section 4.3-1 copyright © 2014, 2012, 2010 pearson education, inc. lecture slides elementary...

Section 4.3-1Copyright © 2014, 2012, 2010 Pearson Education, Inc.

Lecture Slides

Elementary Statistics Twelfth Edition

and the Triola Statistics Series

by Mario F. Triola


Chapter 4Probability

4-1 Review and Preview

4-2 Basic Concepts of Probability

4-3 Addition Rule

4-4 Multiplication Rule: Basics

4-5 Multiplication Rule: Complements and Conditional Probability


Experiments, Outcomes, and Sample Spaces

3

Example #3:Draw a tree diagram for three tosses of a coin. List all outcomes for this experiment in a sample space S.

Solution:Let “H” represent head and “T” represent tail.Therefore, for each experiment, the outcome is either a “H” or “T”.

H

T

H

T

H

T

HHH

1st Selectio

n

2nd Selectio

n

H

T

HT

T

T

H

H

HHT

HTH

HTT

THH

THTTTH

TTT

3rd Selectio

n

S = {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}


Simple and Compound Events

a. Event is a collection of one or more of the outcomes of an experiment. An event could be the entire or portion of a sample space. Therefore, an event can be classified as:

Simple event Compound event

b. A simple event, Ei, consists of one and only one of the final outcomes of an experiment. In Example #3,

c. A compound event consists of more than one outcome of an experiment. It is represented by

A, B, C, D,..., or A1, A2, A3,..., B1, B2, B3,,...

Reconsider Example #3, let A be the event that two of the three tosses will result in heads. Then, event A is given by

4

E1=(HHH), E2=(HHT), E3=(HTH), E4=(HTT), E5=(THH),

E6=(THT), E7=(TTH), and E8=(TTT)

A = {HHH, HHT, HTH, THH}


Simple and Compound Events

5

Example #4: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. 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.

Solution:Let,

D = a defective partG = a good part

The experiment has the following outcomes:DD = both parts are defectiveDG = the 1st part is defective and the 2nd is goodGG = both parts are goodGD = the 1st part is good and the 2nd is defective

a)At least one part is good = {DG, GG, GD} compound event

b)Exactly one part is defective = {DG, GD} compound event

c)The 1st is good and 2nd defective = {GD} simple event

d)At most one part is good = {DD, DG, GD} compound event.


MARGINAL AND CONDITIONAL PROBABILITIES

The following table gives the responses of 2,000 randomly selected adults who were asked whether or not they have shopped on internet.

6

Have shopped Have never shopped

Male 500 700

Female 300 500

Discussion1. The table is a two-way classification of 2,000 adults.2. The table is called contingency table and each box with a

numeric entry is called a cell.3. Each cell gives the frequency of two characteristics:

a) Gender (male or female) andb) Opinion (have shopped or have never shopped).


Marginal and Conditional Probabilities

7

Discussion3. By adding the row totals and column totals, we obtain a new

table.

4. If only one characteristic, “have shopped”, “have never shopped”, “male”, or “female”, is being considered at a time, the probability of each event is called marginal probability or simple probability.

Have shoppedHave never

shoppedTotal

Male 500 700 1200

Female 300 500 800

Total 800 1200 2000



8

The marginal probability or simple probability is a probability of a single event without consideration of any other event.From the table, the marginal probabilities of the characteristics are as follows:

Have shoppedHave never

shoppedTotal

Male 500 700 1200

Female 300 500 800

Total 800 1200 2000

# of adults who have shopped 800 # of males 1200(has shopped) 0.4 (male) 0.6

total number of adults 2000 total number of adults 2000

# of adults who have never shopped(has never shopped)

total number of

P P

P

12000.6

adults 2000# of females 800

(Female) 0.4total number of adults 2000

P



9

Now suppose we want to find the probability that the randomly selected adult has shopped on the internet, assuming that the adult is female. In other words, the event that the adult is female has already occurred. This probability is called conditional probability, and it is written,

P(has shopped | female) and is read as

The probability that the selected adult has shopped on the internet given that the event “female” has already occurred.


