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Chapter 3 Probability 1

Chapter 3 Probability 1. 2 Introduction to Probability

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Page 1: Chapter 3 Probability 1. 2  Introduction to Probability

Chapter 3

Probability

1

Page 2: Chapter 3 Probability 1. 2  Introduction to Probability

2

http://www.learner.org/courses/againstallodds/unitpages/unit18.html

Introduction to Probability

Page 3: Chapter 3 Probability 1. 2  Introduction to Probability

Chapter Outline

• 3.1 Basic Concepts of Probability

• 3.2 Conditional Probability and the Multiplication Rule

• 3.3 The Addition Rule

• 3.4 Additional Topics in Probability and Counting

3

Page 4: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.1

Basic Concepts of Probability

4

Page 5: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.1 Objectives

• Identify the sample space of a probability experiment

• Identify simple events

• Use the Fundamental Counting Principle

• Distinguish among classical probability, empirical probability, and subjective probability

• Determine the probability of the complement of an event

• Use a tree diagram and the Fundamental Counting Principle to find probabilities

5

Page 6: Chapter 3 Probability 1. 2  Introduction to Probability

Probability Experiments

Probability experiment

• An action, or trial, through which specific results (counts, measurements, or responses) are obtained.

Outcome

• The result of a single trial in a probability experiment.

Sample Space

• The set of all possible outcomes of a probability experiment.

Event

• Consists of one or more outcomes and is a subset of the sample space.

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Page 7: Chapter 3 Probability 1. 2  Introduction to Probability

Probability Experiments

• Probability experiment: Roll a die

• Outcome: {3}

• Sample space: {1, 2, 3, 4, 5, 6}

• Event: {Die is even}={2, 4, 6}

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Page 8: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Identifying the Sample Space

A probability experiment consists of tossing a coin and then rolling a six-sided die. Describe the sample space.

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Solution:There are two possible outcomes when tossing a coin: a head (H) or a tail (T). For each of these, there are six possible outcomes when rolling a die: 1, 2, 3, 4, 5, or 6. One way to list outcomes for actions occurring in a sequence is to use a tree diagram.

Page 9: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Identifying the Sample Space

9

Tree diagram:

H1 H2 H3 H4 H5 H6 T1 T2 T3 T4 T5 T6

The sample space has 12 outcomes:{H1, H2, H3, H4, H5, H6, T1, T2, T3, T4, T5, T6}

Page 10: Chapter 3 Probability 1. 2  Introduction to Probability

Simple Events

Simple event

• An event that consists of a single outcome. e.g. “Tossing heads and rolling a 3” {H3}

• An event that consists of more than one outcome is not a simple event. e.g. “Tossing heads and rolling an even number”

{H2, H4, H6}

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Page 11: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Identifying Simple Events

Determine whether the event is simple or not.

• You roll a six-sided die. Event B is rolling at least a 4.

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Solution:Not simple (event B has three outcomes: rolling a 4, a 5, or a 6)

Page 12: Chapter 3 Probability 1. 2  Introduction to Probability

Fundamental Counting Principle

Fundamental Counting Principle

• If one event can occur in m ways and a second event can occur in n ways, the number of ways the two events can occur in sequence is m*n.

• Can be extended for any number of events occurring in sequence.

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Page 13: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Fundamental Counting Principle

You are purchasing a new car. The possible manufacturers, car sizes, and colors are listed.

Manufacturer: Ford, GM, Honda

Car size: compact, midsize

Color: white (W), red (R), black (B), green (G)

How many different ways can you select one manufacturer, one car size, and one color? Use a tree diagram to check your result.

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Page 14: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Fundamental Counting Principle

There are three choices of manufacturers, two car sizes, and four colors.

Using the Fundamental Counting Principle:

3 ∙ 2 ∙ 4 = 24 ways

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Page 15: Chapter 3 Probability 1. 2  Introduction to Probability

Types of Probability

Classical (theoretical) Probability

• Each outcome in a sample space is equally likely.

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Number of outcomes in event E( )

Number of outcomes in sample spaceP E

Page 16: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Classical Probabilities

1. Event A: rolling a 3

2. Event B: rolling a 7

3. Event C: rolling a number less than 5

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Solution:Sample space: {1, 2, 3, 4, 5, 6}

You roll a six-sided die. Find the probability of each event.

