They each hold a ticket to one of the 100 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 their ticket, picking 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 (100th) person to board the plane will sit in their proper seat (#100)?
Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking 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 (100th) person to board the plane will sit in their proper seat (#100)?
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The whole thing stops when someone else sits in Crazy Guy's seat. The chances of that range from 1/99 (First person) to 1/1 for the last guy. So:
1/99 + Summation from 98 to 2 of ( 1/ n(n+1) )
This returns 0.5 or 50%.
Also, the answer remains 50% no matter how many people are added to the line!
1/99 + Summation from 98 to 2 of ( 1/ n(n+1) )
This returns 0.5 or 50%.
Also, the answer remains 50% no matter how many people are added to the line!