, A Solution Manual for:
A First Course In Probability: Seventh Edition
by Sheldon M. Ross.
John L. Weatherwax∗
September 4, 2007
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, 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,
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 #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.
• seat # N and the last or N-th passenger will be unable to sit in their assigned seat.
• any seat before the N-th and after the k-th. Where the k-th passenger’s seat is taken
by a crazy passenger from the previous step. In this case there are N −1 −(k + 1) + 1 =
N − k − 1 “middle” seat choices.
If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select
from that the last passenger sits in his seat. From the argument above we have a recursive
structure give by
N −1
1 1 1 X
p(N, 1) = (1) + (0) + p(N − k, 1)
N N N k=2
N −1
1 1 X
= + p(N − k, 1) .
N N k=2
where the first term is where the first passenger picks the first seat (where the N will sit
correctly with probability one), the second term is when the first passenger sits in the N-th
seat (where the N will sit correctly with probability zero), and the remaining terms represent
the first passenger sitting at position k, which will then require repeating this problem with
the k-th passenger choosing among N − k + 1 seats.
To solve this recursion relation we consider some special cases and then apply the principle
of mathematical induction to prove it. Lets take N = 2. Then there are only two possible
arraignments of passengers (1, 2) and (2, 1) of which one (the first) corresponds to the second
passenger sitting in his assigned seat. This gives
1
p(2, 1) = .
2
, If N = 3, then from the 3! = 6 possible choices for seating arraignments
(1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1)
Only
(1, 2, 3) (2, 1, 3) (3, 2, 1)
correspond to admissible seating arraignments for this problem so we see that
3 1
p(3, 1) = = .
6 2
1
If we hypothesis that p(N, 1) = 2
for all N, placing this assumption into the recursive
formulation above gives
N −1
1 1 X1 1
p(N, 1) = + = .
N N k=2 2 2
Verifying that indeed this constant value satisfies our recursion relationship.
A First Course In Probability: Seventh Edition
by Sheldon M. Ross.
John L. Weatherwax∗
September 4, 2007
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, 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,
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 #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.
• seat # N and the last or N-th passenger will be unable to sit in their assigned seat.
• any seat before the N-th and after the k-th. Where the k-th passenger’s seat is taken
by a crazy passenger from the previous step. In this case there are N −1 −(k + 1) + 1 =
N − k − 1 “middle” seat choices.
If we let p(n, 1) be the probability that given one crazy passenger and n total seats to select
from that the last passenger sits in his seat. From the argument above we have a recursive
structure give by
N −1
1 1 1 X
p(N, 1) = (1) + (0) + p(N − k, 1)
N N N k=2
N −1
1 1 X
= + p(N − k, 1) .
N N k=2
where the first term is where the first passenger picks the first seat (where the N will sit
correctly with probability one), the second term is when the first passenger sits in the N-th
seat (where the N will sit correctly with probability zero), and the remaining terms represent
the first passenger sitting at position k, which will then require repeating this problem with
the k-th passenger choosing among N − k + 1 seats.
To solve this recursion relation we consider some special cases and then apply the principle
of mathematical induction to prove it. Lets take N = 2. Then there are only two possible
arraignments of passengers (1, 2) and (2, 1) of which one (the first) corresponds to the second
passenger sitting in his assigned seat. This gives
1
p(2, 1) = .
2
, If N = 3, then from the 3! = 6 possible choices for seating arraignments
(1, 2, 3) (1, 3, 2) (2, 3, 1) (2, 1, 3) (3, 1, 2) (3, 2, 1)
Only
(1, 2, 3) (2, 1, 3) (3, 2, 1)
correspond to admissible seating arraignments for this problem so we see that
3 1
p(3, 1) = = .
6 2
1
If we hypothesis that p(N, 1) = 2
for all N, placing this assumption into the recursive
formulation above gives
N −1
1 1 X1 1
p(N, 1) = + = .
N N k=2 2 2
Verifying that indeed this constant value satisfies our recursion relationship.