Page 17: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Classical Probabilities

1. Event A: rolling a 3 Event A = {3}

17

1( 3) 0.167

6P rolling a

2. Event B: rolling a 7 Event B= { } (7 is not in the sample space)

0( 7) 0

6P rolling a

3. Event C: rolling a number less than 5

Event C = {1, 2, 3, 4}4

( 5) 0.6676

P rolling a number less than

Page 18: Chapter 3 Probability 1. 2  Introduction to Probability

Types of Probability

Empirical (statistical) Probability

• Based on observations obtained from probability experiments.

• Relative frequency of an event.

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Frequency of event E( )

Total frequency

fP E

n

Page 19: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Empirical Probabilities

A company is conducting an online survey of randomly selected individuals to determine if traffic congestion is a problem in their community. So far, 320 people have responded to the survey. What is the probability that the next person that responds to the survey says that traffic congestion is a serious problem in their community?

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Response Number of times, f

Serious problem 123

Moderate problem 115

Not a problem 82

Σf = 320

Page 20: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Empirical Probabilities

20

Response Number of times, f

Serious problem 123

Moderate problem 115

Not a problem 82

Σf = 320

event frequency

123( ) 0.384

320

fP Serious problem

n

Page 21: Chapter 3 Probability 1. 2  Introduction to Probability

Law of Large Numbers

Law of Large Numbers

• As an experiment is repeated over and over, the empirical probability of an event approaches the theoretical (actual) probability of the event.

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Page 22: Chapter 3 Probability 1. 2  Introduction to Probability

Types of Probability

Subjective Probability

• Intuition, educated guesses, and estimates.

• e.g. A doctor may feel a patient has a 90% chance of a full recovery.

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Page 23: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Classifying Types of Probability

Classify the statement as an example of classical, empirical, or subjective probability.

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Solution:

Subjective probability (most likely an educated guess)

1. The probability that you will be married by age 30 is 0.50.

Page 24: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Classifying Types of Probability

Classify the statement as an example of classical, empirical, or subjective probability.

24

Solution:Empirical probability (most likely based on a survey)

2. The probability that a voter chosen at random will vote Republican is 0.45.

Page 25: Chapter 3 Probability 1. 2  Introduction to Probability

3. The probability of winning a 1000-ticket raffle with one ticket is .

Example: Classifying Types of Probability

Classify the statement as an example of classical, empirical, or subjective probability.

25

1

1000

Solution:Classical probability (equally likely outcomes)

Page 26: Chapter 3 Probability 1. 2  Introduction to Probability

Range of Probabilities Rule

Range of probabilities rule

• The probability of an event E is between 0 and 1, inclusive.

• 0 ≤ P(E) ≤ 1

26

[ ]0 0.5 1

Impossible UnlikelyEven

chance Likely Certain

Page 27: Chapter 3 Probability 1. 2  Introduction to Probability

Complementary Events

Complement of event E

• The set of all outcomes in a sample space that are not included in event E.

• Denoted E ′ (E prime)

• P(E ′) + P(E) = 1• P(E) = 1 – P(E ′)• P(E ′) = 1 – P(E)

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E ′E

Page 28: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Probability of the Complement of an Event

You survey a sample of 1000 employees at a company and record the age of each. Find the probability of randomly choosing an employee who is not between 25 and 34 years old.

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Employee ages Frequency, f

15 to 24 54

25 to 34 366

35 to 44 233

45 to 54 180

55 to 64 125

65 and over 42

Σf = 1000

Page 29: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Probability of the Complement of an Event

• Use empirical probability to find P(age 25 to 34)

29

Employee ages Frequency, f

15 to 24 54

25 to 34 366

35 to 44 233

45 to 54 180

55 to 64 125

65 and over 42

Σf = 1000

366( 25 34) 0.366

1000

fP age to

n

• Use the complement rule366

( 25 34) 11000

6340.634

1000

P age is not to

Page 30: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Probability Using a Tree Diagram

A probability experiment consists of tossing a coin and spinning the spinner shown. The spinner is equally likely to land on each number. Use a tree diagram to find the probability of tossing a tail and spinning an odd number.

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Page 31: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Probability Using a Tree Diagram

Tree Diagram:

31

H T

1 2 3 4 5 76 8 1 2 3 4 5 76 8

H1 H2 H3 H4 H5 H6 H7 H8 T1 T2 T3 T4 T5 T6 T7 T8

P(tossing a tail and spinning an odd number) = 4 1

0.2516 4

Page 32: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Probability Using the Fundamental Counting Principle

Your college identification number consists of 8 digits. Each digit can be 0 through 9 and each digit can be repeated. What is the probability of getting your college identification number when randomly generating eight digits?

