Chapter 6 Probability. Chance experiment – any activity or situation in which there is uncertainty about which of two or more plausible outcomes will.

Chapter 6

Probability

Chance experiment – any activity or situation in which there is uncertainty about which of two or more plausible outcomes will result.

Suppose two six-sided die is rolled and they both land on sixes.

Or a coin is flipped and it lands on heads.

Or record the color of the next 20 cars to pass an intersection.

These would be examples of chance experiments.

Sample space - the collection of all possible outcomes of a chance experiment

Suppose a six-sided die is rolled. The possible outcomes are that the die could land with 1 dot up or 2, 3, 4, 5, or 6 dots up.

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

This would be an example of a sample space.

“S” stands for sample space. We use set notation to list the outcomes of the sample

space.

The sum of the probabilities of the

outcomes in the sample space equals

ONE.

Suppose two coins are flipped. The sample space would be:

S = {HH, HT, TH, TT}Where H = heads and T = tails

H

T

H

T

H

T

We can also use a tree diagram to represent a sample space.

HTWe follow the branches out to show an outcome.

Event - any collection of outcomes (subset) from the sample space of a chance experiment

Suppose a six-sided die is rolled. The outcome that the die would land on an even number would be

E = {2, 4, 6}

This would be an example of an event.

We typically use capital letters to denote an event.

Complement - Consists of all outcomes that are not in the event

Suppose a six-sided die is rolled. The event that the die would land on an even number would be

E = {2, 4, 6}

What would the event be that is the die NOT landing on an even number?

EC = {1, 3, 5}

This is an example of complementary events.

The superscript “C” stands for

complement

E’ and E also denote the complement of E

The sum of the probabilities of complementary events equals

ONE.

These complementary events can be shown on a Venn Diagram.

E = {2, 4, 6} and EC = {1, 3, 5}

Let the rectangle represent the sample space.

Let the circle represent event E.

Let the shaded area represent event not E.

Suppose a six-sided die is rolled. The event that the die would land on an even number would be E = {2, 4, 6}

The event that the die would land on a prime number would be P = {2, 3, 5}

What would be the event E or P happening?

E or P = {2, 3, 4, 5, 6}

This is an example of the union of two events.

The union of A or B - consists of all outcomes that are in at least one of the two events, that is, in A or in B or in both. BAB or A

This symbol means “union”

Consider a marriage or union of two

people – when two

people marry, what do they do with their

possessions ?

The bride takes all her stuff & the

groom takes all his stuff &

they put it together!

And live happily ever after!This is similar to the

union of A and B.All of A and all of B are put together!

Let’s revisit rolling a die and getting an even or a prime number . . .

E or P = {2, 3, 4, 5, 6}Another way to represent this is with a Venn Diagram.

Even number

24

6

Prime number

3

5

1

E or P would be any number in either

circle.

Why is the number 1 outside the

circles?

Suppose a six-sided die is rolled. The event that the die would land on an even number would be E = {2, 4, 6}

The event that the die would land on a prime number would be P = {2, 3, 5}

What would be the event E and P happening?

E and P = {2}

This is an example of the intersection of two events.

The intersection of A and B - consists of all outcomes that are in both of the events

BAB and A

This symbol means

“intersection”

Let’s revisit rolling a die and getting an even or a prime number . . .

E and P = {2}To represent this with a Venn Diagram:

24

6

3

5

1

E and P would be ONLY the middle part that the circles have in

common

Mutually exclusive (or disjoint) events -two events have no outcomes in common; two events that NEVER happen simultaneously

Suppose a six-sided die is rolled. Consider the following 2 events:

A = {2} B = {6}

On a single die roll, is it possible for A and B to happen at the same time?

These events are mutually exclusive.

A Venn Diagram for the roll of a six-sided die and the following two events:

A = {2} B = {6}

2

4

6

3

51

A and B are mutually exclusive (disjoint) since they have no

outcomes in common

The intersection of A and B is empty!

Practice with Venn DiagramsOn the following four slides you will find Venn Diagrams representing the students at your school.Some students are enrolled in Statistics, some in Calculus, and some in Computer Science.