10


General Statement Suppose A and B are two events, then the conditional probability of A given B is written as ,

P(A|B).Again from the contingent table,

Number of males who have never shopped 700

P Male | has never shopped 0.58Total number of adults who have never shopped 1200


MUTUALLY EXCLUSIVE EVENTS

Definition Mutually exclusive events are events that do not have any outcome in

common.

Ex. Events for rolling a die: A = an even number is observed B = an odd number is observe C = a number less than five is observed

Mutually Exclusive Mutually nonexclusive event

Most importantly, the occurrence of one event prevents the occurrence of the other mutually exclusive events.

12


Mutually Exclusive Events

• For example, the outcomes of tossing a coin are mutually exclusive because both Head and Tail outcomes could not occur at the same time. The occurrence of Head prevents occurrence of Tail to occur.

13

H

T

H

T

H

TTT

TH

HT

HH


Mutually Exclusive Events

Example #13

14

Solution:The experiment involves tossing a coin twice. The sample space S = {HH, HT, TT, TH} where H = Head and T = Tail. The events are: A = {HH, HT, TH} & B = {TT}A and B are mutually exclusive. They do not have any outcome in common.

Are A and B mutually exclusive? Explain why or why not.

Define the following two events for two tosses of a coin:A = at least one head

B = both tails are obtained

Example #12 There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” mutually exclusive? Explain why or why not.

SolutionPed Non-Ped Totals

Male 5 80 85

Female 20 55 75

The events “female” and “pediatrician” are not mutually exclusive because a physician could be a female and a pediatrician as shown above.


INDEPENDENT VERSUS DEPENDENT EVENTS

15

Definition 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 if

either P(A | B) = P(A) or P(B | A) = P(B).

If P(A | B) = P(A) is true, then P(B | A) = P(B) is also true. If P(A | B) = P(A) is false, then P(B | A) = P(B) is also false.

If the occurrence of one event affects the probability of the other,

then we say that the events are dependent. In other words, two events are dependent if

either P(A | B) ≠ P(A) or P(B | A) ≠ P(B).


Independent Versus Dependent Events

16

General Statement 1. Two events are either mutually exclusive or independent.

a. Mutually exclusive events are dependentb. Independent events are never mutually exclusive

2. Dependent events may or may not be mutually exclusive.

Example #14There are 160 practicing physicians in a city. Of them, 75 are female and 25 are pediatricians. Of the 75 female, 20 are pediatricians. Are the events “female” and “pediatrician” independent? Explain why or why not.

Solution

Ped Non-Ped Totals

Male 5 80 85

Female 20 55 75

Totals 25 135 160

75P female 0.47

16020

P female | Pediatrician 0.8025

Since P(female | Pediatrician) ≠ P(female), then the events are not independent.



17

Example #15:Two donut bakers baked 1000 donut holes. Baker A baked 600 donuts, of which 450 were sold and the remaining were discarded. Baker B baked the remaining donuts, of which 100 were discarded. The events are “Baker A”, “Baker B”, “sold donuts”, and “discarded donuts”. Prepare a contingent table for this experiment. Are the events “Baker A” and “sold donuts” independent? Explain why or why not.

Solution

Sold Donut

Discarded Donuts

Totals

Baker A 450 150 600

Baker B 300 100 400

Totals 750 250 1000

600P Baker A 0.60

1000450

P Baker A| Sold Donuts 0.60750

Since P(Baker A | sold donut) = P(Baker A), then the events are independent because the occurrence of event “sold donuts” does affect the probability of event “Baker A”.

S

250/1000

A

B

A

B

S | A

D750/1000

450/750

300/750

150/250

100/250

S | B

D | A

D | B


18


Example #15.1:A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15}

a. Are events A and B mutually exclusive?

b. Are event A and B independent events?

Because the two probabilities are not the same, the two events are not independent. Also, we know that mutually exclusive events are always dependent.

yes

0)/(4.010

4)( BAPAP


COMPLEMENTARY EVENTS

Definition

1. The complement of event A, denoted by Ā and is read as “A bar” or “A complement,” is the event that includes all the outcomes for an experiment that are not in A.