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Page 33: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Probability Using the Fundamental Counting Principle

• Each digit can be repeated

• There are 10 choices for each of the 8 digits

• Using the Fundamental Counting Principle, there are

10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10 ∙ 10

= 108 = 100,000,000 possible identification numbers

• Only one of those numbers corresponds to your ID number

33

1

100,000,000P(your ID number) =

Page 34: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.1 Summary

• Identified the sample space of a probability experiment

• Identified simple events

• Used the Fundamental Counting Principle

• Distinguished among classical probability, empirical probability, and subjective probability

• Determined the probability of the complement of an event

• Used a tree diagram and the Fundamental Counting Principle to find probabilities

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Page 35: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.2

Conditional Probability and the Multiplication Rule

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Page 36: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.2 Objectives

• Determine conditional probabilities

• Distinguish between independent and dependent events

• Use the Multiplication Rule to find the probability of two events occurring in sequence

• Use the Multiplication Rule to find conditional probabilities

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Page 37: Chapter 3 Probability 1. 2  Introduction to Probability

Conditional Probability

Conditional Probability

• The probability of an event occurring, given that another event has already occurred

• Denoted P(B | A) (read “probability of B, given A”)

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Page 38: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Conditional Probabilities

Two cards are selected in sequence from a standard deck. Find the probability that the second card is a queen, given that the first card is a king. (Assume that the king is not replaced.)

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Solution:Because the first card is a king and is not replaced, the remaining deck has 51 cards, 4 of which are queens.

4( | ) (2 |1 ) 0.078

51nd stP B A P card is a Queen card is a King

Page 39: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Conditional Probabilities

The table shows the results of a study in which researchers examined a child’s IQ and the presence of a specific gene in the child. Find the probability that a child has a high IQ, given that the child has the gene.

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Gene Present

Gene not present Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102

Page 40: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Conditional Probabilities

There are 72 children who have the gene. So, the sample space consists of these 72 children.

40

33( | ) ( | ) 0.458

72P B A P high IQ gene present

Of these, 33 have a high IQ.

Gene Present

Gene not present Total

High IQ 33 19 52

Normal IQ 39 11 50

Total 72 30 102

Page 41: Chapter 3 Probability 1. 2  Introduction to Probability

Independent and Dependent Events

Independent events

• The occurrence of one of the events does not affect the probability of the occurrence of the other event

• P(B | A) = P(B) or P(A | B) = P(A)

• Events that are not independent are dependent

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Page 42: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Independent and Dependent Events

1. Selecting a king from a standard deck (A), not replacing it, and then selecting a queen from the deck (B).

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4( | ) (2 |1 )

51nd stP B A P card is a Queen card is a King

4( ) ( )

52P B P Queen

Dependent (the occurrence of A changes the probability of the occurrence of B)

Solution:

Decide whether the events are independent or dependent.

Page 43: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Independent and Dependent Events

Decide whether the events are independent or dependent.

2. Tossing a coin and getting a head (A), and then rolling a six-sided die and obtaining a 6 (B).

43

1( | ) ( 6 | )

6P B A P rolling a head on coin

1( ) ( 6)

6P B P rolling a

Independent (the occurrence of A does not change the probability of the occurrence of B)

Solution:

Page 44: Chapter 3 Probability 1. 2  Introduction to Probability

The Multiplication Rule

Multiplication rule for the probability of A and B

• The probability that two events A and B will occur in sequence is P(A and B) = P(A) ∙ P(B | A)

• For independent events the rule can be simplified to P(A and B) = P(A) ∙ P(B) Can be extended for any number of independent

events

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Page 45: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule

Two cards are selected, without replacing the first card, from a standard deck. Find the probability of selecting a king and then selecting a queen.

45

Solution:Because the first card is not replaced, the events are dependent.

( ) ( ) ( | )

4 4

52 5116

0.0062652

P K and Q P K P Q K

Page 46: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule

A coin is tossed and a die is rolled. Find the probability of getting a head and then rolling a 6.

46

Solution:The outcome of the coin does not affect the probability of rolling a 6 on the die. These two events are independent.

( 6) ( ) (6)

1 1

2 61

0.08312

P H and P H P

Page 47: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule

The probability that a particular knee surgery is successful is 0.85. Find the probability that three knee surgeries are successful.

