Old guy, unhelpful staff

Level pending

There are n people namely 1st, 2nd,...i^th,..nth persons standing for boarding in a line in that order. The first person who is an old person enters the airplane and realizes that he has lost his ticket, thus, occupies a random seat. Now, the 2nd person enters and if his seat is vacant, he will occupy his own seat. Otherwise, he will sit in a vacant seat. What is the probability that the (n-2)th person occupies his own seat?


The answer is 0.75.

Vishnu Kadiri by Vishnu Kadiri , 19, India

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1 solution

Vishnu Kadiri
Jun 6, 2020

Let us slightly modify the problem. If the 1st person is sitting in, say, 5th person's seat, when the 5th person comes, they talk and the 1st person leaves, letting the 5th person getting his own seat. This situation is similar to the problem, as there is a person who is going in search for a vacant seat, just that the identities of the persons is not the same in the original problem and is always the 1st person in the modified situation. Now let us calculate the probability that k t h k^{th} person gets his seat(doesn't need to talk with the 1st person to get his seat back). Essentially, if the 1st person sits in his seat, or k t h k^{th} seat or i t h i^{th} seat where i > k i>k , the fate of the k t h k^{th} person getting his seat or not is decided. So, number of actually available choices for the 1st person till the k t h k^{th} boards the plane is n − k + 2 n-k+2 . k t h k^{th} guy gets his seat if the 1st person chooses one of n − k + 1 n-k+1 seats. Hence, probability is n − k + 1 n − k + 2 \frac {n-k+1}{n-k+2}

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