1 introduction to probability chapter 6. 2 introduction in this chapter we discuss the likelihood of...

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Introduction to Probability

Introduction to Probability

Chapter 6

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Introduction

• In this chapter we discuss the likelihood of occurrence for events with uncertain outcomes.

• We define a scale to measure and describe the chance that different outcomes of an uncertain event will take place.

• This ‘measure of uncertainty’ called “Probability” serves as the basis for the analysis and results discussed at the rest of this course.

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• Examples Experiment Outcomes

• Flip a coin Heads and Tails• The marks of a statistics test Numbers between

0 and 100• The time to assemble Non-negative

numbersa computer

6.1 Assigning probabilities to Events

• Random experiment– a random experiment is a process or course of action,

whose outcome is uncertain.

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• Because of the random nature of the experiment, repeated experiments may result in different outcomes. Consequently, one cannot refer to the experiment outcome in certain terms, only in terms of the probability of each outcome to occur.

• To determine the probabilities we need to define and list the possible outcomes first. Such a list should be – exhaustive (list of all the possible outcomes).– the outcomes are mutually exclusive (outcome do not overlap).

• A list of outcomes that meet the two conditions above, is called a sample space.


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Sample Space: S = {O1, O2,…,Ok}

Sample Spacea sample space of a random experimentis a list of all possible outcomes of the experiment. The outcomes must be mutually exclusive and exhaustive.

Simple eventsThe individual outcomes are called simple events. Simple events cannot be further decomposed into constituent outcomes.

EventAn event is any collectionof one or more simple events

O1 O2

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Our objective is to determine P(A), the probability that event A will occur.

Our objective is to determine P(A), the probability that event A will occur.


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Example 1: Build the sample space for the two random experiments described below

The sample space: {B, R}– Case 2: Two balls are randomly selected from an urn that

contains blue and red balls. An outcome is considered the colors of the two balls drawn.

The sample space: {BB, BR, RB, RR}

– Case 1: A ball is randomly selected from an urn containing blue and red balls. An outcome is considered the ball color.

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• Example 2: Build the sample space for the random experiment defined by the random selection of two numbers from the set 1, 2, 3, 4, 5 (a number cannot be selected twice). An outcome is defined by the sum of the two numbers.

The sample space: {3, 4, 5, 6, 7, 8, 9}

1+2

1+3

1+4, 2+3;

1+5, 2+4

2+5, 3+4

3+5

4+5

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• Given a sample space S={O1,O2,…,Ok}, the following characteristics for the probability P(Oi) of the experimental outcome Oi must hold:

n

1i1)iP(O2.

ieachfor1)iP(O01.

n

1i1)iP(O2.

ieachfor1)iP(O01.

• The probability of an event: The probability P(A) of event A is the sum of the probabilities assigned to the simple events contained in A.


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Approaches to Assigning Probabilities and interpretation of Probability

• Approaches– The classical approach– The relative frequency approach– The subjective approach

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Events and their Probabilities

• An event is a collection of sample points.– Example 3: A dice is rolled once. The following events are

defined. Specify the sample points that belong to each event. Event A = the number facing up is 6 Event B = The number facing up is odd Event C = The number facing up is not greater than 4

Solution: A = {6}; B = {1, 3, 5}; C = {1, 2, 3)

– Example 4: A dice is rolled once and the outcome is 3.Which event takes place?

Solution: Event B and event C take place because the outcome ‘3’ belongs to both events.

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Probability of an Event

Example 5: A dice is rolled once. Each number is equally likely to face up. Find the probabilities of the following events:

Event A = the number facing up is 6

Event B = The number facing up is odd

Event C = The number facing up is not greater than 4

Solution:

P(A) = P{6} = 1/6;P(B) = P(1 or 3 or 5) = 3/6P(C) = P(1 or 2 or 3) = 3/6

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Example 6: A dice is rolled twice: Define the following events:D = The numbers facing up are the same and evenE = The sum of the two numbers facing up is 5.Define the sample space and calculate the probabilities of events D and E.

SolutionThe sample space: S = {(1,1); (1,2);….(6,5); (6,6)}P(D) = P{(2,2) or (4,4) or (6,6)} = 3/36 P(E) = P{(1,4) or (2,3) or (3,2) or (4,1)} = 4/36.

Note: there are 36 points in the sample space.

Probability of an Event

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6.2 Joint, Marginal, and Conditional Probability

• Determining the probability of an event involves the counting process of its simple events. Often this may become quite cumbersome.

• Understanding relationships among events may reduce the computational effort involved in determining the probability of combined events.

• Combined events and there probabilities are discussed next

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Intersection and Joint Probability

• The intersection of event A and B is the event that occurs when both A and B occur.