For the next four slides, indicate what relationships the shaded regions represent. (use complement, intersection, and union)

Calculus or Computer Science

Statistics Calculus

Computer Science

(Statistics or Computer Science) and not Calculus

Statistics Calculus

Computer Science

Com Sci

Statistics and Computer Science and not Calculus

Statistics Calculus

Computer Science

Statistics and not (Computer Science or Calculus)

Statistics Calculus

Computer Science

What is Probability?

Three different approaches to probability

The Classical Approach

When the outcomes in a sample space are equally likely, the probability of an event E, denoted by P(E), is the ratio of the number of outcomes favorable to E to the total number of outcomes in the sample space.

outcomes totaloutcomes favorable

)( EP

Examples: flipping a coin, rolling a die, etc.

On some football teams, the honor of calling the toss at the beginning of the football game is determined by random selection.Suppose this week a member of the 11-player offensive team will be selected to call the toss. There are five interior linemen on the offensive team.If event L is defined as the event that an interior linemen is selected to call the toss, what is probability of L?

P(L) = 5/11

Consider an archer shooting arrows at a target.

The probability of getting a bulls’ eye should be the ratio of the area of the inner circle to the area of the entire target.

What if a very experienced archer were shooting the arrows? Would the probability of a bull’s eye still be the same?The classical approach doesn’t

work for every situation.

The Relative Frequency ApproachThe probability of event E, denoted by P(E), is defined to be the value approached by the relative frequency of occurrence of E in a very long series of trials of a chance experiment. Thus, if the number of trials is quite large,

trials of numberoccurs E times of number

)( EP

Consider flipping a coin and recording the relative frequency of heads.

When the number of coin flips is small, there is a lot of variability in the relative frequency of “heads” (as shown in this graph). What do you notice in the graph at the right?

Consider flipping a coin and recording the relative frequency of heads.

The graph at the right shows the relative frequency when the coin is flipped a large number of times. What do you notice in this graph at the right?

Law of Large Numbers

As the number of repetitions of a chance experiment increase, the chance that the relative frequency of occurrence for an event will differ from the true probability by more than any small number approaches 0.

OR in other words, after a large number of trials, the relative

frequency approaches the true probability.

Notice how the relative frequency of heads approaches ½

the larger the number of trials!

The Subjective Approach

Probability can be interpreted as a personal measure of the strength of belief that a particular outcome will occur.

Example: An airline passenger may report that her probability of being placed on standby (denied a seat) due to overbooking is 0.1. She arrived at this through personal experience and observation of events.

The problem with a subjective approach is that different people could assign different

probabilities to the same outcome based on their subjective viewpoints.

Probability Rules!

Fundamental Properties of Probability

Property 1. Legitimate ValuesFor any event E, 0 < P(E) < 1

Property 2. Sample spaceIf S is the sample space, P(S) = 1

Properties Continued . . .

Property 3. AdditionIf two events E and F are

disjoint, P(E or F) = P(E) + P(F)

Property 4. ComplementFor any event E, P(E) + P(not E) = 1

Probabilities of Equally Likely Outcomes

Consider an experiment that can result in any one of N possible outcomes. Denote the simple events by O1, O2, …, ON. If these simple events are equally likely to occur, then

1.

2. For any event E,

NE

EP in outcomes of number

)(

NO, P,

NO, P

NOP N

1)(

1)(

1)( 21

Suppose you roll a six-sided die once. Let E be the event that you roll an even number.

P(E) = P(2 or 4 or 6) = 3/6Number of outcomes in E Over N

Addition Rule for Disjoint EventsIf events E1, E2, . . ., Ek are disjoint (mutually exclusive) events, then

P(E1 or E2 or . . . or Ek) =

P(E1) + P(E2) + . . . + P(Ek)

In words, the probability that any of these k disjoint events occurs is the sum of the probabilities of the individual events.

A large auto center sells cars made by many different manufacturers. Three of these are Honda, Nissan, and Toyota. Consider a chance experiment that consist of observing the make of the next car sold. Suppose that P(H) = 0.25, P(N) = 0.18, P(T) = 0.14.

