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4.1 Experiment, Outcomes, and Sample Space 4.2 Calculating Probability 4.3 Counting Rule
4.4 Marginal and Conditional Probabilities
Case Study 4–1 Facing a Crowd Isn’t Easy 4.5 Mutually Exclusive Events 4.6 Independent versus
Dependent Events 4.7 Complementary Events 4.8 Intersection of Events and
the Multiplication Rule Case Study 4–2 Baseball
Players have “Slumps”
and “Streaks”
4.9 Union of Events and the Addition Rule
Uses and Misuses Glossary
Key Formulas
Chapter
4
PROBABILITY
“P
lease stand up in front of the class and give your oral report on describing data using statistical methods.” Does this request to speak in front of people in public make you panic? If so, what is the likelihood, or probability, that you are a woman? Among professionals, 33% of women and 11% of men fear public speaking. (See Case Study 4–1.) What is probability, and how is it determined?We often make statements about probability. For example, a weather forecaster may pre- dict that there is an 80% chance of rain tomorrow. A health news reporter may state that a smoker has a much greater chance of getting cancer than a nonsmoker does. A col- lege student may ask an instructor about the chances of passing a course or getting an A if he or she did not do well on the midterm examination.
Probability, which measures the likelihood that an event will occur, is an important part of statistics. It is the basis of inferential statistics, which will be introduced in later chapters. In inferential statistics, we make decisions under conditions of uncertainty.
Probability theory is used to evaluate the uncertainty involved in those decisions. For example, estimating next year’s sales for a company is based on many assumptions, some of which may happen to be true and others may not. Probability theory will help
us make decisions under such conditions of imperfect information and uncertainty.
Combining probability and probability distributions (which are discussed in Chapters 5 through 7) with descriptive statistics will help us make decisions about populations based on information obtained from samples. This chapter presents the basic concepts of probability and the rules for computing probability.
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Chapter 4 ProbabilitySupplementary Exercises Self-Review Test Mini-Projects
Technology Instruction Technology Assignment
4.1 EXPERIMENT, OUTCOMES, AND SAMPLE SPACE
Quality control inspector Jack Cook of Tennis Products Company picks up a tennis ball from the production line to check whether it is good or defective. Cook’s act of inspecting a ten- nis ball is an example of a statistical experiment. The result of his inspection will be that the ball is either “good” or “defective.” Each of these two observations is called an outcome (also called a basic or final outcome) of the experiment, and these outcomes taken together constitute the sample space for this experiment.
Definition
Experiment, Outcomes, and Sample Space
An experiment is a process that, when performed, results in one and only one of many observations. These observations are called the outcomes of the experiment. The collection of all outcomes for an experiment is called a sample space.
A sample space is denoted by S. The sample space for the example of inspecting a tennis ball is written as
The elements of a sample space are called sample points.
Table 4.1 lists some examples of experiments, their outcomes, and their sample spaces.
S
5good, defective6
Table 4.1 Examples of Experiments, Outcomes, and Sample Spaces
Experiment Outcomes Sample Space
Toss a coin once Head, Tail S {Head, Tail}
Roll a die once 1, 2, 3, 4, 5, 6 S {1, 2, 3, 4, 5, 6}
Toss a coin twice HH, HT, TH, TT S {HH, HT, TH, TT}
Play lottery Win, Lose S {Win, Lose}
Take a test Pass, Fail S {Pass, Fail}
Select a student Male, Female S {Male, Female}
The sample space for an experiment can also be illustrated by drawing either a Venn di-
agram or a tree diagram. A Venn diagram is a picture (a closed geometric shape such as a
rectangle, a square, or a circle) that depicts all the possible outcomes for an experiment. In
a tree diagram, each outcome is represented by a branch of the tree. Venn and tree diagrams
help us understand probability concepts by presenting them visually. Examples 4–1 through
4–3 describe how to draw these diagrams for statistical experiments.
EXAMPLE 4–1
Draw the Venn and tree diagrams for the experiment of tossing a coin once.
Solution This experiment has two possible outcomes: head and tail. Consequently, the sample space is given by
To draw a Venn diagram for this example, we draw a rectangle and mark two points inside this rectangle that represent the two outcomes, head and tail. The rectangle is labeled S because it represents the sample space (see Figure 4.1a). To draw a tree diagram, we draw two branches starting at the same point, one representing the head and the second representing the tail. The two final outcomes are listed at the ends of the branches (see Figure 4.1b).
S
5H, T6 where H Head and T Tail
4.1 Experiment, Outcomes, and Sample Space
135
(a) (b)
Head
Tail T
H S
H Outcomes
T
Drawing Venn and tree diagrams: one toss of a coin.
Drawing Venn and tree diagrams: two tosses of a coin.
Figure 4.1 (a) Venn dia- gram and (b) tree diagram for one toss of a coin.
EXAMPLE 4–2
Draw the Venn and tree diagrams for the experiment of tossing a coin twice.
Solution This experiment can be split into two parts: the first toss and the second toss.
Suppose the first time the coin is tossed we obtain a head. Then, on the second toss, we can still obtain a head or a tail. This gives us two outcomes: HH (head on both tosses) and
HT (head on the first toss and tail on the second toss). Now suppose we observe a tail onthe first toss. Again, either a head or a tail can occur on the second toss, giving the re- maining two outcomes: TH (tail on the first toss and head on the second toss) and T T (tail on both tosses). Thus the sample space for two tosses of a coin is
The Venn and tree diagrams are given in Figure 4.2. Both these diagrams show the sample space for this experiment.
S
5HH, HT, TH, T
T6
(a) (b)
S
HT HH
TT TH
HH
HT H
T
TH
TT H
T First toss Second
toss
Final outcomes
H
T
Figure 4.2 (a) Venn diagram and (b) tree diagram for two tosses of a coin.
EXAMPLE 4–3
Suppose we randomly select two persons from the members of a club and observe whether the person selected each time is a man or a woman. Write all the outcomes for this exper- iment. Draw the Venn and tree diagrams for this experiment.