A BC

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• The intersection of events A and B is denoted by “A and B” ( ).

• The probability of the intersection of A and B is called also the joint probability of A and B = P(A and B).

BA


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• Example 9– A potential investor examined the relationship

between the performance of mutual funds and the school the fund manager earned his/her MBA.

– The following table describes the joint probabilities.


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• Example 9 – continued – The joint probability of

[Mutual fund outperforms…] and [Top 20 MBA …] = .11– The joint probability of

[Mutual fund outperform…] and […not top 20 MBA …] = .06

Probabilities of Joint Events

Mutual fund outperforms the market

(B1)

Mutual fund doesn’t outperform the market

(B2)

Top 20 MBA program (A1) .11 .29Not top 20 MBA program (A2) .06 .54

P(A1 and B1)

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• Example 9 – continued – The joint probability of

[Mutual fund outperform…] and […Top 20 MBA …] = .11– The joint probability of

[Mutual fund outperform…] and […not from a top 20 …] = .06

Probabilities of Joint Events

Mutual fund outperforms the market

(B1)

Mutual fund doesn’t outperform the market

(B2)


P(A2 and B1)

P(A1 and B1)

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Marginal Probabilities

To better understand the concept of marginal probabilities, observe first the following demonstration

A BC

Event C intersect with both event A and B

Let us separate event C into two sub event: “A and

C”, and “B and C”

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The intersection events “A and C”, and “B and C” are indicated as two triangles, and for clarity are separated for a moment (click).

A A and C BB and C

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As the two intersection events are brought back to there original location, it becomes clear (we hope) that the probability of event C can be calculated as the sum of the two joint probabilities.

A BC

P(C) = P(A and C) + P(B and C))

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• Applying this concept to the table of joint probabilities, we notice that marginal probabilities are determining by adding joint probabilities across rows and columns. Click to continue.

• These probabilities are computed by adding across the rows and down the columns and appear in the margins of the table. Watch.

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Mutual fund outperforms the market (B1)

Mutual fund doesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Top 20 MBA program (A1)

Not top 20 MBA program (A2)

Marginal Probability P(Bj)

+ = P(A1)


P(A1 and B1) P(A1 and B2)

P(A2 and B1) + P(A2 and B2) = P(A2)

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Marginal Prob.

P(Ai)

Top 20 MBA program (A1) .11 .29 .40Not top 20 MBA program (A2) .06 .54 .60Marginal Probability P(Bj)

+ =

+ =

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Marginal Prob.

P(Ai)

Top 20 MBA program (A1) .40Not top 20 MBA program (A2) .60Marginal Probability P(Bj)

P(A1 and B1)+

P(A2 and B1 = P(B1)

P(A1 and B2)+

P(A2 and B2 = P(B2)

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Mutual funddoesn’t outperform the market (B2)

Marginal Prob.

P(Ai)

Top 20 MBA program (A1) .11 .29 .40Not top 20 MBA program (A2) .06 .54 .60Marginal Probability P(Bj) .17 .83

+ +

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Conditional Probability

• Frequently, information about the occurrence of event A changes the probability of event B.

Here is a short demonstration of one possible such situation

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S

B

The sample space is S.

Now, assume event ‘A’ takesplace first followed by event B.Note that event B is containedin event A.

The probability event B takes place is roughly:

{The area of B} {The area of S}.

The probability event B takes place given ‘A’ took place is roughly:

{The area of B} {The area of A}.

Explanation:Since it is known ‘A’ took place, we can reduce the sample space from ‘S’ to ‘A’.

Obviously P(B) is not equal to P(B given A)

AThis is event B

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• The probability of an event given the information about the occurrence of another event is called conditional probability.

• Specifically, the conditional probability of event B given that event A has occurred is calculated as follows:

P(A and B)P(A)

P(B|A) =

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• Example 13– Find the conditional probability that a randomly selected fund

is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market.

• Solution




Marginal Prob.

P(Ai)


.11 .29 .40


.06 .54 .60


.17 .83

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Marginal Prob.

P(Ai)


.11 .29 .40


.06 .54 .60


.17 .83

• Example 13 - continued– Find the conditional probability that a randomly selected fund

is managed by a “Top 20 MBA Program graduate”, given that it did not outperform the market.

• Solution


This informationreduces the relevantsample space to the 83% of event B2.

.83

P(A1 and B2) = .29 .29

P(B2) = .83

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• Example 13 (Solution – continued)P(A1|B2) = P(A1 and B2) = .29 = .3949

P(B2) .83


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• Before the new information becomes available the probability for the occurrence of A1 is

P(A1) = 0.40

• After the new information becomes available P(A1) changes to

P(A1 given B2) = .3949

• Since the occurrence of B2 has changed the probability of A1, the two events are related and are called “dependent events”.

