Instructor’s Solutions Manual
Version October 14, 20171
Probability, A Lively Introduction
Henk Tijms
Cambridge University Press, 2017
1
Despite careful checking, there is non-negligible probability that there are errors in
answers. If anything should be corrected, please let me know be sending an email to
.
, 1
Chapter 1
1.1 Imagine a blue and a red die. Take as sample space the set of the
ordered pairs (b, r), where b is the number shown on the blue and r
is the number shown on the red die. Each of the 36 elements (b, r) is
equally likely. There are 2 × 3 × 3 = 18 elements for which exactly
one component is odd. Thus the probability that the sum of the two
dice is odd equals 18 1
36 = 2 . There are 3 × 3 = 9 elements for which
both components are odd. Thus the probability that the product of
9
the two numbers rolled is odd equals 36 = 14 . Alternatively, you can
obtain the probabilities by using as sample space the set consisting of
the four equiprobable elements (odd, odd), (odd, even), (even, even),
and (even, odd).
1.2 Label the plumbers as 1 and 2. Take as sample space the set of all
possible sequences of ones and twos to the length 3, where a one stands
for plumber 1 and a two for plumber 2. The sample space has 23 = 8
equiprobable outcomes. There are 2 outcomes with three ones or three
twos. The sought probability is 82 = 14 .
1.3 Label the 10 letters of “randomness” as 1 to 10. Take as sample space
the set of all permutations of the numbers 1 to 10. All 10! outcomes are
equally likely. There are 3 × 2 × 8! outcomes that begin and end with
a vowel and there are 8 × 3! × 7! outcomes in which the three vowels
are adjacent to each other. The probabilities are 3 × 2 × 8!/10! = 1/15
and 8 × 3! × 7!/10! = 1/15.
1.4 Take as sample space the set of all possible sequences of zeros and ones
to the length 4, where a zero stands for male gender and an one for
female gender. The sample space has 24 = 16 equiprobable outcomes.
There are 8 outcomes with exactly three zeros or exactly three ones
and 6 outcomes with exactly two zeros. Hence the probability of three
8
puppies of one gender and one of the other is 16 . The probability of
6
two puppies of each gender is 16 .
1.5 Take as sample space the set of all unordered samples of m different
n
numbers. The sample space has m equiprobable elements. There
n−1
are m−1 samples that contain the largest number. The probability
n−1
n
of getting the largest number is m−1 / m = m n . Alternatively, you
can take as sample space the set of all n! permutations of the integers
1 to n. There are m × (n − 1)! permutations for which the number n
,2
is in one of the first m positions.
Note: More generally, the probability that the largest r numbers
n−r n
are among the m numbers picked is given by both m−r / m and
m
r r!(n − r)!/n!.
1.6 Take as sample space the set of all possible combinations of two persons
6
who do the dishes. The sample space has 2 = 15 equally likely
outcomes. The number of outcomes consisting of two boys is 32 =
3
3. The sought probability is 15 = 51 .Alternatively, using an ordered
sample space consisting of all 6! possible ordering of the six people
and imagining that the first two people in the ordering have to do the
dishes, the sought probability can be calculated as 3×2×4!
6! = 15
1.7 Imagine that the balls are labeled as 1, . . . , n. It is no restriction to
assume that the two winning balls have the labels 1 and 2. Take as
sample space the set of all n! permutations of 1, . . . , n. For any k,
the number of permutations having either 1 or 2 on the kth place is
(n − 1)! + (n − 1)!. Thus, the probability that the kth person picks a
winning ball is (n−1)!+(n−1)!
n! = n2 for each k.
1.8 Take as sample space the set of all ordered pairs (i, j) : i, j = 1, . . . , 6,
where i is the number rolled by player A and j is the number rolled by
player B. The sample space has 36 equally likely outcomes. The num-
ber of winning outcomes for player B is 9 + 11 = 20. The probability
of player A winning is 16 4
36 = 9 .
1.9 Take as sample space the set of all unordered samples of six differ-
ent numbers from the numbers 1 to 42. The sample space has 42 6
equiprobable outcomes. There are 41
5 outcomes with the number 10.
Thus the probability of getting the number 10 is 41
42 6
= 42 . The
probability
42that each of the six numbers picked is 20 or more is equal
23
to = 0.0192. Alternatively, the probabilities can be calcu-
lated by using the sample space consisting of all ordered arrangement
of the numbers 1 to 42, where the numbers in the first six positions
6
are the lotto numbers. This leads the calculations (6 × 41!)/42! = 42
23
and ( 6 × 6! × 36!)/42! = 0.0192 for the sought probabilities.
1.10 Take as (unordered) sample space all possible combinations of two
candidates to receive a cup of tea from the waiter. The sample space
5
has 2 = 10 equally likely outcomes. The number of combinations
of two people each getting the cup of tea they ordered is 1. The
, 3
1
sought probability is 10 . Alternatively, using an ordered sample space
consisting of all possible orderings of the five people and imagining
that the first two people in the ordering get a cup of tea from the
waiter, the probability can be calculated as 2×1×3!
