A Solution Manual for:
A First Course In Probability
by Sheldon M. Ross.
John L. Weatherwax∗
February 7, 2012
Introduction
Here you’ll find some notes that I wrote up as I worked through this excellent book. I’ve
workedhardtomakethesenotesasgoodasIcan, butIhavenoillus ionsthattheyareperfect.
If you feel that that there is a better way to accomplish or explain a n exercise or derivation
presented in these notes; or that one or more of the explanations is unclear, incomplete,
or misleading, please tell me. If you find an error of any kind – technic al, grammatical,
typographical, whatever – please tell me that, too. I’ll gladly add to the acknowledgments
in later printings the name of the first person to bring each problem t o my attention.
Acknowledgements
Special thanks to (most recent comments are listed first): Mark C hamness, Dale Peterson,
DougEdmunds, MarleneMiller, JohnWilliams(several contributionsto chapter4), Timothy
Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, W aldo Arriagada, Atul
Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for h elping improve these notes
and solutions. It should be noted that Marlene Miller made several he lpful suggestions
on most of the material in Chapter 3. Her algebraic use of event “se t” notation to solve
probability problems has opened my eyes to this powerful technique . It is a tool that I wish
to become more proficient with.
All comments (no matter how small) are much appreciated. In fact, if you find these notes
useful I would appreciate a contribution in the form of a solution to a problem that is not yet
∗wax@alum.mit.edu
1 worked in these notes. Sort of a “take a penny, leave a penny” typ e of approach. Remember:
pay it forward.
Miscellaneous Problems
The Crazy Passenger Problem
The following is known as the “crazy passenger problem” and is state d as follows. A line of
100 airline passengers is waiting to boardthe plane. They each hold a t icket 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 th e other passengers are
quite normal, and will go to their proper seat unless it is already occup ied. If it is occupied,
they will then find a free seat to sit in, at random. What is the probab ility 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 become s very difficult. We
begin by considering the following cases if the first passenger sits in s eat 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 ca nnot sit there (in fact he
will end up in seat #1). If he sits in any of the 98 seats betweenseats #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 numb erkenters 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 sameoptions 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 h e can sit in seat #100
in which case the passenger #100 is blocked, or finally he can sit in one o f the seats between
seatkand seat #99. The only difference is that this k-th passenger has fewer choices for
the “middle” seats. This kpassenger effectively becomes a new “crazy” passenger.
From this argument we begin to see a recursive structure. To fully s pecify this recursive
structure lets generalize this problem a bit an assume that there ar eNtotal 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 #Nand 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 a reN−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 ntotal seats to select
from that the last passenger sits in his seat. From the argument ab ove we have a recursive
structure give by
p(N,1) =1
N(1)+1
N(0)+1
NN−1/summationdisplay
k=2p(N−k,1)
=1
N+1
NN−1/summationdisplay
k=2p(N−k,1).
where the first term is where the first passenger picks the first se at (where the Nwill sit
correctly with probability one), the second term is when the first pa ssenger sits in the N-th
seat (where the Nwill sit correctly with probability zero), and theremaining terms repr esent
the first passenger sitting at position k, which will then require repeating this problem with
thek-th passenger choosing among N−k+1 seats.
To solve this recursion relation we consider some special cases and t hen apply the principle
of mathematical induction to prove it. Lets take N= 2. Then there are only two possible
arrangements ofpassengers (1 ,2)and(2 ,1) of which one (the first) corresponds to the second
passenger sitting in his assigned seat. This gives
p(2,1) =1
2.
IfN= 3, then from the 3! = 6 possible choices for seating arrangements
(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 arrangements for this problem so we see that
p(3,1) =3
6=1
2.
If we hypothesis that p(N,1) =1
2for allN, placing this assumption into the recursive
formulation above gives
p(N,1) =1
N+1
NN−1/summationdisplay
k=21
2=1
2.
Verifying that indeed this constant value satisfies our recursion re lationship.
