Math 227 Elementary Statistics

Bluman 5th edition

2

CHAPTER 4

Probability and Counting Rules

3

• Determine sample spaces and find the

probability of an event using classical

probability or empirical probability.

• Find the probability of compound events using

the addition rules.

• Find the probability of compound events using

the multiplication rules.

• Find the conditional probability of an event.

Objectives

4

• Determine the number of outcomes of a sequence of

events using a tree diagram.

• Find the total number of outcomes in a sequence of

events using the fundamental counting rule.

• Find the number of ways r objects can be selected

from n objects using the permutation rule.

• Find the number of ways r objects can be selected

from n objects without regard to order using the

combination rule.

• Find the probability of an event using the counting

rules.

Objectives (cont.)

5

• Probability as a general concept can be defined as the

chance of an event occurring. In addition to being used in

games of chance, probability is used in the fields of

insurance, investments, and weather forecasting, and in

various areas.

• Rules such as the fundamental counting rule, combination

rule and permutation rules allow us to count the number of

ways in which events can occur.

• Counting rules and probability rules can be used together to

solve a wide variety of problems.

Introduction

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Section 4.1 Sample Spaces and

Probability

I. Basic Concepts

• A probability experiment is a chance process that leads to

well-defined results called outcomes.

• An outcome is the result of a single trial of a probability

experiment.

• A sample space is the set of all possible outcomes of a

probability experiment.

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Sample Spaces and Events

Examples of some sample space:

Experiment Sample Space

Toss one coin H(head), T(tail)

Roll a die 1, 2, 3, 4, 5, 6

Answer a true/false question True, false

Toss two coins H-H, T-T, H-T, T-H

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Sample Spaces and Events

• An event consists of a set of outcomes of a

probability experiment.

Simple event - an event with one outcome

Compound event - an event with two or

more outcomes.

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Basic Concepts (cont.)

• Equally likely events are events that have

the same probability of occurring.

• A tree diagram is a device consisting of line

segments emanating from a starting point

and also from the outcome point. It is used

to determine all possible outcomes of a

probability experiment.

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Example: A tree diagram to find the sample space

for tossing two coins.

Outcomes

H H H

H

T H T

H T H

T

T T T

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Example 1 : A die is tossed one time.

(a) List the elements of the sample space S.

(b) List the elements of the event consisting of a number that is greater than 4.

S = {1, 2, 3, 4, 5, 6}

Event A: x > 4

A = {5, 6}

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Example 2 :

A coin is tossed twice. List the elements of the sample space S, and list the

elements of the event consisting of at least one head.

H

T

H

T

H

T

S = {HH, HT, TH, TT}

Event A : At least one head (H)

A = {HT, TH, HH}

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Example 3 :

A randomly selected citizen is interviewed and the following information is

recorded: employment status and level of education. The symbols for

employment status are Y = employed and N = unemployed , and the symbols

for level of education are 1 = did not complete high school, 2 = completed

high school but did not complete college, and 3 = completed college.

List the elements of sample space, and list the elements of the following events.

(a) Did not complete high school

Y

N

1

2

3

1

2

3

S = {Y1, Y2, Y3, N1, N2, N3}

(b) Is unemployed

{Y1, N1} {N1, N2, N3}

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II. Calculating Probabilities

A. Empirical Probability

(Relative Frequency Approximation of Probability)

• Empirical probability relies on actual experience

to determine the likelihood of outcomes.

• Given a frequency distribution, the probability of

an event being in a given class is:

P Ef

n( )

frequency for the class

total frequencies in the distribution

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Example 1 : The age distribution of employees for this college is shown below:

Age # of employees

Under 20 25

20 – 29 48

30 – 39 32

40 – 49 15

50 and over 10

If an employee is selected at random, find the probability that he or she is in the

following age groups.

(a) Between 30 and 39 years of age

(b) Under 20 or over 49 years of age

n = 130

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Example 2 :

During a sale at men’s store, 16 white sweaters, 3 red sweaters, 9 blue

sweaters, and 7 yellow sweaters were purchased. If a customer is selected at

random, find the probability that he bought a sweater that was not white.

