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24
Chapter 2
Probability
Exercises
s
FIGURE 2.4
Venn diagram for AB
FIGURE 2.6
25
s
Venn diagram for
mutually exclusive
sets A and 8
two sets A and B, where A is the set of points in the left-hand circle and B is the set
of points in the right-hand circle. The set A U B is the shaded region consisting of
all points inside either circle (or both). The key word for expressing the union of two
sets is or (meaning A or B or both).
The intersection of A and B, denoted by A n B or by A B, is the set of all points in
both A and B. The Venn diagram of Figure 2.4 shows two sets A and B, with A n B
consisting of the points in the shaded region where the two sets overlap. The key word
for expressing intersections is and (meaning A and B simultaneously).
If A is a subset of S, then the complement of A, denoted by A, is the set of points
that are in S but not in A. Figure 2.5 is a Venn diagram illustrating that the shaded
area in S but not in A is A. Note that AU A= S.
Two sets, A and B, are said to be disjoint, or mutually exclusive, if An B = 0. That
is , mutually exclusive sets have no points in common . The Venn diagram in Figure 2.6
illustrates two sets A and B that are mutually exclusive. Referring to Figure 2.5, it is
easy to see that, for any set A, A and A are mutually exclusive.
Consider the die-tossing problem of Section 2.2 and letS denote the set of all possible numerical observations for a single toss of a die. That is, S = {1, 2, 3, 4 , 5, 6) .
Let A = {1, 2}, B = {I, 3), and C = {2, 4 , 6). Then AU B = {1, 2, 3}, An B = {I),
and A= {3, 4, 5, 6} . Also, note that B and Care mutually exclusive, whereas A and
Care not.
We will not attempt a thorough review of set algebra, but we mention four equalities
of considerable importance. These are the distributive laws, given by
An (B U C) = (A n B) U (A n C),
AU (B n C) = (A U B) n (AU C),
and DeMorgan 'slaws:
(A
and
(A
u B) = An B.
In the next section we will proceed with an elementary discussion of probability
theory.
Exercises
2.1
Suppose a family contains two children of different ages, and we are interested in the gender
of these children. Let F denote that a child is female and M that the child is male and let a
pair such as F M denote that the older child is female and the younger is male. There are four
points in the set S of possible observations:
S
= {FF,
FM, MF, MM}.
Let A denote the subset of possibilities containing no males; B, the subset containing two
males; and C , the subset containing at least one male. List the elements of A, B, C, An B,
AU B, An C, AU C , B n C, B U C , and C n B.
s
FIGURE 2.5
Venn diagram for A
n B) = Au B
2.2
Suppose that A and B are two events. Write expressions involving unions, intersections, and
complements that describe the following:
a
Both events occur.
b At least one occurs.
c
A
Neither occurs.
d Exactly one occurs.
2.3
Draw Venn diagrams to verify DeMorgan 's laws. That is, for any two sets A and B, (AU B)=
An Band (A n B) = Au 8.
- -- ~
2.4
a
DEFINITION 2.1
A =(A n B) U (A n B).
Refer to Exercise 2.4. Use the identities A = A n S and S = B U Band a distributive law to
prove that
a
l.7
c
A then A = B U (A
n B) .
When an experiment is performed, it can result in one or more outcomes, which
are called events. In our discussions, events will be denoted by capital letters. If the
experiment consists of counting the number of bacteria in a portion of food, some
events of interest could be
Further, show that (A n B) and (A n B) are mutually exclusive and therefore that A is the
union of two mutually exclusive sets, (A n B) and (A n B).
d Also show that Band (A n B) are mutually exclusive and if B C A, A is the union of two
mutually exclusive sets, B and (A n B).
A : Exactly 110 bacteria are present
Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote
the set of all possible pairs that can be observed. [These pairs can be listed, for example, by
letting (2, 3) denote that a 2 was observed on the first die and a 3 on the second.]
a
B: More than 200 bacteria are present
C : The number of bacteria present is between 100 and 300.
Some events associated with a single toss of a balanced die are these:
Define the following subsets of S:
A: Observe an odd number.
B: Observe a number less than 5.
C: Observe a 2 or a 3.
£ 1 : Observe a 1.
£2 : Observe a 2.
£ 3: Observe a 3.
£4: Observe a 4.
Es : Observe a 5.
£6: Observe a 6.
A: The number on the second die is even .
B : The sum of the two numbers is even.
C: At least one number in the pair is odd.
b List the points in A , C, A n B , A n B , A U B, and A n C.
2.7
A group of five applicants for a pair of identical jobs consists of three men and two women. The
employer is to select two of the five applicants for the jobs. LetS denote the set of all possible
outcomes for the employer's selection. Let A denote the subset of outcomes corresponding t~
the selection of two men and B the subset corresponding to the selection of at least one woman.
List the outcomes in A, B , AU B, A n B, and A n B. (Denote the different men and women
by M 1, M2 , M 3 and W1 , W2 , respectively.)
2.8
From a survey of 60 students attending a university, it was found that 9 were living off campus,
36 were undergraduates, and 3 were undergraduates living off campus. Find the number of
these students who were
a
You can see that there is a distinct difference among some of the events associated
with the die-tossing experiment For example, if you observe event A (an odd number),
at the same time you will have observed £ 1, £ 3, or £ 5. Thus, event A, which can be
decomposed into three other events, is called a compound event. In contrast, the events
E,, £ 2, £3, £ 4, Es, and £ 6 cannot be decomposed and are called simple events. A
simple event can happen only in one way, whereas a compound event can happen in
more than one distinct way.
undergraduates, were living off campus, or both.
b undergraduates living on campus.
c
graduate students living on campus.
2.4 A Probabilistic Model for an Experiment:
The Discrete Case
In Section 2.2 we referred to the die-tossing experiment when we observed the number
appearing on the upper face. We will use the term experiment to include observations
obtained from completely uncontrollable situations (such as observations on the daily
price of a particular stock) as well as those made under controlled laboratory conditions. We have the following definition:
,
An experiment is the process by which an observation is made.
Examples of experiments include coin and die tossing, measuring the IQ score of an
individual , or determining the number of bacteria per cubic centimeter in a portion
of processed food .
A= (A n B) U (A n B).
b If B c
2.6
Case
If A and B are two sets, draw Venn diagrams to verify the following:
b If B c A then A = B U (A n B ).
2.5
... vu"' our an t:xpenment: I he lJiscrete
Certain concepts from set theory are useful for expressing the relationships between
various events associated with an experiment Because sets are collections of points,
we associate a distinct point, called a sample point, with each and every simple event
associated with an experiment
DE FIN IT I 0 N 2.2
A simple event is an event that cannot be decomposed. Each simple event
co1responds to one and only one sample point. The letter E with a subscript
will be used to denote a simple event or the corresponding sample point
Thus, we can think of a simple event as a set consisting of a single point-namely,
the single sample point associated with the event
32
Chapter 2
Probability
Exercises
e Because A= Es U £ 9 U £ 10 , Axiom 3 implies that
P(A)
=
P(Es)
+ P(£9) + P(£10) = 3/10.
Opinion
•
The next section contains an axiomatic description of the method for calculating
P (A) that we just used.
Before we proceed, let us note that there are experiments for which the sample space
is not countable and hence is not discrete. Suppose, for example, that the experiment
consists of measuring the blood glucose level of a diabetic patient. The sample space
for this experiment would contain an interval of real numbers, and any such interval
contains an uncountable number of values. Thus, the sample space is not discrete.
Situations like the latter will be discussed in Chapter 4. The remainder of this chapter
is devoted to developing methods for calculating the probabilities of events defined
on discrete sample spaces.
Proportion
Every person's blood type is A, B, AB, or 0. In addition, each individual either has the
Rhesus (Rh) factor(+) or does not(-). A medical technician records a person's blood type
and Rh factor. List the sample space for this experiment.
2.10
The proportions of blood phenotypes, A, 8 , AB, and 0, in the population of all Caucasians in
the United States are approximately .41, .I 0, .04, and .45, respectively. A single Caucasian is
chosen at random from the population.
.12
Suppose that one American is selected and his or her opinion is recorded.
a
What are the simple events for this experiment?
b Are the simple events that you gave in part (a) all equally likely? If not, what are the
probabilities that should be assigned to each?
c What is the probability that the person selected finds it at least somewhat likely that the Air
Force is withholding information about intelligent life on other planets?
2.14
A survey classified a large number of adults according to whether they were diagnosed as
needing eyeglasses to correct their reading vision and whether they use eyeglasses when reading.
The proportions falling into the four resulting categories are given in the following table:
Uses Eyeglasses
for Reading
Needs glasses
Yes
No
Yes
No
.44
.02
.14
.40
If a single adult is selected from the large group, find the probabilities of the events defined
below. The adult
a
needs glasses.
a List the sample space for this experiment.
b needs glasses but does not use them.
b Make use of the information given above to assign probabilities to each of the simple events.
c
c
2.11
.24
.24
.40
Very likely
Somewhat likely
Unlikely
Other
Exercises
2.9
What is the probability that the person chosen at random has either type A or type AB
blood?
2.15
A sample space consists of five simple events, £ 1, £ 2, £ 3, £ 4, and £ 5.
a
33
If P(£ 1) = P(£ 2 ) = 0.15. P(£ 3 ) = 0.4, and P(£4 ) = 2P(£ 5 ), find the probabilities of
£4 and £ 5.
uses glasses whether the glasses are needed or not.
An oil prospecting firm hits oil or gas on I 0% of its drillings. If the firm drills two wells,
the four possible simple events and three of their associated probabilities are as given in the
accompanying table. Find the probability that the company will hit oil or gas
a on the first drilling and miss on the second.
b on at least one of the two drillings.
