© The McGraw-Hill Companies, Inc., 2000 5-1 Chapter 5 Probability.

© The McGraw-Hill Companies, Inc., 2000

5-15-1

Chapter 5Chapter 5

Probability Probability


5-25-2 OutlineOutline

5-1 Introduction 5-2 Sample Spaces and Probability 5-3 The Addition Rules for

Probability 5-4 The Multiplication Rules and

Conditional Probability


5-35-3 ObjectivesObjectives

Determine Sample Spaces and find the probability of an event using classical probability.

Find the probability of an event using empirical probability.

Find the probability of compound events using the addition rules.


5-45-4 ObjectivesObjectives

Find the probability of compound events using the multiplication rules.

Find the conditional probability of an event.


5-55-5 5-2 Sample Spaces and Probability5-2 Sample Spaces and Probability

A probability experimentprobability experiment is a process that leads to well-defined results called outcomes.

An outcomeoutcome is the result of a single trial of a probability experiment.

NOTE:NOTE: A tree diagram can be used as a systematic way to find all possible outcomes of a probability experiment.


5-65-65-2 Tree Diagram for Tossing Two Coins5-2 Tree Diagram for Tossing Two Coins

First Toss

H

T

H

T H

T

Second Toss


5-75-7 5-2 Sample Spaces -5-2 Sample Spaces - Examples

EXPERIMENT SAMPLE SPACE

Toss one coin H, T

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

Answer a true-false question

True, False

Toss two coins HH, HT, TH, TT


5-85-85-2 Formula for Classical 5-2 Formula for Classical

ProbabilityProbability

Classical probability assumes that all outcomes in the sample space are equally likely to occur.

That is, equally likelyequally likely events are events that have the same probability of occurring.


5-95-95-2 Formula for Classical 5-2 Formula for Classical


( ) = ( )

( )

,

.

The probability of any event E is

number of outcomes in E

total number of outcomes in the sample space

This probability is denoted by

P En E

n S

This probability is called classical probability

and it uses the sample space S

.

.


5-105-10 5-2 Classical Probability -5-2 Classical Probability - Examples

For a card drawn from an ordinary deck, find the probability of getting (a) a queen (b) a 6 of clubs (c) a 3 or a diamond.

Solution:Solution: (a) Since there are 4 queens and 52 cards, P(queen) = 4/52 = 1/13P(queen) = 4/52 = 1/13.

(b) Since there is only one 6 of clubs, then P(6 of clubs) = 1/52P(6 of clubs) = 1/52.



(c) There are four 3s and 13 diamonds, but the 3 of diamonds is counted twice in the listing. Hence there are only 16 possibilities of drawing a 3 or a diamond, thus P(3 or diamond) = 16/52 = 4/13P(3 or diamond) = 16/52 = 4/13.



When a single die is rolled, find the probability of getting a 9.

Solution:Solution: Since the sample space is 1, 2, 3, 4, 5, and 6, it is impossible to get a 9. Hence, P(9) = 0/6 = 0P(9) = 0/6 = 0.

NOTE:NOTE: The sum of the probabilities of all outcomes in a sample space is one.


5-135-13 5-2 Complement of an Event5-2 Complement of an Event

T h e c o m p le m e n t o f a n e v e n t E is th e se t o f o u tc o m e s in th e

sa m p le sp a c e th a t a re n o t in c lu d e d in th e o u tc o m e s

o f e v e n t E T h e c o m p le m e n t o f E is d e n o te d b y E E b a r

. ( ) .

E

E


5-145-145-2 Complement of an Event - 5-2 Complement of an Event - Example

Find the complement of each event. Rolling a die and getting a 4. Solution:Solution: Getting a 1, 2, 3, 5, or 6. Selecting a letter of the alphabet and

getting a vowel. Solution:Solution: Getting a consonant

(assume y is a consonant).


5-155-155-2 Complement of an Event -5-2 Complement of an Event - Example

Selecting a day of the week and getting a weekday.

Solution:Solution: Getting Saturday or Sunday.

Selecting a one-child family and getting a boy.

Solution:Solution: Getting a girl.


5-165-16 5-2 Rule for Complementary Event5-2 Rule for Complementary Event

P E P E

or

P E P E

or

P E P E

( ) ( ) 1

1

1

( ) = ( )

( ) + ( ) = .


