Problem source

On the basis of an interview, the $N$ candidates for admission to a college
are ranked in order according to their  mathematical potential. The candidates are interviewed 
in random order (that is, each possible order is equally likely).
\begin{questionparts}
\item Find the probability that the best amongst the first $n$ candidates
interviewed is the best overall.
\item Find the probability that the best amongst the first $n$ candidates
interviewed is  the best or second best overall.
\end{questionparts}
Verify your answers for the case $N=4$, $n=2$ by listing the possibilities.

Solution source

\begin{questionparts}
\item The probability the best person falls in the first $n$ is $\frac{n}{N}$

\item The probability the best two people do not fall in the first $n$ candidates is 

\begin{align*}
&& 1-P &= \frac{\binom{N-2}{n}}{\binom{N}{n}} \\
&&&= \frac{(N-2)(N-3)\cdots(N-2-n+1)}{n!} \frac{n!}{N(N-1)(N-2) \cdots (N-n+1)} \\
&&&= \frac{(N-n)(N-n-1)}{N(N-1)} \\
\Rightarrow && P &= 1-  \frac{(N-n)(N-n-1)}{N(N-1)} \\
&&&= \frac{N(N-1) - N(N-1)+n(N-n-1)+Nn}{N(N-1)} \\
&&&= \frac{n(2N-n-1)}{N(N-1)}
\end{align*}
\end{questionparts}

If $N = 4, n = 2$ the possibilities are, the best candidate can be first $3!$ ways, or second $3!$ ways, which is $\frac{12}{24} = \frac{1}{2} = \frac{2}{4} = \frac{n}{N}$ so our formula works.

In the case neither of the best two candidates are in the first half, the possibilities are $3412, 3421, 4312, 4321$, ie $\frac{4}{24} = \frac16$ chance, so the probability they are selected in the first $n$ is $\frac56$. our formula says it should be $\frac{2 \cdot (2 \cdot 4 - 2 - 1)}{4 \cdot 3} = \frac{2 \cdot 5}{4 \cdot 3} = \frac56$ as desired.