Solution Let us denote the selection of a man by M and that of a woman by W. We can compare the selection of two persons to two tosses of a coin. Just as each toss of a coin can result in one of two outcomes, head or tail, each selection from the members of this club can result in one of two outcomes, man or woman. As we can see from the Venn and tree diagrams of Figure 4.3, there are four final outcomes: MM, MW, WM, WW. Hence, the sample space is written as
S
5MM, MW, WM, WW6
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Chapter 4 Probability(a) (b)
S
MM
WM
MW
WW
M
W
M
W M
W
MM
MW WM
WW First
selection
Second selection
Final outcomes
Figure 4.3 (a) Venn dia- gram and (b) tree diagram for selecting two persons.
4.1.1 Simple and Compound Events
An event consists of one or more of the outcomes of an experiment.
Definition
Event
An event is a collection of one or more of the outcomes of an experiment.
An event may be a simple event or a compound event. A simple event is also called an
elementary event, and a compound event is also called a composite event.Simple Event
Each of the final outcomes for an experiment is called a simple event. In other words, a sim- ple event includes one and only one outcome. Usually simple events are denoted by E
1, E
2,
E3, and so forth. However, we can denote them by any of the other letters, too—that is, by
A, B, C, and so forth.Definition
Simple Event
An event that includes one and only one of the (final) outcomes for an exper- iment is called a simple event and is usually denoted by E
i.
Example 4–4 describes simple events.
Drawing Venn and tree diagrams: two selections.
EXAMPLE 4–4
Reconsider Example 4–3 on selecting two persons from the members of a club and ob- serving whether the person selected each time is a man or a woman. Each of the final four outcomes (MM, MW, WM, and WW ) for this experiment is a simple event. These four events can be denoted by E
1, E
2, E
3, and E
4, respectively. Thus,
E1
1MM2, E
21MW2, E
31WM2, and E
41WW2
4.1 Experiment, Outcomes, and Sample Space
137
Illustrating simple events.
Compound Event
A compound event consists of more than one outcome.
Definition
Compound Event
A compound event is a collection of more than one outcome for an experiment.
Compound events are denoted by A, B, C, D . . . or by A
1, A
2, A
3. . . , B
1, B
2, B
3. . . , and so forth. Examples 4–5 and 4–6 describe compound events.
EXAMPLE 4–5
Reconsider Example 4–3 on selecting two persons from the members of a club and ob- serving whether the person selected each time is a man or a woman. Let A be the event that at most one man is selected. Event A will occur if either no man or one man is selected.
Hence, the event A is given by
Because event A contains more than one outcome, it is a compound event. The Venn diagram in Figure 4.4 gives a graphic presentation of compound event A.
A
5MW, WM, WW6
S A
WW MM MW
WM
Figure 4.4 Venn dia- gram for event A.
Illustrating a compound event: two selections.
Illustrating simple and compound events: two selections.
EXAMPLE 4–6
In a group of people, some are in favor of genetic engineering and others are against it.
Two persons are selected at random from this group and asked whether they are in favor of or against genetic engineering. How many distinct outcomes are possible? Draw a Venn diagram and a tree diagram for this experiment. List all the outcomes included in each of the following events and mention whether they are simple or compound events.
(a) Both persons are in favor of genetic engineering.
(b) At most one person is against genetic engineering.
(c) Exactly one person is in favor of genetic engineering.
Solution Let
A a person is against genetic engineering
F a person is in favor of genetic engineering
This experiment has the following four outcomes:
The Venn and tree diagrams in Figure 4.5 show these four outcomes.
(a) The event “both persons are in favor of genetic engineering” will occur if FF is ob- tained. Thus,
Because this event includes only one of the final four outcomes, it is a simple event.
(b) The event “at most one person is against genetic engineering” will occur if either none or one of the persons selected is against genetic engineering. Consequently,
Because this event includes more than one outcome, it is a compound event.
At most one person is against genetic engineering 5FF, FA, AF6 Both persons are in favor of genetic engineering 5FF6 AA both persons are against genetic engineering AF the first person is against and the second is in favor FA the first person is in favor and the second is against FF both persons are in favor of genetic engineering
138
Chapter 4 Probability(c) The event “exactly one person is in favor of genetic engineering” will occur if one of the two persons selected is in favor and the other is against genetic engineering. Hence, it includes the following two outcomes:
Because this event includes more than one outcome, it is a compound event.
Exactly one person is in favor of genetic engineering 5FA, AF6
(a) (b)
S
Venn diagram
Tree diagram
FA FF
AA AF
FF
FA F
A
AF
AA F
A F
A First person
Second person
Final outcomes
Figure 4.5 Venn and tree diagrams.
EXERCISES
CONCEPTS AND PROCEDURES
4.1 Define the following terms: experiment, outcome, sample space, simple event, and compound event.
4.2 List the simple events for each of the following statistical experiments in a sample space S.
a. One roll of a die b. Three tosses of a coin
c. One toss of a coin and one roll of a die
4.3 A box contains three items that are labeled A, B, and C. Two items are selected at random (without replacement) from this box. List all the possible outcomes for this experiment. Write the sample space S.
APPLICATIONS
4.4 Two students are randomly selected from a statistics class, and it is observed whether or not they suffer from math anxiety. How many total outcomes are possible? Draw a tree diagram for this experiment. Draw a Venn diagram.
4.5 In a group of adults, some are computer literate, and the others are computer illiterate. If two adults are randomly selected from this group, how many total outcomes are possible? Draw a tree diagram for this experiment.
4.6 A test contains two multiple-choice questions. If a student makes a random guess to answer each question, how many outcomes are possible? Depict all these outcomes in a Venn diagram. Also draw a tree diagram for this experiment. (Hint: Consider two outcomes for each question—either the answer is correct or it is wrong.)
4.7 A box contains a certain number of computer parts, a few of which are defective. Two parts are selected at random from this box and inspected to determine if they are good or defective. How many total outcomes are possible? Draw a tree diagram for this experiment.
4.8 In a group of people, some are in favor of a tax increase on rich people to reduce the federal deficit and others are against it. (Assume that there is no other outcome such as “no opinion” and “do not know.”) Three persons are selected at random from this group and their opinions in favor or against raising such taxes are noted. How many total outcomes are possible? Write these outcomes in a sample space S. Draw a tree diagram for this experiment.
4.9 Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S.