= 3P(£ 2 ) = 0.3, find the probabilities of the remaining simple events if you
know that the remaining simple events are equally probable.
b If P(£ 1)
2.12
A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The
experiment consists of observing the movement of a single vehicle through the intersection.
a List the sample space for this experiment.
b Assuming that all sample points are equally likely, find the probability that the vehicle turns.
2.13
Americans can be quite suspicious, especially when it comes to government conspiracies. On
the question of whether the U.S. Air Force has withheld proof of the existence of intelligent
life on other planets, the proportions of Americans with varying opinions are given in the
table.
2.16
Simple
Event
Outcome of
First Drilling
Outcome of
Second Drilling
Probability
£1
£2
£3
£4
Hit (oil or gas)
Hit
Miss
Miss
Hit (oil or gas)
Miss
Hit
Miss
.01
?
.09
.81
Of the volunteers coming into a blood center, I in 3 have o + blood, I in 15 have o - , 1 in 3
have A+ , and I in 16 have A- . The name of one person who previously has donated blood is
34
Chapter 2
2.5
Probability
selected from the records of the center. What is the probability that the person selected has
2.17
type
b
type 0 blood?
c
type A blood?
d
neither type A nor type 0 blood?
Hydraulic landing assemblies coming from an aircraft rework facility are each inspected for
defects. Historical records indicate that 8% have defects in shafts only, 6% have defects in
bushings only, and 2% have defects in both shafts and bushings. One of the hydraulic assemblies
is selected randomly. What is the probability that the assembly has
a
2.18
a shaft or bushing defect?
c
exactly one of the two types of defects?
d
neither type of defect?
c
d
iii
iv
v
List the sample points for this experiment.
Assign a reasonable probability to each sample point. (Are the sample points equally
likely?)
Let A denote the event that exactly one head is observed and B the event that at least one
head is observed. List the sample points in both A and B.
From your answer to part (c), find P(A), P(B), P(A
n B) , P(A
U B), and P(A U B).
*2.21
a
List the sample points in this experiment of ordering paper on two successive days.
b
Assume the vendors are selected at random each day and assign a probability to each sample
point.
Let A denote the event that the same vendor gets both orders and B the event that V2 gets at
least one order. Find P(A), P(B) , P(A U B) , and P(A n B) by summing the probabilities
of the sample points in these events.
The following game was played on a popular television show. The host showed a contestant
three large curtains. Behind one of the curtains was a nice prize (maybe a new car) and behind
the other two curtains were worthless prizes (duds). The contestant was asked to choose one
curtain. If the curtains are identified by their prizes, they could be labeled G , D 1, and D2 (Good
1
Prize, Dudl, and Dud2). Thus, the sample space for the contestants choice isS = { G, D1, D2}.
a
b
If the contestant has no idea which curtains hide the various prizes and selects a curtain at
random, assign reasonable probabilities to the simple events and calculate the probability
that the contestant selects the curtain hiding the nice prize.
Before showing the contestant what was behind the curtain initially chosen, the game show
host would open one of the curtains and show the contestant one of the duds (he could
always do this because he knew the curtain hiding the good prize). He then offered the
I. Exercises preceded by an asterisk are optional.
If the host shows her one of the duds and she switches to the other unopened curtain,
what will be the result if she had initially selected G ?
Answer the question in part (ii) if she had initially selected one of the duds.
If the contestant switches from her initial choice (as the result of being shown one of
the duds) , what is the probability that the contestant wins the good prize?
Which strategy maximizes the contestant's probability of winning the good prize: stay
with the initial choice or switch to the other curtain?
If A and B are events, use the result derived in Exercise 2.5(a) and the Axioms in Definition
2.6 to prove that
P(A) = P(A
*2.22
A business office orders paper supplies from one of three vendors, V1, V2 , or V3 . Orders are to
be placed on two successive days, one order per day. Thus, (V2 , V3 ) might denote that vendor
V2 gets the order on the first day and vendor V3 gets the order on the second day.
c
*2.20
ii
Suppose two balanced coins are tossed and the upper faces are observed.
a
b
2.19
If the contestant choses to stay with her initial choice, she wins the good prize if and
only if she initially chose curtain G. If she stays with her initial choice, what is the
probability that she wins the good prize?
a bushing defect?
b
35
contestant the option of changing from the curtain initially selected to the other remaining
unopened curtain. Which strategy maximizes the contestant's probability of winning the
good prize: stay with the initial choice or switch to the other curtain? In answering the
following sequence of questions, you will discover that, perhaps surprisingly, this question
can be answered by considering only the sample space above and using the probabilities
that you assigned to answer part (a).
o+ blood?
a
Cal culating th e Probability of an Event: The Sample-Point Method
If A and B are events and B
Definition 2.6 to prove that
c
n
B)+ P(A
n
B).
A, use the result derived in Exercise 2.5(b) and the Axioms in
P(A) = P(B)
+ P(A n B).
s
2.23
If A and Bare events and B C A, why is it "obvious" that P(B)
2.24
Use the result in Exercise 2.22 and the Axioms in Definition 2.6 to prove the "obvious" result
in Exercise 2.23.
P(A)?
2.5 Calculating the Probability of an Event:
The Sample-Point Method
Finding the probability of an event defined on a sample space that contains a finite or
denumerable (countably infinite) set of sample points can be approached in two ways,
the sample-point and the event-composition methods. Both methods use the sample
space model, but they differ in the sequence of steps necessary to obtain a solution
and in the tools that are used. Separation of the two procedures may not be palatable
to the unity-seeking theorist, but it can be extremely useful to a beginner attempting to
find the probability of an event. In this section we consider the sample-point method.
The event-composition method requires additional results and will be presented in
Section 2.9.
38
Chapter 2
Probability
EXAMPLE 2.4
Solution
Exercises
succinctly. Hence, our next topic concerns some elementary results in combinatorial analysis and their application to the sample-point approach for the solution of
probability problems.
The odds are two to one that, when A and B play tennis, A wins. Suppose that A and
B play two matches . What is the probability that A wins at least one match ?
1. The experiment consists of observing the winner (A or B) for each of two
matches. Let A B denote the event that player A wins the first match and player
B wins the second.
2. The sample space for the experiment consists of four sample points:
£1: AA,
£2: AB,
£3: BA,
P(£ 1 ) = 4/9,
P(£2)
= 2/9,
P(£3)
Exercises
2.25
A single car is randomly selected from among all of those registered at a local tag agency.
What do you think of the following claim? "A ll cars are either Yolkswagens or they are not.
Therefore, the probability is l/2 that the car selected is a Volkswagen."
2.26
According to Webster 's New Collegiate Diclionary, a divining rod is "a forked rod believed
to indicate [divine] the presence of water or minerals by dipping downward when held over a
vein." To test the claims of a divining rod expert, skeptics bury four cans in the ground, two
empty and two filled with water. The expert is led to the four cans and told that two contain
water. He uses the divining rod to test each of the four cans and decide which two contain water.
£4: BB
3. Because A has a better chance of winning any match, it does not seem appropriate to assign equal probabilities to these sample points. As you will see in
Section 2.9, under certain conditions it is reasonable to make the following
assignment of probabilities:
= 2/9,
P(£4) = l / 9.
Notice that, even though the probabilities assigned to the simple events are not
all equal, P(E;) ~ 0, fori= 1, 2, 3, 4, and
P(E;) = 1.
4. The event of interest is that A wins at least one game. Thus, if we denote the
event of interest as C, it is easily seen that
a
List the sample space for this experiment.
b
If the divining rod is completely useless for locating water, what is the probability that the
expert will correctly identify (by guessing) both of the cans containing water?
Ls
C
=
2.27
In Exercise 2.12 we considered a si tuation where cars entering an intersection each could turn
right, turn left, or go straight. An experiment consists of observing two vehicles moving through
the intersection.
£1 U £2 U £3.
a
How many sample points are there in the sample space? List them.
b
Assuming that all sample points are equally likely, what is the probability that at least one
car turns left?
c
Again assuming equally likely sample points, what is the probability that at most one
vehicle turns?
5. Finally,
P(C) = P(£ 1)
39
+ P(£ 2 ) + P(£ 3 )
= 4/9
+ 2/9 + 2/9 =
8/9.
•
The sample-point method for solving a probability problem is direct and powerful
and in some respects is a bulldozer approach. It can be applied to find the probability of
any event defined over a sample space containing a finite or countable set of sample
points, but it is not resistant to human error. Common errors include incorrectly
diagnosing the nature of a simple event and failing to list all the sample points in
S. A second complication occurs because many sample spaces contain a very large
number of sample points and a complete itemization is tedious and time consuming
and might be practically impossible.
Fortunately, many sample spaces generated by experimental data contain subsets
of sample points that are equiprobable. (The sample spaces for Examples 2.1, 2.2,
and 2.3 possess this property.) When this occurs, we need not list the points but may
simply count the number in each subset. If such counting methods are inapplicable,
an orderly method should be used to list the sample points (notice the listing schemes
for Examples 2.1, 2.2, and 2.3). The listing of large numbers of sample points can be
accomplished by using a computer.
Tools that reduce the effort and error associated with the sample-point approach
for finding the probability of an event include orderliness, a computer, and the mathematical theory of counting, called combinatorial analysis. Computer programming
and applications form a topic for separate study. The mathematical theory of combi-
2.28
2.29
Four equally qualified people apply for two identical positions in a company. One and only
one applicant is a member of a minority group. The positions are filled by choosi ng two of the
applicants at random.
a
List the possible outcomes for this experiment.
b
Assign reasonable probabilities to the sample points.
c
Find the probability that the applicant from the minority group is selected for a position.
