Soluções ROSS
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Soluções ROSS


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A Solution Manual for:
A First Course In Probability
by Sheldon M. Ross.
John L. Weatherwax\u2217
February 26, 2012
Introduction
Here you\u2019ll find some notes that I wrote up as I worked through this excellent book. I\u2019ve
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 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 \u2013 technical, grammatical,
typographical, whatever \u2013 please tell me that, too. I\u2019ll gladly add to the acknowledgments
in later printings the name of the first person to bring each problem to my attention.
Acknowledgments
Special thanks to (most recent comments are listed first): Ralph Lozano, Vincent Leung,
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 Marlene Miller
made several helpful suggestions on most of the material in Chapter 3. Her algebraic use
of event \u201cset\u201d 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
\u2217
wax@alum.mit.edu
1
worked in these notes. Sort of a \u201ctake a penny, leave a penny\u201d type of approach. Remember:
pay it forward.
Miscellaneous Problems
The Crazy Passenger Problem
The following is known as the \u201ccrazy passenger problem\u201d 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\u2019s 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
the remaining passengers will be in their correct seats and certainly the #100\u2019th 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\u22121 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 \u201cmiddle\u201d seats. This k passenger effectively becomes a new \u201ccrazy\u201d 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
\u2022 seat #1 and the last or N -th passenger will then be able to sit in their assigned seat,
since all intermediate passenger\u2019s seats are unoccupied.
\u2022 seat # N and the last or N -th passenger will be unable to sit in their assigned seat.
\u2022 any seat before the N -th and after the k-th. Where the k-th passenger\u2019s seat is taken
by a crazy passenger from the previous step. In this case there are N\u22121\u2212(k+1)+1 =
N \u2212 k \u2212 1 \u201cmiddle\u201d 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
p(N, 1) =
1
N
(1) +
1
N
(0) +
1
N
N\u22121\u2211
k=2
p(N \u2212 k, 1)
=
1
N
+
1
N
N\u22121\u2211
k=2
p(N \u2212 k, 1) .
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 \u2212 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
arrangements of passengers (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
.
If N = 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
2
for all N , placing this assumption into the recursive
formulation above gives
p(N, 1) =
1
N
+
1
N
N\u22121\u2211
k=2
1
2
=
1
2
.
Verifying that indeed this constant value satisfies our recursion relationship.
Chapter 1 (Combinatorial Analysis)
Chapter 1: Problems
Problem 1 (counting license plates)
Part (a): In each of the first two places we can put any of the 26 letters giving 262 possible
letter combinations for the first two characters. Since the five other characters in the license
plate must be numbers, we have 105 possible five digit letters their specification giving a
total of
262 · 105 = 67600000 ,
total license plates.
Part (b): If we can\u2019t repeat a letter or a number in the specification of a license plate then
the number of license plates becomes
26 · 25 · 10 · 9 · 8 · 7 · 6 = 19656000 ,
total license plates.
Problem 2 (counting die rolls)
We have six possible outcomes for each of the die rolls giving 64 = 1296 possible total
outcomes for all four rolls.
Problem 3 (assigning workers to jobs)
Since each job is different and each worker is unique we have 20! different pairings.
Problem 4 (creating a band)
If each boy can play each instrument we can have 4! = 24 ordering. If Jay and Jack can
play only two instruments then we will assign the instruments they play first with 2! possible
orderings. The other two boys can be assigned the remaining instruments in 2! ways and
thus we have
2! · 2! = 4 ,
possible unique band assignments.
Lucas Lima
Realce
Lucas Lima
Realce
Problem 5 (counting telephone area codes)
In the first specification of this problem we can have 9 \u2212 2 + 1 = 8 possible choices for the
first digit in an area code. For the second digit there are two possible choices. For the third
digit there are 9 possible choices.