Chapter 6wellsmat.startlogic.com/sitebuildercontent/sitebuilderfiles/apstat... · Example –Page...

Chapter 6

Probability

The Study of Randomness

Lesson 6-1

The Idea of Probability

Randomness

Behavior is random if, while individual outcomes are uncertain, for a large number of repetitions, outcomes are regularly distributed.

Example: If I roll a die once, I can’t predict with certainty what number it will land on, but if I roll sixty times, I can expect it to land on 1 ten times, 2 ten times, 3 ten times, etc.

Probability Models

The probability of an outcome is the proportion of times the outcome would occur for a large number of repetitions.

Example: The probability of a die landing on 4 is the proportion of times a die lands on 4 for a large number of repetitions.

Chance Behavior is unpredictable in the short run but has a regular and predictable pattern in the long run.

Tossing a Coin

Example – Page 335, #6.10

A recent opinion poll showed that about 73% of marriedwomen agree that their husbands do at least their fair shareof household chores. Suppose that this is exactly true. Choosing a married woman at random that has probability0.73 of getting one who agrees that her husband does hisshare.

A). Simulate drawing 20 women, then 80 women, then320. What proportion agree in each case? We expect(but because of chance variation we can’t be sure) thatproportion will be closer to 0.73 in longer runs of trials.


A). Use Randint and let 0 – Agree and 1 – disagree.

12 women disagreed and8 women agreed

80.40

20


A). Use Randint and let 0 – Agree and 1 – disagree.


440.55

80

Sample 80 Sample 320


1530.49

320


B). Simulate drawing 20 women 10 times and record thepercents in each trial who agreed. Then simulate drawing 320 women 10 times and again record the10 percents. Which set of 10 results is less variable?We expect the results of 320 trials to be more predictable (less variable) than the results of 20 trials.That is “long-run regularity” showing itself.


20 Women 320 Women

Trail 1 0.50 .51

Trial 2 .70 .54

Trial 3 .54 .55

Trial 4 .55 .53

Trial 5 .45 .54

Trial 6 .55 .52

Trial 7 .45 .48

Trial 8 .50 .50

Trail 9 .35 .46

Trial 10 .65 .45


Probability is a measure of how likely an event is to occur.Match one of the probabilities that follow with each statement about an event. (The probability is usually amuch more exact measure of likelihood than is the verbalstatement.)


0, 0.01, 0.03, 0.6, 0.99, 1

A). This event is impossible. It can never occur.

B). This event is certain. It will occur on every trial ofthe random phenomenon.

C). This event is very unlikely, but it will occur once ina while in a long sequence of trials.

D). This event will occur more often than not.

0

1

0.01

0.6

Lesson 6-2, Part 1

Probability Models

Addition/Complements

Sample Space (S)

The set of all possible outcomes of an event is the sample space S of the event.

Example: For the event “roll a die and observe what number it lands on” the sample space contains all possible numbers the die could land on.

1,2,3,4,5,6S

Event

An event is an outcome (or a set of outcomes) from a sample space

Example 1: When flipping three coins, an event may be getting the result HTH. In this case, the event is one outcome from the sample space.

Example 2: When flipping three coins, an event may be getting two tails. In this case, the event is a set of outcomes (HTT, TTH, THT) from the sample space

An event is usually denoted by a capital letter.

Example: Call getting two tails event A.

The probability of event A is denoted P(A).

Identifying Outcomes

Tree Diagrams

Multiplication Rule

One task a number of ways and a second task b number of ways, then both tasks can be done a x b number of ways.

Tree Diagram Toss a coin and rolling a dice

Possible outcomes

2 6 12

With or Without Replacement

With Replacement

Assume you are drawing index cards with the digits 1 – 10.You are picking 3 cards.

10 10 10 1000

Without Replacement

10 9 8 720

different ways to arrange a 3-digit number


For each of the following, use a tree diagram or the multiplication principal to determine the number of outcomes in the sample space. Then write the samplespace using set notation.

A). Toss 2 coins.

2 2 4

S HH HT TT TH, , ,


B). Toss 3 coins.

2 2 2 8

S HHH HHT HTT TTH HTH TTT THT THH, , , , , , ,


C). Toss 4 coins.

