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English Pages vi+230 [236] Year 1965
Table of contents :
Preface......Page 5
Contents......Page 7
1.1 RANDOM EXPERIMENTS......Page 9
1.2 EMPIRICAL BASIS OF PROBABILITY......Page 11
1.3 SIMPLE EVENTS AND THEIR PROBABILITIES......Page 17
1.4 DEFINITION OF PROBABILITY MODEL......Page 22
1.5 UNIFORM PROBABILITY MODELS......Page 27
1.6 THE ALGEBRA OF EVENTS......Page 31
1.7 SOME LAWS OF PROBABILITY......Page 37
2.1 A MODEL FOR SAMPLING......Page 41
2.2 THE NUMBER OF SAMPLES......Page 47
2.3 ORDERED SAMPLING......Page 52
2.4 SOME FORMULAS FOR SAMPLING......Page 57
3.1 PRODUCT MODELS FOR TWO-PART EXPERIMENTS......Page 61
3.2 REALISM OF PRODUCT MODELS......Page 66
3.3 BINOMIAL TRIALS......Page 74
3.4 THE USE OF RANDOM NUMBERS TO DRAW SAMPLES......Page 79
4.1 THE CONCEPT OF CONDITIONAL PROBABILITY......Page 83
4.2 INDEPENDENCE......Page 89
4.3 TWO-STAGE EXPERIMENTS......Page 96
4.4 PROPERTIES OF TWO-STAGE MODELS; BAYES' LAW......Page 103
4.5 APPLICATIONS OF PROBABILITY TO GENETICS......Page 108
4.6 MARRIAGE OF RELATIVES......Page 115
4.7 PERSONAL PROBABILITY......Page 118
5.1 DEFINITIONS......Page 124
5.2 DISTRIBUTION OF A RANDOM VARIABLE......Page 128
5.3 EXPECTATION......Page 132
5.4 PROPERTIES OF EXPECTATION......Page 136
5.5 LAWS OF EXPECTATION......Page 140
5.6 VARIANCE......Page 146
6.1 THE BINOMIAL DISTRIBUTION......Page 154
6.2 THE HYPERGEOMETRIC DISTRIBUTION......Page 160
6.3 STANDARD UNITS......Page 166
6.4 THE NORMAL CURVE AND THE CENTRAL LIMIT THEOREM......Page 172
6.5 THE NORMAL APPROXIMATION......Page 176
6.6 THE POISSON APPROXIMATION FOR np = 1......Page 181
6.7 THE POISSON APPROXIMATION: GENERAL CASE......Page 184
6.8 THE UNIFORM AND MATCHING DISTRIBUTIONS......Page 189
6.9 THE LAW OF LARGE NUMBERS......Page 193
6.10 SEQUENTIAL STOPPING......Page 197
7.1 JOINT DISTRIBUTIONS......Page 202
7.2 COVARIANCE AND CORRELATION......Page 208
7.3 THE MULTINOMIAL DISTRIBUTION......Page 215
TABLES......Page 221
INDEX......Page 229
INDEX OF EXAMPLES......Page 235
ELEMENTS OF FINITE PROBABILITY
HOLDEN-DAY SERIES IN PROBABILITY AND STATISTICS E. L. Lehmann, Edi/or
ELEMENTS OF FINITE PROBABILITY
J. L. Hodges, Jr. and E. L. Lehmann University of California, Berkeley
1965
HOLDEN-DAY, INC. San Francisco, London, Amsterdam
© Copyright 1964, 1965, by HOLDEN-DAY, INC. 728 Montgomery Street, San Francisco
All rights reserved. No part of this book may be reproduced in any form, by mimeograph or any other means, without permission in writing from the publisher. Printed in the United States of America. Library of Congress Catalog Card Number: 64-66419
Preface
After publication of our book Basic Concepts of Probability and Statistics, it was suggested to us that "Part I. Probability' be made available sepa'
rately, to serve as a text for a
one quarter or one semester course in prob-
ability at the precalculus level.