4.10 Refer to Exercise 4.4. List all the outcomes included in each of the following events. Indicate which are simple and which are compound events.
a. Both students suffer from math anxiety.
b. Exactly one student suffers from math anxiety.
c. The first student does not suffer and the second suffers from math anxiety.
d. None of the students suffers from math anxiety.
4.11 Refer to Exercise 4.5. List all the outcomes included in each of the following events. Indicate which are simple and which are compound events.
a. One person is computer literate and the other is not.
b. At least one person is computer literate.
c. Not more than one person is computer literate.
d. The first person is computer literate and the second is not.
4.12 Refer to Exercise 4.6. List all the outcomes included in each of the following events and mention which are simple and which are compound events.
a. Both answers are correct.
b. At most one answer is wrong.
c. The first answer is correct and the second is wrong.
d. Exactly one answer is wrong.
4.13 Refer to Exercise 4.7. List all the outcomes included in each of the following events. Indicate which are simple and which are compound events.
a. At least one part is good.
b. Exactly one part is defective.
c. The first part is good and the second is defective.
d. At most one part is good.
4.14 Refer to Exercise 4.8. List all the outcomes included in each of the following events and mention which are simple and which are compound events.
a. At most one person is against a tax increase on rich people.
b. Exactly two persons are in favor of a tax increase on rich people.
c. At least one person is against a tax increase on rich people.
d. More than one person is against a tax increase on rich people.
4.1 Experiment, Outcomes, and Sample Space
139
140
Chapter 4 Probability4.2 CALCULATING PROBABILITY
Probability, which gives the likelihood of occurrence of an event, is denoted by P. The prob- ability that a simple event E
iwill occur is denoted by P(E
i), and the probability that a com- pound event A will occur is denoted by P(A).
Definition
Probability Probability is a numerical measure of the likelihood that a specific event will
occur.
First Property of Probability
0 P1A2 1 0 P1E
i2 1
Second Property of Probability
Thus, for an experiment:
P1E
i2 P1E
12 P1E
22 P1E
32 # # # 1
Two Properties of Probability
1. The probability of an event always lies in the range 0 to 1.
Whether it is a simple or a compound event, the probability of an event is never less than 0 or greater than 1. Using mathematical notation, we can write this property as follows.
An event that cannot occur has zero probability; such an event is called an impossible event. An event that is certain to occur has a probability equal to 1 and is called a sure event.
That is,
2. The sum of the probabilities of all simple events (or final outcomes) for an experi- ment, denoted by P(E
i), is always 1.
For a sure event C: P 1C2 1 For an impossible event M: P 1M2 0
From this property, for the experiment of one toss of a coin,
For the experiment of two tosses of a coin,
For one game of football by a professional team,
P
1win2 P1loss2 P1tie2 1
P1HH 2 P1HT 2 P1TH 2 P1T
T2 1
P
1H 2 P1T 2 1
4.2.1 Three Conceptual Approaches to Probability
The three conceptual approaches to probability are (1) classical probability, (2) the relative frequency concept of probability, and (3) the subjective probability concept. These three con- cepts are explained next.
Classical Probability
Many times, various outcomes for an experiment may have the same probability of occurrence.
Such outcomes are called equally likely outcomes. The classical probability rule is applied to compute the probabilities of events for an experiment in which all outcomes are equally likely.
4.2 Calculating Probability
141
Definition
Equally Likely Outcomes
Two or more outcomes (or events) that have the same probability of occurrence are said to be equally likely outcomes (or events).
According to the classical probability rule, the probability of a simple event is equal to 1 divided by the total number of outcomes for the experiment. This is obvious because the sum of the probabilities of all final outcomes for an experiment is 1, and all the final outcomes are equally likely. In contrast, the probability of a compound event A is equal to the number of outcomes favorable to event A divided by the total number of outcomes for the experiment.
Classical Probability Rule to Find Probability
P 1A2 Number of outcomes favorable to A Total number of outcomes for the experiment
P 1E
i2 1
Total number of outcomes for the experiment
Examples 4–7 through 4–9 illustrate how probabilities of events are calculated using the clas- sical probability rule.
EXAMPLE 4–7
Find the probability of obtaining a head and the probability of obtaining a tail for one toss of a coin.
Solution The two outcomes, head and tail, are equally likely outcomes. Therefore,
1Similarly,
P
1tail2 1 2 .50
P
1head2 1
Total number of outcomes 1 2 .50
1If the final answer for the probability of an event does not terminate within three decimal places, usually it is rounded to four decimal places.
Calculating the probability of a simple event.
EXAMPLE 4–8
Find the probability of obtaining an even number in one roll of a die.
Solution This experiment has a total of six outcomes: 1, 2, 3, 4, 5, and 6. All these out- comes are equally likely. Let A be an event that an even number is observed on the die.
Event A includes three outcomes: 2, 4, and 6; that is,
If any one of these three numbers is obtained, event A is said to occur. Hence,
EXAMPLE 4–9
In a group of 500 women, 80 have played golf at least once. Suppose one of these 500 women is randomly selected. What is the probability that she has played golf at least once?
Solution Because the selection is to be made randomly, each of the 500 women has the same probability of being selected. Consequently this experiment has a total of 500 equally likely outcomes. Eighty of these 500 outcomes are included in the event that the selected woman has played golf at least once. Hence,
Relative Frequency Concept of Probability Suppose we want to calculate the following probabilities:
1. The probability that the next car that comes out of an auto factory is a “lemon”
2. The probability that a randomly selected family owns a home 3. The probability that a randomly selected woman has never smoked
4. The probability that an 80-year-old person will live for at least 1 more year 5. The probability that the tossing of an unbalanced coin will result in a head
6. The probability that a randomly selected person owns a Sport-Utility Vehicle (SUV) These probabilities cannot be computed using the classical probability rule because the var- ious outcomes for the corresponding experiments are not equally likely. For example, the next car manufactured at an auto factory may or may not be a lemon. The two outcomes, “it is a lemon” and “it is not a lemon,” are not equally likely. If they were, then (approximately) half the cars manufactured by this company would be lemons, and this might prove disas- trous to the survival of the firm.
