TEST BANK FOR A First Course in Probability 8th Edition By Sheldon Ross

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 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 – technical, 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 to my attention.
Acknowledgements
Special thanks to (most recent comments are listed first): Mark Chamness, Dale Peterson,
Doug Edmunds, Marlene Miller, JohnWilliams (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 “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 form of a solution to a problem that is not yet
[email protected]
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 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
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
p(N, 1) =
1
N
(1) +
1
N
(0) +
1
N
NX−1
k=2
p(N − k, 1)
=
1
N
+
1
N
NX−1
k=2
p(N − 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 − 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
NX−1
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’t 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.
Problem 5 (counting telephone area codes)
In the first specification of this problem we can have 9 − 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. So in total we have
8 · 2 · 9 = 144 ,
possible area codes. In the second specification of this problem, if we must start our area
codes with the digit “four” we will only have 2 · 9 = 18 area codes.
Problem 6 (counting kittens)
The traveler would meet 74 = 2401 kittens.
Problem 7 (arranging boys and girls)
Part (a): Since we assume that each person is unique, the total number of ordering is given
by 6! = 720.
Part (b): We have 3! orderings of each group of the three boys and girls. Since we can put
these groups of boys and girls in 2! different ways (either the boys first or the girls first) we
have
(2!) · (3!) · (3!) = 2 · 6 · 6 = 72 ,
possible orderings.
Part (c): If the boys must sit together we have 3! = 6 ways to arrange the block of boys.
This block of boys can be placed either at the ends or in between any of the individual 3!
orderings of the girls. This gives four locations where our block of boys can be placed we
have
4 · (3!) · (3!) = 144 ,
possible orderings.
Part (d): The only way that no two people of the same sex can sit together is to have the
two groups interleaved. Now there are 3! ways to arrange each group of girls and boys, and
to interleave we have two different choices for interleaving. For example with three boys and
girls we could have
g1b1g2b2g3b3 vs. b1g1b2g2b3g3 ,
thus we have
2 · 3! · 3! = 2 · 62 = 72 ,
possible arrangements.
Problem 8 (counting arrangements of letters)
Part (a): Since “Fluke” has five unique letters we have 5! = 120 possible arrangements.
Part (b): Since “Propose” has seven letters of which four (the “o”’s and the “p”’s) repeat
we have
7!
2! · 2!
= 1260 ,
arrangements.
Part (c): Now “Mississippi” has eleven characters with the “i” repeated four times, the “s”
repeated four times and the “p” repeated two times, so we have
11!
4! · 4! · 2!
= 34650 ,
possible rearranges.
Part (d): “Arrange” has seven characters with a double “a” and a double “r” so it has
7!
2! · 2!
= 1260 ,
different arrangements.
Problem 9 (counting colored blocks)
Assuming each block is unique we have 12! arrangements, but since the six black and the
four red blocks are not distinguishable we have
12!
6! · 4!
= 27720 ,
possible arrangements.
Problem 10 (seating people in a row)
Part (a): We have 8! = 40320 possible seating arrangements.
Part (b): We have 6! ways to place the people (not including A and B). We have 2! ways
to order A and B. Once the pair of A and B is determined, they can be placed in between
any ordering of the other six. For example, any of the “x”’s in the expression below could
be replaced with the A B pair
x P1 x P2 x P3 x P4 x P5x P6 x .
Giving seven possible locations for the A,B pair. Thus the total number of orderings is given
by
2! · 6! · 7 = 10080 .
Part (c): To place the men and women according to the given rules, the men and women
must be interleaved. We have 4! ways to arrange the men and 4! ways to arrange the
women. We can start our sequence of eight people with a woman or a man (giving two
possible choices). We thus have
2 · 4! · 4! = 1152 ,
possible arrangements.
Part (d): Since the five men must sit next to each other their ordering can be specified in
5! = 120 ways. This block of men can be placed in between any of the three women, or at
the end of the block of women, who can be ordered in 3! ways. Since there are four positions
we can place the block of men we have
5! · 4 · 3! = 2880 ,
possible arrangements.
Part (e): The four couple have 2! orderings within each pair, and then 4! orderings of the
pairs giving a total of
(2!)4 · 4! = 384 ,
total orderings.
Problem 11 (counting arrangements of books)
Part (a): We have (3 + 2 + 1)! = 6! = 720 arrangements.
Part (b): The mathematics books can be arranged in 2! ways and the novels in 3! ways.
Then the block ordering of mathematics, novels, and chemistry books can be arranged in 3!
ways resulting in
(3!) · (2!) · (3!) = 72 ,
possible arrangements.
Part (c): The number of ways to arrange the novels is given by 3! = 6 and the other three
books can be arranged in 3! ways with the blocks of novels in any of the four positions in
between giving
4 · (3!) · (3!) = 144 ,
possible arrangements.
Problem 12 (counting awards)
Part (a): We have 30 students to choose from for the first award, and 30 students to choose
from for the second award, etc. So the total number of different outcomes is given by
305 = 24300000
Part (b): We have 30 students to choose from for the first award, 29 students to choose
from for the second award, etc. So the total number of different outcomes is given by
30 · 29 · 28 · 27 · 26 = 17100720
Problem 13 (counting handshakes)
With 20 people the number of pairs is given by