The
present book, the result of these
suggestions, contains all the material of Part I of the earlier book, supple-
mented by two new
A
sections at the end of Chapter
6.
careful treatment of probability without calculus
attention
is
restricted to finite probability models,
senting experiments with a finite
number
is
i.e.,
of outcomes.
basic notions of probability, such as probability model,
possible only
if
to models repre-
Fortunately, the
random
variable,
expectation and variance, are in fact most easily introduced with
finite
Furthermore, many interesting and important applications lead to finite models, such as the binomial, hypergeometric and sampling models. More than most branches of mathematics, probability theory retains its models.
connections with the empirical is
phenomena from which
it
grew.
While
it
possible to present probability in a purely abstract manner, divorced
from its empirical origins, the subject thereby loses much of its vitality and interest. On the other hand, if the theory is developed in its empirical context, it becomes essential to make clear the distinction between the real world and its mathematical representation. Our treatment is therefore centered on the notion of a probability model and emphasizes its relation to, and distinction from, the underlying random experiment which it represents.
One aspect of this approach is a careful treatment of the independence assumption in building probability models. Many complex experiments are made up of simpler parts, and it is frequently possible to build models for such experiments by first building simple models for the parts, and then combining them into a "product model' for the experiment as a whole. Particular attention is devoted to the realism of this procedure, and in general to the empirical meaning of the independence assumption, which '
so often
is
applied invalidly.
While basically we have included only concepts that can be denned and can be proved without calculus, there is one exception: we give approximations to certain probabilities whose calculation we have explained in principle, but whose actual computation would be too laboriThus, in particular, we discuss the normal approximation to the ous. binomial and hypergeometric distributions and the Poisson approximation Since the limit theoto the binomial and Poisson-binomial distributions. rems which underlie these approximations require advanced analytical techniques and are hence not available to us, we give instead numerical results that
illustrations to develop
The book
some
feeling for their accuracy.
some of which provide simple and results of the text.* Six of at the end of the book as Tables
contains over 400 problems,
exercises while others extend the ideas
the more important tables are collected
A-F.
A
decimal numbering system
is used for sections, formulas, examples, Thus, Section 5.4 is the fourth section of Chapter 5. To simplify the writing, within Chapter 5 itself we omit the chapter and refer to this section simply as Section 4. Similarly Example However, within Section 5.4 5.4.2 is the second example in Section 5.4. we omit the chapter and section and refer to the example as Example 2; in other sections of Chapter 5 we omit the chapter and refer to the example
tables, problems,
as
Example
and
so forth.
4.2. J.
L.
Hodges, Jr.
E. L. Berkeley
January, 1965
*
Instructors
may
obtain answer books by writing to the publisher.
Lehmann
Contents
l.
1. 2. 3. 4. 5. 6. 7.
PROBABILITY MODELS
Random experiment Empirical basis of probability . Simple events and their probabilities . Definition of a probability model . Uniform probability models The algebra of events Some laws of probability 2.
1.
2. 3. 4.
16 21
25
31
35 41 46 51
PRODUCT MODELS
1. Product models for two-part experiments 2. Realism of product models. 3. Binomial trials 4. The use of random numbers to draw samples 4.
5 11
SAMPLING
A model for sampling The number of samples Ordered sampling Some formulas for sampling 3.
3
55 60 68 73
CONDITIONAL PROBABILITY
1. The concept of conditional probability 2. Independence 3. Two-stage experiments . 4. Properties of two-stage models; Bayes' Jaw
77 83
90 97
CONTENTS 102
5.
Applications of probability to genetic
6.
Marriage of relatives
109
7.
Personal probability
112
5.
RANDOM VARIABLES 118
1.
Definitions
2.
Distribution of a
3.
Expectation
126
4.
Properties of expectation
130
5.
Laws
134
6.
Variance
140
7.
Laws
143
random
of expectation
of variance
6.
1.
2. 3.
4. 5.