Although the various outcomes for each of these experiments are not equally likely, each of these experiments can be performed again and again to generate data. In such cases, to calculate probabilities, we either use past data or generate new data by performing the ex- periment a large number of times. The relative frequency of an event is used as an approxi- mation for the probability of that event. This method of assigning a probability to an event is called the relative frequency concept of probability. Because relative frequencies are de- termined by performing an experiment, the probabilities calculated using relative frequencies
P
1selected woman has played golf at least once2 80 500 .16
P1A2 Number of outcomes included in A
Total number of outcomes 3 6 .50
A52, 4, 66
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Chapter 4 ProbabilityCalculating the probability of a compound event.
Calculating the probability of a compound event.
Examples 4–10 and 4–11 illustrate how the probabilities of events are approximated using the relative frequencies.
EXAMPLE 4–10
Ten of the 500 randomly selected cars manufactured at a certain auto factory are found to be lemons. Assuming that the lemons are manufactured randomly, what is the probability that the next car manufactured at this auto factory is a lemon?
Solution Let n denote the total number of cars in the sample and f the number of lemons in n. Then,
Using the relative frequency concept of probability, we obtain
This probability is actually the relative frequency of lemons in 500 cars. Table 4.2 lists the frequency and relative frequency distributions for this example.
P
1next car is a lemon2
f n10
500 .02
n500 and f 10
4.2 Calculating Probability
143
may change almost each time an experiment is repeated. For example, every time a new sam- ple of 500 cars is selected from the production line of an auto factory, the number of lemons in those 500 cars is expected to be different. However, the variation in the percentage of lemons will be small if the sample size is large. Note that if we are considering the population, the relative frequency will give an exact probability.
Using Relative Frequency as an Approximation of Probability
If an experiment is repeated n times and an event A is observed f times, then, according to the relative frequency concept of probability:
P
1A2
f nTable 4.2 Frequency and Relative Frequency Distributions for the Sample of Cars
Car f Relative Frequency
Good 490 490500 .98
Lemon 10 10500 .02
n 500 Sum 1.00
The column of relative frequencies in Table 4.2 is used as the column of approximate probabilities. Thus, from the relative frequency column,
P 1next car is a good car2 .98 P 1next car is a lemon2 .02
Approximating probability by relative frequency:
sample data.
Note that relative frequencies are not probabilities but approximate probabilities.
However, if the experiment is repeated again and again, this approximate probability of an outcome obtained from the relative frequency will approach the actual probability of that outcome. This is called the Law of Large Numbers.
144
Chapter 4 ProbabilityDefinition
Law of Large Numbers
If an experiment is repeated again and again, the probability of an event obtained from the relative frequency approaches the actual or theoretical probability.
EXAMPLE 4–11
Allison wants to determine the probability that a randomly selected family from New York State owns a home. How can she determine this probability?
Solution There are two outcomes for a randomly selected family from New York State:
“This family owns a home” and “this family does not own a home.” These two events are not equally likely. (Note that these two outcomes will be equally likely if exactly half of the families in New York State own homes and exactly half do not own homes.) Hence, the classical probability rule cannot be applied. However, we can repeat this experiment again and again. In other words, we can select a sample of families from New York State and ob- serve whether or not each of them owns a home. Hence, we will use the relative frequency approach to probability.
Suppose Allison selects a random sample of 1000 families from New York State and observes that 670 of them own homes and 330 do not own homes. Then,
Consequently,
Again, note that .670 is just an approximation of the probability that a randomly selected family from New York State owns a home. Every time Allison repeats this experiment she may obtain a different probability for this event. However, because the sample size (n 1000) in this example is large, the variation is expected to be very small.
P
1a randomly selected family owns a home2
fn
670
1000 .670 f number of families who own homes 670
n sample size 1000
Subjective Probability
Many times we face experiments that neither have equally likely outcomes nor can be re- peated to generate data. In such cases, we cannot compute the probabilities of events using the classical probability rule or the relative frequency concept. For example, consider the following probabilities of events:
1. The probability that Carol, who is taking statistics, will earn an A in this course 2. The probability that the Dow Jones Industrial Average will be higher at the end of the
next trading day
Approximating probability by relative frequency.
3. The probability that the Miami Dolphins will win the Super Bowl next season 4. The probability that Joe will lose the lawsuit he has filed against his landlord
Neither the classical probability rule nor the relative frequency concept of probability can be applied to calculate probabilities for these examples. All these examples belong to experiments that have neither equally likely outcomes nor the potential of being re- peated. For example, Carol, who is taking statistics, will take the test (or tests) only once, and based on that she will either earn an A or not. The two events “she will earn an A”
and “she will not earn an A” are not equally likely. The probability assigned to an event in such cases is called subjective probability. It is based on the individual’s own judg- ment, experience, information, and belief. Carol may assign a high probability to the event that she will earn an A in statistics, whereas her instructor may assign a low probability to the same event.
4.2 Calculating Probability
145
Definition
Subjective Probability Subjective probability is the probability assigned to an event based on
subjective judgment, experience, information, and belief.
Subjective probability is assigned arbitrarily. It is usually influenced by the biases, preferences, and experience of the person assigning the probability.
EXERCISES
CONCEPTS AND PROCEDURES
4.15 Briefly explain the two properties of probability.
4.16 Briefly describe an impossible event and a sure event. What is the probability of the occurrence of each of these two events?
4.17 Briefly explain the three approaches to probability. Give one example of each approach.
4.18 Briefly explain for what kind of experiments we use the classical approach to calculate probabilities of events and for what kind of experiments we use the relative frequency approach.
4.19 Which of the following values cannot be probabilities of events and why?
15 .97 .55 1.56 53 0.0 27 1.0
4.20 Which of the following values cannot be probabilities of events and why?
.46 23 .09 1.42 .96 94 14 .02
APPLICATIONS
4.21 Suppose a randomly selected passenger is about to go through the metal detector at the JFK New York airport. Consider the following two outcomes: The passenger sets off the metal detector, and the passenger does not set off the metal detector. Are these two outcomes equally likely? Explain why or why not. If you are to find the probability of these two outcomes, would you use the classical approach or the relative frequency approach? Explain why.