Two additional jurors are needed to complete a jury for a criminal trial. There are six prospective
jurors, two women and four men. Two jurors are randomly selected from the six available.
a
2.30
b
Define the experiment and describe one sample point. Assume that you need describe only
the two jurors chosen and not the order in which they were selected.
List the sample space associated with this experiment.
c
What is the probability that both of the jurors selected are women?
Three imported wines are to be ranked from lowest to highest by a purported wine expert. That
is, one wine will be identified as best, another as second best, and the remaining wine as worst.
a
b
Describe one sample point for this experiment.
List the sample space.
c
Assume that the "expert" really knows nothing about wine and randomly assigns ranks to
the three wines. One of ·
- ........................
~~,
2.6
2.31
A boxcar contains six complex electronic systems. Two of the six are to be randomly selected
for thorough testing and then classified as defective or not defective.
2.32
a
If two of the six systems are actually defective, find the probability that at least one of the
two systems tested will be defective. Find the probability that both are defective.
b
If four of the six systems are actually defective, find the probabilities indicated in part (a).
a
List the possibilities for preference arrangements among the four customers (that is, list
the sample space) .
b
Assign probabilities to the sample points.
c
Let A denote the event that all four customers prefer the same style. Find P(A).
List the points in the sample space.
b
Identify the simple events in each of the following events:
8:
At least two had incomes exceeding $43 ,318.
Exactly two had incomes exceeding $43,318.
C:
Exactly one had income less than or equal to $43 ,318.
A:
c
a3
am
bl
b3
'·I I I I:TI
The first result from combinatorial analysis that we present, often called the mn
rule, is stated as follows:
THEOREM 2.1
With m elements a 1, a2, ... , am and n elements b 1, b 2 , ••• , b11 , it is possible
to form mn = m x n pairs containing one element from each group.
Proof
Verification of the theorem can be seen by observing the rectangular table in
Figure 2.9. There is one square in the table for each a;, b j pair and hence a total
of m x n squares.
Make use of the given interpretation for the median to assign probabilities to the simple
events and find P(A) , P(B), and P(C).
The mn rule can be extended to any number of sets. Given three sets of elementsa I, a2, ... , am; b 1, b2, ... , h11 ; and c 1 , c 2 , ••• , cp-the number of distinct triplets
Patients aniving at a hospital outpatient clinic can select one of three stations for service.
Suppose that physicians are assigned randomly to the stations and that the patients therefore
have no station preference. Three patients anive at the clinic and their selection of stations is
observed.
a
{/ 2
b2
The Bureau of the Census reports that the median family income for all families in the United
States during the year 2003 was $43,318. That is, half of all American families had incomes
exceeding this amount, and half had incomes equal to or below this amount. Suppose that four
families are surveyed and that each one reveals whether its income exceeded $43,318 in 2003.
a
{/I
41
(a;, bi)
A retailer sells only two styles of stereo consoles, and experience shows that these are in equal
demand. Four customers in succession come into the store to order stereos. The retailer is
interested in their preferences.
2.33
2.34
FIGURE 2.9
Table indicating the
number of pairs
Tools for Counting Sample Points
containing one element from each set is equal to mnp. The proof of the theorem for
three sets involves two applications of Theorem 2.1. We think of the first set as an
(a;, b j) pair and unite each of these pairs with elements of the third set, c 1, c2 , ..• , c p·
Theorem 2.1 implies that there are mn pairs (a;, b j). Because there are p elements
c1, c2, ... , cp, another application of Theorem 2.1 implies that there are (mn)(p) =
mnp triplets a;bjck.
List the sample points for the experiment.
b Let A be the event that each station receives a patient. List the sample points in A.
c Make a reasonable assignment of probabilities to the sample points and find P(A).
2.6 Tools for Counting Sample Points
This section presents some useful results from the theory of combinatorial analysis
and illustrates their application to the sample-point method for finding the probability
of an event. In many cases, these results enable you to count the total number of
sample points in the sample space S and in an event of interest, thereby providing
a confirmation of your listing of simple events. When the number of simple events
in a sample space is very large and manual enumeration of every sample point is
tedious or even impossible, counting the number of points in the sample space and in
the event of interest may be the only efficient way to calculate the probability of an
event. Indeed, if a sample space contains N equiprobable sample points and an event
A contains exactly na sample points, it is easily seen that P(A) = nafN. _
EXAMPLE 2.5
An experiment involves tossing a pair of dice and observing the numbers on the upper
faces. Find the number of sample points in S, the sample space for the experiment.
Solution
A sample point for this experiment can be represented symbolically as an ordered
pair of numbers representing the outcomes on the first and second die, respectively.
Thus, (4, 5) denotes the event that the uppermost face on the first die was a 4 and on
the second die, a 5. The sample space S consists of the set of all possible pairs (x, y ),
where x andy are both integers between I and 6.
The first die can result in one of six numbers. These represent a 1 , a 2 , ... ,
a6. Likewise, the second die can fall in one of six ways, and these correspond to
b1, b2, ... , b6. Then m = n = 6 and the total number of sample points in S is
mn = (6)(6) = 36.
•
'-"al-'"'"
L
nooa0111ty
Exercises
Exercises
2.35
An airline has six Aights from New York to California and seven Aights from California to
Hawaii per day. If the Aights are to be made on separate days, how many different Aight
arrangements can the airline offer from New York to Hawaii?
2.36
An assembly operation in a manufacturing plant requires three steps that can be performed in
any sequence. How many different ways can the assembly be performed?
2.37
A businesswoman in Philadelphia is preparing an itinerary for a visit to six major cities. The
distance traveled, and hence the cost of the trip, will depend on the order in which she plans
her route.
2.38
2.39
2.40
a
How many different itineraries (and trip costs) are possible?
b
If the businesswoman randomly selects one of the possible itineraries and Denver and San
Francisco are two of the cities that she plans to visit, what is the probability that she will
visit Denver before San Francisco?
2.46
Ten teams are playing in a basketball tournament. In the first round , the teams are randomly
assigned to games I, 2, 3, 4 and 5. In how many ways can the teams be assigned to the games?
*2.47
Refer to Exercise 2.46. lf2n teams are to be assigned to games I, 2 , ... , n , in how many ways
can the teams be assigned to the games?
b
Find the probability that the sum of the numbers appearing on the dice is equal to 7.
2.49
Students attending the University of Florida can select from 130 major areas of study. A
student's major is identified in the registrar's records with a two-or three-letter code (for
example, statistics majors are identified by STA, math majors by MS) . Some students opt for
a double major and complete the requirements for both of the major areas before graduation.
The registrar was asked to consider assigning these double majors a distinct two- or three-letter
code so that they could be identified through the student records ' system.
How many autos would a dealer have to stock if he included one for each style-enginetransmission combination?
b
How many would a distribution center have to carry if all colors of cars were stocked for
each combination in part (a)?
a
What is the maximum number of possible double majors available to University of Florida
students?
b
If any two- or three-letter code is available to identify majors or double majors, how many
major codes are available?
c
How many major codes are required to identify students who have either a single major or
a double major?
d
Are there enough major codes available to identify all single and double majors at the
University of Florida?
2.50
Probability played a role in the rigging of the April 24, 1980, Pennsylvania state lottery (Los
Angeles Times, September 8, I 980). To determine each digit of the three-digit winning number,
each of the numbers 0, I, 2, ... , 9 is placed on a Ping-Pong ball, the ten balls are blown into
a compartment, and the number selected for the digit is the one on the ball that Aoats to the
top of the machine. To alter the odds, the conspirators injected a liquid into all balls used in
the game except those numbered 4 and 6, making it almost certain that the lighter balls would
be selected and determine the digits in the winning number. Then they bought lottery tickets
bearing the potential winning numbers. How many potential winning numbers were there (666
was the eventual winner)?
2.51
A brand of automobile comes in five different styles, with four types of engines, with two types
of transmissions, and in eight colors.
a
what is the coefficient of x 5 y 3 ? What is the coefficient of
If we wish to expand (x
x 3y 5?
An experiment consists of tossing a pair of dice.
Use the combinatorial theorems to determine the number of sample points in the sample
spaceS.
+ y )8 ,
2.48
An upscale restaurant offers a special fixe prix menu in which, for a fixed dinner cost, a diner can
select from four appetizers, three salads, four entrees, and five desserts . How many different
dinners are available if a dinner consists of one appetizer, one salad, one entree, and one
dessert?
a
49
2.41
How many different seven-digit telephone numbers can be formed if the first digit cannot be
zero?
A local fraternity is conducting a rafAe where 50 tickets are to be sold-one per customer.
There are three prizes to be awarded. If the four organizers of the raffle each buy one ticket,
what is the probability that the four organizers win
2.42
A personnel director for a corporation has hired ten new engineers. If three (distinctly different)
positions are open at a Cleveland plant, in how many ways can she fill the positions?
a
all of the prizes?
b
exactly two of the prizes?
c
exactly one of the prizes?
d
none of the prizes?
2.43
2.44
2.45
A Aeet of nine taxis is to be dispatched to three airports in such a way that three go to airport
A, five go to airport B , and one goes to airport C. In how many distinct ways can this be
accomplished?
2.52
An experimenter wishes to investigate the effect of three variables-pressure, temperature,
and the type of catalyst-on the yield in a refining process. If the experimenter intends to
use three settings each for temperature and pressure and two types of catalysts, how many
experimental runs will have to be conducted if he wishes to run all possible combinations of
pressure, temperature, and types of catalysts?