6. 7. 8. 9.
10.
122
variable
SPECIAL DISTRIBUTIONS
The binomial distribution The hypergeometric distribution
148
Standard units The normal curve and the central limit theorem The normal approximation The Poisson approximation for np = 1
160
The Poisson approximation: general case The uniform and matching distributions The law of large numbers
178
Sequential stopping
191
7.
154
166 170
175 183
187
MULTIVARIATE DISTRIBUTIONS
1.
Joint distributions
196
2.
Co variance and correlation The multinomial distribution
202 209
3.
TABLES A.
B-C. D. E. F.
Number
of combinations Binomial distributions Square roots Normal approximation Poisson approximation
216 217 219 220 221
INDEX
223
EXAMPLE INDEX
229
CHAPTER
1
PROBABILITY MODELS
1.1
RANDOM EXPERIMENTS
The theory
of probability is a mathematical discipline which has found important applications in many different fields of human activity. It has extended the scope of scientific method, making it applicable to experiments whose results are not completely determined by the experimental
conditions.
The agreement among
most scientific on the fact that the experiments on which the theories are based will yield essentially the same results when they are repeated. When a scientist announces a discovery, other scienscientists regarding the validity of
theories rests to a considerable extent
tists in different
Sometimes the turns out to
same ment
in the is
parts of the world can verify his findings for themselves.
results of
mean
two
two workers appear to
disagree, but this usually
that the experimental conditions were not quite the
cases.
If
the same results are obtained
repeated under the same conditions,
we may say
when an
experi-
that the result
is
determined by the conditions, or that the experiment is deterministic. It is the deterministic nature of science that permits the use of scientific theory for predicting what will be observed under specified conditions. However, there are also experiments whose results vary, in spite of all Familiar examples efforts to keep the experimental conditions constant. are provided by gambling games: throwing dice, tossing pennies, dealing from a shuffled deck of cards can all be thought of as "experiments" with unpredictable results. More important and interesting instances occur in many fields. For example, seeds that are apparently identical will produce plants of differing height, and repeated weighings of the same object with a chemical balance will show slight variations. A machine which sews together two pieces of material, occasionally for no apparent reason
—
—
will
miss a
stitch.
If
we
are willing to stretch our idea of "experi-
PROBABILITY MODELS
4
ment," length of
life
may
be considered a variable experimental
since people living under similar conditions will die at different
predictable ages.
[CHAP.
1
result,
and un-
We shall refer to experiments that are not deterministic,
and thus do not always yield the same result when repeated under the same conditions, as random experiments. Probability theory (and the theory of statistics, which is based on it) are the branches of mathematics that have been developed to deal with random experiments. Let us now consider two random experiments in more detail. As the first
we take the experiment of throwing a die. This is one of random experiments and one with which most people are acquainted. In fact, probability theory had its beginnings in
example,
the simplest personally
the study of dice games.
Example
Throwing a die. Suppose we take a die, shake it vigorously and throw it against a vertical board so that it bounces onto a table. When the die comes to rest, we observe as the experimental result the number, say X, of points on the upper face. The experiment is not deterministic: the result X may be any of the six numbers 1, 2, 3, 4, 5, or 6, and no one can predict which of the values will be obtained on any partic1
.
in a dice cup,
ular performance of the experiment.
We may make every effort to control
by always placing the die in number of times, always throwing it against the same spot on the backboard, and so In spite of all such efforts, the result will remain variable and on.
or standardize the experimental conditions,
the cup in the same position, always shaking the cup the same
unpredictable.
Example 2.
Measuring a distance. It between two points a few miles apart. distance by use of a surveyor's chain.
is
desired to determine the distance
A
surveying party measures the
found in practice that the measured distance will not be exactly the same if two parties do the job, In spite of or even if the same party does the job on consecutive days. the best efforts to measure precisely, small differences will accumulate and the final measurement will vary from one performance of the experiment It is
to another.