47

Solution:The probability that each knee surgery is successful is 0.85. The chance for success for one surgery is independent of the chances for the other surgeries.

P(3 surgeries are successful) = (0.85)(0.85)(0.85) ≈ 0.614

Page 48: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule

Find the probability that none of the three knee surgeries is successful.

48

Solution:Because the probability of success for one surgery is 0.85. The probability of failure for one surgery is 1 – 0.85 = 0.15

P(none of the 3 surgeries is successful) = (0.15)(0.15)(0.15) ≈ 0.003

Page 49: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule

Find the probability that at least one of the three knee surgeries is successful.

49

Solution:“At least one” means one or more. The complement to the event “at least one successful” is the event “none are successful.” Using the complement rule

P(at least 1 is successful) = 1 – P(none are successful)≈ 1 – 0.003= 0.997

Page 50: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule to Find Probabilities

More than 15,000 U.S. medical school seniors applied to residency programs in 2007. Of those, 93% were matched to a residency position. Seventy-four percent of the seniors matched to a residency position were matched to one of their top two choices. Medical students electronically rank the residency programs in their order of preference and program directors across the United States do the same. The term “match” refers to the process where a student’s preference list and a program director’s preference list overlap, resulting in the placement of the student for a residency position. (Source: National Resident Matching Program)

50

(continued)

Page 51: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule to Find Probabilities

1. Find the probability that a randomly selected senior was matched a residency position and it was one of the senior’s top two choices.

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Solution:A = {matched to residency position}B = {matched to one of two top choices}

P(A) = 0.93 and P(B | A) = 0.74

P(A and B) = P(A)∙P(B | A) = (0.93)(0.74) ≈ 0.688dependent events

Page 52: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Multiplication Rule to Find Probabilities

2. Find the probability that a randomly selected senior that was matched to a residency position did not get matched with one of the senior’s top two choices.

52

Solution:Use the complement:

P(B′ | A) = 1 – P(B | A)

= 1 – 0.74 = 0.26

Page 53: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.2 Summary

• Determined conditional probabilities

• Distinguished between independent and dependent events

• Used the Multiplication Rule to find the probability of two events occurring in sequence

• Used the Multiplication Rule to find conditional probabilities

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Page 54: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.3

Addition Rule

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Page 55: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.3 Objectives

• Determine if two events are mutually exclusive

• Use the Addition Rule to find the probability of two events

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Page 56: Chapter 3 Probability 1. 2  Introduction to Probability

Mutually Exclusive Events

Mutually exclusive

• Two events A and B cannot occur at the same time

56

AB A B

A and B are mutually exclusive

A and B are not mutually exclusive

Page 57: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Mutually Exclusive Events

Decide if the events are mutually exclusive.

Event A: Roll a 3 on a die.

Event B: Roll a 4 on a die.

57

Solution:Mutually exclusive (The first event has one outcome, a 3. The second event also has one outcome, a 4. These outcomes cannot occur at the same time.)

Page 58: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Mutually Exclusive Events

Decide if the events are mutually exclusive.

Event A: Randomly select a male student.

Event B: Randomly select a nursing major.

58

Solution:Not mutually exclusive (The student can be a male nursing major.)

Page 59: Chapter 3 Probability 1. 2  Introduction to Probability

The Addition Rule

Addition rule for the probability of A or B

• The probability that events A or B will occur is P(A or B) = P(A) + P(B) – P(A and B)

• For mutually exclusive events A and B, the rule can be simplified to P(A or B) = P(A) + P(B) Can be extended to any number of mutually

exclusive events

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Page 60: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Addition Rule

You select a card from a standard deck. Find the probability that the card is a 4 or an ace.

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Solution:The events are mutually exclusive (if the card is a 4, it cannot be an ace)

(4 ) (4) ( )

4 4

52 528

0.15452

P or ace P P ace

4♣4♥

4♦4♠ A♣

A♥

A♦

A♠

44 other cards

Deck of 52 Cards

Page 61: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Addition Rule

You roll a die. Find the probability of rolling a number less than 3 or rolling an odd number.

61

Solution:The events are not mutually exclusive (1 is an outcome of both events)

Odd

5

3 1

2

4 6

Less than three

Roll a Die

Page 62: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Using the Addition Rule

62

( 3 )

( 3) ( ) ( 3 )

2 3 1 40.667

6 6 6 6

P less than or odd

P less than P odd P less than and odd

Odd

5

3 1

2

4 6

Less than three

Roll a Die

Page 63: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Addition Rule

The frequency distribution shows the volume of sales (in dollars) and the number of months a sales representative reached each sales level during the past three years. If this sales pattern continues, what is the probability that the sales representative will sell between $75,000 and $124,999 next month?