2. Therefore, complementary events are always mutually exclusive.

3. Two complementary events, combined together, includes all the outcomes of the experiment.

19

P A P(A) 1 P A 1 P(A)

and

P A 1 P(A)


Complementary Events

20

Example #15 – Problem

Let A be the event that a number less than 3 is obtained if we roll a die once. What is the probability of A? What is the complementary event of A, and what is its probability?

Solution

2P A 0.33

6A ={a number 3}

4P A 1 P(A) 0.67 or P A 0.67

6


21

Complementary EventsExample #15.1:

A statistical experiment has 10 equally likely outcomes that are denoted by 10, 11, 12, 13, 14, 15, 16, 17, 18, 19. Let event A = { 10, 12, 14, 16} and event B = {11, 13, 15}

Solution

What are the complements of event A and B, respectively, and their probabilities?

10

6)(},19,18,17,15,13,11{ APA

7.10

7)(},19,18,17,16,14,12,10{ BPB


INTERSECTION OF EVENTS AND THE MULTIPLICATION RULE

Suppose an experiment resulted in a sample space described as,

S = {1, 2, 3, 4, 5, 6, 7, 8}

Also, three events from the experiment are define as consisting of the following outcomes:

A = {1, 2, 3, 4}

B = {3, 4, 5, 6}

C = {5, 6, 7, 8}

22

1. You can see that the Events A and B are not mutually exclusive because they have two common outcomes, 3 and 4.

2. Likewise, Events B and C are not mutually exclusive because they have two outcomes, 5 and 6, in common.

3. Given two Events, A and B, we can say that the intersection of Events A and B is the collection of all outcomes that are common to both A and B. It can be written as,

A and BA B, or simply AB


Multiplication Rule

23

The probability of the intersection of Events A and B is called the joint probability and is define as the product of the marginal and conditional probabilities. Joint probability is written as,

P(A B) = P(A) P(B|A) or P(A B) = P(B) P(A|B)

From the two formulas, we can calculate the conditional probabilities as,

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

P(A) P(B)


Multiplication Rule for Independent Events

24

Recall, P(A B) = P(A) P(B|A) or P(A B) = P(B) P(A|B)

This is true for dependent events if the probability of one is affected by the occurrence of the other event. In other words,

P(A|B) ≠ P(A) and P(B|A)≠ P(B).

For independent events, the occurrence of one event does not affect the probability of the other. Therefore,

P(A|B) = P(A) and P(B|A) = P(B).

Hence, we can rewrite the formula for calculating probability of intersection of two independent events as,

P(A B) = P(A) P(B) Note: You can extend the multiplication rule to calculate the joint probability of as many events as you want.


Joint Probability of Mutually Exclusive Events

25

We know from discussion of mutually exclusive events that mutually exclusive events have no common outcomes. Therefore, they do not have an intersection. In this case, we write the intersection of two or more mutually exclusive events as

P(AB) = 0

Find the joint probability of A and B for the following:a. P(B) = 0.59 and P(A|B) = 0.77b. P(A) = 0.28 and P(B|A) = 0.35

Example #17 Solution

a. P(AB)= P(B) P(A|B) = (0.59)(0.77) = 0.4543b. P(AB)= P(A) P(B|A) = (0.28)(0.35) = 0.098

Find the joint probability for the following three independent events:a. P(A) = 0.49, P(B) = 0.67, P(C) = .75b. P(A) = 0.71, P(B) = 0.34, P(C) = 0.45

Example #18 Solution

a. P(ABC)= P(A)P(B)P(C) = (.49)(.67)(.75) = 0.2462b. P(ABC)= P(A)P(B)P(C) = (.71)(.34)(.45) = 0.1086


Intersection of Events and Multiplication Rule

26

The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate.

Example #19Solution

Graduate Did not Graduate

Totals

Male 126 55 181

Female 133 32 165

Totals 259 87 346

a. If one of these players is selected at random, find the following probabilities:

i. P(female and graduate)ii. P(male and did not graduate)

b. Find P(graduate and did not graduate). Is this probability zero? If yes, why?

133P(female and graduate) = 0.3844

346P(F G) = P(F)P(G|F)

165 1330.3844

346 165

55P(male and did not graduate) = 0.1590

346P(M not G) = P(M)P( not G|M)

181 55346 1810.1590

P(graduate and did not graduate) = 0, because these events are mutually exclusive and you could not have someone that is both “a graduate” and “a non graduate”.