How
is it
on indeterminacy? by an empirical observation: while the result on
possible for a scientific theory to be based
The paradox
is
resolved
any particular performance
of such
an experiment cannot be predicted,
a long sequence of performances, taken together, reveals a stability that
can serve as the basis for quite precise predictions. The property of longrun stability lies at the root of the ideas of probability, and we shall examine it in more detail in the next section.
EMPIRICAL BASIS OF PROBABILITY
1.2]
EMPIRICAL BASIS OF PROBABILITY
1.2
To
obtain some idea about the behavior of results in a sequence of repetirandom experiment, we shall now consider some specific examples.
tions of a
Example
Throwing a
1.
throws of a
number
value of the
Figure
die.
In this experiment,
die.
X of points showing,
n=10
(a)
shows certain
1
we were not
results of 5000
interested in the actual
but only in whether
(b) tt=50
1.0
(c)
1.0
1.0
X was less 7i=250
r
•
-OS 0.5
•
—
•
•
0.5-
0.5 •
0
•
•
•••
•
-
•
.
• •
•
•
1
-
•
1
1
A
1
i
1
20
10
Number
three
(X ^
2. We shall employ the same terminology in the model. as is
the
number
of points
X
Definition.
The complement
of
an event
E is the event E which
consists
just of those simple events in 8 that do not belong to E. It follows from this definition that the sets E and E have no common member, and that between them they contain all simple events of the
event set
8.
The
situation
Figure the simple events in 8,
namely
e 3 , eb
e h e2
,
e4
,
is
illustrated in Figure
1.
and
1,
where
E
consists of
Complementary events
E
consists of the remaining simple events
.
An interesting special case is the complement of the event set 8 itself. By definition, 8 consists of all simple events of the model under discussion, so that 8
is
a set without members.
This set
is
known
as the empty
set.
THE ALGEBRA OF EVENTS
1.6]
27
we think in terms of results, 8 corresponds to a result that cannot occur, = 7 when rolling a die. such as the result Related to any two given results R, S is the result which occurs when both the result R and the result S occur, and which will be called the result If
X
U
R
For example, if with two-child families R denotes the result is a boy and S that the second child is a boy, then "R and S" denotes the result that both children are boys. Similarly, if R indicates that a card drawn from a deck is a heart while S indicates that it is a face card, then the result "R and S" will occur if a heart face card is drawn. Suppose the results R, S are represented in the model by the events E, F respectively. What event in the model then corresponds to the result "R and S" ? To answer this question notice that the result U R and *S" occurs whenever the experiment yields a simple result, which is a simple Therefore the event in the model corresponding result both of R and of S. to "R and S" will consist of those simple events that belong both to E and The situation is illustrated in Figure 2a, where E = {e h e 3 e4 ee} to F.
and S."
that the
first
child
,
,
(a)
Figure
(b)
2.
Intersection and union
corresponds to the result R consisting of the simple results r h r3 n, r 6 and where F = {e 2 e 4 e*, e-i) corresponds to the result S consisting of the simple Here "R and £" will occur whenever the experiment results r 2 r 4 r 6 r 7 ,
,
,
.
,
yields one of the simple results r 4 or r6
.
The event corresponding
"R and S" consists therefore of the simple events common part or intersection of the two events E, F.
result
Definition.
which consists It
may
The
intersection
(E and F)
of
e4 , e6
-
to the
It is the
two events E, F is the event both to E and to F.
of all simple events that belong
of course
common.
,
,
,
The
happen that two events E, F have no simple events and F) is then the empty set.
intersection (E
in
PROBABILITY MODELS
28 Definition.
Two
belongs to both
E
events E,
and
F
[CHAP.
are said to be exclusive
if
1
no simple event
to F.
The results corresponding to two exclusive events will be such that not both of them can happen on the same trial of the random experiment. For example, when throwing a die, the results > 4 are ^ 2 and exclusive, but the results "X is even" and "X is divisible by 3" are not, = 6. Again, for two-child families, the results since both occur when "first child is a boy" and "both children are girls" are exclusive, but the results "first child is a boy" and "second child is a boy" are not, since both
X
X
X
children
may
The
be boys.
trated in Figure
case of two exclusive events
E
and
F
is illus-
3.