2 2 2 2 16

, , , , , ,

, , , , , ,

, , ,

HHHH HHHT HHTH HTHH THHH HHTT

S HTHT HTTH THTH TTHH THHT HTTT

THTT TTHT TTTH TTTT


Suppose you select a card from a standard deck of 52playing cards. In how many ways can the selected card be:

A). A red card?

26

B). A heart?

13

C). A queen and a heart?

1

E). A queen that is not a heart?

3

D). A queen or a heart?

16

Probability Rules

The probability of any event is between 0 and 1.

A probability of 0 indicates the event will never occur.

A probability of 1 indicates the event will always occur

0 ≤ P(A) ≤ 1

If S is the sample space, then P(S) = 1.

The sum of the probabilities of all possible outcomes must be 1


The New York Times (August 21, 1989) reported a poll that interviewed a random sample of 1025 women. The married women in the sample were asked whether their husbands did their fair share of household chores. Here are the results:

Outcome Probability

Does more than his fair share 0.12

Does his fair share 0.61

Does less than his fair share ?

These proportions are probabilities for the random phenomenon of choosing a married women at random and asking her opinion.


Outcome Probability




A). What must be the probability that women chosen saysthat her husband does less than his fair share? Why?

All three probabilities must add up to 1

1 (0.12 0.61) 0.27


Outcome Probability




B). The event “I think my husband does at least his fairshare” contains the first two outcomes. What is theprobability.

(0.12 0.61) 0.73

Complement

The complement of an event A, denoted by Ac , is the set of outcomes that are not in A. P(Ac) = 1 – P(A)

Example: When flipping two coins, the probability of getting two heads is 0.25. The probability of not getting two heads is

1 – 0.25 = 0.75

Complement

The set A and its complement

( ) 1P S

A

AC

The sample space S

S

( ) 1 ( )CP A P A

Addition Rule

If events A and B are disjoint if they have no outcomes in common.

Events like this, that can’t occur together, are called disjoint or mutually exclusive.

For two disjoint events A and B, the probability that one or the other occurs is the sum of the probabilities of the two events.

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

Addition Rule

Example: Let event A be rolling a die and landing on an even number, and event B be rolling a die and landing on an odd number. The outcomes for A are {2 , 4, 6} and the

outcomes for B are {1, 3 , 5}. These events are disjoint because they have no outcomes in common.

The probability of A or B (landing on either even or an odd number) P(A or B) = P(A) + P(B)

Disjoint

B

A

S

Two disjoint sets, A and B

We write P(A or B) as P(A B). The symbol means “union,” representing the outcomes in event A or event B (or both).

Lesson 6-2, Part 2

Probability Models

Independence

Independent

Events A and B are independent because the probability A does not change the probability of B.

Example: Roll a yellow die and a red die. Event A is the yellow die landing on an even number, and event B is the red die landing on an odd number.

Multiplication Rule

For two independent events A and B, the probability that both A and B occur is the product of the probabilities of the two events.

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

Multiplication Rule

Example: The probability that a yellow die lands on an even number and the red die lands on an odd number is:

P(A and B) = P(A) · P(B) = ½ x ½ = ¼

If events A and B are independent,

then their complements, Ac and Bc are also independent, and

Ac independent of B

Intersection

A

B

A and B

Two sets A and B are not disjoint.

We write P(A and B) as P(A B). The symbol means “intersection,” representing the outcomes that are in both event A and event B.


Example 6.10 (page 345) states that the first digits of numbers in legitimate records often follow a distribution know as Benford’s Law. Here is the distribution:

1st Digit 1 2 3 4 5 6 7 8 9

Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046

P(A) = P(first digit is 1) = 0.301P(B) = P(first digit is 6 or greater) = 0.222P(C) = P(first digit is odd) = 0.609P(D) = P(first digit is less than 4) = 0.602


1st Digit 1 2 3 4 5 6 7 8 9

Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046


A). P(D) = P(1, 2, or 3) = 0.301 + 0.176 + 0.125 = 0.602

B). P B D( ) P B P D( ) ( ) 0.222 0.602 0.824

C). CP D( ) P D1 ( ) 1 0.602 0.398


1st Digit 1 2 3 4 5 6 7 8 9

Probability 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046


D).

E).

