Probability of Sitting in Correct Seat n passengers board a plane with n seats. Each passenger has an assigned seat. The first passenger forgets and takes a seat at random. Every subsequent passenger sits in their assigned seat unless it is already taken, at which point they take a random seat. What is the probability (in terms of n) that:
Every passenger sits in the correct seat?
At least one passenger sits in the correct seat?
Exactly one passenger sits in the correct seat?
 A: Every passenger sits in the correct seat?
This only happens if the first passenger choses his own seat by accident, the probability of which is $1/n$.
At least one passenger sits in the correct seat?
This is the complementary event of noone being in their correct seat. For that to happen passenger 1 must pick the seat of passenger 2, passsenger 2 must pick the seat of passenger 3, etc. The probability of that is $1/n\cdot1/(n-1)\cdot\ldots\cdot1/1=1/(n!)$. The complementary event has probability $1-1/(n!)$
Exactly one passenger sits in the correct seat?
This happens if every passenger picks the seat of the next passenger except for one, which picks the seat of the second next passenger. The first passenger cannot be in his correct seats.
Probability of the $i+1$th passenger being in their correct seat: 
$$1/n\cdot 1/(n-1)\cdot ...\cdot 1/(n-i+1)\cdot 1/(n-i-1)\cdot ...\cdot 1/1=(n-i)/n!$$
It can be anyone of passenger $1$ to $n-1$, yielding a total probability of
\begin{align*}
\sum_{i=1}^{n-1} (n-i)/(n!)&=1/(n!)\sum_{i=1}^{n-1} n-i=1/(n!)\cdot(n(n-1)-1/2\cdot(n-1)n)\\
&=1/2\cdot 1/(n!)\cdot (n-1)n=1/2\cdot 1/((n-2)!)
\end{align*}
An extra question for you: What is the probability of exactly one passenger being in the wrong seat?