Figure
3.
Exclusive events
Another result related to the given results R, S is the result "R or S" which occurs when either the result R occurs or the result S occurs, or both If for example R denotes the result that the first child of a twooccur. child family is a boy while S denotes that the second child is a boy, then "R or S" will occur if at least one of the children is a boy. Similarly, if R denotes that a card drawn from a deck is a heart, while S denotes that it is a face card, then
"R or S" will occur if the card when it is a heart face card.
is
a heart or a face card,
including the case
What event in the model corresponds to the result "R or S" ? To answer this question, notice that the result U R or S" occurs whenever the experiment yields one of the simple results of R or of S or both. Therefore the event in the model corresponding to "R or S" will consist of those simple events belonging to in Figure 2b,
E or to F or E = {e h e
where as before
both. 3,
e4
,
e6}
The
F =
consisting of the simple results n, r 3 r4 r 6 while ,
,
situation
illustrated
is
corresponds to the result
,
{e 2 , e if
e&,
R
e 7 } corre-
sponds to the result S consisting of the simple results r2 n, r 6 r7 Here "R or S" will occur whenever the experiment yields one of the simple The composite event corresponding to "R or S" results n, r 2 r 3 r 4 r 6 r 7 It is obtained by consists therefore of the simple events e h e 2 e h e 4 e & e 7 combining or uniting the simple events of E with those of F, producing the set (E or F) enclosed by the dashed line in Figure 2b. ,
,
,
,
,
.
,
,
,
.
,
.
THE ALGEBRA OF EVENTS
1.6]
Definition.
The union (E
or F) of
29
two events E, F
is
the event which
consists of all simple events that belong to E, or to F, or to both
Example
1.
E
5.3, let
Two
E and F.
In the model for throwing two dice of Example "first die shows one point," so that E row of tableau (5.2). Similarly, let F correspond to
dice.
correspond to the result
consists of the first
"second die shows one point," consisting of the first column of (5.2). or F) consists of the eleven simple events in the top and left mar-
Then (E
gins of (5.2),
On
point."
and corresponds
(1, 1)
Since
By
result "at least one die
shows one
to the results "both dice
show one point."
Consider an event E and its complement have no simple event in common, they are exclusive. the definition of complement, any simple event not in E belongs to E,
Example T$.
and corresponds to the
the other hand, (E and F) consists only of the simple event
2.
E
Complementation.
and
E
so that every simple event belongs either to
union of
E and E is the entire event set 8:
The reader should be warned
E or to E.
(E or E)
of a confusion that
=
It follows that the 8.
sometimes
arises
due to
connotations of the word "and" in other contexts. From the use of "and" as a substitute for "plus," it might be thought that (E and F) should represent the set obtained by putting together or uniting the sets E and F. However, in the algebra of events (E and F) corresponds to the result il R and S" which occurs only if both R and S occur, and thus (E and F) is the common part or intersection of E and F. F) is used for (E or F) while (E O F) In many books the notation (E
U
is
used for (E and F).
PROBLEMS 1.
An
of the
experiment consists of ten tosses with a coin. of each of the following events:
(i)
at least six heads
(ii)
(iii)
2.
most six heads no heads. at
In Problem 5.6 give a verbal description of the complements of the events
described in parts 3.
Give a verbal description
complement
(i),
(ii),
(v),
and
(viii).
In Problem 5.9 give verbal descriptions of the complements of the events
described in parts (i)-(v). 4.
In Problem 5.6 give verbal descriptions of the intersections of each of the follow-
ing pairs of events: (i)
the events
(i)
(ii)
the events
(i)
and and
(ii)
(iii)
the events
(ii)
(iii)
(iv)
the events
(i)
and (iii) and (iv)
:
:
PROBABILITY MODELS
30
and (v) and (vi) the events (ii) and (vi) events
(i)
(viii)
the events
(i)
(ix)
the events (iv) and
(vii)
(x)
the events (vi) and
(vii).