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B. Classical Probability

• Classical probability uses sample spaces to

determine the numerical probability that an

event will happen.

• Classical probability assumes that all outcomes

in the sample space are equally likely to occur.

( ) Number of outcomes in E( )

( ) total number of outcomes in the sample space

n EP E

n S

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Example 1 :

A statistics class contains 14 males and 20 females. A student is to be

selected by chance and the gender of the student recorded. (Ref: General

Statistics by Chase/Bown, 4th Ed.)

(a) Give a sample space S for the experiment.

S = {M, F}

(b) Is each outcome equally likely? Explain.

P(M) ≠ P(F) . It is not equally likely because we have more females than males.

(c) Assign probabilities to each outcome.

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Example 2 :

Two dice are tossed. Find the probability that the sum of two dice is greater than 8?

S = { (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6)

(2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)

(3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6)

(5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6)

(4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6)

(6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) }

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Example set of 52 playing cards; 13 of each suit clubs,

diamonds, hearts, and spades

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Example 3 : If one card is drawn from a deck, find the probability of getting

(a) a club

52 cards in a deck, 13 of those are clubs.

(b) a 4 and a club

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Example 4 :

Three equally qualified runners, Mark (M), Bill (B) and Alan (A), run a 100-meter,

sprint, and the order of finish is recorded.

(a) Give a sample space S.

B

A

A

B

M

A

A

M

M

B

B

M

S = {MBA, MAB, BMA, BAM, AMB, ABM}

(b) What is the probability that Mark will finish last?

M

B

A

{BAM, ABM}

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III. Rounding Rule for Probabilities

• Probabilities should be expressed as

reduced fractions or rounded to two or three

decimal places.

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1. The probability of an event E is a number

(either a fraction or decimal) between and

including 0 and 1. This is denoted by:

*Rule 1 states that probabilities cannot be

negative or greater than one.

0 1 P E( )

Probability Rules

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Probability Rules (cont.)

2. If an event E cannot occur (i.e., the event

contains no members in the sample space), the

probability is zero.

3. If an event E is certain, then the probability of E

is 1.

4. The sum of the probabilities of the outcomes in

the sample space is 1.

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Example 1: A probability experiment is conducted. Which

of these can be considered a probability of an

outcome?

a) 2/5 Yes

b) -0.28 No

c) 1.09 No

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Example 2 : Given : S = {E1, E2, E3, E4}

P(E1) = P(E2) = 0.2 and P(E3) = 0.5

Find : P(E4)

P(E1) + P(E2) + P(E3) +P(E4) = 1

0.2 + 0.2 + 0.5 +P(E4) = 1

P(E4) = 1 – 0.9

P(E4) = 0.1

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P E P E( ) ( ) 1

P E P E( ) ( ) 1P E P E( ) ( ), 1

IV. Complementary Events

• The complement of an event E is the set of outcomes in the sample space that are not included in the outcomes of event E. The complement of E is denoted by .

• Rule for Complementary Events

OR

E

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Example 1 : The chance of raining

tomorrow is 70%. What is the probability

that it will not rain tomorrow?

P (No Rain) = 1 – 0.70

= 0.3

30% chance it will not rain tomorrow.

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Mutually Exclusive Events

• Two events are mutually exclusive if they cannot occur at the same time (i.e., they have no outcomes in common).

• The probability of two or more events can be determined by the addition rules.

Section 4.2 The Addition Rules for

Probability

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I. Addition Rules

• Addition Rule 1—When two events A and

B are mutually exclusive, the probability

that A or B will occur is:

P A B P A P B( ) ( ) ( ) or

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Ex1) A single card is drawn from a deck. Find the

probability of selecting a club or a diamond.

P (club or diamond) mutually exclusive

P (club or diamond) = P (club) + P (diamond)

13 13

52 52

26 1

52 2

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Ex2) In a large department store, there are 2 managers, 4

department heads, 16 clerks, and 4 stock persons. If a

person is selected at random, find the probability that the

person is either a clerk or a manager.

