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07 Feb 2021, 23:48
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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. What is 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?
A. \(\frac{1}{n^3}\)
B. \(\frac{3n-3}{n^3}\)
C. \(\frac{3n-2}{n^3}\)
D. \(\frac{3n+1}{n^3}\)
E. \(\frac{28}{n^3}\)
Are You Up For the Challenge: 700 Level Questions
A. \(\frac{1}{n^3}\)
B. \(\frac{3n-3}{n^3}\)
C. \(\frac{3n-2}{n^3}\)
D. \(\frac{3n+1}{n^3}\)
E. \(\frac{28}{n^3}\)
Are You Up For the Challenge: 700 Level Questions
08 Feb 2021, 00:56
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Bunuel
So a particular candidate can be chosen in two ways..
(1) All three place him/her first
Probability that each place the same candidate first = \(\frac{1}{n}*\frac{1}{n}*\frac{1}{n}=\frac{1}{n^3}\)
(2) Exactly two place him/her first
Probability that exactly 2 place the same candidate first = \(\frac{1}{n}*\frac{1}{n}*\frac{n-1}{n}=\frac{n-1}{n^3}\)
But the different candidate can be chosen by any of the three members, so \(\frac{n-1}{n^3}*3\)
Total P = \(\frac{1}{n^3}+\frac{n-1}{n^3}*3=\frac{1+3n-3}{n^3}=\frac{3n-2}{n^3}\)
C
04 Aug 2024, 22:40
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Harsha
Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com
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Harsha
Passionate about all things GMAT | Exploring the GMAT space | My website: gmatanchor.com
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