Dependent Events

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Independent Events

• Two events A and B are said to be independent if

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

P(B|A) = P(B)

• That is: The probability of one event is not affected by the occurrence of the other event.

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Dependent and independent events

• Example 13 – continued– We have already seen the dependency between A1

and B2.

– Let us check A2 and B2.• P(B2) = .83

• P(B2|A2)=P(B2 and A2)/P(A2) = .54/.60 = .90

– Conclusion: A2 and B2 are dependent.

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Mutually Exclusive Events

• Two events are said to be mutually exclusive if the occurrence of one precludes the occurrence of the other one.

• If A and B are mutually exclusive, by definition, the probability of their intersection is equal to zero.

• Example: When rolling a dice once the event “The number facing up is 6” and the event “The number facing up is odd” are mutually exclusive.

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Union

• The union event of A and B is the event that occurs when either A or B or both occur.

• It is denoted “A or B” (A U B).

A BC

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Union

• Example 9 – continued Calculating P(A or B))– Determine the probability that a randomly selected

fund outperforms the market or the manager graduated from a top 20 MBA Program.

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• SolutionMutual fund outperforms the market (B1)



A1 or B1 occurs whenever either: A1 and B1 occurs,

A1 and B2 occurs,

A2 and B1 occurs.

P(A1 or B1) = P(A1 and B1) + P(A1 and B2) + P(A2 and B1) = .11 +.29 + .06 = .46

Union

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6.3 Probability Rules and Trees

• We present more methods to determine the probability of the intersection and the union of two events.

• Three rules assist us in determining the probability of complex events.– The complement rule – The multiplication rule– The addition rule

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The complement of event A (denoted by AC) is the event that occurs when event A does not occur.

ACA

Complement of an Event

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Computing Probability using the Complement

• The probability of event A can be calculated using the probability of its complement event by the complement rule

P(A) = 1 - P(AC)P(A) = 1 - P(AC)

Why is this true? Because A and AC consist of all the simple events in the sample space. Therefore, P(A) + P(AC) = 1

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• Example 14What is the probability that two female students will be selected at random to participate in a certain research project, from a class of seven males and three female students?

• Solution– Define the events:

A – the first student selected is a femaleB – the second student selected is a female

– P(A and B) = P(A)P(B|A) = (3/10)(2/9) = 6/90 = .067

Multiplication Rule

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• Example 15What is the probability that a female student will be selected at random in each of two classes of seven males and three female students?

• Solution– Define the events:

A – the student selected in one class is a femaleB – the student selected in the other class is a female

– P(A and B) = P(A)P(B) = (3/10)(3/10) = 9/100 = .09

Multiplication Rule

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For any two events A and B

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

A

B

P(B) =5/13

P(A) =6/13+

P(A and B) =3/13

_

P(A or B) = 8/13

Skip

Addition Rule

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When A and B are mutually exclusive,

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

Addition Rule for Mutually Exclusive Events

A

P(A and B) = 0

B

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• Example 11– The circulation departments of two newspapers in a

large city report that 22% of the city’s households subscribe to the Sun, 35% subscribe to the Post, and 6% subscribe to both.

– What proportion of the city’s household subscribe to either newspaper?

Addition Rule

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• Solution– Define the following events:

• A = the household subscribes to the Sun• B = the household subscribes to the Post

– Calculate the probabilityP(A or B) = P(A) + P(B) – P(A and B) = .22+.35 -.06 = .51

Addition Rule

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Addition Rule

• Example 12 (repeat of example 9) Calculating P(A or B) using the addition ruleDetermine the probability that a randomly selected fund outperforms the market or the manager graduated from a top 20 MBA Program.

• SolutionP(A1 or B1) = P(A1)+P(B1)-P(A1 and B1)= .4 + .17 - .11 = .46


Mutual funddoesn’t outperform the market (B2)

Marginal Prob.P(Ai)


.11 .29 .40


.06 .54 .60


.17 .83

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Addition, Multiplication, and the complementary rules

• Example• A local photocopy shop has three black-and white

(BW) copy machines and two color copiers (CC). It is known that a BW is down 10% of the time for repairs and a CC is down 20%. Assume machines break down independently.

– What proportion of the time a customer cannot run a color photocopy job?

Answer: The probability both CCs are down is (.2)(.2) = .04

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• Example - Continued– If a customer wants both a color copy and a black and

white copy now, what is the probability he can complete the two jobs?

Answer:Each job can be completed if at least one machine of each type is operational.

• P(At least one CC is working) = 1 – P(Both CCs are down) = 1 – (.2)2 = .96.