P (clerk or manager) mutually exclusive

P (clerk or manager) = P (clerk) + P (manager)

16 2

26 26

18 9

26 13

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I. Addition Rules (cont.)

• Addition Rule 2—If A and B are not

mutually exclusive, then:

P A B P A P B P A B( ) ( ) ( ) ( ) or and

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Example1 : A single card is drawn from a deck. Find

the probability of selecting a jack or a black card.

P (Jack or Black) = P (Jack) + P (Black) – P (Jack and Black)

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Example 2 :

In a certain geographic region, newspapers are classified as being published

daily morning, daily evening, and weekly. Some have a comics section and

other do not. The distribution is shown here.

Having comics Section Morning Evening Weekly

Yes 2 3 1

No 3 4 2

If a newspaper is selected at random, find these probabilities.

(b) The newspaper is a daily morning publication or has comics.

S = 2 + 3 + 1 + 3 + 4 +2 = 15

(a) The newspaper is weekly publication.

P (Daily morning or has comics) = P (Daily morning) + P (Has comics)

– P (Daily morning and has comics)

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(c) The newspaper is published weekly or does not have comics.

P (Weekly or No comics) = P (Weekly) + P (No comics)

– P (Weekly and no comics)

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Section 4.3 The Multiplication

Rules and Conditional Probability

Independent Events Two events A and B are independent if the

fact that A occurs does not affect the probability of B occurring.

When the outcome or occurrence of the first event affects the outcome or occurrence of the second event in such a way that the probability is changed, the events are said to be dependent.

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I. Multiplication Rules

• The multiplication rules can be used to find

the probability of two or more events that

occur in sequence.

• Multiplication Rule 1—When two events are

independent, the probability of both

occurring is:

P A B P A P B( ) ( ) ( ) and

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Example 1 : If 36% of college students are overweight, find

the probability that if three college students are selected

at random, all will be overweight.

P (1st overweight and 2nd overweight and 3rd overweight)

= 0.36 · 0.36 · 0.36

= 0.046656

≈ 0.047

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Example 2 :

If 25% of U.S. federal prison inmates are not U.S. citizens, find the probability that

two randomly selected federal prison inmates will be U.S. citizens.

P (1st U.S. citizen and 2nd U.S. citizen) = 0.75 · 0.75

= 0.5625

≈ .563

Example 3 :

Suppose the probability of remaining with a particular company 10 years or longer is

1/6. A man and a woman start work at the company on the same day.

(a) What is the probability that the man will work there less than 10 years?

(b) What is the probability that both the man and woman will work there less

than 10 years?

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Example 4 :

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 0.95; by device B, 0.98; and by both

devices, 0.94.

(a) If smoke is present, find the probability that the smoke will be detected by

device A or device B or by both devices.

(b) Find the probability that the smoke will not be detected.

P ( A or B) = P (A ) + P (B) – P (A and B)

= 0.95 + 0.98 – 0.94

= 0.99

1 – 0.99

= 0.01

P (Neither ) =

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Example 5 :

If you make random guesses for four multiple-choice test questions

(each with five possible answers), what is the probability of getting at

least one correct?

P (All wrong) = P (1st wrong & 2nd wrong & 3rd wrong & 4th wrong )

P (At least 1 correct) = 1 – P (All wrong)

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Example 6 :

There are 2000 voters in a town. Consider the experiment of randomly selecting a

voter to be interviewed. The event A consists of being in favor of more stringent

building codes; the event B consists of having lived in the town less than 10

years. The following table gives the number of voters in various categories. (Ref:

General statistics by Chase/Bown, 4th Ed.)

Favor more Do not favor more

Stringent codes stringent codes

Less than 10 years

At least 10 years

100 700

1000 200

Find each of the following:

(a)

(b)

(c)

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Multiplication Rules (cont.)

Multiplication Rule 2—When two events are

dependent (e.g. the outcome of event A

affects the outcomes of event B), the

probability of both occurring is:

P A B P A P B A( ) ( ) ( | ) and

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Example 1 :

A box has 5 red balls and 2 white balls. If two balls are randomly selected (one

after the other), what is the probability that they both are red?