P C D( ) P P(1) (3) 0.301 0.125 0.426P(1 and 3)

P B C( ) P(7 or 9) P P(7) (9) 0.058 0.046 0.104

Example – Page 354, #28

An automobile manufacturer buys computer chips froma supplier. The supplier sends a shipment containing5% defective chips. Each chip chosen from this shipmenthas probability 0.05 of being defective, and eachautomobile uses 12 chips selected independently. Whatis the probability that all 12 chips in a car will work properly?

P(all chips work properly)

12 12

1 .05 0.95 0.5404

= 1 – P(defective chips)


Choose at random a U.S. resident at least 25 years of age. Weare interested in the events

A = {The person chosen completed 4 years of college}B = {The person chosen is 55 years old or older}


Government data recorded in table 4.6 on page 241 allowsus to assign probabilities to these events.


A). Explain why P(A) = 0.256 P A44845

( ) 0.256175230


B). Find P(B). P B56008

( ) 0.320175230


C). Find the probability that the person is at least 55 years oldand had 4 years of college education, P(A and B). Areevents A and B independent?


105960.0604

175230 P(A and B)

A and B are not independent because P(A and B) P A P B( ) ( )

0.0604 0.256(0.320)

0.0604 0.08192


An athlete suspected of having used steroids is given two tests that operate independently of each other. Test A has a probability 0.9 of being positive if steroids have been used. Test B has probability 0.8 of being positive if steroids have been used. What is the probability that neither test is positive if steroids have used?

P(neither test is positive)

(1 0.9)(1 0.8)

C CP A P B

0.02

Lesson 6-3, Part 1

General Probability Rules

Basic Properties of Probability

0 ≤ P(A) ≤ 1 for any event A P(S) = 1 Complement Rule

For any event A, P(Ac) = 1 – P(A)

Addition Rule Events A and B are disjoint

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

Multiplication Rule Events A and B are independent

P(A and B) = P(A B) = P(A) · P(B)

Addition Rule – Not Disjoint

The union of two or more events that at least one of those occurs.

The addition Rule for the Union to Two Events:

P(A or B) = P(A) + P(B) – P(A and B)

P(A B) = P(A) +P(B) – P(A B)

Mutually Exclusive

Two events are mutually exclusive:

If the P(A and B) =0

Addition Rule – Not Disjoint

P(A)

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

P P(A B) = P(A) +P(B) – P(A B)

Two events are mutually exclusive if the P(A and B) = 0

Example 1Addition Rule – Not Disjoint

The accompanying table represents the blood groups and RH types of 100 people.

RH Types of

100 People

Group

A B AB O Total

RH

Positive 35 8 4 39 86

Negative 5 2 1 6 14

Total 40 10 5 45 100

Based on the table what is the probability a person willhave group B blood or be RH Positive?

Example 1Addition Rule – Not Disjoint

RH Types of

100 People

Group

A B AB O Total

RH

Positive 35 8 4 39 86

Negative 5 2 1 6 14

Total 40 10 5 45 100

P(B Blood or RH +) = P(B Blood) + P(RH +) – P(B Blood and RH +)

10 86 80.88

100 100 100

Example 2 Addition Rule – Mutually Exclusive

A study of credit card fraud was conducted by Master Card International, and the table below is based on the results”

Method of Fraud Number

Stolen Card 243

Counterfeit Card 85

Mail/Phone Order 52

Other 46

Total 426

Based on the table, what is the probability that fraud resulted from a stolen card or a mail/phone order ?

Example 2 Addition Rule – Mutually Exclusive

Method of Fraud Number

Stolen Card 243

Counterfeit Card 85

Mail/Phone Order 52

Other 46

Total 426

P(stolen card) + P(Mail/Phone) – P(Stolen and Mail/Phone)

243 520 0.69

426 426

P(stolen or Mail/Phone)=


Call a household prosperous if its income exceeds $100,000. Call the household educated if the household completed college. Select an American household at random, and let A be the event that the selected household is prosperous and B the event that it is educated. According the Census Bureau, P(A) = 0.134, P(B) = 0.254, and the joint probabilitythat a household is both prosperous and educated is P(A and B) = 0.080. What is the probability P(A or B) that the household selected is either prosperous or educated?