4.22 Thirty-two persons have applied for a security guard position with a company. Of them, 7 have previous experience in this area and 25 do not. Suppose one applicant is selected at random. Consider the following two events: This applicant has previous experience, and this applicant does not have pre- vious experience. If you are to find the probabilities of these two events, would you use the classical approach or the relative frequency approach? Explain why.
4.23 The president of a company has a hunch that there is a .80 probability that the company will be successful in marketing a new brand of ice cream. Is this a case of classical, relative frequency, or sub- jective probability? Explain why.
4.24 The coach of a college football team thinks there is a .75 probability that the team will win the national championship this year. Is this a case of classical, relative frequency, or subjective probability?
Explain why.
4.25 A hat contains 40 marbles. Of them, 18 are red and 22 are green. If one marble is randomly selected out of this hat, what is the probability that this marble is
a. red b. green?
4.26 A die is rolled once. What is the probability that a. a number less than 5 is obtained
b. a number 3 to 6 is obtained?
4.27 A random sample of 2000 adults showed that 860 of them have shopped at least once on the Internet. What is the (approximate) probability that a randomly selected adult has shopped on the Internet?
4.28 In a statistics class of 42 students, 28 have volunteered for community service in the past. Find the probability that a randomly selected student from this class has volunteered for community service in the past.
4.29 In a group of 50 executives, 29 have a type A personality. If one executive is selected at random from this group, what is the probability that this executive has a type A personality?
4.30 Out of the 3000 families who live in an apartment complex in New York City, 600 paid no in- come tax last year. What is the probability that a randomly selected family from these 3000 families paid income tax last year?
4.31 A multiple-choice question on a test has five answers. If Dianne chooses one answer based on
“pure guess,” what is the probability that her answer is a. correct? b.
wrong?
Do these two probabilities add up to 1.0? If yes, why?
4.32 A university has a total of 320 professors and 105 of them are women. What is the probability that a randomly selected professor from this university is a
a. woman? b.man?
Do these two probabilities add up to 1.0? If yes, why?
4.33 A company that plans to hire one new employee has prepared a final list of six candidates, all of whom are equally qualified. Four of these six candidates are women. If the company decides to select at random one person out of these six candidates, what is the probability that this person will be a woman? What is the probability that this person will be a man? Do these two probabilities add up to 1.0? If yes, why?
4.34 A sample of 500 large companies showed that 120 of them offer free psychiatric help to their employees who suffer from psychological problems. If one company is selected at random from this sample, what is the probability that this company offers free psychiatric help to its employees who suf- fer from psychological problems? What is the probability that this company does not offer free psy- chiatric help to its employees who suffer from psychological problems? Do these two probabilities add up to 1.0? If yes, why?
4.35 A sample of 400 large companies showed that 130 of them offer free health fitness centers to their employees on the company premises. If one company is selected at random from this sample, what is the probability that this company offers a free health fitness center to its employees on the company premises? What is the probability that this company does not offer a free health fit- ness center to its employees on the company premises? Do these two probabilities add up to 1.0?
If yes, why?
4.36 In a large city, 15,000 workers lost their jobs last year. Of them, 7400 lost their jobs because their companies closed down or moved, 4600 lost their jobs due to insufficient work, and the remainder lost
146
Chapter 4 Probabilitytheir jobs because their positions were abolished. If one of these 15,000 workers is selected at random, find the probability that this worker lost his or her job
a. because the company closed down or moved b. due to insufficient work
c. because the position was abolished Do these probabilities add up to 1.0? If so, why?
4.37 A sample of 820 adults showed that 80 of them had no credit cards, 116 had one card each, 94 had two cards each, 77 had three cards each, 43 had four cards each, and 410 had five or more cards each. Write the frequency distribution table for the number of credit cards an adult possesses. Calculate the relative frequencies for all categories. Suppose one adult is randomly selected from these 820 adults.
Find the probability that this adult has
a. three credit cards b. five or more credit cards
4.38 In a sample of 500 families, 90 have a yearly income of less than $30,000, 270 have a yearly in- come of $30,000 to $70,000, and the remaining families have a yearly income of more than $70,000.
Write the frequency distribution table for this problem. Calculate the relative frequencies for all classes.
Suppose one family is randomly selected from these 500 families. Find the probability that this family has a yearly income of
a. less than $30,000 b. more than $70,000
4.39 Suppose you want to find the (approximate) probability that a randomly selected family from Los Angeles earns more than $75,000 a year. How would you find this probability? What procedure would you use? Explain briefly.
4.40 Suppose you have a loaded die and you want to find the (approximate) probabilities of different outcomes for this die. How would you find these probabilities? What procedure would you use? Explain briefly.
4.3 COUNTING RULE
The experiments dealt with so far in this chapter have had only a few outcomes, which were easy to list. However, for experiments with a large number of outcomes, it may not be easy to list all outcomes. In such cases, we may use the counting rule to find the total number of outcomes.
4.3 Counting Rule
147
Counting Rule to Find Total Outcomes
If an experiment consists of three steps and if the first step can result in m outcomes, the second step in n outcomes, and the third step in k outcomes, then
Total outcomes for the experiment m n k
The counting rule can easily be extended to apply to an experiment that has fewer or more than three steps.
EXAMPLE 4–12
Suppose we toss a coin three times. This experiment has three steps: the first toss, the second toss, and the third toss. Each step has two outcomes: a head and a tail. Thus,
The eight outcomes for this experiment are HHH, HHT, HTH, HTT, THH, THT, TTH, and
TTT.Total outcomes for three tosses of a coin 2 2 2 8
Applying the counting rule:
3 steps.
EXAMPLE 4–13
A prospective car buyer can choose between a fixed and a variable interest rate and can also choose a payment period of 36 months, 48 months, or 60 months. How many total outcomes are possible?
Solution This experiment is made up of two steps: choosing an interest rate and selecting a loan payment period. There are two outcomes (a fixed or a variable interest rate) for the first step and three outcomes (a payment period of 36 months, 48 months, or 60 months) for the second step. Hence,
Total outcomes 2 3 6
148
Chapter 4 ProbabilityEXAMPLE 4–14
A National Football League team will play 16 games during a regular season. Each game can result in one of three outcomes: a win, a loss, or a tie. The total possible outcomes for 16 games are calculated as follows:
One of the 43,046,721 possible outcomes is all 16 wins.
