<s>[INST] A simple probability problem 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.
 [/INST] ${3\choose2} \cdot 1/n \cdot 1/n \cdot (1 - 1/n)$ --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:
$$ \array{
&   & 3 \cdot 1/n^2 \cdot (n-1)/n + 1/n^3 & = \\
& = & 3(n-1)/n^3 + 1/n^3 & = \\
& = & (3n - 3)/n^3 + 1/n^3 & = \\
& = & (3n - 3 + 1)/n^3 & = \\
& = & (3n - 2)/n^3
}$$
</s>