2.53
Five firms, F1, F2, ... , Fs, each offer bids on three separate contracts, C 1 , C 2 , and C 3 . Any one
firm will be awarded at most one contract. The contracts are quite different, so an assignment
of C 1 to F 1 , say, is to be distinguished from an assignment of C2 to F 1 •
Refer to Exercise 2.43. Assume that taxis are allocated to airports at random.
a
If exactly one of the taxis is in need of repair, what is the probability that it is dispatched
to airport C?
b
If exactly three of the taxis are in need of repair, what is the probability that every airport
receives one of the taxis requiring repairs?
Suppose that we wish to expand (x
+ y + z) 17 • What is the coefficient of x 2 y 5 z 10?
50
Chapter 2
2.7
Probabi Iity
a
How many sample points are there altogether in this experiment involving assignment of
contracts to the firms? (No need to list them all.)
b
Under the assumption of equally likely sample points, find the probability that F3 is awarded
a contract.
2.54
A group of three undergraduate and five graduate students are available to fill certain student government posts. If four students are to be randomly selected from this group, find the
probability that exactly two undergraduates will be among the four chosen.
2.55
A study is to be conducted in a hospital to determine the attitudes of nurses toward various
administrative procedures. A sample of I0 nurses is to be selected from a total of the 90 nurses
employed by the hospital.
2.56
2.57
2.58
2.59
2.60
2.61
a
How many different samples of 10 nurses can be selected?
b
Twenty of the 90 nurses are male. If I0 nurses are randomly selected from those employed
by the hospital, what is the probability that the sample of ten will include exactly 4 male
(and 6 female) nurses?
A student prepares for an exam by studying a list of ten problems. She can solve six of them.
For the exam, the instructor selects five problems at random from the ten on the list given
to the students. What is the probability that the student can solve all five problems on the
exam?
2.64
A balanced die is tossed six times , and the number on the uppermost face is recorded each
time. What is the probability that the numbers recorded are I, 2, 3, 4, 5, and 6 in any order?
2.65
Refer to Exercise 2.64. Suppose that the die has been altered so that the faces are 1, 2, 3, 4, 5,
and 5. If the die is tossed five times, what is the probability that the numbers recorded are 1, 2,
3, 4, and 5 in any order?
2.66
Refer to Example 2.10. What is the probability that
2.67
a
an ethnic group member is assigned to each type of job?
b
no ethnic group member is assigned to a type 4 job?
Refer to Example 2.13. Suppose that the number of distributors is M = I0 and that there are
n = 7 orders to be placed. What is the probability that
a
2.68
all of the orders go to different distributors?
*b
distributor I gets exactly two orders and distributor II gets exactly three orders?
*c
distributors I, II, and III get exactly two, three, and one order(s), respectively?
Show that, for any integer n :::_ I ,
C) =
a
3 aces and 2 kings?
a
b
a "full house" (3 cards of one kind, 2 cards of another kind)?
b
(~) = I. Interpret this result.
c
C) =
d
t(
Five cards are dealt from a standard 52-card deck. What is the probability that we draw
a
I ace, 1 two, l three, I four, and I five (this is one way to get a "straight")?
b
any straight?
a
Give an expression for the probability that none share the same birthday.
b
What is the smallest value of 11 so that the probability is at least .5 that at least two people
share a birthday?
Suppose that we ask n randomly selected people whether they share your birthday.
a
Give an expression for the probability that no one shares your birthday (ignore leap years).
b
How many people do we need to select so that the probability is at least .5 that at least one
shares your birthday?
2.62
A manufacturer has nine distinct motors in stock, two of which came from a particular supplier.
The motors must be divided among three production lines, with three motors going to each
line. If the assignment of motors to lines is random, find the probability that both motors from
the particular supplier are assigned to the first line.
2.63
The eight-member Human Relations Advisory Board of Gainesville, Florida, considered
the complaint of a woman who claimed discrimination, based on sex, on the part of a local
*2.70
I. Interpret this result.
(. ~J Interpret this result.
11
i= O
2.69
Refer to Example 2.7. Suppose that we record the birthday for each of n randomly selected
persons.
51
company. The board, composed of five women and three men, voted 5- 3 in favor of the plaintiff,
the five women voting in favor of the plaintiff, the three men against. The attorney representing
the company appealed the board's decision by claiming sex bias on the part of the board members. If there was no sex bias among the board members, it might be reasonable to conjecture
that any group of five board members would be as likely to vote for the complainant as any
other group of five. If this were the case, what is the probability that the vote would split along
sex lines (five women for, three men against)?
Two cards are drawn from a standard 52-card playing deck. What is the probability that the
draw will yield an ace and a face card?
Five cards are dealt from a standard 52-card deck. What is the probability that we draw
Conditional Probability and the Independence of Events
)
= 2". [Hint: Consider the binomial expansion of (x
+ y )"
with x = y = 1.]
l
Prove that
1
("+
" 1).
k ) = (")
k + (k-
Consider the situation where n items are to be partitioned into k < n distinct subsets. The
"* )
multinomial coefficients ( , 1 ,;'.
provide the number of distinct partitions where n 1 items
are in group l, 11 2 are in group 2, ... , nk are in group k. Prove that the total number of distinct
partitions equals k". [Hint: Recall Exercise 2.68(d).]
2.7 Conditional Probability
and the Independence of Events
The probability of an event will sometimes depend upon whether we know that other
events have occurred. For example, Florida sport fishermen are vitally interested
in the probability of rain. The probability of rain on a given day, ignoring the daily
atmospheric conditions or any other events, is the fraction of days in which rain occurs
over a long period of time. This is the unconditional probability of the event "rain on
a given day." Now suppose that we wish to consider the probability of rain tomorrow.
d P(A IA n B)
e P(A n BIA U B)
a Are A and B independent events?
b Are A and C independent events?
2.72
Solution
a To decide whether A and B are independent, we must see whether they satisfy
the conditions of Definition 2.10. In this example, P(A) = l / 2, P(B) = 1/ 2 ,
and P(C) = l / 3. Because A n B = 0, P(AIB) = 0, and it is clear that
P(AiB) -=J P(A). Events A and Bare dependent events.
b Are A and C independent? Note that P(A IC) = 1/ 2 and, as before, P(A) =
l / 2. Therefore, P(AiC) = P(A) , and A and Care independent.
•
EXAMPLE 2.16
Sex
Male (M)
Female (F)
Total
Pass (A)
Fail (A)
24
16
40
36
24
60
60
40
100
a
Are the events A and M independent?
b Are the events A and F independent?
2.73
C: Brand X is ranked second best.
D : Brand X is ranked third best.
If the judge actually has no taste preference and randomly assigns ranks to the
brands, is event A independent of events B, C , and D?
Gregor Mendel was a monk who, in 1865, suggested a theory of inheritance based on the
science of genetics. He identified heterozygous individuals for flower color that had two alleles
(one r = recessive white color allele and one R = dominant red color allele) . When these
individuals were mated, 3/ 4 of the offspring were observed to have red flowers, and I / 4 had
white flowers. The following table summarizes thi s mating; each parent gives one of its alleles
to form the gene of the offspring.
Parent 2
The six equally likely sample points for this experiment are given by
E 1: XYZ , £ 3: YXZ ,
£2: XZY, £4: YZX,
Parent I
£ 5 : ZXY ,
£6: ZYX,
where X Y Z denotes that X is ranked best, Y is second best, and Z is last.
Then A = {£, , E2, Es), B = {£, , £2}, C = {£3, Es), D = {£4 , £6}, and it
follows that
P(A)
= 1/ 2 ,
P(AiB)= P(AnB)
P(B)
=
l,
P(AiC) = 1/ 2 ,
P(AiD)
r
n·
R
Rr
rR
RR
a at least one dominant allele?
b at least one recessive allele?
c
2.7 4
If two events, A and B , are such that P(A)
following:
R
the probability that an offspring has
= 0.
Exercises
r
We assume that each parent is equally likely to give either of the two alleles and that, if either
one or two of the alleles in a pair is dominant (R), the offspring will have red flowers . What is
Thus, events A and C are independent, but events A and B are dependent. Events A
and D are also dependent.
•
2.71
Outcome
Total
Three brands of coffee, X , Y, and Z , are to be ranked according to taste by a judge.
Define the following events :
A: Brand X is preferred to Y.
B: Brand X is ranked best.
Solution
For a certain population of employees, the percentage passing or failing a job competency exam,
listed according to sex , were as shown in the accompanying table. That is, of all the people
taking the exam, 24% were in the male-pass category, 16% were in the male-fail category, and
so forth. An employee is to be selected randomly from this population. Let A be the event that
the employee scores a passing grade on the exam and let M be the event that a male is selected.
one recessive allele, given that the offspring has red flowers?
One hundred adults were interviewed in a telephone survey. Of interest was their opinions
regarding the loan burdens of college students and whether the respondent had a child currently
in college. Their responses are summarized in the table below:
Loan Burden
= .5,
P(B)
= .3, and
P(A
n
B)= .I, find the
Child in College
a
P(A IB)
b
Yes (D)
No (E)
P(BIA)
Total
c
P(A IA U B)
Too High (A)
About Right (B)
Too Little (C)
Total
.20
.41
.61
.09
.21
.30
.01
.08
.09
.30
.70
1.00
2.8
.:>o
Lnapter L
Two Laws of Probability
57
1-'robability
holds, A and B are independent. Show that if any of these equalities hold, the other two also
Which of the following are independent events?
a
b
c
2.75
2.76
2.79
2.80
a
If the first 2 cards are both spades, what is the probability that the next 3 cards are also
spades?
b
If the first 3 cards are all spades, what is the probability that the next 2 cards are also
spades?
c
If the first 4 cards are all spades, what is the probability that the next card is also a spade?
Suppose that A C Band that P(A) > 0 and P(B) > 0. Are A and B independent? Prove your
answer.
