### Crazy guy on the airplane

A line of 1000 airline passengers is waiting to board a plane. They each hold a ticket to one of the 1000 seats on that flight. For convenience, let's say that the nth passenger in line has a ticket for the seat number n. Unfortunately, the first person in line is crazy, and will ignore the seat number on his ticket and picks a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (1000th) person to board the plane will sit in their proper seat (#1000)? Now what is the probability for the nth guy to sit in the nth seat number, if there an 'n' people waiting to board the plane?

Level: Easy to Medium

(Image credit: http://www.hasslefreeclipart.com/clipart_carttravel/page1.html)

Level: Easy to Medium

(Image credit: http://www.hasslefreeclipart.com/clipart_carttravel/page1.html)

## 1 Comments:

Well, we forgot about this ocmpletely didn't we. Anyway here is one wayto solve this. Assume a set of 3 people and apply these rules to it and what do we get?

Let the 3 people be a,b,c

Valid combinations based on the rules are:

abc

bac

cab

cba

None of the others are valid - check them out and you will know why - read the problem

So of these 4 ways the 3 people can sit logically according to the problem only in 2 of them c gets to sit in his designated place.

So p = 2/4 = 1/2 = 50%

Yippeee, I get one point more!

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