P(A or B) = P(A) + P(B) – P(A and B)

= 0.134 + 0.254 – 0.080= 0.308


Consolidated builders has bid on two large construction projects. The company president believes that the probability of winning the first contract (event A) is 0.6, that the probability of winning the second (event B) is 0.5, and that the joint probability of winning both jobs (event {A and B}) is 0.3. What is the probability of event {A or B} that Consolidated will win at least one of the jobs?

P(at least one contract)

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

= 0.6 + 0.5 – 0.3= 0.80


Musical styles other than rock and pop are becoming morepopular. A survey of college students finds that 40% likecountry music, 30% like gospel music, and 10% like both.

A) Make a Venn Diagram with these results.

Country Only0.30

Gospel Only0.2

Both0.1

Neither0.4



B) What percent of college students like country but notgospel?

P(country but not gospel) = P(C) – P(C and G)

0.40 .10

0.30



C) What percent like neither country nor gospel.

P(neither country or gospel) = 1 – P(C or G)

1 [.40 .30 .10]

0.40

= 1 – [P(C) + P(G) – P(C and G)]

Basic Properties of Probability

0 ≤ P(A) ≤ 1 for any event A

P(S) = 1

Complement Rule For any event A, P(Ac) = 1 – P(A)

Addition Rule Events A and B are not disjoint

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

Events A and B are disjoint P(A or B) = P(A B) = P(A) + P(B)

Multiplication Rule Events A and B are independent

P(A and B) = P(A B) = P(A) · P(B)

Conditional Probability Notation

The probability that event B occurs if we know for certain that event A will occur is called conditional probability.

The conditional probability of B given A is: P(B|A) which reads “the probability of event B given

event A has occurred.”

If events A and B are independent, then knowing that event A will occur does not change the probability of B. P(A|B) = P(A) or P(B|A) = P(B)

Conditional ProbabilityIndependent

Example: When flipping a coin twice, what is the probabilityof getting heads on the second flip if the first flip was head?

Event A: getting head on first flipEvent B: getting head on second flip

Events A and B are independent since the outcome of thefirst flip does not change the probability of the second flip.

1( | ) ( )

2P B A P B

Multiplication RuleNot Independent

The intersection of two or more events is that all of those events occur.

The probability that two events, A and B bothoccur is the probability that event A occurs multiplied by the probability that event B alsooccurs – that is by the probability that event B occurs given that event A occurs.

P(A and B) = P(A) · P(B|A)

P(A B) = P(A) · P(B|A)

Conditional Probability

To find the probability of event B given the event A, we restrict our attention to the outcomes in A. We then find in what fraction of those outcomes B also occurred.

P(B|A) =

P(B|A) =

P(A and B)

P(A)

P(A B)

P(A)

ExampleConditional Probability

RH Types of

100 People

Group

A B AB O Total

RH

Positive 35 8 4 39 86

Negative 5 2 1 6 14

Total 40 10 5 45 100

P(RH +|O Blood) =

3939100 0.87

45 45

100

What is the probability that a person will be RH positive giventhey have blood type O?

P(O and RH +)

P(O blood)

ExampleAre the Events Disjoint? Independent?

Police report that 78% of drivers are given a breath test,36% a blood test, and 22% both tests

1. Are giving a DWI suspect a blood test and a breath testmutually exclusive?

2. Are giving the two tests independent?

State the events we’re interest in.

Let A = {suspect is given a breath test}.Let B = {suspect is given a blood test}.


Police report that 78% of drivers are given a breath test,36% a blood test, and 22% both tests

State the events we’re interest in.


State the given probabilities.

P(A) = 0.78P(B) = 0.36P(A and B) = P(A B) = 0.22



P(A) = 0.78P(B) = 0.36P(A and B) = P(A B) = 0.22

1. Are giving a DWI suspect a blood test and a breath testmutually exclusive?

Disjoint events cannot both happen at the same time, socheck to see if P(A and B) = 0.

P(A and B) = 0.22, the events cannot be mutuallyexclusive.

22% of all suspects get both tests, so a breath testand blood test are not disjoint events.

Lesson 6-3, Part 2

General Probability Rules

Conditional Probability



P(A) = 0.78P(B) = 0.36P(A and B) = P(A B) = 0.22

2. Are the two tests independent?

Does getting a breath test change the probability ofgetting a blood test? That is, does P(B|A) = P(B)?

Because the two probabilities are not the same, the twoevents are not independent.