20
2

= 190 .
Problem 14 (counting poker hands)
A deck of cards has four suits with thirteen cards each giving in total 52 cards. From these
52 cards we need to select five to form a poker hand thus we have

52
5

= 2598960 ,
unique poker hands.
Problem 15 (pairings in dancing)
We must first choose five women from ten in

10
5

possible ways, and five men from 12
in

12
5

ways. Once these groups are chosen then we have 5! pairings of the men and
women. Thus in total we will have

10
5

12
5

5! = 252 · 792 · 120 = 23950080 ,
possible pairings.
Problem 16 (forced selling of books)
Part (a): We have to select a subject from three choices. If we choose math we have
6
2

= 15 choices of books to sell. If we choose science we have

7
2

= 21 choices of
books to sell. If we choose economics we have

4
2

= 6 choices of books to sell. Since each
choice is mutually exclusive in total we have 15 + 21 + 6 = 42, possible choices.
Part (b): We must pick two subjects from

3
2

= 3 choices. If we denote the letter “M”
for the choice math the letter “S” for the choice science, and the letter “E” for the choice
economics then the three choices are
(M, S) (M,E) (S,E) .
For each of the choices above we have 6 · 7 + 6 · 4 + 7 · 4 = 94 total choices.
Problem 17 (distributing gifts)
We can choose seven children to give gifts to in

10
7

ways. Once we have chosen the
seven children, the gifts can be distributed in 7! ways. This gives a total of

10
7

· 7! = 604800 ,
possible gift distributions.
Problem 18 (selecting political parties)
We can choose two Republicans from the five total in

5
2

ways, we can choose two
Democrats from the six in

6
2

ways, and finally we can choose three Independents from
the four in

4
3

ways. In total, we will have

5
2

·

6
2

·

4
3

= 600 ,
different committees.
Problem 19 (counting committee’s with constraints)
Part (a): We select three men from six in

6
3

, but since two men won’t serve together
we need to compute the number of these pairings of three men that have the two that won’t
serve together. The number of committees we can form (with these two together) is given
by
2
2

·

4
1

= 4 .
So we have
6
3

− 4 = 16 ,
possible groups of three men. Since we can choose

8
3

= 56 different groups of women,
we have in total 16 · 56 = 896 possible committees.
Part (b): If two women refuse to serve together, then we will have

2
2

·

6
1

groups
with these two women in them from the

8
3

ways to draw three women from eight. Thus
we have
8
3

−

2
2

·

6
1

= 56 − 6 = 50 ,
possible groupings of woman. We can select three men from six in

6
3

= 20 ways. In
total then we have 50 · 20 = 1000 committees.
Part (c): We have

8
3

·

6
3

total committees, and

1
1

·

7
2

·

1
1

·

5
2

= 210 ,
committees containing the man and women who refuse to serve together. So we have

8
3

·

6
3

−

1
1

·

7
2

·

1
1

·

5
2

= 1120 − 210 = 910 ,
total committees.
Problem 20 (counting the number of possible parties)
Part (a): There are a total of

8
5

possible groups of friends that could attend (assuming
no feuds). We have

2
2

·

6
3

sets with our two feuding friends in them, giving

8
5

−

2
2

·

6
3

= 36
possible groups of friends
Part (b): If two fiends must attend together we have that