(a) With replacement → independent case

(b) Without replacement → dependent case

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Example 2 :

Three cards are drawn from a deck without replacement. Find the probability

that all are Jacks.

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Multiplication Rules (cont.)

• The conditional probability of an event B in

relationship to an event A is the probability

that event B occurs after event A has

already occurred. The notation for

conditional probability is P(B|A).

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Formula for Conditional

Probability

• The probability that the second event B occurs

given that the first event A has occurred can be

found dividing the probability that both events

occurred by the probability that the first event has

occurred. The formula is:

P B AP A B

P A( | )

( )

and

( )

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Example 1: Two fair dice are tossed. Consider the following events.

A = sum is 7 or more,

B = sum is even,

and C = a match (both numbers are the same).

A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3)

(4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5)

(5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) }

B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5)

(4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) }

C = { (1,1) (2,2) (3,3) (4,4) (5,5) (6,6) }

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(a) P (A and B) =

A = { (1,6) (2,5) (2,6) (3,4) (3,5) (3, 6) (4, 3)

(4,4) (4,5) (4,6) (5,2) (5,3) (5, 4) (5, 5)

(5,6) (6,1) (6,2) (6,3) (6,4) (6, 5) (6, 6) }

B = { (1,1) (1,3) (1,5) (2,2) (2,4) (2,6) (3,1) (3,3) (3,5)

(4,2) (4,4) (4,6) (5,1) (5,3) (5,5) (6,2) (6,4) (6,6) }

(b) P (A or B) = P (A) + P (B) – P (A and B)

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(c) P (A | B) =

(d) P (B | C) =

Example 2 :

At a local Country Club, 65% of the members play bridge and swim, and 72%

play bridge. If a member is selected at random, find the probability that the

member swims, given that the member plays bridge.

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Example 3 :

Eighty students in a school cafeteria were asked if they favored a ban on

smoking in the cafeteria. The results of the survey are shown in the table.

Class Favor Oppose No opinion

Freshman 15 27 8

Sophomore 23 5 2

If a student is selected at random, find these probabilities.

(a) The student is a freshman or favors the ban.

P (Freshman or Favor ban) = P (Freshman) + P (Favor ban)

– P (Freshman and Favor ban)

(b) Given that the student favors the ban, the student is a sophomore.

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Section 4.4 Counting Rules

I. The Fundamental Counting Rule

In a sequence of n events in which the first one has k1

possibilities and the second event has k2 and the third

has k3, and so forth, the total number of possibilities of

the sequence will be:

Note: “And” in this case means to multiply.

k k k kn1 2 3

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Example 1 : Two dice are tossed. How many outcomes are in S.

6 ·

How many outcomes for Task 1? How many outcome for Task 2?

Example 2 :

A password consists of two letters followed by one digit. How many different

passwords can be created? (Note: Repetitions are allowed)

1st letter → 26 outcomes (A – Z)

2nd letter → 26 outcomes (A – Z)

One digit → 10 outcomes (0-9)

26 · 26 · 10 = 6,760 ways

6 = 36 outcomes

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Example 3 :

Suppose four digits are to be randomly selected (with repetitions

allowed). (Note: the set of digits is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}.) If the

first digit must be a 2 and repetitions are permitted, how many

different 4-digit can be made?

1st digit → has to be a 2 → 1 outcome

2nd digit → any number (0 – 9) →10 outcomes

3rd digit → any number (0 – 9)→10 outcomes

4th digit → any number (0 – 9) →10 outcomes

1 · 10 · 10 · 10 = 1,000 ways

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II. Permutations and Combinations

• Factorial Notation

5! = 5·4·3·2·1

7! = 7·6·5·4·3·2·1

In general,

n! = n · (n-1) ·(n-2) ·(n-3) ···· 1

0!=1

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Permutation

• A permutation is an arrangement of n objects in

a specific order.

The arrangement of n objects in a specific order

using r objects at a time is called a permutation

of n objects taking r objects at a time.

Note: Order does matter.

It is written as nPr, and the formula is:

n rPn

n r

!

( )!

