Solution Manual for A First Course in
Probability 10th Edition / All Chapters Full
Complete
1.1 Introduction
1.2 The Basic Principle of Counting
1.3 Permutations
1.4 Combinations
1.5 Multinomial Coefficients
1.6 The Number of Integer Solutions of Equations
Here is a typical problem of interest involving probability: A communication system is
to consist of 𝑛 seemingly identical antennas that are to be lined up in a linear order.
1 of 835
, The resulting system will then be able to receive all incoming signals and will be
called functional as long as no two consecutive antennas are defective. If it turns out
that exactly 𝑚 of the 𝑛 antennas are defective, what is the probability that the
resulting system will be functional? For instance, in the special case where 𝑛 = 4 and
𝑚 = 2, there are 6 possible system configurations, namely,
0 1 1 0
0 1 0 1
1 0 1 0
0 0 1 1
1 0 0 1
1 1 0 0
where 1 means that the antenna is working and 0 that it is defective. Because the
resulting system will be functional in the first 3 arrangements and not functional in the
remaining 3, it seems reasonable to take 3 = 1 as the desired probability. In the case
6 2
of general 𝑛 and 𝑚, we could compute the probability that the system is functional in
a similar fashion. That is, we could count the number of configurations that result in
the system’s being functional and then divide by the total number of all possible
configurations.
From the preceding discussion, we see that it would be useful to have an effective
method for counting the number of ways that things can occur. In fact, many
problems in probability theory can be solved simply by counting the number of
different ways that a certain event can occur. The mathematical theory of counting is
formally known as combinatorial analysis.
The basic principle of counting will be fundamental to all our work. Loosely put, it
states that if one experiment can result in any of 𝑚 possible outcomes and if another
experiment can result in any of 𝑛 possible outcomes, then there are mn possible
outcomes of the two experiments.
The basic principle of counting
Suppose that two experiments are to be performed. Then if experiment 1 can
result in any one of 𝑚 possible outcomes and if, for each outcome of
experiment 1, there are 𝑛 possible outcomes of experiment 2, then together
there are mn possible outcomes of the two experiments.
Proof of the Basic Principle: The basic principle may be proven by enumerating all
the possible outcomes of the two experiments; that is,
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, (1, 1), (1, 2), . . ., (1, 𝑛)
(2, 1), (2, 2), . . ., (2, 𝑛)
⋮
(𝑚, 1), (𝑚, 2), . . ., (𝑚, 𝑛)
where we say that the outcome is (𝑖, 𝑗) if experiment 1 results in its 𝑖th possible
outcome and experiment 2 then results in its 𝑗th possible outcome. Hence, the set of
possible outcomes consists of 𝑚 rows, each containing 𝑛 elements. This proves the
result.
Example 2a
A small community consists of 10 women, each of whom has 3 children. If one
woman and one of her children are to be chosen as mother and child of the year,
how many different choices are possible?
Solution
By regarding the choice of the woman as the outcome of the first experiment and
the subsequent choice of one of her children as the outcome of the second
experiment, we see from the basic principle that there are 10 × 3 = 30 possible
choices.
When there are more than two experiments to be performed, the basic principle can
be generalized.
The generalized basic principle of counting
If 𝑟 experiments that are to be performed are such that the first one may result
in any of 𝑛1 possible outcomes; and if, for each of these 𝑛1 possible outcomes,
there are 𝑛2 possible outcomes of the second experiment; and if, for each of
the possible outcomes of the first two experiments, there are 𝑛3 possible
outcomes of the third experiment; and if, then there is a total of 𝑛1 ⋅ 𝑛2⋯𝑛r
possible outcomes of the 𝑟 experiments.
Example 2b
A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors,
and 2 seniors. A subcommittee of 4, consisting of 1 person from each class, is to
be chosen. How many different subcommittees are possible?
Solution
We may regard the choice of a subcommittee as the combined outcome of the
four separate experiments of choosing a single representative from each of the
classes. It then follows from the generalized version of the basic principle that
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, there are 3 × 4 × 5 × 2 = 120 possible subcommittees.
Example 2c
How many different 7-place license plates are possible if the first 3 places are to
be occupied by letters and the final 4 by numbers?
Solution
By the generalized version of the basic principle, the answer is
26 ⋅ 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 175,760,000.
Example 2d
How many functions defined on 𝑛 points are possible if each functional value is
either 0 or 1?
Solution
Let the points be 1,2, . . . ,𝑛. Since 𝑓(𝑖) must be either 0 or 1 for each 𝑖 = 1,2, . . . ,𝑛,
it follows that there are 2n possible functions.
Example 2e
In Example 2c , how many license plates would be possible if repetition
among letters or numbers were prohibited?
Solution
In this case, there would be 26 ⋅ 25 ⋅ 24 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 = 78,624,000 possible
license plates.
How many different ordered arrangements of the letters 𝑎, 𝑏, and 𝑐 are possible? By
direct enumeration we see that there are 6, namely, abc, acb, bac, bca, cab, and
cba. Each arrangement is known as a permutation. Thus, there are 6 possible
permutations of a set of 3 objects. This result could also have been obtained from
the basic principle, since the first object in the permutation can be any of the 3, the
second object in the permutation can then be chosen from any of the remaining 2,
and the third object in the permutation is then the remaining 1. Thus, there are
3 ⋅ 2 ⋅ 1 = 6 possible permutations.