2
2

6
3

if the do attend
the party together and

6
5

if they don’t attend at all, giving a total of

2
2

6
3

+

6
5

= 26 .
Problem 21 (number of paths on a grid)
From the hint given that we must take four steps to the right and three steps up, we can
think of any possible path as an arraignment of the letters ”U” for up and “R” for right.
For example the string
U U U RRRR,
would first step up three times and then right four times. Thus our problem becomes one of
counting the number of unique arrangements of three “U”’s and four “R”’s, which is given
by
7!
4! · 3!
= 35 .
Problem 22 (paths on a grid through a specific point)
One can think of the problem of going through a specific point (say P) as counting the
number of paths from the start A to P and then counting the number of paths from P to
the end B. To go from A to P (where P occupies the (2, 2) position in our grid) we are
looking for the number of possible unique arrangements of two “U”’s and two “R”’s, which
is given by
4!
2! · 2!
= 6 ,
possible paths. The number of paths from the point P to the point B is equivalent to the
number of different arrangements of two “R”’s and one “U” which is given by
3!
2! · 1!
= 3 .
From the basic principle of counting then we have 6 · 3 = 18 total paths.
Problem 23 (assignments to beds)
Assuming that twins sleeping in different bed in the same room counts as a different arraignment,
we have (2!) · (2!) · (2!) = 8 possible assignments of each set of twins to a room. Since
there are 3! ways to assign the pair of twins to individual rooms we have 6 · 8 = 48 possible
assignments.
Problem 24 (practice with the binomial expansion)
This is given by
(3x2 + y)5 =
X5
k=0

5
k

(3x2)ky5−k .
Problem 25 (bridge hands)
We have 52! unique permutations, but since the different arrangements of cards within a
given hand do not matter we have
52!
(13!)4 ,
possible bridge hands.
Problem 26 (practice with the multinomial expansion)
This is given by the multinomial expansion
(x1 + 2x2 + 3x3)4 =
X
n1+n2+n3=4

4
n1 , n2 , n3

xn1
1 (2x2)n2(3x3)n3
The number of terms in the above summation is given by

4 + 3 − 1
3 − 1

=

6
2

=
6 · 5
2
= 15 .
Problem 27 (counting committees)
This is given by the multinomial coefficient

12
3 , 4 , 5

= 27720
Problem 28 (divisions of teachers)
If we decide to send n1 teachers to school one and n2 teachers to school two, etc. then the
total number of unique assignments of (n1, n2, n3, n4) number of teachers to the four schools
is given by
8
n1 , n2 , n3 , n4

.
Since we want the total number of divisions, we must sum this result for all possible combinations
of ni, or
X
n1+n2+n3+n4=8

8
n1 , n2 , n3 , n4

= (1 + 1 + 1 + 1)8 = 65536 ,
possible divisions.
If each school must receive two in each school, then we are looking for

8
2 , 2 , 2 , 2

=
8!
(2!)4 = 2520 ,
orderings.
Problem 29 (dividing weight lifters)
We have 10! possible permutations of all weight lifters but the permutations of individual
countries (contained within this number) are irrelevant. Thus we can have
10!
3! · 4! · 2! · 1!
=

10
3 , 4 , 2 , 1

= 12600 ,
possible divisions. If the united states has one competitor in the top three and two in
the bottom three. We have

3
1

possible positions for the US member in the first three
positions and

3
2

possible positions for the two US members in the bottom three positions,
giving a total of
3
1

3
2

= 3 · 3 = 9 ,
combinations of US members in the positions specified. We also have to place the other countries
participants in the remaining 10 − 3 = 7 positions. This can be done in

7
4 , 2 , 1

=
7!
4!·2!·1! = 105 ways. So in total then we have 9 · 105 = 945 ways to position the participants.
Problem 30 (seating delegates in a row)
If the French and English delegates are to be seated next to each other, they can be can be
placed in 2! ways. Then this pair constitutes a new “object” which we can place anywhere
among the remaining eight people, i.e. there are 9! arrangements of the eight remaining
people and the French and English pair. Thus we have 2·9! = 725760 possible combinations.
Since in some of these the Russian and US delegates are next to each other, this number
over counts the true number we are looking for by 2 · 28! = 161280 (the first two is for the
number of arrangements of the French and English pair). Combining these two criterion we
have
2 · (9!) − 4 · (8!) = 564480 .
Problem 31 (distributing blackboards)
Let xi be the number of black boards given to school i, where i = 1, 2, 3, 4. Then we must
have
P
i xi = 8, with xi 0. The number of solutions to an equation like this is given by

8 + 4 − 1
4 − 1

=

11
3

= 165 .
If each school must have at least one blackboard then the constraints change to xi 1 and
the number of such equations is give by

8 − 1
4 − 1

=

7
3

= 35 .
Problem 32 (distributing people)
Assuming that the elevator operator can only tell the number of people getting off at each
floor, we let xi equal the number of people getting off at floor i, where i = 1, 2, 3, 4, 5, 6.
Then the constraint that all people are off at the sixth floor means that
P
i xi = 8, with
xi 0. This has

n + r − 1
r − 1

=

8 + 6 − 1
6 − 1

=

13
5

= 1287 ,
possible distribution people. If we have five men and three women, let mi and wi be the
number of men and women that get off at floor i. We can solve this problem as the combination
of two problems. That of tracking the men that get off on floor i and that of tracking
the women that get off on floor i. Thus we must