3
1643,046,721
Total outcomes 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3
4.4 MARGINAL AND CONDITIONAL PROBABILITIES
Suppose all 100 employees of a company were asked whether they are in favor of or against paying high salaries to CEOs of U.S. companies. Table 4.3 gives a two-way classification of the responses of these 100 employees.
Table 4.3 Two-Way Classification of Employee Responses
In Favor Against
Male 15 45
Female 4 36
Table 4.3 shows the distribution of 100 employees based on two variables or character- istics: gender (male or female) and opinion (in favor or against). Such a table is called a con-
tingency table. In Table 4.3, each box that contains a number is called a cell. Notice thatthere are four cells. Each cell gives the frequency for two characteristics. For example, 15 employees in this group possess two characteristics: “male” and “in favor of paying high salaries to CEOs.” We can interpret the numbers in other cells the same way.
Applying the counting rule:
2 steps.
Applying the counting rule:
16 steps.
Suppose one employee is selected at random from these 100 employees. This employee may be classified either on the basis of gender alone or on the basis of opinion. If only one characteristic is considered at a time, the employee selected can be a male, a female, in favor, or against. The probability of each of these four characteristics or events is called marginal probability or simple probability. These probabilities are called marginal probabilities be- cause they are calculated by dividing the corresponding row margins (totals for the rows) or column margins (totals for the columns) by the grand total.
4.4 Marginal and Conditional Probabilities
149
Table 4.4 Two-Way Classification of Employee Responses with Totals
In Favor Against Total
Male 15 45 60
Female 4 36 40
Total 19 81 100
Definition
Marginal Probability Marginal probability is the probability of a single event without con-
sideration of any other event. Marginal probability is also called simple probability.
For Table 4.4, the four marginal probabilities are calculated as follows:
As we can observe, the probability that a male will be selected is obtained by dividing the total of the row labeled “Male” (60) by the grand total (100). Similarly,
These four marginal probabilities are shown along the right side and along the bottom of Table 4.5.
P 1against2 81 100 .81
P 1in favor2 19 100 .19
P 1female2 40 100 .40
P
1male2 Number of males
Total number of employees 60 100 .60
Table 4.5 Listing the Marginal Probabilities
In Favor Against
(A) (B) Total
Male (M ) 15 45 60
Female (F ) 4 36 40
Total 19 81 100
P(A) 19100 P(B) 81100
.19 .81
P(M) 60100 .60 P(F ) 40100 .40
Now suppose that one employee is selected at random from these 100 employees. Fur- thermore, assume it is known that this (selected) employee is a male. In other words, the
By adding the row totals and the column totals to Table 4.3, we write Table 4.4.
event that the employee selected is a male has already occurred. What is the probability that the employee selected is in favor of paying high salaries to CEOs? This probability is written as follows:
150
Chapter 4 ProbabilityRead as “given”
P(in favor | male)
The event whose probability This event has
is to be determined already occurred
T c c
This probability, P(in favor 0 male), is called the conditional probability of “in favor.”
It is read as “the probability that the employee selected is in favor given that this employee is a male.”
Definition
Conditional Probability Conditional probability is the probability that an event will occur
given that another event has already occurred. If A and B are two events, then the conditional probability of A given B is written as
and read as “the probability of A given that B has already occurred.”
P
1A 0 B2
EXAMPLE 4–15
Compute the conditional probability P(in favor 0 male) for the data on 100 employees given in Table 4.4.
Solution The probability P(in favor 0 male) is the conditional probability that a ran- domly selected employee is in favor given that this employee is a male. It is known that the event “male” has already occurred. Based on the information that the employee selected is a male, we can infer that the employee selected must be one of the 60 males and, hence, must belong to the first row of Table 4.4. Therefore, we are concerned only with the first row of that table.
In Favor Against Total
Male 15 45 60
Males who Total number
are in favor of males
The required conditional probability is calculated as follows:
As we can observe from this computation of conditional probability, the total number of males (the event that has already occurred) is written in the denominator and the num- ber of males who are in favor (the event whose probability we are to find) is written in the numerator. Note that we are considering the row of the event that has already occurred. The tree diagram in Figure 4.6 illustrates this example.
P
1in favor 0 male2 Number of males who are in favor Total number of males 15
60 .25
¡ ¡
Calculating the conditional probability: two-way table.
EXAMPLE 4–16
For the data of Table 4.4, calculate the conditional probability that a randomly selected em- ployee is a female given that this employee is in favor of paying high salaries to CEOs.
Solution We are to compute the probability, P(female 0 in favor). Because it is known that the employee selected is in favor of paying high salaries to CEOs, this employee must belong to the first column (the column labeled “in favor”) and must be one of the 19 employees who are in favor.
In Favor 15
4 — Females who are in favor
19 — Total number of employees who are in favor
Hence, the required probability is
The tree diagram in Figure 4.7 illustrates this example.
4
19 .2105
P 1female 0 in favor2 Number of females who are in favor Total number of employees who are in favor
4.4 Marginal and Conditional Probabilities
151
FavorsMale
We are to find the probability of this event
Male This event has already occurred
15/60 Against
Male 45/60
FavorsFemale Female
40/100 60/100
Against4/40
Female 36/40
Required probability
Figure 4.6 Tree diagram.
MakeFavors
We are to find the probability of this event Favors
This event has already occurred
15/19 Female
Favors 4/19
MaleAgainst Against
81/100 19/100
45/81 Female
Against 36/81
Required probability
Figure 4.7 Tree diagram.
Calculating the conditional probability: two-way table.
152
Case Study 4–1 illustrates the conditional probabilities of professional women and men having fear of public speaking.
4.5 MUTUALLY EXCLUSIVE EVENTS
Events that cannot occur together are called mutually exclusive events. Such events do not have any common outcomes. If two or more events are mutually exclusive, then at most one of them will occur every time we repeat the experiment. Thus the occurrence of one event excludes the occurrence of the other event or events.
