- #1
Mogarrr
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Homework Statement
An employer is about to hire one new employee from a group of N candidates, whose future potential can be rated on a scale from 1 to N. The employer proceeds according to the following rules:
(a) Each candidate is seen in succession (in random order) and a decision is made whether to hire the candidate.
(b) Having rejected m-1 candidates (m>1), the employer can hire the mth candidate only if the mth candidate is better than the previous m-1.
Suppose a candidate is hired on the ith trial. What is the probability that the best candidate was hired.
Homework Equations
I have the solution, however I'm not entirely convinced that it's correct.
This is most easily seen by doing each possibility. Let P(i) = probability that the candidate is hired on the ith trial is best. then
[itex]P(1)=\frac 1N[/itex], [itex]P(2)=\frac 1{N-1}[/itex], ... , [itex]P(i)=\frac 1{N-i+1}[/itex], ... , P(N)=1.
The Attempt at a Solution
As a counterexample, suppose that there are 5 candidates, with an ordering of 'future potential' (a bit redundant eh) corresponding to an arbitrary number assignment, i.e. candidate number one has a future potential of one and so forth.
Now Suppose that a candidate is hired on the 2nd trial. We're assuming that each outcome is equally likely, as per (a). Thus we can use the standard method, [itex] P(E)= \frac {number of outcomes favorable to event E}{total number of possible outcomes}[/itex].
Here are the possibilities enumerated as ordered pairs: (1,2), (1,3), (2,3), (1,4), (2,4), (3,4), (1,5), (2,5), (3,5) and (4,5). Note that ordered pairs like (3,2) are not considered, as they would violate (b).
Now the number of total possible outcomes is 10, and the number favorable to the best candidate is 4. However, [itex]\frac 4{10} \neq \frac 14[/itex].
Am I right here? Please help.