63

Sales volume ($) Months

0–24,999 3

25,000–49,999 5

50,000–74,999 6

75,000–99,999 7

100,000–124,999 9

125,000–149,999 2

150,000–174,999 3

175,000–199,999 1

Page 64: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Using the Addition Rule

• A = monthly sales between $75,000 and $99,999

• B = monthly sales between $100,000 and $124,999

• A and B are mutually exclusive

64

Sales volume ($) Months

0–24,999 3

25,000–49,999 5

50,000–74,999 6

75,000–99,999 7

100,000–124,999 9

125,000–149,999 2

150,000–174,999 3

175,000–199,999 1

( ) ( ) ( )

7 9

36 3616

0.44436

P A or B P A P B

Page 65: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Addition Rule

A blood bank catalogs the types of blood given by donors during the last five days. A donor is selected at random. Find the probability the donor has type O or type A blood.

65

Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

Page 66: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Using the Addition Rule

66

The events are mutually exclusive (a donor cannot have type O blood and type A blood)

Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

( ) ( ) ( )

184 164

409 409348

0.851409

P type O or type A P type O P type A

Page 67: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Using the Addition Rule

Find the probability the donor has type B or is Rh-negative.

67

Solution:

The events are not mutually exclusive (a donor can have type B blood and be Rh-negative)

Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

Page 68: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Using the Addition Rule

68

Type O Type A Type B Type AB Total

Rh-Positive 156 139 37 12 344

Rh-Negative 28 25 8 4 65

Total 184 164 45 16 409

( )

( ) ( ) ( )

45 65 8 1020.249

409 409 409 409

P type B or Rh neg

P type B P Rh neg P type B and Rh neg

Page 69: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.3 Summary

• Determined if two events are mutually exclusive

• Used the Addition Rule to find the probability of two events

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Page 70: Chapter 3 Probability 1. 2  Introduction to Probability

Car is behind door #

P(C=c) Player selects door #

P(S=s) Host opens door #

P(H=h) P(branch) StayW=carL=goat

Total P(W)

Switch Total P(W)

C=1 1/3

S=1 1/3H=2 1/2 1/18 W

1/18+1/18+1/18+1/18+1/18+1/18=6/18

L

1/9+1/9+1/9+1/9+1/9+1/9=6/9

H=3 1/2 1/18 W L

S=2 1/3 H=3 1 1/9 L W

S=3 1/3 H=2 1 1/9 L W

C=2 1/3

S=1 1/3 H=3 1 1/9 L W

S=2 1/3H=1 1/2 1/18 W L

H=3 1/2 1/18 W L

S=3 1/3 H=1 1 1/9 L W

C=3 1/3

S=1 1/3 H=2 1 1/9 L W

S=2 1/3 H=1 1 1/9 L W

S=3 1/3H=1 1/2 1/18 W L

H=2 1/2 1/18 W L

The calculations for the 3-door problem:

Page 71: Chapter 3 Probability 1. 2  Introduction to Probability

71

Probability Models

http://www.learner.org/courses/againstallodds/unitpages/unit19.html

Page 72: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.4

Additional Topics in Probability and Counting

72

Page 73: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.4 Objectives

• Determine the number of ways a group of objects can be arranged in order

• Determine the number of ways to choose several objects from a group without regard to order

• Use the counting principles to find probabilities

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Page 74: Chapter 3 Probability 1. 2  Introduction to Probability

Permutations

Permutation

• An ordered arrangement of objects

• The number of different permutations of n distinct objects is n! (n factorial) n! = n∙(n – 1)∙(n – 2)∙(n – 3)∙ ∙ ∙3∙2 ∙1 0! = 1 Examples:

• 6! = 6∙5∙4∙3∙2∙1 = 720

• 4! = 4∙3∙2∙1 = 24

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Page 75: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Permutation of n Objects

The objective of a 9 x 9 Sudoku number puzzle is to fill the grid so that each row, each column, and each 3 x 3 grid contain the digits 1 to 9. How many different ways can the first row of a blank 9 x 9 Sudoku grid be filled?