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n rCn

n r r

!

( )! !

Combinations

• A selection of distinct objects without regard to order is called a combination.

(Order does NOT matter!)

• The number of combinations of r objects selected from n objects is denoted nCr and is given by the formula:

Note: Combinations are always less than permutations for the

same n and r.

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Example 1 :

(a) 5! 5 · 4 · 3 · 2 · 1 = 120

(b) 25P8

(c) 12P4

(d) 25C8

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Example 2 :

A television news director wishes to use three news stories on an evening

show. One story will be the lead story, one will be the second story, and the

last will be a closing story. If the director has a total of eight stories to choose

from, how many possible ways can the program be set up?

Example 3 :

If a person can select 3 presents from 10 presents, how many different

combinations are there?

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Example 4 :

Your family vacation involves a cross-country air flight, a rental car, and a hotel

stay in Boston. If you can choose from four major air carriers, five car rental

agencies, and three major hotel chains, how many options are available for your

vacation accommodations?

Multiplication Rule

4 · 5 · 3 = 60 ways

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Section 4.5 Probability and

Counting Rules

I. Basic Concepts

By using the fundamental counting rule, the

permutation rules, and the combination rule,

the probability of outcomes of many

experiments can be computed.

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Example 1 :

A combination lock consists of 26 letters of the alphabet. If a three-letter

combination is needed, find the probability that the combination will consist of the

first two letters AB in that order. The same letter can be used more than once.

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Example 2 :

Five cards are selected from a 52-card deck for a poker hand. (A poker hand

consists of 5 cards dealt in any order.)

(a) How many outcomes are in the sample space?

(b) A royal flush is a hand that contains that A, K, Q, J, 10, all in the same suit.

How many ways are there to get a royal flush?

A, K, Q, J, 10 - ♠

A, K, Q, J, 10 - ♣

A, K, Q, J, 10 -♥

A, K, Q, J, 10 - ♦ n(A) = 4 combinations

(c) What is the probability of being dealt a royal flush?

Example 4-52: Committee Selection

A store has 6 TV Graphic magazines and 8 Newstime

magazines on the counter. If two customers purchased a

magazine, find the probability that one of each magazine

was purchased.

TV Graphic: One magazine of the 6 magazines

Newstime: One magazine of the 8 magazines

Total: Two magazines of the 14 magazines

66 Bluman, Chapter 4

6 1 8 1

14 2

C C

C

6 8 48

91 91

67

Summary

• The three types of probability are classical,

empirical, and subjective.

• Classical probability uses sample spaces.

• Empirical probability uses frequency

distributions and is based on observations.

• In subjective probability, the researcher

makes an educated guess about the chance

of an event occurring.

68

Summary (cont)

• An event consists of one or more outcomes

of a probability experiment.

• Two events are said to be mutually exclusive

if they cannot occur at the same time.

• Events can also be classified as

independent or dependent.

• If events are independent, whether or not the

first event occurs does not affect the

probability of the next event occurring.

69

Summary (cont.)

• If the probability of the second event

occurring is changed by the occurrence of

the first event, then the events are

dependent.

• The complement of an event is the set of

outcomes in the sample space that are not

included in the outcomes of the event itself.

• Complementary events are mutually

exclusive.

70

Summary

The number of ways a sequence of n events can occur; if the first event can

occur in k1 ways, the second event can occur in k2 ways, etc.

(Multiplication rule)

The arrangement of n objects in a specific order using r objects at a time

(Permutation rule)

The number of combinations of r objects selected from n objects (order is not important)

(Combination rule)

n rPn

n r

!

( )!

n rCn

n r r

!

( )! !

k k k kn1 2 3

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Conclusions

• Probability can be defined as the chance of

an event occurring. It can be used to

quantify what the “odds” are that a specific

event will occur. Some examples of how

probability is used everyday would be

weather forecasting, “75% chance of snow”

or for setting insurance rates.

72

Conclusions

• A tree diagram can be used when a list of all

possible outcomes is necessary. When only

the total number of outcomes is needed, the

multiplication rule, the permutation rule, and

the combination rule can be used.