weatherwax_ross_solutions

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

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A Solution Manualfor: A FirstCourse In Probability bySheldon M. Ross. John L. Weatherwax December 16, 2011 Introduction Acknowledgements Special thanks to (most recent comments are listed first): John Williams (several contri- butions to chapter 4), Timothy Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, Waldo Arriagada, Marlene Miller, Atul Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for helping improve these notes and solutions. It should be noted that Mar- lene Miller made several helpful suggestions to the problems in Chapter 3. All comments were (and are) much appreciated. 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, let’s 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, * [email protected] 1

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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 the remaining passengers will be in their correct seats and certainly the #100’th 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 passenger’s seats are unoccupied.
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