(v) the (vi) (vii)
[CHAP.
the events
(iii)
and
1
(vi)
In Problem 5.6 determine for each of the following pairs whether they are
5.
exclusive (i)
(ii)
(iii)
(iv)
and (iii) and (iii) the events (iii) and (v) the events (iv) and (v) the events
(i)
the events
(ii)
(v)
the events (iv) and
(vi)
the events (v) and
(vi)
the events (v) and
(vii)
(vii) (viii)
the events
(vi)
(vi)
and
(vii).
In Problem 5.9 pick out three pairs from the events described in parts
6.
(i)-(vi)
that are exclusive. 7.
Give verbal descriptions of the unions of each of the pairs of events of Problem 4.
8.
In Problem
give verbal descriptions of the unions of each of the following
5.9,
pairs of events (i)
the events
(i)
(ii)
the events
(ii)
For each and G.
9.
F,
If E,
(i)
If
(ii)
E,
If E,
(iii)
From a
tion;
if
Let
etc.
E
of the following statements
class of ten,
first
first
an instructor
student does not
Suppose that the event
Is it true that then
JDraw diagrams any sets E and F (i)
(ii)
(Ei
selects a student to
first
know
all
E,
answer a certain ques-
F
student knows the answer, and
the
the answer but the second one does.
F
broken up into exclusive pieces E h
E
2,
.
.
.
,Ea
can be written as
and F) or {E2 and F) or
to illustrate the facts
E or F) = E&ndF (E and F) = E or F. (
set £ is
any event
F =
(1)
12.
correct for
E and F exclusive?
11.
for
it is
student cannot answer, he selects one of the remaining students;
denote the event that the
event that the
Are
determine whether
F are exclusive, and F, G are exclusive, then E, G are exclusive, F are exclusive, and E, G are exclusive, then E, (F or G) are exclusive, (F or G) are exclusive, then E, F are exclusive.
10.
the
and (iii) and (iii).
.
.
.
(Ea and F)?
(known as De Morgan's Laws)
that,
.
SOME LAWS OF PROBABILITY
1.7]
31
SOME LAWS OF PROBABILITY
1.7
In the preceding section we have denned three operations on events: complementation, intersection, and union. We shall now consider how probabilities behave under union and complementation. The probabilities of intersections will be studied in Chapters 3 and 4. A law connecting the probability of the union (E or F) with the probabilities of E and of F is obtained most easily in the special case that E and F are exclusive. If R and S are two exclusive results, then not both can happen on the same trial of the experiment, and therefore in a sequence of trials
KR or S) = For
example, in
an evening
player will be the
diamond
sum
number
of the
f(R or S)
=
f(R)
#08).
number
of red flushes dealt to a
of heart flushes
and the number
we
trials
+ /(£)
see that
R,
if
S
are exclusive.
Since probability is supposed to represent long-run frequency, gests that in the model we should have the equation
P(E
(1)
The proof Figure so
if
E,
F
this sug-
are exclusive.
may
be illustrated on
of the five simple events e h e h e 6 e 6 e 8 ,
,
,
definition of the probability of a composite event (Section 4),
right side
or F)
may
just
P(E)
= P(eO
+
P(e 8 )
+
P(e B )
+
P(e«)
+
Pfo).
be written as
[P(e 8 ) is
P(F)
There (E or F) consists
P(E
which
+
= P(E)
of this addition law for exclusive events
6.3.
by the
The
or F)
of
Dividing both sides of the displayed
flushes that he receives.
equation by the total number of
+
#(«)
of poker, the
+
+
+
P(*)
P(cg)]
+
[P(ei)
It is clear that the
P(F).
+
P(ee)]
argument would hold
in
general.
Example
1.