Suppose now that we have 𝑛 objects. Reasoning similar to that we have just
used for the 3 letters then shows that there are
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Probability 10th Edition / All Chapters Full
Complete
1.1 Introduction
1.2 The Basic Principle of Counting
1.3 Permutations
1.4 Combinations
1.5 Multinomial Coefficients
1.6 The Number of Integer Solutions of Equations
Here is a typical problem of interest involving probability: A communication system is
to consist of 𝑛 seemingly identical antennas that are to be lined up in a linear order.
1 of 835
, The resulting system will then be able to receive all incoming signals and will be
called functional as long as no two consecutive antennas are defective. If it turns out
that exactly 𝑚 of the 𝑛 antennas are defective, what is the probability that the
resulting system will be functional? For instance, in the special case where 𝑛 = 4 and
𝑚 = 2, there are 6 possible system configurations, namely,
0 1 1 0
0 1 0 1
1 0 1 0
0 0 1 1
1 0 0 1
1 1 0 0
where 1 means that the antenna is working and 0 that it is defective. Because the
resulting system will be functional in the first 3 arrangements and not functional in the
remaining 3, it seems reasonable to take 3 = 1 as the desired probability. In the case
6 2
of general 𝑛 and 𝑚, we could compute the probability that the system is functional in
a similar fashion. That is, we could count the number of configurations that result in
the system’s being functional and then divide by the total number of all possible
configurations.
From the preceding discussion, we see that it would be useful to have an effective
method for counting the number of ways that things can occur. In fact, many
problems in probability theory can be solved simply by counting the number of
different ways that a certain event can occur. The mathematical theory of counting is
formally known as combinatorial analysis.
The basic principle of counting will be fundamental to all our work. Loosely put, it
states that if one experiment can result in any of 𝑚 possible outcomes and if another
experiment can result in any of 𝑛 possible outcomes, then there are mn possible
outcomes of the two experiments.
The basic principle of counting
Suppose that two experiments are to be performed. Then if experiment 1 can
result in any one of 𝑚 possible outcomes and if, for each outcome of
experiment 1, there are 𝑛 possible outcomes of experiment 2, then together
there are mn possible outcomes of the two experiments.
Proof of the Basic Principle: The basic principle may be proven by enumerating all
the possible outcomes of the two experiments; that is,
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, (1, 1), (1, 2), . . ., (1, 𝑛)
(2, 1), (2, 2), . . ., (2, 𝑛)
⋮
(𝑚, 1), (𝑚, 2), . . ., (𝑚, 𝑛)
where we say that the outcome is (𝑖, 𝑗) if experiment 1 results in its 𝑖th possible
outcome and experiment 2 then results in its 𝑗th possible outcome. Hence, the set of
possible outcomes consists of 𝑚 rows, each containing 𝑛 elements. This proves the
result.
Example 2a
A small community consists of 10 women, each of whom has 3 children. If one
woman and one of her children are to be chosen as mother and child of the year,
how many different choices are possible?
Solution
By regarding the choice of the woman as the outcome of the first experiment and
the subsequent choice of one of her children as the outcome of the second
experiment, we see from the basic principle that there are 10 × 3 = 30 possible
choices.
When there are more than two experiments to be performed, the basic principle can
be generalized.
The generalized basic principle of counting
If 𝑟 experiments that are to be performed are such that the first one may result
in any of 𝑛1 possible outcomes; and if, for each of these 𝑛1 possible outcomes,
there are 𝑛2 possible outcomes of the second experiment; and if, for each of
the possible outcomes of the first two experiments, there are 𝑛3 possible
outcomes of the third experiment; and if, then there is a total of 𝑛1 ⋅ 𝑛2⋯𝑛r
possible outcomes of the 𝑟 experiments.
Example 2b
A college planning committee consists of 3 freshmen, 4 sophomores, 5 juniors,
and 2 seniors. A subcommittee of 4, consisting of 1 person from each class, is to
be chosen. How many different subcommittees are possible?
Solution
We may regard the choice of a subcommittee as the combined outcome of the
four separate experiments of choosing a single representative from each of the
classes. It then follows from the generalized version of the basic principle that
3 of 835
, there are 3 × 4 × 5 × 2 = 120 possible subcommittees.
Example 2c
How many different 7-place license plates are possible if the first 3 places are to
be occupied by letters and the final 4 by numbers?
Solution
By the generalized version of the basic principle, the answer is
26 ⋅ 26 ⋅ 26 ⋅ 10 ⋅ 10 ⋅ 10 ⋅ 10 = 175,760,000.
Example 2d
How many functions defined on 𝑛 points are possible if each functional value is
either 0 or 1?
Solution
Let the points be 1,2, . . . ,𝑛. Since 𝑓(𝑖) must be either 0 or 1 for each 𝑖 = 1,2, . . . ,𝑛,
it follows that there are 2n possible functions.
Example 2e
In Example 2c , how many license plates would be possible if repetition
among letters or numbers were prohibited?
Solution
In this case, there would be 26 ⋅ 25 ⋅ 24 ⋅ 10 ⋅ 9 ⋅ 8 ⋅ 7 = 78,624,000 possible
license plates.
How many different ordered arrangements of the letters 𝑎, 𝑏, and 𝑐 are possible? By
direct enumeration we see that there are 6, namely, abc, acb, bac, bca, cab, and
cba. Each arrangement is known as a permutation. Thus, there are 6 possible
permutations of a set of 3 objects. This result could also have been obtained from
the basic principle, since the first object in the permutation can be any of the 3, the
second object in the permutation can then be chosen from any of the remaining 2,
and the third object in the permutation is then the remaining 1. Thus, there are
3 ⋅ 2 ⋅ 1 = 6 possible permutations.
Suppose now that we have 𝑛 objects. Reasoning similar to that we have just
used for the 3 letters then shows that there are
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