27

The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents.

Example #20 < High Sch (D)

High Sch (E)

> High Sch (F)

Better off (A)

140 450 420

Same as (B)

60 250 110

Worse off (C)

200 300 70

a. Suppose one adult is selected at random from these 2000 adults. Find the following probabilities:

i. P(better off and high school)ii. P(more than high school and

worse off)b. Find the joint probability of the events

“worse off” and “better off.” Is this probability zero? Explain why or why not.



28

Solution

< High Sch (D)

High Sch (E)

> High Sch (F)

Totals

Better off (A)

140 450 420 1010

Same as (B)

60 250 110 420

Worse off (C)

200 300 70 570

Totals 400 1000 600 2000

450P(A E) = 0.225

2000P(A E) = P(A)P(E|A)

1010 4500.225

2000 1010

70P(F ) = 0.035

2000P(F C) = P(F)P(C|F)

600 700.035

2000 600

C

P(C = 0Mutually exclusive events.

A


UNION OF EVENTS AND THE ADDITION RULE

Definition

Let a sample space, S, consist of all outcomes in Events A and B. Then the union of the two events is the collection of all outcomes that belong to either A or B or to both A and B. This is denoted by

A B or just A or B

29

Addition RuleAddition rule is the method for calculating the probability of the union of events. It is defined as,

P(A B) = P(A)+P(B)-P(A B)

S

BA

S

BA

Essentially, we calculate the probability of union of events by:1. Adding the probability of each event and2. Subtract the probability of the intersection of the events from result in (1).


Addition Rule for Mutually Exclusive Events

Let re-examine the formula for calculating the probability of union of events.

P(A B) = P(A)+P(B)-P(A B)

However, we have said that for mutually exclusive events,

P(A B) = 0

Then, for mutually exclusive events, the union of two events is,

P(A B) = P(A)+P(B)

30


Union of Events and the Addition Rule

31

The following table gives two way classification of all basketball players at a state university who began their college careers between 2001 and 2005, based on gender and whether or not they graduate.

Example #24Example #24 - Solution

Graduate (C)

Did not Graduate (D)

Totals

Male (A)

126 55 181

Female (B)

133 32 165

Totals 259 87 346If one of these players is selected at random, find the following probabilities:a. P(female or did not graduate)b. P(graduate or male)

P(B D) = P(B) + P(D) - P(B )165 87 32

0.6358346 346 346

D

P(C ) = P(C) + P(A) - P(C )259 181 126

0.9075346 346 346

A A



32

The following table gives two way classification of the responses based on the education levels of the persons included in the survey and whether they are financially better off, the same as, or worse off than their parents.

Example #25 < High Sch (D)

High Sch (E)

> High Sch (F)

Better off (A)

140 450 420

Same as (B)

60 250 110

Worse off (C)

200 300 70

Suppose one adult is selected at random from these 2000 adults. Find the following probabilities:i. P(better off or high school)ii. P(more than high school or worse off)iii. P(better off or worse off)



33

Solution

< High Sch (D)

High Sch (E)

> High Sch (F)

Totals

Better off (A)

140 450 420 1010

Same as (B)

60 250 110 420

Worse off (C)

200 300 70 570

Totals 400 1000 600 2000

a. P(A E) = P(A) + P(E) - P(A E) 1010 1000 450

= 0.782000 2000 2000

b. P(F ) = P(F) + P(C) - P(F ) 600 570 70

= 0.552000 2000 2000

C C

c. P(A ) = P(A) + P(C) 1010 570

= 0.792000 2000

C



34

The probability of a student getting an A grade in an economics class is 0.24 and that of getting a B grade is 0.28. What is the probability that a randomly selected student from this class will get an A or a B in this class? Explain why the probability is not equal to 1.0.

Example #26

P(A) = 0.24P(B) = 0.28

Then, the probability of getting an A or B is,

P(A or B) = P(A) + P(B) = 0.24 + 0.28 = 0.52

The probability is not equal to 1.0 because the student can get a grade of C, D, or F.

Solution

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