Are these disjoint events?

P(H or N or T) =

P(not (H or N or T)) =

yes

.25 + .18+ .14 = .57

1 - .57 = .43

Sometimes the knowledge that one event has occurred changes our assessment of the likelihood that another event occurs.

Consider the following example:

Suppose that 0.1% of all the individuals in a population have a certain disease. The presence of the disease is not discernable from appearances, but there is a screening test for the disease.

Let D = the event that a person has the disease

P(D) = .001

Disease example continued . . .

Suppose that 0.1% of all the individuals in a population have a certain disease.

80% of those with positive test results actually have the disease.20% of those with positive test results actually do NOT have the disease (false positive)

Let P = the event that a person tests positive for the disease

P(D|P) = 0.80This is an example of conditional probability.

Knowing that event P, the person tested

positive, has occurred, changes the probability of event D, the person

having the disease, from 0.001 to 0.80.Read:

Probability that a person has the disease “GIVEN” the person tests

positive

Conditional Probability

A probability that takes into account a given condition has occurred

P(A)B)P(A

P(B|A)

The article “Chances Are You Know Someone with a Tattoo, and He’s Not a Sailor” (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50. The accompanying data are consistent with summary values given in the article.

At Least One Tattoo

No Tattoo Totals

Age 18-29 18 32 50

Age 30-50 6 44 50

Totals 24 76 100

Assuming these data are representative of adult Americans and that an adult is selected at random, use the given information to estimate the following probabilities.

Tattoo Example Continued . . .

At Least One Tattoo

No Tattoo Totals

Age 18-29 18 32 50

Age 30-50 6 44 50

Totals 24 76 100

What is the probability that a randomly selected adult has a tattoo?

P(tattoo) =

24/100 = 0.24

Tattoo Example Continued . . .

At Least One Tattoo

No Tattoo Totals

Age 18-29 18 32 50

Age 30-50 6 44 50

Totals 24 76 100

What is the probability that a randomly selected adult has a tattoo if they are between 18 and 29 years old?

P(tattoo|age 18-29) =

18/50 = 0.36

This is a condition!How many adults in

the sample are ages 18-29?

How many adults in the sample are ages 18-29

AND have a tattoo?

Tattoo Example Continued . . .

At Least One Tattoo

No Tattoo Totals

Age 18-29 18 32 50

Age 30-50 6 44 50

Totals 24 76 100

What is the probability that a randomly selected adult is between 18 and 29 years old if they have a tattoo?

P(age 18-29|tattoo) =

18/24 = 0.75

This is a condition!

How many adults in

the sample have a tattoo?

How many adults in the sample are ages 18-29

AND have a tattoo?

Sometimes the knowledge that one event has occurred does NOT change our assessment of the likelihood that another event occurs.Consider the genetic trait, hitch hiker’s thumb, which is the ability to bend the last joint of the thumb back at an angle of 60° or more.

Whether or not an offspring has hitch hiker’s thumb is determined by two random events: which gene is contributed by the father and which gene is contributed by the mother. Which gene is contributed by the father does NOT affect which gene is contributed by the mother

These are independent events.

Let’s consider a bank that offers different types of loans:

The bank offers both adjustable-rate and fixed-rate loans on single-family dwellings, condominiums and multifamily dwellings. The following table, called a joint-probability table, displays probabilities based upon the bank’s long-run loaning practices.

Single Family

Condo Multifamily Total

Adjustable .40 .21 .09 .70

Fixed .10 .09 .11 .30

Total .50 .30 .20

P(Adjustable loan) =

.70

Bank Loan’s Continued . . .

Single Family

Condo Multifamily Total

Adjustable .40 .21 .09 .70

Fixed .10 .09 .11 .30

Total .50 .30 .20

P(Adjustable loan) =

.70

P(Adjustable loan|Condo) =.21/.30 = .70Knowing that the loan is for a condominium does not change the probability that it is an adjustable-rate loan. Therefore, the event that a randomly selected loan is adjustable and the event that a randomly selected loan is for a condo are independent.