Two
Suppose that
dice.
in
throwing two dice, the 36 simple If T denotes the sum of the
events displayed in (5.2) are equally likely. points on the
The
two
dice, then, as
shown
probabilities of the other values of
Table t
P{T =
t)
1.
in
T
Example
5.3,
P{T =
7)
=
?%.
are as follows (Problem 5.4).
Probabilities of values of
T
2
3
4
5
6
7
8
9
10
11
12
&
&
A
&
A
A
A
A
A
A
&
PROBABILITY MODELS
32
Let us
now by
divisible 10,
= P(E)
P(G)
event set
TH oi
S, it
sum T
will
be
if
+ P(F) law
special case of the addition
complement
that the
and only if T is either equal to 5 or to the exclusive events E: T = 5 and F: T = 10. Hence
Since this occurs
G is the union of
A
G
find the probability of the event 5.
[CHAP. 1
(1) is
E
Since i? and
E.
A + A = A-
=
obtained by taking for
and
are exclusive
union
their
F
the
is
the
follows that
+
P(E)
P(E)
= P(E
=
E) = P(S)
or
1.
This establishes the important law of complementation
=
P(E)
(2)
As we that
shall see later,
-
1
P(E).
the main usefulness
of this relation resides in the fact
often easier to calculate the probability that something does not
it is
happen than that it does happen. As an illustration of (2), consider once more the model
of
Example
suppose we wish to determine the probability of the event E: the of the points of
E is the
showing on the two dice is three or more. The complement E T = 2, which has probability -jV Hence
event
:
P(E)
The concept
1-^=35.
=
of exclusive events extends readily to
The events E h E2 event belongs to more than one Definition.
The addition law
.
,
.
of
.
.
.)
=
P(E2 )
F
no simple
manner:
+
.
if
E,
if
them.
+
P(ft)
more than two events.
are said to be exclusive
.
also extends in the obvious
P(Ei or E* or
If
and
1
sum T
.
.
Eh E
2,
.
.
.
are exclusive.
are not exclusive, law (1) will in general not be correct, as
be seen by examining Figure 6.2b.
P(E) = P( ei ) P(F)
+
Pfe)
+
=
may
Here P(e 4 ) P(e 4 )
+ +
P(e 6 ) P(e 6 )
+
P(e 2 )
+
P(e 7 )
so that
P(E)
+
= Pfe)
P(P)
+
P(e 3 )
+
2P(e 4 )
+
2P(e 6 )
+
P(e 2 )
+
P(e7 )
while
P(E
or F)
=
P{e,)
+
P(*)
+
P(e 4 )
+
P(e»)
+
P(e2)
+
P(e7 ).
+
P(P) but only once in Since P(e 4 ) and P(e 6 ) occur twice in P{E) P(P) is greater than P{E or P) (except P(E or P), it follows that P(E)
+
in the trivial case
A
correct law
when both P(e 4 ) and
must allow
P(ee) are equal to zero).
for the double counting of the probabilities of
those simple events which belong both to
E
and
to F, that
is
to the inter-
SOME LAWS OF PROBABILITY
1.7]
section
(E and F).
33
In Figure 6.2a the event (E and F) consists of
e4
and
that
e 6 , so
F)
=
P(e 4 )
+
= P(E)
+
P(F)
- P(#
P(E and
P(ee).
It follows that
P{E
(4)
or F)
This addition law holds for
all
whether or not they are excluin Figure 6.3), (E and F) will be
sive. If they happen to be exclusive (as empty and have probability 0, in which case
Example two
Two
1.
dice (continued).
F
dice the events E,
point'
'
and "second
die
The
(4)
reduces to
In Example
6.1
corresponding to the results
P(F)
fa,
= &,
likely, it is
we considered If
for
shows one
the 36 simple
seen that
P(E and
and
(1).
"first die
shows one point" respectively.
events displayed in (5.2) are equally
P(E) =
and F).
