1
$\begingroup$

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.

My reasoning:

  • There are $n \cdot (n-1) \cdot (n-2)$ many ways the interviewers put a different person at the first position.

  • So, $n^3 - n \cdot (n-1) \cdot (n-2)$ ways that at least two interviewers will put the same person at the first position

  • So, $\frac{n^3 - n \cdot (n-1) \cdot (n-2)}{n^3} = \frac{3n^2-2n}{n^3}$ is the desired probability

But the answer given is $\frac{3n-2}{n^3}$

Where is the problem in my reasoning? thanks.

$\endgroup$
  • $\begingroup$ The question is ambiguous and the difference in answers seems to hinge on that ambiguity: is it asking the chance that some candidate will be accepted or the chance that a particular candidate will be accepted? Because it's impossible for more than one candidate to be accepted, the probability of the former must be $n$ times the probability of the latter. $\endgroup$ – whuber Mar 3 at 21:17
3
$\begingroup$

The probability that the second interviewer puts a different person at the first position is

$$\frac{n-1}{n}\,.$$

The probability that the third interviewer puts a different person at the first position is

$$\frac{n-2}{n}\,.$$

Hence, the probability that there are three different persons at the first position is

$$p(\text{all different})=\frac{n-1}{n}\cdot \frac{n-2}{n} = \frac{n^2-3n+2}{n^2} \,.$$

The contrary of three different persons at the first position is not three different persons at the first position (i.e., at least two times the same person is at position one). Therefore, we calculate

$$p(\text{at least two identical}) = p(\lnot\text{all different}) = 1 - p(\text{all different}) =\\1-\frac{n^2-3n+2}{n^2} = \frac{3n-2}{n^2}\,.$$

| cite | improve this answer | |
$\endgroup$
  • 2
    $\begingroup$ Hi Sven, you have found exactly the same answer with me, but the given answer in the text book is different, note the $n^3$ at the denominator. $\endgroup$ – Sanyo Mn Sep 6 '17 at 11:53
  • 1
    $\begingroup$ Might still be a typo. $\endgroup$ – bayerj Sep 6 '17 at 11:54
  • 1
    $\begingroup$ You can understand the question as relating to a specific candidate being selected and not just any candidate. In this case you have to multiply your result by 1/n. $\endgroup$ – Zahava Kor Sep 6 '17 at 16:34
1
$\begingroup$

${3\choose2} \cdot 1/n \cdot 1/n \cdot (1 - 1/n)$ --this represents two out of the three committee members selecting the same individual times the third committee member selecting a different person.

${3\choose3} \cdot 1/n \cdot 1/n \cdot 1/n$ --this represents all three committee members selecting the same individual.

Add the two up:

$$ \array{ & & 3 \cdot 1/n^2 \cdot (n-1)/n + 1/n^3 & = \\ & = & 3(n-1)/n^3 + 1/n^3 & = \\ & = & (3n - 3)/n^3 + 1/n^3 & = \\ & = & (3n - 3 + 1)/n^3 & = \\ & = & (3n - 2)/n^3 }$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Welcome to CV. I believe you obtain a different answer than the existing one because you interpret the question differently. $\endgroup$ – whuber Mar 3 at 21:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.