Independent Events

Two events are independent if knowing that one will occur (or has occurred) does not change the probability that the other occurs.

Two events, E and F, are said to be independent if P(E|F) = P(E).

If P(E|F) = P(E), it is also true that P(F|E) = P(F).

If two events are not independent, they are said to be dependent

events.

Two events E and F are independent, if and only if,

P(F) P(E)F) P(E

Multiplication Rule for Two Independent Events

P(H+ from mom AND H+ from dad) =

Hitch Hiker’s Thumb Revisited

Suppose that there is a 0.10 probability that a parent will pass along the hitch hiker’s thumb gene to their offspring. What is the probability that a child will have a hitch hiker’s thumb?

Since these are independent events, we just multiply the

probabilities together.

0.1 × 0.1 = 0.01

This would happen if the mother contributes a hitch hiker’s gene

(H+) AND if the father contributes a hitch hiker’s gene

(H+).

Events E1, E2, . . ., Ek are independent if knowledge that some number of the events have occurred does not change the probabilities that any particular one or more of the other events occurred.

This relationship remains valid if one or more of the events are replaced by their complement (not E).

)(...)()()...( 2121 kk EPEPEPEEEP

Multiplication Rule for k Independent Events

Suppose that a desktop computer system consist of a monitor, a mouse, a keyboard, the computer processor itself, and storage devices such as a disk drive. Most computer system problems due to manufacturer defects occur soon in the system’s lifetime. Purchasers of new computer systems are advised to turn their computers on as soon as they are purchased and then to let the computer run for a few hours to see if any problems occur.LetE1 = event that a newly purchased monitor is not defective

E2 = event that a newly purchased mouse is not defective

E3 = event that a newly purchased disk drive is not defective

E4 = event that a newly purchased processor is not defective

Suppose the four events are independent with

P(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99

LetE1 = event that a newly purchased monitor is not defective

E2 = event that a newly purchased mouse is not defective

E3 = event that a newly purchased disk drive is not defective

E4 = event that a newly purchased processor is not defective

Suppose the four events are independent withP(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99

What is the probability that none of these components are defective?

)( 4321 EEEEP(.98)(.98)(.94)(.99) = .89

In the long run, 89% of such systems will run properly when tested shortly after

purchase.

LetE1 = event that a newly purchased monitor is not defective

E2 = event that a newly purchased mouse is not defective

E3 = event that a newly purchased disk drive is not defective

E4 = event that a newly purchased processor is not defective

Suppose the four events are independent withP(E1) = P(E2) = .98 P(E3) = .94 P(E4) = .99

What is the probability that all these components will run properly except the monitor?

)( 4321 EEEEP C

(.02)(.98)(.94)(.99) = .018

Suppose I will pick two cards from a standard deck. This can be done two ways:

1)Pick a card at random, replace the card, then pick a second card

2) Pick a card at random, do NOT replace, then pick a second card.

If I pick two cards from a standard deck without replacement, what is the probability that I select two spades?Are the events E1 = first card is a spade and E2 = second card is a spade independent?NOP(E1 and E2) =

P(E1) × P(E2|E1) =

Sampling with replacement – the events are typically independent

events.

Sampling without replacement – the events are typically dependent

events.

Probability of a spade given I drew a spade on the first card.

171

5112

41

Suppose the manufacturer of a certain brand of light bulbs made 10,000 of these bulbs and 500 are defective. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that both bulbs are defective?

Are the events E1 = the first bulb is defective and E2 = the second bulb is defective independent?

What would be the probability of selecting a defective light bulb? 500/10,000 = .05

To answer this question, let’s explore the probabilities of these

two events?

Light Bulbs Continued . . .

What would be the probability of selecting a defective light bulb?

Having selected one defective bulb, what is the probability of selecting another without replacement?

500/10,000 = .05

499/9999 = .0499

These values are so close to each other that when rounded to three decimal places they are both .050. For all

practical purposes, we can treat them as being independent.