75

Solution:The number of permutations is

9!= 9∙8∙7∙6∙5∙4∙3∙2∙1 = 362,880 ways

Page 76: Chapter 3 Probability 1. 2  Introduction to Probability

Permutations

Permutation of n objects taken r at a time

• The number of different permutations of n distinct objects taken r at a time

76

!

( )!n r

nP

n r

■ where r ≤ n

Page 77: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding nPr

Find the number of ways of forming three-digit codes in which no digit is repeated.

77

Solution:• You need to select 3 digits from a group of 10• n = 10, r = 3

10 3

10! 10!

(10 3)! 7!

10 9 8 7 6 5 4 3 2 1

7 6 5 4 3 2 1

720 ways

P

Page 78: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding nPr

Forty-three race cars started the 2007 Daytona 500. How many ways can the cars finish first, second, and third?

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Solution:• You need to select 3 cars from a group of 43• n = 43, r = 3

43 3

43! 43!

(43 3)! 40!

43 42 41

74,046 ways

P

Page 79: Chapter 3 Probability 1. 2  Introduction to Probability

Distinguishable Permutations

Distinguishable Permutations

• The number of distinguishable permutations of n objects where n1 are of one type, n2 are of another type, and so on

79

1 2 3

!

! ! ! !k

n

n n n n ■

where n1 + n2 + n3 +∙∙∙+ nk = n

Page 80: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Distinguishable Permutations

A building contractor is planning to develop a subdivision that consists of 6 one-story houses, 4 two-story houses, and 2 split-level houses. In how many distinguishable ways can the houses be arranged?

80

Solution:• There are 12 houses in the subdivision• n = 12, n1 = 6, n2 = 4, n3 = 2

12!

6! 4! 2!13,860 distinguishable ways

Page 81: Chapter 3 Probability 1. 2  Introduction to Probability

Combinations

Combination of n objects taken r at a time

• A selection of r objects from a group of n objects without regard to order

81

!

( )! !n r

nC

n r r

Page 82: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Combinations

A state’s department of transportation plans to develop a new section of interstate highway and receives 16 bids for the project. The state plans to hire four of the bidding companies. How many different combinations of four companies can be selected from the 16 bidding companies?

82

Solution:• You need to select 4 companies from a group of 16• n = 16, r = 4• Order is not important

Page 83: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Combinations

83

16 4

16!

(16 4)!4!

16!

12!4!16 15 14 13 12!

12! 4 3 2 11820 different combinations

C

Page 84: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Probabilities

A student advisory board consists of 17 members. Three members serve as the board’s chair, secretary, and webmaster. Each member is equally likely to serve any of the positions. What is the probability of selecting at random the three members that hold each position?

84

Page 85: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Probabilities

• There is only one favorable outcome

• There are

ways the three positions can be filled

85

17 3

17!

(17 3)!

17!17 16 15 4080

14!

P

1( 3 ) 0.0002

4080P selecting the members

Page 86: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Probabilities

You have 11 letters consisting of one M, four Is, four Ss, and two Ps. If the letters are randomly arranged in order, what is the probability that the arrangement spells the word Mississippi?

86

Page 87: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Probabilities

• There is only one favorable outcome

• There are

distinguishable permutations of the given letters

87

11!34,650

1! 4! 4! 2!

1( ) 0.000029

34650P Mississippi

11 letters with 1,4,4, and 2 like letters

Page 88: Chapter 3 Probability 1. 2  Introduction to Probability

Example: Finding Probabilities

A food manufacturer is analyzing a sample of 400 corn kernels for the presence of a toxin. In this sample, three kernels have dangerously high levels of the toxin. If four kernels are randomly selected from the sample, what is the probability that exactly one kernel contains a dangerously high level of the toxin?

88

Page 89: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Probabilities

• The possible number of ways of choosing one toxic kernel out of three toxic kernels is

3C1 = 3

• The possible number of ways of choosing three nontoxic kernels from 397 nontoxic kernels is

397C3 = 10,349,790

• Using the Multiplication Rule, the number of ways of choosing one toxic kernel and three nontoxic kernels is

3C1 ∙ 397C3 = 3 ∙ 10,349,790 3 = 31,049,37089

Page 90: Chapter 3 Probability 1. 2  Introduction to Probability

Solution: Finding Probabilities

• The number of possible ways of choosing 4 kernels from 400 kernels is

400C4 = 1,050,739,900

• The probability of selecting exactly 1 toxic kernel is

90

3 1 397 3

400 4

(1 )

31,049,3700.0296

1,050,739,900

C CP toxic kernel

C

Page 91: Chapter 3 Probability 1. 2  Introduction to Probability

Section 3.4 Summary

• Determined the number of ways a group of objects can be arranged in order

• Determined the number of ways to choose several objects from a group without regard to order

• Used the counting principles to find probabilities

91

Page 92: Chapter 3 Probability 1. 2  Introduction to Probability
Page 93: Chapter 3 Probability 1. 2  Introduction to Probability

Barnett/Ziegler/Byleen Finite Mathematics 12e

Page 94: Chapter 3 Probability 1. 2  Introduction to Probability

Three kings!