FACING A CROWD ISN’T EASY
4–1
CA SE S TU D Y
The above chart shows the percentage of professional women and men who fear public speaking. These percentages can be written as conditional probabilities as follows. Suppose one professional is selected at random. Then, given that this person is a female, the probability is .35 that she has a public speak- ing fear. On the other hand, if this selected person is a male, this probability is only .11. These probabilities can be written as follows:
Note that these are approximate probabilities because the data given in the chart are based on a sample survey
P1has fear of public speaking 0 male2 .11 P1has fear of public speaking 0 female2 .35
Source: USA TODAY, May 20, 2002. Copyright © 2002, USA TODAY. Chart reproduced with permission.
Definition
Mutually Exclusive Events
Events that cannot occur together are said to be mutually exclu-
sive events.By Mary M. Kershaw and Bob Laird, USA TODAY
Professionals who fear public speaking:
Source: Sims Wyeth & Co.
Women 35%
Men 35%
USA TODAY Snapshots
RFacing
a crowd
isn't easy
For any experiment, the final outcomes are always mutually exclusive because one and only one of these outcomes is expected to occur in one repetition of the experiment. For ex- ample, consider tossing a coin twice. This experiment has four outcomes: HH, HT, TH, and
TT. These outcomes are mutually exclusive because one and only one of them will occurwhen we toss this coin twice.
EXAMPLE 4–17
Consider the following events for one roll of a die:
Are events A and B mutually exclusive? Are events A and C mutually exclusive?
Solution Figures 4.8 and 4.9 show the diagrams of events A and B and events A and
C, respectively.As we can observe from the definitions of events A and B and from Figure 4.8, events
A and B have no common element. For one roll of a die, only one of the two events, Aand B, can happen. Hence, these are two mutually exclusive events. We can observe from the definitions of events A and C and from Figure 4.9 that events A and C have two com- mon outcomes: 2-spot and 4-spot. Thus, if we roll a die and obtain either a 2-spot or a 4-spot, then A and C happen at the same time. Hence, events A and C are not mutually exclusive.
C a number less than 5 is observed 51, 2, 3, 46 B an odd number is observed 51, 3, 56 A an even number is observed 52, 4, 66
4.5 Mutually Exclusive Events
153
A
B S
4 6 2 5
3 1
Figure 4.8 Mutually exclusive events A and B.
A
C S
6 2 3 5 1
4
Figure 4.9 Mutually nonex- clusive events A and C.
EXAMPLE 4–18
Consider the following two events for a randomly selected adult:
Are events Y and N mutually exclusive?
Solution Note that event Y consists of all adults who have shopped at least once on the Internet, and event N includes all adults who have never shopped on the Internet. These two events are illustrated in the Venn diagram in Figure 4.10.
N this adult has never shopped on the Internet Y this adult has shopped at least once on the Internet
Illustrating mutually exclusive and mutually nonexclusive events.
Illustrating mutually exclusive events.
As we can observe from the definitions of events Y and N and from Figure 4.10, events
Y and N have no common outcome. They represent two distinct sets of adults: the ones whohave shopped at least once on the Internet and the ones who have never shopped on the Internet. Hence, these two events are mutually exclusive.
4.6 INDEPENDENT VERSUS DEPENDENT EVENTS
In the case of two independent events, the occurrence of one event does not change the probability of the occurrence of the other event.
154
Chapter 4 ProbabilityN
S Y Figure 4.10 Mutually exclusive events Y and N.
Definition
Independent Events
Two events are said to be independent if the occurrence of one does not affect the probability of the occurrence of the other. In other words, A and B are independent
events ifeither P1A 0 B2 P1A2 or P1B 0 A2 P1B2
It can be shown that if one of these two conditions is true, then the second will also be true, and if one is not true, then the second will also not be true.
If the occurrence of one event affects the probability of the occurrence of the other event, then the two events are said to be dependent events. In probability notation, the two events are dependent if either
EXAMPLE 4–19
Refer to the information on 100 employees given in Table 4.4 in Section 4.4. Are events
“female (F)” and “in favor (A)” independent?
Solution Events F and A will be independent if
Otherwise they will be dependent.
Using the information given in Table 4.4, we compute the following two probabilities:
Because these two probabilities are not equal, the two events are dependent. Here, de- pendence of events means that the percentages of males who are in favor of and against paying high salaries to CEOs are different from the percentages of females who are in favor and against.
In this example, the dependence of A and F can also be proved by showing that the probabilities P(A) and
P1A 0 F2 are not equal.
P
1F2 40 100 .40 and P1F 0 A2 4 19 .2105
P
1F2 P1F 0 A2
P1A 0 B2 P1A2 or P1B 0 A2 P1B2.
Illustrating two dependent events: two-way table.
EXAMPLE 4–20
A box contains a total of 100 CDs that were manufactured on two machines. Of them, 60 were manufactured on Machine I. Of the total CDs, 15 are defective. Of the 60 CDs that were manufactured on Machine I, 9 are defective. Let D be the event that a randomly se- lected CD is defective, and let A be the event that a randomly selected CD was manufactured on Machine I. Are events D and A independent?
Solution From the given information,
Hence,
Consequently, the two events, D and A, are independent.
Independence, in this example, means that the probability of any CD being defective is the same, .15, irrespective of the machine on which it is manufactured. In other words, the two machines are producing the same percentage of defective CDs. For example, 9 of the 60 CDs manufactured on Machine I are defective and 6 of the 40 CDs manufactured on Machine II are defective. Thus, for each of the two machines, 15% of the CDs produced are defective.
Actually, using the given information, we can prepare Table 4.6. The numbers in the shaded cells are given to us. The remaining numbers are calculated by doing some arithmetic manipulations.
P
1D2 P1D 0 A2
P
1D2 15 100 .15 and P1D 0 A2 9 60 .15
4.6 Independent versus Dependent Events
155
Using this table, we can find the following probabilities:
Because these two probabilities are the same, the two events are independent.
Two Important Observations
We can make two important observations about mutually exclusive, independent, and de- pendent events.
1. Two events are either mutually exclusive or independent.
2a. Mutually exclusive events are always dependent.
b. Independent events are never mutually exclusive.
2. Dependent events may or may not be mutually exclusive.
P 1D 0 A2 9 60 .15
P 1D2 15 100 .15
2The exception to this rule occurs when at least one of the two events has a zero probability.