5!
1
= 10 .
1.11 Label the nine socks as s1 , . . . , s9 . The probability model in which the
order of selection of the socks is considered relevant has a sample space
with 9 × 8 = 72 equiprobable outcomes (si , sj ). There are 4 × 5 = 20
outcomes for which the first sock chosen is black and the second is
white, and there are 5 × 4 = 20 outcomes for which the first sock is
white and the second is black. The sought probability is 40/72 = 5/9.
The probability model in which the order of selection of the socks is
9
not considered relevant has a sample space with 2 = 36 equiprobable
outcomes. The number of outcomes for which the socks have different
5 4
colors is 1 × 1 = 20, yielding the same value 20/36 = 5/9 for the
sought probability.
1.12 This problem can be solved by using either an ordered sample space or
an unordered sample space. Label the ten letters of the word Cincin-
nati as 1, 2, . . . , 10. As ordered sample space, take the set of all or-
dered pairs (i1 , i2 ), where i1 is the label of the first letter dropped and
i2 is the label of the second letter dropped. This sample space has
10 × 9 = 90 equally likely outcomes. Let A be the event that the two
letters dropped are the same. Noting that in the word Cincinnati the
letter c occurs two times and the letters
i and n each
occur three times,
it follows that there are 22 × 2! + 32 × 2! + 32 × 2! = 14 outcomes
14 7
leading to the event A. Hence P (A) = 90 = 45 . An unordered sample
space can also be used . This sample space consists of all possible sets
of two differently labeled letters from the ten letters of Cincinnati.
This sample space has 10 2 = 45 equally likely outcomes. The number
of outcomes for which the two labeled letters in the set represent the
same letter is 22 + 32 + 32 = 7. This gives the same value 45 7
for the
probability that the two letters dropped are the same.
1.13 Take as sample space the set of all unordered pairs of two distinct
cards. The sample space has 52 2 equally likely outcomes. There are
1 51 3 12
1 × 1 = 51 outcomes with the ten of hearts, and 1 × 1 = 36
outcomes with hearts and aten but not the ten of hearts. The sought
probability is (51 + 36)/ 52
2 = 0.0656.
1.14 Represent the words chance and choice by chanCe and choiCe. Take as
Version October 14, 20171
Probability, A Lively Introduction
Henk Tijms
Cambridge University Press, 2017
1
Despite careful checking, there is non-negligible probability that there are errors in
answers. If anything should be corrected, please let me know be sending an email to
.
, 1
Chapter 1
1.1 Imagine a blue and a red die. Take as sample space the set of the
ordered pairs (b, r), where b is the number shown on the blue and r
is the number shown on the red die. Each of the 36 elements (b, r) is
equally likely. There are 2 × 3 × 3 = 18 elements for which exactly
one component is odd. Thus the probability that the sum of the two
dice is odd equals 18 1
36 = 2 . There are 3 × 3 = 9 elements for which
both components are odd. Thus the probability that the product of
9
the two numbers rolled is odd equals 36 = 14 . Alternatively, you can
obtain the probabilities by using as sample space the set consisting of
the four equiprobable elements (odd, odd), (odd, even), (even, even),
and (even, odd).
1.2 Label the plumbers as 1 and 2. Take as sample space the set of all
possible sequences of ones and twos to the length 3, where a one stands
for plumber 1 and a two for plumber 2. The sample space has 23 = 8
equiprobable outcomes. There are 2 outcomes with three ones or three
twos. The sought probability is 82 = 14 .
1.3 Label the 10 letters of “randomness” as 1 to 10. Take as sample space
the set of all permutations of the numbers 1 to 10. All 10! outcomes are
equally likely. There are 3 × 2 × 8! outcomes that begin and end with
a vowel and there are 8 × 3! × 7! outcomes in which the three vowels
are adjacent to each other. The probabilities are 3 × 2 × 8!/10! = 1/15
and 8 × 3! × 7!/10! = 1/15.
1.4 Take as sample space the set of all possible sequences of zeros and ones
to the length 4, where a zero stands for male gender and an one for
female gender. The sample space has 24 = 16 equiprobable outcomes.
There are 8 outcomes with exactly three zeros or exactly three ones
and 6 outcomes with exactly two zeros. Hence the probability of three
8
puppies of one gender and one of the other is 16 . The probability of
6
two puppies of each gender is 16 .
1.5 Take as sample space the set of all unordered samples of m different
n
numbers. The sample space has m equiprobable elements. There
n−1
are m−1 samples that contain the largest number. The probability
n−1
n
of getting the largest number is m−1 / m = m n . Alternatively, you
can take as sample space the set of all n! permutations of the integers
1 to n. There are m × (n − 1)! permutations for which the number n
,2
is in one of the first m positions.