Page 95: Chapter 3 Probability 1. 2  Introduction to Probability

95

Probability of an Earlier Event Given a Later Event

A survey of middle-aged men reveals that 28% of them are balding at the crown of their head. Moreover, it is known that such men have an 18% probability of suffering a heart attack in the next 10 years. Men who are not balding in this way have an 11% probability of a heart attack. If a middle-aged man is randomly chosen, what is the probability that he is balding, given that he suffered a heart attack? See tree diagram on next slide.

Page 96: Chapter 3 Probability 1. 2  Introduction to Probability

96

Tree Diagram

We want P(B|H) = probability that he is balding, given that he suffered a heart attack.

Balding 0.28

Not balding 0.72

heart attack 0.18

heart attack 0.11

no heart attack 0.82

no heart attack 0.89

Page 97: Chapter 3 Probability 1. 2  Introduction to Probability

97

Derivation of Bayes’ Formula

( )( )

( )

p B HP

pH

HB

( ) (( ) )p B H p NBp HH

( ) ( ) ( ) ( ) ( )p H p B p H B p NB p H NB

( ) ( ) ( ) (

( )( )

)

p B HP B H

p B p H B p NB p H NB

Page 98: Chapter 3 Probability 1. 2  Introduction to Probability

98

Solution of Problem

( )

( ) ( )(

))

( ( ) ( )p B p H

p B p H B

B p NB pp H

H NBB

( ) ( ) ( ) (

( )( )

)

p B HP B H

p B p H B p NB p H NB

0.28(0.18)

0.28 (0.

0.72

18)(

(0)

.11)p B H = 0.389

Page 99: Chapter 3 Probability 1. 2  Introduction to Probability

99

Another Method of Solution

Balding 0.28

Not balding 0.72

heart attack 0.18

heart attack 0.11

no heart attack 0.82

no heart attack 0.89

(0.28)(0.18)=0.0504

(0.72)(0.11)=0.0792

Another way to look at this problem is that we know the person had a heart attack, so he has followed one of the two red paths. The sample space has been reduced to these paths. So theprobability that he is balding is

( )

( ) ( )

P B H

P B H P NB H

which is 0.0504/(0.0504 + 0.0792) = 0.389

Page 100: Chapter 3 Probability 1. 2  Introduction to Probability

100

Summary of Tree Method of Solution

You do not need to memorize Bayes’ formula. In practice, it is usually easier to draw a probability tree and use the following:

Let U1, U2,…Un be n mutually exclusive events whose union is the sample space S. Let E be an arbitrary event in S such that P(E) 0. Then

product of branch probabilities leading to E through U1

P(U1|E) = sum of all branch products leading to E

Similar results hold for U2, U3,…Un

Page 101: Chapter 3 Probability 1. 2  Introduction to Probability

Slide 4- 101© 2012 Pearson Education, Inc.

Elementary Statistics:

Picturing the World

Fifth Edition

by Larson and Farber

Chapter 3: Probability

Page 102: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

How many 4-letter television call signs are possible, if each sign must start with either a K or a W?

A. 456,976

B. 35,152

C. 16

D. 104

Slide 3- 102

Page 103: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

How many 4-letter television call signs are possible, if each sign must start with either a K or a W?

A. 456,976

B. 35,152

C. 16

D. 104

Slide 3- 103

1.26.26.26 1.26.26.26 35152

Page 104: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The spinner shown is spun one time. Find the probability the spinner lands on blue.

A. 0.375

B. 0.5

C. 0.125

D. 0.25

Slide 3- 104

Page 105: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The spinner shown is spun one time. Find the probability the spinner lands on blue.

A. 0.375

B. 0.5

C. 0.125

D. 0.25

Slide 3- 105

Page 106: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The bar graph shows the cell phone provider for students in a class. One of these students is chosen at random. Find the probability that their provider is not AT&T.