Illustrating two independent events.
Table 4.6 Two-Way Classification Table
Defective Good
(D) (G) Total
Machine I (A) 9 51 60
Machine II (B) 6 34 40
Total 15 85 100
4.7 COMPLEMENTARY EVENTS
Two mutually exclusive events that taken together include all the outcomes for an experiment are called complementary events. Note that two complementary events are always mutually exclusive.
156
Chapter 4 ProbabilityDefinition
Complementary Events
The complement of event A, denoted by and read as “A bar” or “A complement,” is the event that includes all the outcomes for an experiment that are not in A.
A
Events A and are complements of each other. The Venn diagram in Figure 4.11 shows the complementary events A and
Because two complementary events, taken together, include all the outcomes for an ex- periment and because the sum of the probabilities of all outcomes is 1, it is obvious that
P
1A2 P1A2 1
A.A
A S
A –
Figure 4.11 Venn diagram of two complementary events.
From this equation we can deduce that
Thus, if we know the probability of an event, we can find the probability of its com- plementary event by subtracting the given probability from 1.
EXAMPLE 4–21
In a group of 2000 taxpayers, 400 have been audited by the IRS at least once. If one tax- payer is randomly selected from this group, what are the two complementary events for this experiment, and what are their probabilities?
Solution The two complementary events for this experiment are
Note that here event A includes the 400 taxpayers who have been audited by the IRS at least once, and includes the 1600 taxpayers who have never been audited by the IRS.
Hence, the probabilities of events A and are
As we can observe, the sum of these two probabilities is 1. Figure 4.12 shows a Venn diagram for this example.
P
1A2 400 2000 .20 and P1A2 1600 2000 .80
A A
A the selected taxpayer has never been audited by the IRS A the selected taxpayer has been audited by the IRS at least once
P
1A2 1 P1A2 and P1A2 1 P1A2
Calculating probabilities of complementary events.
EXAMPLE 4–22
In a group of 5000 adults, 3500 are in favor of stricter gun control laws, 1200 are against such laws, and 300 have no opinion. One adult is randomly selected from this group. Let
A be the event that this adult is in favor of stricter gun control laws. What is the comple-mentary event of A? What are the probabilities of the two events?
Solution The two complementary events for this experiment are
Note that here event includes 1500 adults who either are against stricter gun control laws or have no opinion. Also notice that events A and are complements of each other. Be- cause 3500 adults in the group favor stricter gun control laws and 1500 either are against stricter gun control laws or have no opinion, the probabilities of events A and are
As we can observe, the sum of these two probabilities is 1. Also, once we find P(A), we can find the probability of as
Figure 4.13 shows a Venn diagram for this example.
P
1A2 1 P1A2 1 .70 .30
P1A2
P
1A2 3500 5000 .70 and P1A2 1500 5000 .30
A A
A
A the selected adult either is against such laws or has no opinion A the selected adult is in favor of stricter gun control laws
4.7 Complementary Events
157
A S
A –
Figure 4.12 Venn diagram.
A S
A –
Figure 4.13 Venn diagram.
EXERCISES
CONCEPTS AND PROCEDURES
4.41 Briefly explain the difference between the marginal and conditional probabilities of events. Give one example of each.
4.42 What is meant by two mutually exclusive events? Give one example of two mutually exclusive events and another example of two mutually nonexclusive events.
4.43 Briefly explain the meaning of independent and dependent events. Suppose A and B are two events. What formula can you use to prove whether A and B are independent or dependent?
4.44 What is the complement of an event? What is the sum of the probabilities of two complementary events?
Calculating probabilities of complementary events.
4.45 How many different outcomes are possible for four rolls of a die?
4.46 How many different outcomes are possible for 10 tosses of a coin?
4.47 A statistical experiment has eight equally likely outcomes that are denoted by 1, 2, 3, 4, 5, 6, 7, and 8. Let event A {2, 5, 7} and event B {2, 4, 8}.
a. Are events A and B mutually exclusive events?
b. Are events A and B independent events?
c. What are the complements of events A and B, respectively, and their probabilities?
4.48 A statistical experiment has 10 equally likely outcomes that are denoted by 1, 2, 3, 4, 5, 6, 7, 8, 9, and 10. Let event A {3, 4, 6, 9} and event B {1, 2, 5}.
a. Are events A and B mutually exclusive events?
b. Are events A and B independent events?
c. What are the complements of events A and B, respectively, and their probabilities?
APPLICATIONS
4.49 A small ice cream shop has 10 flavors of ice cream and 5 kinds of toppings for its sundaes. How many different selections of one flavor of ice cream and one kind of topping are possible?
4.50 A man just bought 4 suits, 8 shirts, and 12 ties. All of these suits, shirts, and ties coordinate with each other. If he is to randomly select one suit, one shirt, and one tie to wear on a certain day, how many different outcomes (selections) are possible?
4.51 A restaurant menu has four kinds of soups, eight kinds of main courses, five kinds of desserts, and six kinds of drinks. If a customer randomly selects one item from each of these four categories, how many different outcomes are possible?
4.52 A student is to select three courses for next semester. If this student decides to randomly select one course from each of eight economics courses, six mathematics courses, and five computer courses, how many different outcomes are possible?
4.53 A sample of 2000 adults were asked whether or not they have ever shopped on the Internet. The following table gives a two-way classification of the responses.
Have Shopped Have Never Shopped
Male 300 900
Female 200 600
a. If one adult is selected at random from these 2000 adults, find the probability that this adult i. has never shopped on the Internet
ii. is a male
iii. has shopped on the Internet given that this adult is a female iv. is a male given that this adult has never shopped on the Internet
b. Are the events “male” and “female” mutually exclusive? What about the events “have shopped”
and “male”? Why or why not?
c. Are the events “female” and “have shopped” independent? Why or why not?
4.54 The following table gives a two-way classification, based on gender and employment status, of the civilian labor force age 16 to 24 years as of July 2002. The numbers in the table are in thousands.
Employed Unemployed
Male 11,232 1615
Female 10,353 1435
Source: U.S. Bureau of Labor Statistics, August 13, 2002.