Note: More generally, the probability that the largest r numbers
n−r n
are among the m numbers picked is given by both m−r / m and
m
r r!(n − r)!/n!.
1.6 Take as sample space the set of all possible combinations of two persons
6
who do the dishes. The sample space has 2 = 15 equally likely
outcomes. The number of outcomes consisting of two boys is 32 =
3
3. The sought probability is 15 = 51 .Alternatively, using an ordered
sample space consisting of all 6! possible ordering of the six people
and imagining that the first two people in the ordering have to do the
dishes, the sought probability can be calculated as 3×2×4!
6! = 15
1.7 Imagine that the balls are labeled as 1, . . . , n. It is no restriction to
assume that the two winning balls have the labels 1 and 2. Take as
sample space the set of all n! permutations of 1, . . . , n. For any k,
the number of permutations having either 1 or 2 on the kth place is
(n − 1)! + (n − 1)!. Thus, the probability that the kth person picks a
winning ball is (n−1)!+(n−1)!
n! = n2 for each k.
1.8 Take as sample space the set of all ordered pairs (i, j) : i, j = 1, . . . , 6,
where i is the number rolled by player A and j is the number rolled by
player B. The sample space has 36 equally likely outcomes. The num-
ber of winning outcomes for player B is 9 + 11 = 20. The probability
of player A winning is 16 4
36 = 9 .
1.9 Take as sample space the set of all unordered samples of six differ-
ent numbers from the numbers 1 to 42. The sample space has 42 6
equiprobable outcomes. There are 41
5 outcomes with the number 10.
Thus the probability of getting the number 10 is 41
42 6
= 42 . The
probability
42that each of the six numbers picked is 20 or more is equal
23
to = 0.0192. Alternatively, the probabilities can be calcu-
lated by using the sample space consisting of all ordered arrangement
of the numbers 1 to 42, where the numbers in the first six positions
6
are the lotto numbers. This leads the calculations (6 × 41!)/42! = 42
23
and ( 6 × 6! × 36!)/42! = 0.0192 for the sought probabilities.
1.10 Take as (unordered) sample space all possible combinations of two
candidates to receive a cup of tea from the waiter. The sample space
5
has 2 = 10 equally likely outcomes. The number of combinations
of two people each getting the cup of tea they ordered is 1. The
, 3
1
sought probability is 10 . Alternatively, using an ordered sample space
consisting of all possible orderings of the five people and imagining
that the first two people in the ordering get a cup of tea from the
waiter, the probability can be calculated as 2×1×3!
5!
1
= 10 .
1.11 Label the nine socks as s1 , . . . , s9 . The probability model in which the
order of selection of the socks is considered relevant has a sample space
with 9 × 8 = 72 equiprobable outcomes (si , sj ). There are 4 × 5 = 20
outcomes for which the first sock chosen is black and the second is
white, and there are 5 × 4 = 20 outcomes for which the first sock is
white and the second is black. The sought probability is 40/72 = 5/9.
The probability model in which the order of selection of the socks is
9
not considered relevant has a sample space with 2 = 36 equiprobable
outcomes. The number of outcomes for which the socks have different
5 4
colors is 1 × 1 = 20, yielding the same value 20/36 = 5/9 for the
sought probability.
1.12 This problem can be solved by using either an ordered sample space or
an unordered sample space. Label the ten letters of the word Cincin-
nati as 1, 2, . . . , 10. As ordered sample space, take the set of all or-
dered pairs (i1 , i2 ), where i1 is the label of the first letter dropped and
i2 is the label of the second letter dropped. This sample space has
10 × 9 = 90 equally likely outcomes. Let A be the event that the two
letters dropped are the same. Noting that in the word Cincinnati the
letter c occurs two times and the letters
i and n each
occur three times,
it follows that there are 22 × 2! + 32 × 2! + 32 × 2! = 14 outcomes
14 7
leading to the event A. Hence P (A) = 90 = 45 . An unordered sample
space can also be used . This sample space consists of all possible sets
of two differently labeled letters from the ten letters of Cincinnati.
This sample space has 10 2 = 45 equally likely outcomes. The number
of outcomes for which the two labeled letters in the set represent the
same letter is 22 + 32 + 32 = 7. This gives the same value 45 7
for the
probability that the two letters dropped are the same.
1.13 Take as sample space the set of all unordered pairs of two distinct
cards. The sample space has 52 2 equally likely outcomes. There are
1 51 3 12
1 × 1 = 51 outcomes with the ten of hearts, and 1 × 1 = 36
outcomes with hearts and aten but not the ten of hearts. The sought
probability is (51 + 36)/ 52
2 = 0.0656.
1.14 Represent the words chance and choice by chanCe and choiCe. Take as
