Probability: A Lively Introduction 1st Edition – Henk Tijms – (Solutions Manual & Lecture Slides)

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Version October 14, 20171




Probability, A Lively Introduction



Henk Tijms


Cambridge University Press, 2017




1
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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
18
dice is odd equals 36 = 12 . 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 = 41 . 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 = 41 .

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

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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 = 15 .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! = 51

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

42
that 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

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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 a
ten 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

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sample space the set of all possible pairs (l1 , l2 ), where l1 is an element
from the word chanCe and l2 is an element from the word choiCe.
By distinguishing between c and C, the sample space has 6 × 6 = 36
equally likely outcomes. The number of outcomes for which the two
chosen letters represent the same letter is 4 + 1 + 1 = 6. The sought
probability is 61 .
1.15 Take as sample space the set of all sequences (i1 , . . . , ik ), where ik is the
number shown on the kth roll of the die. Each element of the sample
space is equally likely. The explanation is that there is P a one-to-one
10
correspondence between the elements (i1 ,P . . . , ik ) with k=1 ik = s
and the elements (7 − i1 , . . . , 7 − ik ) with 10 k=1 (7 − i k ) = 70 − s.
1.16 Take as ordered sample space the set of all sequences (i1 , . . . , i12 ),
where ik is the number rolled by the kth die. The sample space has
612 equally likely outcomes. The number
of outcomes
in which each
12 10 8 6 4




number appears exactly two times is 2 × 2 × 2 × 2 × 2 =
12!/26 . The sought probability is 2612!
×612
= 0.0034.
1.17 Take as sample space the
set
of
all possible

16 samples of three residents.
This leads to the value 41 41 3 = 55 for the sought probability.

1.18 Take as sample space the set of all ordered arrangements of 10 people,
where the people in the first five positions form group 1 and the other
five people form group 2. The sample space has 10! equally likely
elements. The number of elements for which your two friends and you
together are in the same group is 5 × 4 × 3 × 7! + 5 × 4 × 3 × 7!. The
sought probability is 120×7! 1
10! = 6 . Alternatively, the probability can be
(3)(7)+(3)(7)
calculated as 3 2 10 0 5 = 61 , using as sample space the set of all
(5)
possible combinations of five people for the first group. A third way
( 5) + ( 5)
to calculate the probability is 3 10 3 = 61 , using as sample space the
(3)
set of all possible combinations of three positions for the three friends.
1.19 Take as sample space the set of the 9! possible orderings of the nine
books. The subjects mathematics, physics and chemistry can be or-
dered in 3 × 2 × 1 = 6 ways and so the number of favorable orderings is
6 × 4! × 3! × 2!. The sought probability is (6 × 4! × 3! × 2!)/9! = 1/210.
1.20 The sample space is Ω = {(i, j, k) : i, j, k = 0, 1}, where the three
components corresponds to the outcomes of the three individual tosses
of the three friends. Here 0 means heads and 1 means tails. Each

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element of the sample space gets assigned a probability of 81 . Let A
denote the event that one of the three friends pays for all the three
tickets. The set A is given by A = Ω\{(0, 0, 0), (1, 1, 1)} and consists
of six elements. The sought probability is P (A) = 68 .

1.21 Label the eleven letters of the word Mississippi as 1, 2, . . . , 11 and take
as sample space the set of the 1111 possible ordered sequences of eleven
numbers from 1, . . . , 11. The four positions for a number representing
i, the four positions for a number representing s, the two positions
for a number representing p, and the
one
position for the number
representing m can be chosen in 11 7 3


4 × 4 × 2 ways. Therefore the
number of outcomes in which all letters of the word Mississippi are
11 7 3 4 4 2




represented is 4 × 4 × 2 × 4 × 4 × 2 . Dividing this number by
1111 gives the value 0.0318 for the sought probability.

1.22 One pair is a hand with the pattern aabcd, where a, b, c and d are
from distinct kinds of cards. There are 13 kinds and four of each kind
in a standard deck of 52 cards. The probability of getting one pair is
13

4
12

4
3
1 2
52

3 1
= 0.4226.
5

Two pair is a hand with the pattern aabbc, where a, b and c are from
distinct kinds of cards. The probability of getting two pair is
13 4 4 11 4






2 2 2
52

1 1
= 0.0475.
5


1.23 Take as sample space the set of all possible combinations of two apart-
ments from the 56 apartments. These two apartment represent the
56


vacant apartments. The sample space has 2 equiprobable elements.
The
number of elements with no vacant  56apartment
 on
the top floor is
48 48 56


2 . Thus the sought probability is 2 − 2 / 2 = 0.2675. Al-
ternatively, using a sample space made up of all permutations of the
56 apartments, the probability can be calculated as 1 − 48×47×54! 56! =
0.2675.