( ) 0.22( | ) 0.28

( ) 0.78

P A BP B A

P A

( ) 0.36P B


Overall, 36% of the drivers get blood tests, but only 22%of those get a breath test too. Since suspects who get a breath test are less likely to have a blood test, the two events are not independent.


Choose an adult American woman at random. Table 6.1 describes the population from which we draw. Use the information in that table to answer the following questions.

A). What is the probability that the women chosen is65 years old or older?

B). What is the conditional probability that the women chosen is married, given that she is 65 or over?

C). How many women are both married and in the over 65age group? What is the probability that the women wechoose is a married women at least 65 years old?


A). P( 65 years old or older) =18669

0.18103870


B). P(M|65) =P(65 and M)

P(65)

82708270103870 0.443

18669 18669

103870


C). P(M and 65) = 8270

0.08103870


Choose an employed person at random. Let A be the event that theperson chosen is a women and B the event that the person holds amanagerial or professional job. Government data tell us that P(A) = 0.46and the probability of managerial and professional jobs among womenis P(B|A) = 0.32. Find the probability that a randomly chosen employedperson is a women holding a managerial or professional position.

P(Women and Managerial/Professional Position)

P(A and B) = P(A) · P(B|A) = 0.46 0.32 0.1472


A poker player holds a flush when all 5 cards in the hand belongto the same suit. We will find the probability of a flush when 5cards are dealt. Remember that a deck contains 52 cards, 13 ofeach suit, and that when the deck is well shuffled, each card dealtis equally likely to be any of those that remain in the deck.


A). We will concentrate on spades. What is the probability that the firstcard dealt is a spade? What is the conditional probability that the second card is a spade, given that the first is a spade?

P(1st card Spade) = 13

0.2552

P(2nd card spade|1st card spade) =12

0.235351


B). Continue to count the remaining cards to find the conditional probability of a spade on the third, fourth, and fifth card, giventhat all previous cards are spades.

P(3rd card spade|2nd card spade) =

P(4th card spade|3rd card spade) =

P(5th card spade|4th card spade) =

110.22

50

100.2041

49

90.1875

48


C). The probability of being dealt 5 spades is the product of the five probabilities you have found? Why? What is this probability?

P(5 spades) = 0.25 0.2353 0.22 0.2041 0.1875 0.0004952

The product of these conditional probabilities givesthe probability of a flush in spades by the extendedmultiplication rule: We must draw a spade then another,and then a third, fourth, and fifth.


D). The probability of being dealt 5 hearts, or 5 diamonds or 5 clubs isthe same as the probability of being dealt 5 spades. What isthe probability of being dealt a flush?

P(flush) = (0.0004952)4 0.001981


Enzyme immunoassay (EIA) tests are used to screen blood specimens for the presence of antibodies to HIV, the virus that causes AIDS. Antibodies indicate the presence of the virus. The test is quite accurate but is not always correct. Here are approximate probabilities of positive and negative EIA outcomes when the blood tested does and does notactually contain antibodies to HIV.

Test result

Positive Negative

Antibodies Present 0.9985 0.0015

Antibodies Absent 0.006 0.994

Suppose that 1% of a large population carries antibodies to HIV in their blood.


Test result

Positive Negative



A). Draw a tree diagram for selecting a person from thispopulation (outcomes: antibodies present or absent) and for testing his or her blood (outcomes: EIA positiveor negative).


Test result

Positive Negative



Subject

AntibodyPresent

AntibodyAbsent

EIA +

EIA –

EIA +

EIA –

0.01

0.99

0.9985

0.0015

0.006

0.994


Subject

AntibodyPresent

AntibodyAbsent

EIA +

EIA –

EIA +

EIA –

0.01

0.99

0.9985

0.0015

0.006

0.994

B). What is the probability that the EIA is positive for an randomlychosen person from this population?

P(Test +) = P(antibody and Test +) + P(no antibody and Test +) =

(0.01)(0.9985) (0.99)(0.006) 0.016


Subject

AntibodyPresent

AntibodyAbsent

EIA +

EIA –

EIA +

EIA –

0.01

0.99

0.9985

0.0015

0.006

0.994

C). What is the probability that a person has the antibody given that the EIA test is positive?

P(Antibody|Test +) = P(Test + and Antibody)

P(Test +)

(0.9985)(0.01)0.624

0.016

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