2.81
If P(A) > 0, P(B) > 0, and P(A) < P(AIB) , show that P(B) < P(BIA).
2.82
Suppose that A
2.83
If A and Bare mutually exclusive events and P(B) > 0, show that
c
Band that P(A) > 0 and P(B) > 0. Show that P(BIA)
an unsatisfactory plumbing job, given that the plumber was A.
a satisfactory plumbing job, given that the plumber was A.
P(A)
P(AIA U B)= P(A)
P(B)
+
2.8 Two Laws of Probability
A study of the posttreatment behavior of a large number of drug abusers suggests that the
likelihood of conviction within a two-year period after treatment may depend upon the offenders
education. The proportions of the total number of cases falling in four education-conviction
categories are shown in the following table:
Education
I 0 years or more
9 years or less
Total
The following two laws give the probabilities of unions and intersections of events.
As such, they play an important role in the event-composition approach to the solution
of probability problems.
THEOREM 2.5
Convicted
Not Convicted
Total
. 10
.27
.37
.30
.33
.63
.40
.60
1.00
A: The offender has I 0 or more years of education.
B: The offender is convicted within two years after completion of treatment.
Find the following:
P(A).
d
P(A U B).
e
f
P(A).
P(A U B).
g
P(A
h
P(AIB).
The Multiplicative Law of Probability The probability of the intersection of
two events A and B is
P(A
n
n
B).
B).
P(BIA).
In the definition of the independence of two events, you were given three equalities to check:
P(AIB) = P(A) or P(BIA) = P(B) or P(AnB) = P(A)P(B).lfanyoneoftheseequalities
P(A)P(B\A)
If A and B are independent, then
P(A
Proof
n B)=
P(A)P(B).
The multiplicative law follows directly from Definition 2.9, the definition of
conditional probability.
Notice that the multiplicative Jaw can be extended to find the probability of the
intersection of any number of events. Thus, twice applying Theorem 2.5, we obtain
P(B).
P(A
n B)=
= P(B)P(A\B) .
Suppose that a single offender is selected from the treatment program. Define the events:
a
b
c
= 1 and P(AIB) =
P(A) / P(B).
A survey of consumers in a particular community showed that 10% were dissatisfied with
plumbing jobs done in their homes. Half the complaints dealt with plumber A, who does 40%
of the plumbing jobs in the town. Find the probability that a consumer will obtain
Status within 2 Years
after Treatment
2.78
Suppose that A and B are mutually exclusive events, with P(A) > 0 and P(B) < 1. Are A
and B independent? Prove your answer.
C and D
Cards are dealt, one at a time, from a standard 52-card deck.
a
b
2.77
hold.
A and D
Band D
P(A
n B n C)=
P[(A
n B) n C]
= P(A
= P(A)P(B\A)P(C\A
n B)P(C\A n B)
n B).
The probability of the intersection of any number of, say, k events can be obtained in
the same manner:
P(At
n A2 n A3 n · · · n Ak) =
P(At)P(A2\At)P(A3\At
· · · P(Ak!At
n A2)
n A2 n · · · n Ak-J).
The additive law of probability gives the probability of the union of two events.
~
.. u,...,c:,
L
rrooa0111ty
Exercises
THEOREM 2.6
The Additive Law of Probability The probability of the union of two events
A and B is
P(A U B)= P(A)
+
P(B)- P(A
If A and Bare mutually exclusive events, P(A
P(A U B) = P(A)
Proof
P(A) = 1 - P(A).
n B).
and
+ P(B).
P(B) = P(A n B)+ P(A
As we will see in Section 2.9, it is sometimes easier to calculate P (A) than to
calculate P(A). In such cases, it is easier to find P(A) by the relationship P(A) =
1 - P(A) than to find P(A) directly.
=
P(A)
+
P(B)- P(A
Exercises
n B).
The equality given on the right implies that PeA n B) = P(B) - P(A n B).
Substituting this expression for P(A n B) into the expression for P(A U B)
given in the left-hand equation of the preceding pair, we obtain the desired
result:
P(A U B)
n B).
The probability of the union of three events can be obtained by making use of
Theorem 2.6. Observe that
2.84
+ P(B U C)-
P[A
n (B
Suppose that A and Bare two events such that P(A)
a
b
c
d
2.87
Is it possible that P(A
n B)
n B)
B are also independent. Are A and
=
.8 and P(B)
=
B
.7.
= .I? Why or why not?
What is the smallest possible value for P(A
Is it possible that P(A
=I 0 but P(A z n A 3) = 0,
+ P(A 2 ) + P(A 3) - 2P(A 1 n A z).
2.86
n B)?
= .77? Why or why not?
What is the largest possible value for P(A
n B)?
Suppose that A and Bare two events such that P(A)
+ P(B)
> I.
a What is the smallest possible value for P(A n B)?
b What is the largest possible value for
2.88
P(A
n
B)?
Suppose that A and Bare two events such that P(A) = .6 and P(B) = .3.
a Is it possible that P (A n B) = .I? Why or why not?
b What is the smallest possible value for P(A n B)?
because (A n B) n (A n C) = A n B n C.
Another useful result expressing the relationship between the probability of an
event and its complement is immediately available from the axioms of probability.
c
Is it possible that P(A
n B) = .7? Why or why not?
d What is the largest possible value for P(A n B)?
2.89
FIGURE 2.10
Venn diagram for the
union of A and B
P(A 1 n A 3)
If A and B are independent events, show that A and
independent?
= P(A)
=
=
2.85
U C)]
+ P(B) + P(C)- P(B n C)- P[(A n B) U (An C)]
P(A) + P(B) + P(C)- P(B n C)- P(A n B)- P(A n C)
+ P(A n B n C)
If A 1, A 2 , and A 3 are three events and P(A 1 n A 2 )
show that
P(at least one A;)= P(A 1)
P(A U B U C)= P[A U (B U C)]
= P(A)
Observe that S = A U A. Because A and A are mutually exclusive events, it
follows that P(S) = P(A) + P(A). Therefore, P(A) + P(A) = I and the
result follows.
Proof
n B) = 0 and
Notice that A U B = A U (4 n B), where A and (An B) are mutually
exclusive events. Further, B = (An B) U (A n B), where (An B) and (An B)
are mutually exclusive events. Then, by Axiom 3,
+ P(A n B)
If A is an event, then
THEOREM 2.7
The proof of the additive law can be followed by inspecting the Venn diagram
in Figure 2.1 0.
P(A U B) = P(A)
59
Suppose that A and Bare two events such that P(A)
+ P(B)
< I.
a What is the smallest possible value for P(A n B)?
b What is the largest possible value for P(A n B)?
2.90
Suppose that there is a I in 50 chance of injury on a single skydiving attempt.
a If we assume that the outcomes of different jumps are independent, what is the probability
A
8
that a skydiver is injured if she jumps twice?
b A friend claims if there is a I in 50 chance of injury on a single jump then there is a 100%
chance of injury if a skydiver jumps 50 times. Is your friend correct? Why?
2.91
Can A an 8 be mutually exclusive if P(A) = .4 and P(B) = .7? If P(A) = .4 and P(B) = .3?
Why?
2.92
A policy requiring all hospital employees to take lie detector tests may reduce losses due to theft,
but some employees regard such tests as a violation of their rights. Past experience indicates
that lie detectors have accuracy rates that vary from 92% to 99%. 2 To gain some insight into the
risks that employees face when taking a lie detector test, suppose that the probability is .05 that
a lie detector concludes that a person is lying who, in fact, is telling the truth and suppose
that any pair of tests are independent. What is the probability that a machine will conclude
that
a
L..------1.
2.98
2. 94
A smoke detector system uses two devices, A and B. If smoke is present, the probability that
it will be detected by device A is .95; by device 8, .90; and by both devices, .88.
a
If smoke is present, find the probability that the smoke will be detected by either device A
or B or both devices.
Two events A and B are such that P(A)
following:
a
P(A
= .2, P(B) = .3,
and P(A U B)
2.99
.4. Find the
2.96
a
1-b -a
-_-b1
2.101
Articles coming through an inspection line are visually inspected by two successive inspectors.
When a defective article comes through the inspection line, the probability that it gets by the
first inspector is .I. The second inspector will "miss" five out of ten of the defective items that
get past the first inspector. What is the probability that a defective item gets by both inspectors?
2.102
Diseases I and li are prevalent among people in a certain population. It is assumed that 10% of
the population will contract disease I sometime during their lifetime, 15 % will contract disease
II eventually, and 3% will contract both diseases .
= .5 and P(B) = .2, find the following:
a
P(A U B)
Consider the following portion of an electric circuit with three relays. Current will flow from
point a to point b if there is at least one closed path when the relays are activated. The relays
may malfunction and not close when activated. Suppose that the relays act independently of
one another and close properly when activated, with a probability of .9.
a What is the probability that current will flow when the relays are activated?
b Given that current flowed when the relays were activated, what is the probability that relay
I functioned?
2. Source: Copyright© 1980 Sentinel Communications Co. All rights reserved.
Find the probability that a randomly chosen person from this population will contract at
least one disease.
b P(A nB)
c P(A UB)
2.97
Suppose that A and B are independent events such that the probability that neither occurs is a
Show that Theorem 2.6, the additive law of probability, holds for conditional probabilities.
That is, if A, B, and Care events such that P(C) > 0, prove that P(A U BIC) = P(AIC) +
P(BIC)- P(A n BIC). [Hint: Make use of the distributive law (AUB)nC = (AnC)U(BnC).l
P(AIB)
If A and B are independent events with P (A)
•B
*2.1 00
n B)
P(AnB)
~
and the probability of B is b. Show that P(A) = -
b P(AUB)
c
d
{J)
with the probability of flow in the parallel system shown.
b Find the probability that the smoke will be undetected.