F
events E,
F)
=
fa.
probability of the event (E or F) corresponding to the result that at
one of the dice shows one point is then, by (4), equal to fa + fa — This can also be seen directly from the fact that the event -g-J. fa (E or F) consists of 1 1 simple events each having probability fa It is important to keep in mind that law (3) requires the assumption that the events are exclusive. When they are not, it is necessary to correct for multiple counting, as is done in law (4) for the case of two events. The formulas for nonexclusive events become rapidly more complicated least
=
as the
number
of events
is
increased (see Problem 12).
PROBLEMS 1.
In Problem
scribed in parts 2.
5.6, find (i),
In Problem
(ii),
the probabilities of the complements of the events de-
(v),
5.9, find
and
(vii).
the probabilities of the complements of the events de-
scribed in parts (i)-(v). 3.
In Problem
5.8,
use the law of complementation to find the probabilities of the
following events. (i)
(ii)
4.
At At
least one of the least
two
digits is greater
one of the two digits
In Example
1,
is
than zero,
even.
use the results of Table
1
and the addition law
(3) to find
the
probabilities of the following events: (i)
(ii)
(hi)
T ^ 10 T < 5 T is divisible by
3.
be the number of boys. Make a similar table to Table 1 showing the probabilities of B taking on its various possible values 0, L 2, and 3. 5.
In Problem 5.6 let
B
,,
PROBABILITY MODELS
34 Use the table
6.
of the preceding
[CHAP.
problem and the addition law
(3) to find
1
the
probabilities of the following events: (i) (ii)
(iii)
B^
2
B > B ^
2 1.
A
manufacturer of fruit drinks wishes to know whether a new formula (say B) has greater consumer appeal than his old formula (say A) Each of four customers is presented with two glasses, one prepared from each formula, and is asked to state which he prefers. The results may be represented by a sequence of four letters; thus ABAB would mean that the first and third customers preferred A, while the second and fourth preferred B. Suppose that the 2 4 = 16 possible outcomes of the experiment are equally likely, as would be reasonable if the two formulas are If S denotes the number of consumers preferring the in fact equally attractive. new formula, make a table similar to Table 1, showing the probabilities of S taking on its various possible values. 7.
.
Use the table
8.
of the preceding
problem and the addition law
(3) to find
the
probabilities of the following events.
At least two of the customers prefer the new product. At most two of the customers prefer the new product, More than two of the customers prefer the new product.
(i) (ii)
(iii)
Let S consist of the six simple events
9.
=
P{ez)
.2,
=
P(e 4 )
.25,
E=
=
P(e 5 )
{e h e 2 },
.3,
F =
eh
=
P(««)
.
{e 2 e 3 e 4} ,
.
.
,
let
P(eO
=
P(e 2 )
=
.1,
If
G =
,
and
e6,
,
.05.
{e 2 e s e b } ,
,
find
(ii)
P(E) P(F)
(iii)
P(E
(i)
10. .05,
P(F P(E
(iv)
(v)
Let 8 consist of the six simple events e h = .14, P(e 4 ) = .1, P(e b ) = .2, P(e 6 )
E=
{e h e 4 e 6 } ,
F =
,
{e 3
,
.
.
.
=
,
(i)
(iii)
and left-hand
11.
or G).
e4, e 5 },
let P(ei)
=
.01,
P(e 2)
=
If
G = it is
{e 2 e 5 } ,
true
by computing both the
+ P(F) + P{G) + P(G).
results.
Check equation
(4) for
the events
E and F
Check the following generalization 12. when E, F, G are the events of Problem
P(E
and
sides:
P(E or F) = P(E) P{E or G) = P{E) P{F or G) = P{F)
Explain your
e6,
.5.
check for each of the following equations whether
(ii)
F
or
or F)
P(e 3 )
right-
or G)
or
F
or G)
=
[P{E)
-
+
of
Problem
of equation (4)
9.
by computing both sides
9:
P(F)
[P{E and F)
+ P(G)] + P{E and G) + P(F and