A. 0.3

B. 0.6

C. 0.125

D. 0.4

Slide 3- 106

Page 107: Chapter 3 Probability 1. 2  Introduction to Probability

The bar graph shows the cell phone provider for students in a class. One of these students is chosen at random. Find the probability that their provider is not AT&T.

A. 0.3

B. 0.6

C. 0.125

D. 0.4

Page 108: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

One card is selected at random from a standard deck, then replaced, and a second card is drawn. Find the probability of selecting two face cards.

A. 0.050

B. 0.053

C. 0.038

D. 0.462

Slide 3- 108

Page 109: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

One card is selected at random from a standard deck, then replaced, and a second card is drawn. Find the probability of selecting two face cards.

A. 0.050

B. 0.053

C. 0.038

D. 0.462

Slide 3- 109

Page 110: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

One card is selected at random from a standard deck, not replaced, and then a second card is drawn. Find the probability of selecting two face cards.

A. 0.050

B. 0.053

C. 0.446

D. 0.038

Slide 3- 110

Page 111: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

One card is selected at random from a standard deck, not replaced, and then a second card is drawn. Find the probability of selecting two face cards.

A. 0.050

B. 0.053

C. 0.446

D. 0.038

Slide 3- 111

Page 112: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is male, given that they prefer Onion.

A. 0.333

B. 0.6

C. 0.208

D. 0.556

Slide 3- 112

Tomato Onion Mushroom Total

Male 8 5 2 15

Female 2 4 3 9

Total 10 9 5 24

Page 113: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is male, given that they prefer Onion.

A. 0.333

B. 0.6

C. 0.208

D. 0.556

Slide 3- 113

5( ) 524( | ) 0.556

9( ) 924

p m op m o

p o

Tomato Onion Mushroom Total

Male 8 5 2 15

Female 2 4 3 9

Total 10 9 5 24

Page 114: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

True or False:

The following events are mutually exclusive.

Event A: Being born in California Event B: Watching American Idol

A. True

B. False

Slide 3- 114

Page 115: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

True or False:

The following events are mutually exclusive.

Event A: Being born in California Event B: Watching American Idol

A. True

B. False

Slide 3- 115

They CAN happen at the same time!

Page 116: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefers mushroom.

A. 0.458

B. 0.583

C. 0.125

D. 0.556

Slide 3- 116

Tomato Onion Mushroom Total

Male 8 5 2 15

Female 2 4 3 9

Total 10 9 5 24

Page 117: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

The table shows the favorite pizza topping for a sample of students. One of these students is selected at random. Find the probability the student is female or prefersm mushroom.

A. 0.458

B. 0.583

C. 0.125

D. 0.556

Slide 3- 117

( ) ( ) ( ) ( )

9 5 3 110.458

24 24 24 24

p f msh p f p msh p f msh

Tomato Onion Mushroom Total

Male 8 5 2 15

Female 2 4 3 9

Total 10 9 5 24

Page 118: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded?

A. 45

B. 455

C. 2,730

D. 3,375

Slide 3- 118

Page 119: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

There are 15 dogs entered in a show. How many ways can first, second, and third place be awarded?

A. 45

B. 455

C. 2,730

D. 3,375

Slide 3- 119

Page 120: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

There are 13 students in a club. How many ways can four students be selected to attend a conference?

A. 17,160

B. 52

C. 28,561

D. 715

Slide 3- 120

Page 121: Chapter 3 Probability 1. 2  Introduction to Probability

© 2012 Pearson Education, Inc.

There are 13 students in a club. How many ways can four students be selected to attend a conference?

A. 17,160

B. 52

C. 28,561

D. 715

Slide 3- 121

Page 122: Chapter 3 Probability 1. 2  Introduction to Probability

122

How many Ways 8 people can sit together around a round circular table?

A. 7053

B. 5040

C. 64

D. 8!

Page 123: Chapter 3 Probability 1. 2  Introduction to Probability

123

Total arrangements:The number of ways to arrange n elements in a circle = (n-1)!.Thus, the total number of ways to arrange the 8 people = (8-1)! = 7! = 5040.

How many Ways 8 people can sit together around a round circular table?

A. 7053

B. 5040

C. 64

D. 8!

! number of arrangments= ( 1)!

Fix the first position:

A

1*7*6*5*4*3*2*1 ( 1)!

nTotal n

n

n