2.95
~----'
to b in the series system shown
A•
In a game, a participant is given three attempts to hit a ball. On each try, she either scores a
hit, H, or a miss, M . The game requires that the player must alternate which hand she uses in
successive attempts. That is, if she makes her first attempt with her right hand, she must use
her left hand for the second attempt and her right hand for the third. Her chance of scoring a
hit with her right hand is .7 and with her left hand is .4. Assume that the results of successive
attempts are independent and that she wins the game if she scores at least two hits in a row.
If she makes her first attempt with her right hand, what is the probability that she wins the
game?
3
With relays operating as in Exercise 2.97, compare the probability of current flowing from a
each of three employees is lying when all are telling the truth?
b at least one of the three employees is lying when all are telling the truth?
2.93
B
A
b Find the conditional probability that a randomly chosen person from this population will
contract both diseases, given that he or she has contracted at least one disease.
2.103
Refer to Exercise 2.50. Hours after the rigging of the Pennsylvania state lottery was announced,
Connecticut state lottery officials were stunned to learn that their winning number for the day
was 666 (Los Angeles Times , September 21, 1980).
a All evidence indicates that the Connecticut selection of 666 was due to pure chance. What
b
is the probability that a 666 would be drawn in Connecticut, given that a 666 had been
selected in the April 24, 1980, Pennsylvania lottery?
What is the probability of drawing a 666 in the April 24, 1980, Pennsylvania lottery
(remember, this drawing was rigged) and a 666 in the September 19, 1980, Connecticut
lottery?
2.9
2.104
If A and 8 are two events, prove that P(A n B) 2: 1- P(4)- P(ii). [Note: This is a simplified
version of the 8onferroni inequality.]
2.1 OS
If the probability of injury on each individual parachute jump is .05, use the result in Exercise 2.104 to provide a lower bound for the probability of landing safely on both of two jumps.
2.106
If least
A and.98,
8 are
equally
likely events and we require that the probability of their intersection be
at
what
is P(A)?
2.107
Let A, 8 , and C be events such that P(A) > P(B) and P(C) > 0. Construct an example to
demonstrate that it is possible that P(A/C) < P(B/C).
2.108
If A, B, and C are three events, use two applications of the result in Exercise 2.104 to prove
that P(A n 8 n C) 2: I - P(4)- P(ii)- P(C).
2.109
If A, B, and C are three equally likely events, what is the smallest value for P(A) such that
P(A n B n C) always exceeds 0.95?
2.9
Calculating the Probabi Iity of an Event:
The Event-Composition Method
We learned in Section 2.4 that sets (events) can often be expressed as unions, intersections, or complements of other sets. The event-composition method for calculating
the probability of an event, A, expresses A as a composition involving unions and/or
intersections of other events. The laws of probability are then applied to find P(A).
We will illustrate this method with an example.
Calculating the Probability of an Event: The Event-Composition Method
s
FIGURE 2.11
Venn diagram
for events of
Example 2.17
R
•
EXAMPLE 2.18
In Example 2.7 we considered an experiment wherein the birthdays of 20 randomly
selected persons were recorded. Under certain conditions we found that P(A) =
.5886, where A denotes the event that each person has a different birthday. Let B
denote the event that at least one pair of individuals share a birthday. Find P(B).
Solution
The event B is the set of all sample points in S that are not in A, that is, B = A.
Therefore,
P(B)
=
I - P(A)
=
1 - .5886
=
.4114.
(Most would agree that this probability is surprisingly high!)
EXAMPLE 2.17
Of the voters in a city, 40% are Republicans and 60% are Democrats. Among the
Republicans 70% are in favor of a bond issue, whereas 80% of the Democrats favor
the issue. If a voter is selected at random in the city, what is the probability that he or
she will favor the bond issue?
Solution
Let F denote the event "favor the bond issue," R the event "a Republican is selected,"
.4, P(D)
.6, P(F/R) =
and D the event "a Democrat is selected." Then P(R)
.7, and P(F/D)
.8. Notice that
=
=
P(F) = P[(F
n R)
U (F
n D)]=
P(F
=
n R) + P(F n D)
=
It follows that
= P(F/R)P(R) =
P(F
n R)
P(F
n D)=
(.7)(.4) = .28,
P(F/D)P(D) = (.8)(.6) = .48.
P(F) = .28
+ .48 = .76.
•
Let us refer to Example 2.4, which involves the two tennis players, and let D 1
and D2 denote the events that player A wins the first and second games, respectively. The information given in the example implies that P(D 1) = P(D 2 ) = 2/3.
Further, if we make the assumption that D 1 and D 2 are independent, it follows that
P(D1 n D2) = 2/3 x 2/3 = 4/9. In that example we identified the simple event £ 1 ,
which we denoted AA, as meaning that player A won both games. With the present
notation,
£1 = D1
because (F n R) and (F n D) are mutually exclusive events. Figure 2.11 will help
you visualize the result that F
(F n R) U (F n D). Now
63
n D2,
and thus P(£ 1) = 4/9. The probabilities assigned to the other simple events in
Example 2.4 can be verified in a similar manner.
The event-composition approach will not be successful unless the probabilities of
the events that appear in P(A) (after the additive and multiplicative laws have been
applied) are known. If one or more of these probabilities is unknown, the method fails.
Often it is desirable to form compositions of mutually exclusive or independent events.
Mutually exclusive events simplify the use of the additive law and the multiplicative
law of orobab
PlY to indeJ
Exercises
69
b We leave it to you to show that
?(exactly one match)= P(A 1)
+ P(A 2) + P(A 3)
- 2[P(A 1 n A2)
+ P(A1 n A3) + P(Az n A 3)]
+ 3[P(A 1 n Az n A3))
= (3)(1/3)- (2)(3)(1/6)
+ (3)(1/6) =
1/2.
2.115
A football team has a probability of .75 of winning when playing any of the other four teams
in its conference. If the games are independent, what is the probability the team wins all its
conference games?
2.116
A communications network has a built-in safeguard system against failures. In this system if
line I fails, it is bypassed and line II is used. If line II also fails , it is bypassed and line III is
used. The probability of failure of any one of these three lines is .0 I, and the failures of these
lines are independent events. What is the probability that this system of three lines does not
completely fail?
2.117
A state auto-inspection station has two inspection teams. Team I is lenient and passes all
automobiles of a recent vintage; team 2 rejects all autos on a first inspection because their
" headlights are not properly adjusted." Four unsuspecting drivers take their autos to the station
for inspection on four different days and randomly select one of the two teams.
•
The best way to learn how to solve probability problems is to learn by doing. To
assist you in developing your skills, many exercises are provided at the end of this
section, at the end of the chapter, and in the references.
a
b What is the probability that all four will pass?
Exercises
2.110
2.118
An accident victim will die unless in the next I 0 minutes he receives some type A, Rh-positive
blood, which can be supplied by a single donor. The hospital requires 2 minutes to type a
prospective donor's blood and 2 minutes to complete the transfer of blood. Many untyped
donors are available, and 40% of them have type A, Rh-positive blood. What is the probability
that the accident victim will be saved if only one blood-typing kit is available? Assume that
the typing kit is reusable but can process only one donor at a time.
*2.119
Suppose that two balanced dice are tossed repeatedly and the sum of the two uppermost faces
is determined on each toss. What is the probability that we obtain
Of the items produced daily by a factory, 40% come from line I and 60% from line II. Line I
has a defect rate of 8%, whereas line II has a defect rate of I 0 %. If an item is chosen at random
from the day 's production, find the probability that it will not be defective.
2.111
An advertising agency notices that approximately I in 50 potential buyers of a product sees
a given magazine ad, and I in 5 sees a corresponding ad on television. One in I 00 sees both.
One in 3 actually purchases the product after seeing the ad, 1 in I 0 without seeing it. Wbat is
the probability that a randomly selected potential customer will purchase the product?
2.112
a
Three radar sets, operating independently, are set to detect any aircraft flying through a certain
area. Each set has a probability of .02 of failing to detect a plane in its area. If an aircraft enters
the area, what is the probability that it
a sum of 3 before we obtain a sum of 7?
b a sum of 4 before we obtain a sum of 7?
2.120
a
If all four cars are new and in excellent condition, what is the probability that three of the
four will be rejected?
goes undetected?
Suppose that two defective refrigerators have been included in a shipment of six refrigerators.
The buyer begins to test the six refrigerators one at a time.
b is detected by all three radar sets?
a
2.113
2.114
A lie detector will show a positive reading (indicate a lie) 10% of the time when a person is
telling the truth and 95% of the time when the person is lying. Suppose two people are suspects
in a one-person crime and (for certain) one is guilty and will lie. Assume further that the lie
detector operates independently for the truthful person and the liar. What is the probability that
the detector
a
b
c
shows a positive reading for both suspects?
shows a positive reading for the guilty suspect and a negative reading for the innocent
suspect?
is completely wrong-that is, that it gives a positive reading for the innocent suspect and
a negative reading for the guilty?
d gives a positive reading for either or both of the two suspects?
What is the probability that the last defective refrigerator is found on the fourth test?
b What is the probability that no more than four refrigerators need to be tested to locate both
Consider one of the radar sets of Exercise 2.1 I 2. What is the probability that it will correctly
detect exactly three aircraft before it fails to detect one, if aircraft arrivals are independent
single events occurring at different times?
of the defective refrigerators?
c
2.121
When given that exactly one of the two defective refrigerators has been located in the first
two tests, what is the probability that the remaining defective refrigerator is found in the
third or fourth test?
