined: 24 Jun 2003
Posts: 146
Location: India
Followers: 2

Kudos [?]: 2 [0], given: 0

### Show Tags

06 Nov 2003, 10:05
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
Manager
Joined: 11 Mar 2003
Posts: 54
Location: Chicago
Followers: 1

Kudos [?]: 0 [0], given: 0

### Show Tags

08 Nov 2003, 17:54
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?
Intern
Joined: 30 Oct 2003
Posts: 33
Location: uk
Followers: 0

Kudos [?]: 3 [0], given: 0

### Show Tags

19 Nov 2003, 17:27
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?
SVP
Joined: 30 Oct 2003
Posts: 1790
Location: NewJersey USA
Followers: 6

Kudos [?]: 101 [0], given: 0

### Show Tags

19 Dec 2003, 19:43
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 ?
Intern
Joined: 17 Sep 2003
Posts: 30
Location: Mumbai,India
Followers: 0

Kudos [?]: 0 [0], given: 0

### Show Tags

20 Dec 2003, 00:43
(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.
20 Dec 2003, 00:43
Display posts from previous: Sort by