If a random sample of size n is taken from a population of size N, then the outcomes

of selecting successive items from the population without replacement can be

treated as independent when the sample size n is at most 5% of the population size

N.

Light Bulbs Continued . . .

What is the probability that both bulbs are defective?

Are the selections independent?

defective) eP(defectiv

We can assume independence.

(0.05)(0.05) = .0025

General Rule for Addition

For any two events E and F,

)()()()( FEPFPEPFEP

E F

Since the intersection is added in twice, we subtract out the

intersection.

Musical styles other than rock and pop are becoming more popular. A survey of college students finds that the probability they like country music is .40. The probability that they liked jazz is .30 and that they liked both is .10. What is the probability that they like country or jazz?

.4 + .3 -.1 = .6

)( JazzCountryP

)( FEP Ask yourself, “Are the events mutually exclusive?”

)()( FPEP

Yes No

)()()( FEPFPEP

If independent

)()( FPEP

Here is a process to use when calculating the union of two or more events.

In some problems, the intersection of the two events

is given (see previous example).

In some problems, the intersection of the two events is not given, but we

know that the events are

independent.

Suppose two six-sided dice are rolled (one white and one red). What is the probability that the white die lands on 6 or the red die lands on 1?

Let A = white die landing on 6

B = red die landing on 1

Are A and B disjoint?NO, independent events cannot be disjoint

How can you find

the probability of A and

B?

)()()()( BAPBPAPBAP

3611

61

61

61

61

General Rule for MultiplicationFor any two events E and F,

)()|()( FPFEPFEP

Ask yourself, “ Are these events independent?”

)( BAP

Yes No

)()( BPAP

)|()( ABPAP

Here is a process to use when calculating the intersection of two or more events.

There are seven girls and eight boys in a math class. The teacher selects two students at random to answer questions on the board. What is the probability that both students are girls?

Are these events independent?

2.146

157

)G P(G 21

NO

Light Bulbs Revisited . . .

A certain brand of light bulbs are defective five percent of the time. You randomly pick a package of two such bulbs off the shelf of a store. What is the probability that exactly one bulb is defective?

Let D1 = first light bulb is defective

D2 = second light bulb is defective

= (.05)(.95) + (.95)(.05) = .095

2121defective) one exactly( DDDDPP CC

An electronics store sells DVD players made by one of two brands. Customers can also purchase extended warranties for the DVD player. The following probabilities are given:

Let B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4

If a DVD customer is selected at random, what is the probability that they purchased the extended warranty?

This can happen in one of two ways:1)They purchased the extended warranty and Brand 1 DVD player

OR2) They purchased the extended warranty and Brand 2 DVD player

21 BEBEE

DVD Player Continued . . .

Let B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4

If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? 21 BEBEE

These are disjoint events

21 BEPBEPEP

Use the General Multiplication Rule:

)(|)(| 2211 BPBEPBPBEPEP

DVD Player Continued . . .

Let B1 = event that brand 1 is purchased

B2 = event that brand 2 is purchased

E = event that extended warranty is purchased

P(B1) = .7 P(B2) = .3 P(E|B1) = .2 P(E|B2) = .4

If a DVD customer is selected at random, what is the probability that they purchased the extended warranty? )(|)(| 2211 BPBEPBPBEPEP

P(E) = (.2)(.7) + (.4)(.3) = .26

This is an example of the Law of Total Probabilities.

Law of Total ProbabilitiesIf B1 and B2 are disjoint events with probabilities P(B1) + P(B2) = 1, for any event E

More generally B1, B2, …, Bk are disjoint events with probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E

)()()( 21 BEPBEPEP )()|()()|( 2211 BPBEPBPBEP

)(...)()()( 21 kBEPBEPBEPEP

)()|(...

)()|()()|( 2211

kk BPBEP

BPBEPBPBEP

Bayes Rule (Theorem)• A formula discovered by the

Reverend Thomas Bayes, an English Presbyterian minister, to solve what he called “converse” problems.

Let’s examine the following problem before looking at the formula . . .

