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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
worked hard to make these notes as good as I can, but I have no illusions that they are perfect.
If you feel that that there is a better way to accomplish or explain an exerciseor 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 – technical, grammatical,
in later printings the name of the ﬁrst person to bring each problem to my attention.

AcknowledgementsSpecial thanks to (most recent comments are listed ﬁrst): Mark Chamness, Dale Peterson,
Doug Edmunds, Marlene Miller, John Williams (several contributions to chapter 4), Timothy
Alsobrooks, Konstantinos Stouras, William Howell, Robert Futyma, Waldo Arriagada, Atul
Narang, Andrew Jones, Vincent Frost, and Gerardo Robert for helping improve these notes
and solutions. It should be noted that MarleneMiller made several helpful suggestions
on most of the material in Chapter 3. Her algebraic use of event “set” 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 formof 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” type of approach. Remember:
pay it forward.

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 holda 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 isoccupied,
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 intheir 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 thepassenger 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 inhis 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...