A new secretary has been given n computer passwords, only one of which will permit access
to a computer file. Because the secretary has no idea which password is correct, he chooses
one of the passwords at random and tries it. If the password is incorrect, he discards it and
randomly selects another password from among those remaining, proceeding in this manner
until he finds the correct password.
a
What is the probability that he obtains the correct password on the first try?
b What is the probability that he obtains the correct password on the second try? The third try?
c
A security system has been set up so that if three incorrect passwords are tried before
the correct one, the computer file is locked and access to it denied. If n = 7, what is the
probability that the secretary will gain access to the file?
Exercises
FIGURE 2.13
Tree diagram for
calculations in
Example 2.23. ~A
and~ 8 are
alternative notations
/
[\.\:,';~
A
a What impact would this have on P (A IB) ?
b Suppose that P(A IB) decreased to .12 and all other probabilities remained unchanged. Use
0.0271
the applet Bayes ' Rule as a Tree to re-evaluate P(B /A).
c
B~-AOI729
for A and 8,
respective/ y.
73
How does the answer you obtained in part (b) compare to that obtained in Exercise 2.122?
Are you surprised by this result?
d Assume that all probabilities remain the same except P (A IB). Use the applet and trial and
error to find the value of P(A/B) for which P(B/A)
= .3000.
e If line I produces only defective items but all other probabilities remain unchanged, what
is P(B/A)?
P(BIAJ
A friend expected the answer to part (e) to be I. Explain why, under the conditions of part
(e), P(B /A) =f. 1.
=0.0271 I (0.0271 + 0.0461) =0.3700
(\\J';~
A
0.0461
2.124
A population of voters contains 40% Republicans and 60% Democrats. It is reported that
30% of the Republicans and 70% of the Democrats favor an election issue. A person chosen
at random from this population is found to favor the issue in question . Find the conditional
probability that this person is a Democrat.
2.125
A diagnostic test for a disease is such that it (correctly) detects the disease in 90% of the
individuals who actually have the disease. Also, if a person does not have the disease, the test
will report that he or she does not have it with probability .9. Only 1% of the population has the
disease in question. If a person is chosen at random from the population and the diagnostic test
indicates that she has the disease, what is the conditional probability that she does, in fact, have
the disease? Are you surprised by the answer? Would you call this diagnostic test reliable?
2.126
Applet Exercise Refer to Exercise 2.125. The probability that the test detects the disease given
that the patient has the disease is called the sensitivity of the test. The specificity of the test is the
probability that the test indicates no disease given that the patient is disease free. The positive
predictive value of the test is the probability that the patient has the disease given that the test
~
-B~
0.9,jz;'~
-A
0.7539
Similarly,
P(B) = 0.8
and
P(A/B) = 3(.02)(.98) 2 = .057624.
Note that these conditional probabilities were very easy to calculate. Using the law
of total probability,
P(A)
+ P(A/B)P(B)
= (.135375)(.2) + (.057624)(.8) = .0731742.
=
P(A/B)P(B)
indicates that the disease is present. Tn Exercise 2.125, the disease in question was relatively
rare, occurring with probability .0 I, and the test described has sensitivity = specificity = .90
and positive predictive value = .0833 .
Finally,
P(B/A)
=
P(B n A)
P(A)
P(A/B)P(B)
P(A)
(.135375)(.2)
.0731742
=
a
.37
,
and
In an effort to increase the positive predictive value of the test, the sensitivity was increased
to .95 and the specificity remained at .90, what is the positive predictive value of the
"improved" test?
b Still not satisfied with the positive predictive value of the procedure, the sensitivity of the
test is increased to .999. What is the positive predictive value of the (now twice) modified
test if the specificity stays at .90?
P(B/A) = l - P(B/A) = l- .37 = .63.
Figure 2.13, obtained using the applet Bayes' Rule as a Tree, illustrates the various
steps in the computation of P(B/A) .
•
c
Look carefully at the various numbers that were used to compute the positive predictive
value of the tests. Why are all of the positive predictive values so small? [Hint: Compare
the size of the numerator and the denominator used in the fraction that yields the value of
the positive predictive value. Why is the denominator so (relatively) large?]
d The proportion of individuals with the disease is not subject to our control. If the sensitivity
of the test is .90, is it possible that the positive predictive value of the test can be increased
to a value above .5? How? [Hint: Consider improving the specificity of the test.]
Exercises
e
2.122
Applet Exercise Use the applet Bayes' Rule as a Tree to obtain the results given in Figure 2.13.
2.123
Applet Exercise Refer to Exercise 2.122 and Example 2.23. Suppose that lines 2 through
5 remained the same, but line I was partially repaired and produced a smaller percentage
of defects.
2.127
Based on the results of your calculations in the previous parts, if the disease in question
is relatively rare, how can the positive predictive value of a diagnostic test be significantly
increased?
Applet Exercise Refer to Exercises 2.125 and 2. 126. Suppose now that the disease is not
particularly rare and occurs with probability .4 .
a
If, as in Exercise 2.125, a test has sensitivity
predictive value of the test?
=
specificity
=
selecting the correct answer is .25. If the student correctly answers a question, what is the
probability that the student really knew the correct answer?
.90, what is the positive
b Why is the value of the positive predictive value of the test so much higher that the value
obtained in Exercise 2.125? [Hint: Compare the size of the numerator and the denominator
used in the fraction that yields the value of the positive predictive value.]
2.134
Two methods, A and B, are available for teaching a certain industrial skill. The failure rate is
20% for A and I 0% for B. However, B is more expensive and hence is used only 30% of the
time. (A is used the other 70%. ) A worker was taught the skill by one of the methods but failed
to learn it COITectly. What is the probability that she was taught by method A?
2.135
Of the travelers arriving at a small airport, 60% fly on major airlines, 30% fly on privately
owned planes, and the remainder fly on commercially owned planes not belonging to a major
airline. Of those traveling on major airlines, 50% are traveling for business reasons, whereas
60% of those arriving on private planes and 90% of those arriving on other commercially owned
planes are traveling for business reasons. Suppose that we randomly select one person aniving
at this airport. What is the probability that the person
c If the specificity of the test remains .90, can the sensitivity of the test be adjusted to obtain
a positive predictive value above .87?
d If the sensitivity remains at .90, can the specificity be adjusted to obtain a positive predictive
value above .95? How?
e
2.128
The developers of a diagnostic test want the test to have a high positive predictive value.
Based on your calculations in previous parts of this problem and in Exercise 2.126, is the
value of the specificity more or less critical when developing a test for a rarer disease?
Use Theorem 2.8, the law of total probability, to prove the following:
a
If P(AIB)
b If P(AIC)
2.129
2.130
A study of Georgia residents suggests that those who worked in shipyards during World War II
were subjected to a significantly higher risk of lung cancer (Wall Street Journal, September 21,
1978). 3 It was found that approximately 22% of those persons who had lung cancer worked at
some prior time in a shipyard. In contrast, only 14% of those who had no lung cancer worked·
at some prior time in a shipyard. Suppose that the proportion of all Georgians living during
World War II who have or will have contracted lung cancer is .04%. Find the percentage of
Georgians living during the same period who will contract (or have contracted) lung cancer,
given that they have at some prior time worked in a shipyard.
2.131
The symmetric difference between two events A and B is the set of all sample points that are
in exactly one of the sets and is often denoted A 6. B. Note that A 6. B = (A n B) U (An B).
Prove that P(A 6. B) = P(A) + P(B)- 2P(A n B).
2.132
A plane is missing and is presumed to have equal probability of going down in any of three
regions. If a plane is actually down in region i, let I -a; denote the probability that the plane
will be found upon a search of the ith region, i = I, 2, 3. What is the conditional probability
that the plane is in
region l, given that the search of region I was unsuccessful?
b region 2, given that the search of region I was unsuccessful?
c region 3, given that the search of region I was unsuccessful?
2.133
A student answers a multiple-choice examination question that offers four possible answers.
Suppose the probability that the student knows the answer to the question is .8 and the probability that the student will guess is .2. Assume that if the student guesses, the probability of
3. Source: Wall Srreer Journal , © Dow Jones & Company, Inc. 1981. All rights reserved worldwide.
L
is traveling on business?
b is traveling for business on a privately owned plane?
c arrived on a privately owned plane, given that the person is traveling for business reasons ?
d is traveling on business, given that the person is flying on a commercially owned plane?
> P(BIC) and P(A IC) > P(BIC), then P(A) > P(B).
Males and females are observed to react differently to a given set of circumstances. It has
been observed that 70% of the females react positively to these circumstances, whereas only
40% of males react positively. A group of 20 people, 15 female and 5 male, was subjected
to these circumstances, and the subjects were asked to describe their reactions on a written
questionnaire. A response picked at random from the 20 was negative. What is the probability
that it was that of a male?
a
a
= P(AIB), then A and Bare independent.
2.136
A personnel director has two lists of applicants for jobs. List I contains the names of five
women and two men, whereas list 2 contains the names of two women and six men. A name is
randomly selected from list I and added to list 2. A name is then randomly selected from the
augmented list 2. Given that the name selected is that of a man, what is the probability that a
woman 's name was originally selected from list I?
2.137
Five identical bowls are labeled I, 2, 3, 4, and 5. Bowl i contains i white and 5 - i black
balls, with i = l , 2 , ... , 5. A bowl is randomly selected and two balls are randomly selected
(without replacement) from the contents of the bowl.
a
What is the probability that both balls selected are white?
b Given that both balls selected are white, what is the probability that bowl 3 was selected ?