Lyme’s disease is the leading tick-borne disease in the United States and England. Diagnosis of the disease is difficult and is aided by a test that detects particular antibodies in the blood. The article, “Laboratory Consideration in the Diagnosis and Management of Lyme Borreliosis”, American Journal of Clinical Pathology, 1993, used the following notations:

+ represents a positive result on a blood test- represents a negative result on a blood testL represents the patient actually has LymesLC represents the patient doesn’t have Lymes

The article gave the following probabilities:P(L) = .00207 P(LC) = .99723P(+|L) = .937 P(-|L) = .063P(+|LC) = .03 P(-|LC) = .97

The article gave the following probabilities:P(L) = .00207 P(LC) = .99723P(+|L) = .937 P(-|L) = .063P(+|LC) = .03 P(-|LC) = .97

Bayes’s converse problem poses this question:

“Given that a patient test positive, what is the probability that he or she really has the disease?” written: P(L|+)

Lyme’s Disease Continued . . .

This question is of primary concern in medical diagnosis

problems!

)()|()()|(

)()|(CC LPLPLPLP

LPLP

Lyme’s Disease Continued . . .The article gave the following probabilities:

P(L) = .00207 P(LC) = .99723P(+|L) = .937 P(-|L) = .063P(+|LC) = .03 P(-|LC) = .97

Bayes reasoned as follows:

)()(

)|(

PLP

LP

Using the Law of Total Probabilities, the denominator becomes

P(+|L)P(L) + P(+|LC)P(LC).

Substitute values:

0596.)99793(.03.)00207(.937.

)00207(.937.

Since

we can use P(+|L) × P(L) for the numerator.

)()( LPLP

Bayes Rule (Theorem)

If B1 and B2 are disjoint events with probabilities P(B1) + P(B2) = 1, for any event E

More generally B1, B2, …, Bk are disjoint events with probabilities P(B1) + P(B2) + … + P(Bk) = 1, for any event E

)()|()()|()()|(

)|(2211

111 BPBEPBPBEP

BPBEPEBP

)()|(...)()|()()|()()|(

)|(2211 kk

iii BPBEPBPBEPBPBEP

BPBEPEBP

Estimating Probabilities using Simulation

Simulation1. Design a method that uses a random mechanism

(such as a random number generator or table, tossing a coin or die, etc.) to represent an observation. Be sure that the important characteristics of the actual process are preserved.

2. Generate an observation using the method in step 1 and determine if the outcome of interest has occurred.

3. Repeat step 2 a large number of times4. Calculate the estimated probability by dividing the

number of observations of the outcome of interest by the total number of observations generated.

Simulation provides a means of estimating probabilities when we are unable to determine them analytically or it is impractical to estimate them

empirically by observation.

Suppose that couples who wanted children were to continue having children until a boy was born. Would this change the proportion of boys in the population?

We will use simulation to estimate the proportion of boys in the population if couples were to continue having children until a boy was born.

1)We will use a single-random digit to represent a child, where odd digits represent a male birth and even digits represent a female birth.

2) Select random digits from a random digit table until a male is selected and record the number of boys and girls.

3) Repeat step 2 a large number of times.

Boy Simulation Continued . . .

Below are four rows from the random digit table at the back of our textbook.

Row

6 0 9 3 8 7 6 7 9 9 5 6 2 5 6 5 8 4 2 6 4

7 4 1 0 1 0 2 2 0 4 7 5 1 1 9 4 7 9 7 5 1

8 6 4 7 3 6 3 4 5 1 2 3 1 1 8 0 0 4 8 2 0

9 8 0 2 8 7 9 3 8 4 0 4 2 0 8 9 1 2 3 3 2

Trial 1: girl, boy

Trial 2: boy

Trial 3: girl, boy

Trial 4: girl, boy

Trial 5: boy

Trial 6: boy

Trial 7: boy

Trial 8: girl, girl, boy

Trial 9: girl, boy

Trial 10: girl, girl, girl, girl, girl, girl, boy

Continue this process a large number of times (at least 100 trials).

Calculate the proportion of boys out of the number of children born.

Notice that with only 10 trials, the proportion of boys is 10/22,

which is close to 0.5!