5-175-17 5-2 Empirical Probability5-2 Empirical Probability

The difference between classical and empirical probabilityempirical probability is that classical probability assumes that certain outcomes are equally likely while empirical probability relies on actual experience to determine the probability of an outcome.


5-185-185-2 Formula for Empirical 5-2 Formula for Empirical


Given a frequency distribution

the probability of an event being

in a given class is

P Efrequency for the class

total frequencies in the distribution

f

nThis probability is called the empirical

probability and is based on observation

,

( ) =

.

.


5-195-195-2 Empirical Probability -5-2 Empirical Probability -

Example

In a sample of 50 people, 21 had type O blood, 22 had type A blood, 5 had type B blood, and 2 had AB blood. Set up a frequency distribution.



Example

Type Frequency

ABABO

225221

50 = n



Example

Find the following probabilities for the previous example.

A person has type O blood. Solution:Solution: P(O) = f /n = 21/50. A person has type A or type B blood. Solution:Solution: P(A or B) = 22/50+ 5/50

= 27/50.


5-225-225-3 The Addition Rules for 5-3 The Addition Rules for ProbabilityProbability

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


5-235-235-3 The Addition Rules for 5-3 The Addition Rules for ProbabilityProbability

A B

A and B are mutually exclusive


5-245-24 5-3 Addition Rule 15-3 Addition Rule 1

When two events A and B are mutually exclusive, the probabilitythat A or B will occur is

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


5-255-25 5-3 Addition Rule 1-5-3 Addition Rule 1- Example

At a political rally, there are 20 Republicans (R), 13 Democrats (D), and 6 Independents (I). If a person is selected, find the probability that he or she is either a Democrat or an Independent.

Solution:Solution: P(D or I) = P(D) + P(I) = 13/39 + 6/39 = 19/39.



A day of the week is selected at random. Find the probability that it is a weekend.

Solution:Solution: P(Saturday or Sunday) = P(Saturday) + P(Sunday) = 1/7 + 1/7 = 2/7.



When two events A and B

are not mutually exclusive, the

probabilityy that A or B will

occur is

P A or B P A P B P A and B

( ) ( ) ( ) ( )



A B

A and B (common portion)



In a hospital unit there are eight nurses and five physicians. Seven nurses and three physicians are females. If a staff person is selected, find the probability that the subject is a nurse or a male.

The next slide has the data.


5-305-30 5-3 Addition Rule 2 -5-3 Addition Rule 2 - Example

STAFF FEMALES MALES TOTAL

NURSES 7 1 8

PHYSICIANS 3 2 5

TOTAL 10 3 13

STAFF FEMALES MALES TOTAL

NURSES 7 1 8

PHYSICIANS 3 2 5

TOTAL 10 3 13



Solution:Solution: P(nurse or male) = P(nurse) + P(male) –

P(male nurse) = 8/13 + 3/13 – 1/13 = 10/13.



On New Year’s Eve, the probability that a person driving while intoxicated is 0.32, the probability of a person having a driving accident is 0.09, and the probability of a person having a driving accident while intoxicated is 0.06. What is the probability of a person driving while intoxicated or having a driving accident?



Solution:Solution: P(intoxicated or accident)

= P(intoxicated) + P(accident) – P(intoxicated and accident) = 0.32 + 0.09 – 0.06 = 0.35.


5-345-34

Two events A and B are independentindependent if the fact that A occurs does not affect the probability of B occurring.

Example:Example: Rolling a die and getting a 6, and then rolling another die and getting a 3 are independent events.

5-4 The Multiplication Rules and 5-4 The Multiplication Rules and Conditional Probability Conditional Probability


5-355-35 5-4 Multiplication Rule 15-4 Multiplication Rule 1


are independent the

probability of both

occurring is

P A and B P A P B

,

( ) ( ) ( ).


5-365-365-4 Multiplication Rule 1 -5-4 Multiplication Rule 1 - Example

A card is drawn from a deck and replaced; then a second card is drawn. Find the probability of getting a queen and then an ace.

Solution:Solution: Because these two events are independent (why?), P(queen and ace) = (4/52)(4/52) = 16/2704 = 1/169.



A Harris pole found that 46% of Americans say they suffer great stress at least once a week. If three people are selected at random, find the probability that all three will say that they suffer stress at least once a week.

Solution:Solution: Let S denote stress. Then P(S and S and S) = (0.46)3 = 0.097.