1.24 Imagine that the balls are labeled as 1 to 11, where the white balls get
the labels 1 to 7 and the red balls the labels 8 to 11. Take as sample
space is the set of all possible permutations of 1, 2, . . . , 11. The number
of outcomes in which a red ball appears for the first time at the ith

,6

7


drawing is i−1 × (i − 1)! × 4 × (7 − (i − 1) + 3)! for 1 ≤ i ≤ 8. The
sought probability is
4  
1 X 7 13
× (2k − 1)! × 4 × (7 − (2k − 1) + 3)! = .
11! 2k − 1 33
k=1


1.25 Take as sample space the set of all ordered pairs (i, j), where i is
the first number picked and j is the second number picked. There
are n2 equiprobable outcomes. For r ≤ n + 1, the r − 1 outcomes
(1, r − 1), (2, r − 2), . . . , (r − 1, 1) are the only outcomes (i, j) for which
i + j = r. Thus the probability that the sum of the two numbers
picked is r is r−1 n2
for 2 ≤ r ≤ n + 1. Therefore the probability of
getting a sum s when rolling two dice is s−1 36 for 2 ≤ s ≤ 7. By a
symmetry argument, the probability of getting a sum s is the same as
the probability of getting a sum 14 − s for 7 ≤ s ≤ 12 (opposite faces
of a die always total 7). Thus the probability of rolling a sum s has
the value 14−s−1
36 for 7 ≤ s ≤ 12.

1.26 Take as sample space the interval (0, 1). The outcome x means that
the stick is broken on the point x. The length of the longer piece is
at least three times the length of the shorter piece if x ∈ (0, 14 ) or
x ∈ ( 43 , 1). The sought probability is 41 + 41 = 12 .

1.27 Take as sample space the square {(x, y) : 0 ≤ x, y ≤ a}. The outcome
(x, y) refers to the position of the middle point of the coin. The sought
probability is given by the probability that a randomly chosen point
in the square falls in the subset {(x, y) : d2 ≤ x, y ≤ a − d2 } and is equal
to (a − d)2 /a2 .

1.28 Take as sample space the interval (0, 1). The outcome x means that a
randomly chosen point in (0, 1) is equal to x. The sought probability is
the probability that a randomly chosen point in (0, 1) falls into one of
1
the intervals (0, 12 ) or ( 21 , 12
7 1
). The sought probability is 12 1
+ 12 = 61 .

1.29 This problem can be solved with the model of picking at random a
point inside a rectangle. The rectangle R = {(x, y) : 0 ≤ x ≤ 1, 12 ≤
y ≤ 1} is taken as sample space, where the outcome (x, y) means that
you arrive 60x minutes past 7 a.m. and your friend arrives 60y minutes
past 7 a.m. The probability assigned to each subset of the sample space
is the area of the subset divided by the area of the rectangle R. The
sought probability is P (A), where the set A is the union of the three

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disjoint subsets {(x, y) : 12 < x, y < 12 + 12
1
}, {(x, y) : 21 + 12
1
< x, y < 34 }
3
and {(x, y) : 4 < x, y < 1}. This gives

1 1 1 1 1 1 1 7
P (A) = × + × + × = .
2 12 12 6 6 4 4 36

1.30 Translate the problem into choosing a point at random inside the unit
square {(x, y) : 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}. The probability that the
two persons will meet within 10 minutes of each other is given by the
probability that a point chosen at random in the unit square will fall
inside the shaded region in the figure. The area of the shaded region
is calculated as 1 − 65 × 56 = 0.3056. This gives the desired probability.


5/6




1/6


1/6 5/6



1.31 For q = 1, take the square {(x, y) : −1 < x, y < 1} as sample
space. The sought probability is the probability that a point (x, y)
chosen at random in the square satisfies y ≤ 14 x2 and is equal to
1
R1 1 2
4 (2 + −1 4 x dx) = 0.5417. For the general case, the sought probabil-
ity is (make a picture!):
Z q
1  2 1 2  1 q
2q + x dx = + for 0 < q < 4
4q 2 −q 4 2 24
Z 2√q
1  2 1 2 √  2
2
2q + √
x dx + 2(q − 2 q)q = 1 − √ for q ≥ 4.
4q −2 q 4 3 q

1.32 Take as sample space the set {(x, y) : 0 ≤ x, y ≤ 1}. The outcome
(x, y) means that a randomly chosen point in the unit square is equal to
(x, y). The probability that the Manhattan distance from a randomly
chosen point to the point (0, 0) is no more than a is given by the
probability that the randomly chosen point (x, y) satisfies x + y ≤ a.
The area of the region {(x, y) : 0 ≤ x, y ≤ 1 and x + y ≤ a} is 21 a2 .
This gives the sought probability. By a symmetry argument, this