The "n" passengers for a flight in an airplane with "n" seats have been told their seat numbers. The get on the plane one by one. The first person sits in the wrong seat. Subsequent passengers sit in their assigned seats whenever they find them available, or otherwise in a randomly chosen empty seat. What is the probability that the last passenger finds his seat free?
An interesting problem. Is this from the course or from somewhere else? Basically, as I see it, at any one time there is at most one passenger whose seat is already taken. The question is whether or not that passenger solves the problem by: Being the next to board the plane Happening to choose the seat of someone who is already on the plane If we don't have both of these things happening then we return to where we started, i.e. with one passenger still to board whose seat is already taken. We can view the process as having two end-stages, i.e. the first and last passengers to board, and several intermediate stages, where the next passenger to board could solve the problem and allow the remaining passengers to board in their correct seats. We want the probability that none of the intermediate stages succeeds. We can get this by examining an individual intermediate stage, assuming that there is still a passenger whose seat is already taken and that \(r\) passengers have already boarded and there are \(n-r\) still to board. The probability of the passenger whose seat is already taken being the next to board the plane is \(\frac{1}{n-r}\) and, given that this passenger boards, the probability that he solves the problem by randomly choosing the seat of someone who is already on the plane is \(\frac{1}{n-r}\), since there is only one such seat out of the \(n-r\) remaining. So the probability of the problem being solved at this stage is \(\frac{1}{(n-r)^2}\) and the probability of it not being solved is \[1-\frac{1}{(n-r)^2}=\frac{(n-r+1)(n-r-1)}{(n-r)^2}\]. Now we want the probability that the problem is not solved at any of the \(n-2\) intermediate stages: \[\prod_{r=1}^{n-2}\frac{(n-r+1)(n-r-1)}{(n-r)^2}=\frac{n}{2(n-1)}\] This gives the probability that the last passenger finds his seat already taken. So the probability that the last passenger finds his seat free is: \[1-\frac{n}{2(n-1)}=\frac{n-2}{2(n-1)}\] I've a hunch that this conclusion could have been arrived at rather more simply and I'd be interested to hear from anyone who can demonstrate how.
Actually I got the sequence but couldn't find a general formula for it in terms of "n". This guy has done just that. Thanks td290. That was great explanation! Made it look so easy.
Forgive me Suraj. I just saw that you had posted three questions on the forum - with no explanation of what the problem was and I assumed (incorrectly) that you were just hoping for everyone to do the work for you (based on other students in the past who had done similar things). So it appears my model was incorrect and shall have to be amended...alternatively I could just ask people what they're stuck on...
Hmm.. Maybe I should've mentioned where was I stuck. Actually the thing is, I've nothing to do these days (looking for job). Already cleared 6 exams and I don't want to clear anymore without job. So just reading random stuff on probability now-a-days, where I came across these wicked questions. Had to do something to pass my time I have to say it's quite fun studying when you are not under exam pressure. You learn a lot better compared to when you are preparing for the exam. Just finished a book on Elementary Statistics. It was awesome. Completely changed the way I used to look at Stats.
