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.
 A: 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
$$p(\text{at least two identical}) = p(\lnot\text{all different}) = 1 - p(\text{all different}) =\\1-\frac{n^2-3n+2}{n^2} = \frac{3n-2}{n^2}\,.$$
A: ${3\choose2} \cdot 1/n \cdot 1/n \cdot (1 - 1/n)$ --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:
$$ \array{
&   & 3 \cdot 1/n^2 \cdot (n-1)/n + 1/n^3 & = \\
& = & 3(n-1)/n^3 + 1/n^3 & = \\
& = & (3n - 3)/n^3 + 1/n^3 & = \\
& = & (3n - 3 + 1)/n^3 & = \\
& = & (3n - 2)/n^3
}$$