The probability that a specific medical test will show positive is 0.32. If four people are tested, find the probability that all four will show positive.

Solution:Solution: Let T denote a positive test result. Then P(T and T and T and T) = (0.32)4 = 0.010.


5-395-39

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.

Example:Example: Having high grades and getting a scholarship are dependent events.



5-405-40

The conditional probabilityconditional probability of an event B in relationship to an event A is the probability that an event B occurs after event A has already occurred.

The notation for the conditional probability of B given A is P(B|A).

NOTE:NOTE: This does not mean B A.



5-415-41 5-4 Multiplication Rule 25-4 Multiplication Rule 2


are dependent the

probability of both

occurring is

P A and B P A P B A

,

( ) ( ) ( | ).


5-425-42

In a shipment of 25 microwave ovens, two are defective. If two ovens are randomly selected and tested, find the probability that both are defective if the first one is not replaced after it has been tested.

Solution:Solution: See next slide.

5-4 The Multiplication Rules and 5-4 The Multiplication Rules and Conditional Probability - Conditional Probability - Example


5-435-43

Solution:Solution: Since the events are dependent, P(D1 and D2) = P(D1)P(D2| D1) = (2/25)(1/24) = 2/600 = 1/300.



5-445-44

The WW Insurance Company found that 53% of the residents of a city had homeowner’s insurance with its company. Of these clients, 27% also had automobile insurance with the company. If a resident is selected at random, find the probability that the resident has both homeowner’s and automobile insurance.



5-455-45

Solution:Solution: Since the events are dependent, P(H and A) = P(H)P(A|H) = (0.53)(0.27) = 0.1431.



5-465-46

Box 1 contains two red balls and one blue ball. Box 2 contains three blue balls and one red ball. A coin is tossed. If it falls heads up, box 1 is selected and a ball is drawn. If it falls tails up, box 2 is selected and a ball is drawn. Find the probability of selecting a red ball.



5-475-47

5-4 Tree Diagram for 5-4 Tree Diagram for Example

P(B1) 1/2

Red

Red

Blue

Blue

Box 1

P(B2) 1/2Box 2

P(R|B1) 2/3

P(B|B1) 1/3

P(R|B2) 1/4

P(B|B2) 3/4

(1/2)(2/3)

(1/2)(1/3)

(1/2)(1/4)

(1/2)(3/4)


5-485-48

Solution:Solution: P(red) = (1/2)(2/3) + (1/2)(1/4) = 2/6 + 1/8 = 8/24 + 3/24 = 11/24.



5-495-495-4 Conditional Probability -5-4 Conditional Probability -

FormulaFormula

.

( | ) =( )

( )

The probability that the event B occurs

given that the first event A has occurred can be

found by dividing the probability that both events

occurred by the probability that the first event has

occurred The formula is

P B AP A and B

P A

second

.


5-505-50

The probability that Sam parks in a no-parking zone and gets a parking ticket is 0.06, and the probability that Sam cannot find a legal parking space and has to park in the no-parking zone is 0.2. On Tuesday, Sam arrives at school and has to park in a no-parking zone. Find the probability that he will get a ticket.

5-4 Conditional Probability -5-4 Conditional Probability - Example


5-515-51

Solution:Solution: Let N = parking in a no-parking zone and T = getting a ticket.

Then P(T |N) = [P(N and T) ]/P(N) = 0.06/0.2 = 0.30.



5-525-52

A recent survey asked 100 people if they thought women in the armed forces should be permitted to participate in combat. The results are shown in the table on the next slide.



5-535-535-4 Conditional Probability -5-4 Conditional Probability -

Example

Gender Yes No Total

Male 32 18 50

Female 8 42 50

Total 40 60 100

Gender Yes No Total

Male 32 18 50

Female 8 42 50

Total 40 60 100


5-545-54

Find the probability that the respondent answered “yes” given that the respondent was a female.

Solution:Solution: Let M = respondent was a male; F = respondent was a female; Y = respondent answered “yes”; N = respondent answered “no”.



5-555-55

P(Y|F) = [P( F and Y) ]/P(F) = [8/100]/[50/100] = 4/25.

Find the probability that the respondent was a male, given that the respondent answered “no”.

Solution: P(M|N) = [P(N and M)]/P(N) = [18/100]/[60/100] = 3/10.


© The McGraw-Hill Companies, Inc., 2000 5-1 Chapter 5 Probability.

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