*2.138
2.11
Following is a description of the game of craps. A player rolls two dice and computes the total
of the spots showing. If the player's first toss is a 7 or an I l, the player wins the game. If the
first toss is a 2, 3, or 12, the player loses the game. If the player rolls anything else (4, 5, 6, 8, 9
or I 0) on the first toss, that value becomes the player's point. Tf the player does not win or lose
on the first toss, he tosses the dice repeatedly until he obtains either his point or a 7. He wins
if he tosses his point before tossing a 7 and loses if he tosses a 7 before his point. What is the
probability that the player wins a game of craps? [Hint: Recall Exercise 2.1 19.]
Numerical Events and Random Variables
Events of major interest to the scientist, engineer, or businessperson are those identified by numbers, called numerical events. The research physician is interested in the
event that ten of ten treated patients survive an illness; the businessperson is interested in the event that sales next year will reach $5 million. Let Y denote a variable
to be measured in an experiment. Because the value of Y will vary depending on the
outcome of the experiment, it is called a random variable.
To each point in the sample space we will assign a real number denoting the value
of the variable Y. The value assigned to Y will vary from one sample point to another,
---
FIGURE 2.14
Partitioning 5 into
subsets that define
the events
Y = 0, 1, 2, 3, and 4
Compute the probabilities for each value of Y in Example 2.24.
EX AMPLE 2.25
Solution
The event {Y
= 0} results only from sample point £4 . If the coins are balanced, the
sample points are equally likely; hence,
P(Y = 0) = P(£4) = 1/ 4.
Similarly,
s
P(Y =I)= P(Ez)
but some points may be assigned the same numerical value. Thus, we have defined
a variable that is a function of the sample points in S, and {all sample points where
Y = a} is the numerical event assigned the number a. Indeed, the sample spaceS can
be partitioned into subsets so that points within a subset are all assigned the same value
of Y. These subsets are mutually exclusive since no point is assigned two different
numerical values. The partitioning of Sis symbolically indicated in Figure 2.14 for a
random variable that can assume values 0, I, 2, 3, and 4.
DEFINITION 2.12
EXAMPLE 2.24
Solution
A random variable is a real-valued function for which the domain is a sample
space.
Define an experiment as tossing two coins and observing the results. Let Y equal the
number of heads obtained. Identify the sample points in S, assign a value of Y to
each sample point, and identify the sample points associated with each value of the
random variable Y.
Let H and T represent head and tail, respectively; and let an ordered pair of symbols
identify the outcome for the first and second coins. (Thus, H T implies a head on the
first coin and a tail on the second.) Then the four sample points inS are £ 1: H H, Ez:
HT, £ 3 : T Hand £ 4 : TT. The values of Y assigned to the sample points depend
on the number of heads associated with each point. For E 1 : H H, two heads were
observed, and £ 1 is assigned the value Y = 2. Similarly, we assign the values Y = 1
to £ 2 and £3 and Y = 0 to £ 4. Summarizing, the random variable Y can take three
values, Y = 0, I, and 2, which are events defined by specific collections of sample
points:
{Y = 0} = {£4},
{Y
=
l}
=
{Ez, £3},
{Y = 2} ={Ed.
•
Let y denote an observed value of the random variable Y. Then P (Y = y) is the
sum of the probabilities of the sample points that are assigned the value y .
+
P(£3) = 1/ 2
and
P(Y = 2) = P(£ 1) = l / 4.
•
A more detailed examination of random variables will be undertaken in the next
two chapters.
Exercises
2.139
Refer to Exercise 2.112. Let the random variable Y represent the number of radar sets that
detect a particular aircraft. Compute the probabilities associated with each value of Y.
2.140
Refer to Exercise 2. 120. Let the random variable Y represent the number of defective refrigerators found after three refrigerators have been tested. Compute the probabilities for each
2.141
value of Y.
Refer again to Exercise 2.120. Let the random variable Y represent the number of the test in
which the last defective refrigerator is identified. Compute the probabilities for each value of Y.
2.142
A spinner can land in any of four positions, A , B, C, and D, with equal probability. The
spinner is used twice, and the position is noted each time. Let the random variable Y denote
the number of positions on which the spinner did not land. Compute the probabilities for each
value of Y.
2.12 Random Sampling
As our final topic in this chapter, we move from theory to application and examine
the nature of experiments conducted in statistics. A statistical experiment involves the
observation of a sample selected from a larger body of data, existing or conceptual,
called a population. The measurements in the sample, viewed as observations of the
values of one or more random variables, are then employed to make an inference
about the characteristics of the target population.
How are these inferences made? An exact answer to this question is deferred until
later, but a general observation follows from our discussion in Section 2.2. There we
learned that the probability of the observed sample plays a major role in making an
inference and evaluating the credibility of the inference.
Without belaboring the point, it is clear that the method of sampling will affect
the probability of a particular sample outcome. For example, suppose that a fictitious
uo.:>lliUU[IOns for the random variables that will be discussed in Chapter 3. The numerical events of interest to us appear in a sample, and we will wish to calculate the
probability of an observed sample to make an inference about the target population.
Probability provides both the foundation and the tools for statistical inference, the
objective of statistics.
2.148
A bin contains three components from supplier A, four from supplier 8 , and five from supplier
C. Jffour of the components are randomly selected for testing, what is the probability that each
supplier would have at least one component tested?
2.149
A large group of people is to be checked for two common symptoms of a certain disease. It
is thought that 20% of the people possess symptom A alone, 30% possess symptom B alone,
I 0% possess both symptoms, and the remainder have neither symptom. For one person chosen
at random from this group, find these probabilities:
References and Further Readings
a
b
c
Cramer, H. I 973. The Elements of Probability Theory and Some oflts Applications,
2d ed. Huntington, N.Y.: Krieger.
Feller, W. 1968. An Introduction to Probability Theory and Its Applications, 3d ed.,
vol. 1. New York: Wiley.
- - - . 197 I. An Introduction to Probability Theory and Its Applications, 2d ed.,
vol. 2. New York: Wiley.
2.150
2.144
LetS contain four sample points, £ 1 , £ 2 , £ 3 , and £
4
Refer to Exercise 2.149. Let the random variable Y represent the number of symptoms possessed
by a person chosen at random from the group. Compute the probabilities associated with each
value of Y.
2.152
We know the following about a colormetric method used to test lake water for nitrates. If
water specimens contain nitrates, a solution dropped into the water will cause the specimen to
turn red 95 % of the time. When used on water specimens without nitrates, the solution causes
the water to turn red I 0% of the time (because chemicals other than nitrates are sometimes
present and they also react to the agent). Past experience in a lab indicates that nitrates are
contained in 30% of the water specimens that are sent to the lab for testing. If a water specimen
is randomly selected
Supplementary Exercises
Show that Theorem 2.7 holds for conditional probabilities. That is, if P(B) > 0, then
P(A/B) = I - P(A/B).
A Model for the World Series Two teams
a
of events in S.
c
2.145
2.146
2.147
(;')
2.153
.
Medical case histories indicate that different illnesses may produce identical symptoms. Suppose that a particular set of symptoms, denoted H, occurs only when any one of three illnesses,
I,, fz, or / 3, occurs. Assume that the simultaneous occurrence of more that one of these illnesses
is impossible and that
P(/ 1)
= 2". Use this result to give the total number
P(H/1 1 )
P(/2) = .005 ,
P(/3) = .02.
= .90,
P(H/I2)
= .95 ,
P(H/13) = .75.
Assuming that an ill person exhibits the symptoms, H , what is the probability that the person
has illness / 1?
2.154
Five cards are drawn from a standard 52-card playing deck. What is the probability that all 5
cards will be of the same suit?
Refer to Exercise 2.146. A gambler has been dealt five cards: two aces, one king, one five, and
one 9. He discards the 5 and the 9 and is dealt two more cards. What is the probability that he
ends up with a full house?
= .01 ,
The probabilities of developing the set of symptoms H, given each of these illnesses, are known
to be
Let A and B be the events {£ 1, Ez , £ 3 ) and {£ 2 , £ 4 ) , respectively. Give the sample points
in the following events: AU B , An B , An 8, and AU B.
A patient receiving a yearly physical examination must have 18 checks or tests performed. The
sequence in which the tests are conducted is important because the time lost between tests will
vary depending on the sequence. If an efficiency expert were to study the sequences to find the
one that required the minimum length of time, how many sequences would be included in her
study if all possible sequences were admissible?
from among those sent to the lab, what is the probability that it will turn red when tested?
b and turns red when tested, what is the probability that it actually contains nitrates?
a List all possible events in S (include the null event).
b In Exercise 2.68(d), you showed that :L;'~ 1
The person has both symptoms, given that he has symptom B.
A and B play a series of games until one team
wins four games. We assume that the games are played independently and that the probability
that A wins any game is p. What is the probability that the series lasts exactly five games?
Riordan, J. 2002. Introduction to Combinatorial Analysis. Mineola, N .Y.: Dover
Publications.
2.143
The person has at least one symptom.
*2.151
Meyer, P. L. I 970. Introductory Probability and Statistical Applications, 2d ed.
Reading, Mass.: Addison- Wesley.
Parzen, E. 1992. Modern Probability Theory and Its Applications. New York:
Wi ley-lnterscience.
The person has neither symptom.
2.155
a
A drawer contains 11 = 5 different and distinguishable pairs of socks (a total often socks).
If a person (perhaps in the dark) randomly selects four socks, what is the probability that
there is no matching pair in the sample?
*b
A drawer contains n different and distinguishable pairs of socks (a total of 2n socks). A
person randomly selects 2r of the socks, where 2r < 11. In terms of 11 and r , what is the
probability that there is no matching pair in the sample?
A group of men possesses the three characteristics of being married (A}, having a college
degree (B), and being a citizen of a specified state (C), according to the fractions given in the
accompanying Venn diagram. That is, 5% of the men possess all three charac