weatherwax_ross_solutions - A Solution Manual for: A First...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: A Solution Manual for: A First Course In Probability: Seventh Edition by Sheldon M. Ross. John L. Weatherwax May 26, 2008 Introduction Acknowledgements Special thanks to Vincent Frost and Andrew Jones for helping find and correct various typos in these solutions. Miscellaneous Problems The Crazy Passenger Problem The following is known as the crazy passenger problem and is stated as follows. A line of 100 airline passengers is waiting to board the plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, lets say that the k-th passenger in line has a ticket for the seat number k .) Unfortunately, the first person in line is crazy , and will ignore the seat number on their ticket, picking a random seat to occupy. All 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)? If one tries to solve this problem with conditional probability it becomes very difficult. We begin by considering the following cases if the first passenger sits in seat number 1, then all * 1 the remaining passengers will be in their correct seats and certainly the #100th will also. If he sits in the last seat #100, then certainly the last passenger cannot sit there (in fact he will end up in seat #1). If he sits in any of the 98 seats between seats #1 and #100, say seat k , then all the passengers with seat numbers 2 , 3 ,...,k 1 will have empty seats and be able to sit in their respective seats. When the passenger with seat number k enters he will have as possible seating choices seat #1, one of the seats k + 1 ,k + 2 ,..., 99, or seat #100. Thus the options available to this passenger are the same options available to the first passenger. That is if he sits in seat #1 the remaining passengers with seat labels k +1 ,k +2 ,..., 100 can sit in their assigned seats and passenger #100 can sit in his seat, or he can sit in seat #100 in which case the passenger #100 is blocked, or finally he can sit in one of the seats between seat k and seat #99. The only difference is that this k-th passenger has fewer choices for the middle seats. This k passenger effectively becomes a new crazy passenger. From this argument we begin to see a recursive structure. To fully specify this recursive structure lets generalize this problem a bit an assume that there are N total seats (rather than just 100). Thus at each stage of placing a k-th crazy passenger we can choose from seat #1 and the last or N-th passenger will then be able to sit in their assigned seat, since all intermediate passengers seats are unoccupied....
View Full Document

This note was uploaded on 06/28/2009 for the course STAT 430 taught by Professor Krieger during the Summer '08 term at UPenn.

Page1 / 281

weatherwax_ross_solutions - A Solution Manual for: A First...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online