• P(At least one BW is working) = 1 – (.1)3 = .999• P(At least one CC and at least one BW is working) =

P(At least one CC is working)*P(At least one BW is working) = (.96)(.999).


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• Example - Continued– If a customer wants both a color copy and a black and

white copy now, but this time a CC can perform also a BW job, what is the probability he can complete the two jobs?

Answer:P(The two jobs can be completed) = P(The two CCs are working or one CC and at least one BW are working) = P(The two CCs are working)+P(One CC is working)*P(At least one BW is working) = 2(.8)(.2) + [1 – (.1)3]


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• This is an effective graphical tool that help apply probability rules.

• The events are represented by the tree branches.

• The given probabilities are indicated along these branches too.

Probability Trees

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Probability Trees

• Example 14 revisited (dependent events).– Find the probability of selecting two female students (without

replacement), if there are 3 female students in a class of 10.

First selectionP(F) = 3/10

P( M) = 7/10

P(F|F) = 2/9

Second selection

Second selectionP(F|M) = 3/9

P( M|M) = 6/9

P( M|F) = 7/9

P(FF)=(3/10)(2/9)

P(FM)=(3/10)(7/9)

P(MF)=(7/10)(3/9)

P(MM)=(7/10)(6/9)

Joint probabilitiesFF

MF

MM

FM

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• Example 15 – revisited (independent events)– Find the probability of selecting one female student in each of

two classes if there are 3 female students in a class of 10.

FF

MF

MM

FMFirst selection

P(F) = 3/10

P( M) = 7/10

Second selection

Second selection

P(F|M) = 3/10

P(F|F) = 3/10

P( M|M) =7/10

P( M|F) = 7/10

P(FF)=(3/10)(3/10)

P(FM)=(3/10)(7/10)

P(MF)=(7/10)(3/10)

P(MM)=(7/10)(7/10)

= P(F) =

= P(F) =

= P(M) =

= P(M) =

Probability Trees

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• Example 16 (conditional probabilities)– The pass rate of first-time takers for the bar exam at

a certain jurisdiction is 72%.– Of those who fail, 88% pass their second attempt.– Find the probability that a randomly selected law

school graduate becomes a lawyer (candidates cannot take the exam more than twice).

Probability Trees

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• Solution

Probability Trees

P(Pass1) = .72

P(Fail1 and Pass2)=..28(.88)=.2464

P(Fail1 and Fail2) = (.28)(.12) = .0336

First exam

P(Pass1) = .72

P( Fail1) = .28 Second exam

P(Pass2|Fail1) = .88

P( Pail2|Fail1) = .12

P(Pass) = P(Pass on first exam) + P(Fail on first and Pass on second) = .9664

.72+ .2464

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Baye’s Law

• We use Baye’s law to find the conditional probability of a possible cause for an event that is known to have occurred.

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A

B|A

A CB|A

C

If event B took place,either event A or AC

could have occurred before.

Baye’s law calculates the conditional probability P(A|B) and P(AC|B).

Baye’s Law

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A

B|A

A CB|A

C

Baye’s Law

)B(P)BandA(P

)B|A(P

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• Example 17– Medical tests can produce false-positive or false-negative

results.– A particular test is found to perform as follows:

• Correctly diagnose “Positive” 94% of the time.• Correctly diagnose “Negative” 98% of the time.

– It is known that 4% of men in the general population suffer from the illness.

– What is the probability that a men is suffering from the illness, if the test result were positive?

Baye’s Law

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• Solution– Define the following events

• D = Has the disease • DC = Does not have the disease• PT = Positive test results• NT = Negative test results

– Build a probability tree

Baye’s Law

Called before: A

Called before: AC

Called before: B

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• Solution – Continued– The probabilities provided are:

• P(D) = .04 P(DC) = .96

• P(NT|DC) = .98 P(PT|DC) = .02

• P(PT|D) = .94 P(NT|D)= .06

• The probability to be determined is

)PT|D(P

Baye’s Law

4% of the population has the disease.

The test result is “Negative” 98% of the time given that a patient is not ill.

The test result is “Positive” 94% of the time given that a patient is ill.

The patient is ill given that the testcomes “Positive”.

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P(PT|DC ) = .02

P( NT|D C) = .98

P(PT|D) = .94

P( NT|D) = .06

P(D C) = .96

P(D) = .04

)PT|D(P 6620.0568.0376.

Baye’s Law

Prior probabilities

Likelihoodprobabilities

Posterior probabilities

The prior probability of event ‘D’ was 4%,but because event ‘PT’ occurred theprobability the cause had been ‘D’ increased to 66.2%.

1 introduction to probability chapter 6. 2 introduction in this chapter we discuss the likelihood of...

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