BM6043_Probability

8/3/2019 BM6043_Probability

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Methods of Assigning Probabilities


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Methods to Assigning Probabilities

There are three ways to assign a probability, P(E), to anevent, E, namely:

Classical approach: assigning probabilities based on lawsand rules and present in the form of ratio or fraction.

Relative frequency: assigning probabilities based oncumulated historical data: frequency an event occurred in the

past / total number of opportunities for the event to haveoccurred.

Subjective approach: Assigning probabilities based on theassignors (subjective) judgment.


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Classical Approach

If an experiment has n possible outcomes, this methodwould assign a probability of 1/n to each outcome. It is

necessary to determine the number of possible (collectively

exhaustive) events/outcomes.

Experiment: Rolling adieOutcomes {1, 2, 3, 4, 5, 6}

Probabilities: Each sample point has a 1/6 chance of

occurring.


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Classical Approach

Experiment: Rolling two dice and observing the totalOutcomes: {2, 3, , 12}

Examples:

P(2) = 1/36

P(6) = 5/36

P(10) = 3/36

1 2 3 4 5 6

1 2 3 4 5 6 7

2 3 4 5 6 7 8

3 4 5 6 7 8 9

4 5 6 7 8 9 10

5 6 7 8 9 10 11

6 7 8 9 10 11 12


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Relative Frequency Approach

UMS Computer Shop tracks the number of tablet PC it sellsover a month (30 days):

For example,

10 days out of 30

2 units were sold.

From this we can constructthe probabilities of an event

(i.e. the # of tablet PC sold on a given day)

Tablet PC Sold # of Days

0 1

1 22 10

3 12

4 5


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Relative Frequency Approach

There is a 40% chance UMS Computer will sell 3 desktops

on any given day

Tablet PC Sold # of Days Tablet PC Sold

0 1 1/30 = .03

1 2 2/30 = .07

2 10 10/30 = .33

3 12 12/30 = .404 5 5/30 = .17

= 1.00


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Subjective Approach

Not a scientific approach.

Probability comes from persons intuition or reasoning.

Based on the feelings or insights of the person determiningthe probability.

Examples of subjective probability?

Is it reliable?


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Joint, Marginal, Conditional Probability

Methods to determine probabilities of events that result fromcombining other events in various ways.

There are several types of combinations and relationships

between events:

Complement eventIntersection of events

Union of events

Mutually exclusive events

Dependent and independent events


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Complement of an Event

Thecomplement of event A is defined to be the eventconsisting of all sample points that are not in A.

Complement of A is denoted by Ac (or A')

The Venn diagram below illustrates the concept of acomplement.

P(A) + P(Ac ) = 1 A Ac


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For example, the rectangle stores all the possible tosses of 2dice {(1,1), 1,2), (6,6)} Let A = tosses totaling 7 {(1,6),

(2, 5), (3,4), (4,3), (5,2), (6,1)}

P(Total = 7) + P(Total not equal to 7) = 1

A Ac


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Can you provide examples ofcomplement probability?



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6.12

Intersection of Two Events

The intersection of events A and B is the set of all samplepoints that are in both A and B.

The intersection is denoted: A and B

Thejoint probability of

A and B is the probability of

the intersection of A and B,i.e. P(A and B)

or P(AB)

A B


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6.13


For example, let A = tosses where first toss is 1 {(1,1), (1,2),(1,3), (1,4), (1,5), (1,6)}

and B = tosses where the second toss is 5 {(1,5), (2,5), (3,5),

(4,5), (5,5), (6,5)}

The intersection is {(1,5)} must occur in both A and B

Thejoint probability of

A and B is the probability of

the intersection of A and B,

i.e. P(A and B) = 1/36

A B


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Can you provide examples of jointprobability?


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6.15

Union of Two Events

The union of two events A and B, is the event containing allsample points that are in A or B or both:

Union Probability of A and B is denoted: A or B

or P(AuB)

A B


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6.16

Union of Two Events

For example, let A = tosses where first toss is 1 {(1,1), (1,2), (1,3),(1,4), (1,5), (1,6)}

and B is the tosses that the second toss is 5 {(1,5), (2,5), (3,5), (4,5),

(5,5), (6,5)}

Union of A and B is {(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,5), (3,5),

(4,5), (5,5), (6,5)}

A B


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6.17

Mutually Exclusive Events

When two events aremutually exclusive (that is the twoevents cannot occur together), their joint probability is 0,

hence:

A B

Mutually exclusive; no points in common

Gender of employees: A = Male and B = Female

Manufactured part: A = Okay and B = Defective

f


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6.18

Basic Relationships of Probability

Complement of Event

Intersection of Events

Union of Events

Mutually Exclusive Events

A Ac A B

A BA B

E l 1


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6.19

Example 1

Why are some mutual fund managers more successful thanothers? One possible factor is where the manager earned his

or her MBA. The following table compares mutual fund

performance against the ranking of the university where thefund manager earned their MBA:

Mutual fund outperformsthe market

Mutual fund doesntoutperform the market

Top 20 MBA program .11 .29

Not top 20 MBA program .06 .54

E.g. This is the probability that a mutual fundoutperformsAND the manager was in a top-

20 MBA program; its a joint probability. Probability Matrix

E l 1


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6.20

Example 1

Alternatively, we could introduce shorthand notation torepresent the events:

A1 = Fund manager graduated from a top-20 MBA program

A2 = Fund manager did not graduate from a top-20 MBA programB1 = Fund outperforms the market

B2 = Fund does not outperform the market

B1 B2

A1 .11 .29

A2

.06 .54

E.g. P(A2 and B1) = .06

= the probability a fund outperforms the market

and the manager isnt from a top-20 school.

0.40

0.60

0.17 0.83 1.00

M i l P b biliti


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6.21

Marginal Probabilities

Marginal probabilities are computed by adding across rowsand down columns; that is they are calculated in themargins

of the table:

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

P(B1) = .11 + .06

P(A2) = .06 + .54

whats the probability a fundoutperforms the market?

whats the probability a fund

manager isnt from a top school?

BOTH margins must add to 1

(useful error check)

C diti l P b bilit


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6.

Conditional Probability

Conditional probability is used to determine how two eventsare related; that is, we can determine the probability of one

eventgiven the occurrence of another related event.

Conditional probabilities are written as P(A | B) and read as

the probability of A given B and is calculated as:

B A

How is this differ from

joint probability? Look at the denominator

C diti l P b bilit


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6.23


Again, the probability of an eventgiven that another eventhas occurred is called a conditional probability

Note how A given B and B given A are related

Example: Conditional Probability


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6.24

Example: Conditional Probability

You want to determine what changes in office design wouldimprove productivity. 70% of respondents believed noise

reduction would improve productivity and 67% said

increased office space would improve productivity. Inaddition, 56% respondents believed both would improve

productivity.

A worker is picked randomly and this worker believes noise

reduction would improve productivity. What is the

probability that this worker also believe increased officespace would improve productivity?



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6.25


We want to calculate P(B1 | A1)

Thus, there is a 27.5% chance that that a fund will outperform the market

given that the manager graduated from a top-20 MBA program.

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

Independence


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6.26

Independence

One of the objectives of calculating conditional probability isto determine whether two events are related.

In particular, we would like to know whether they areindependent, that is, if the probability of one event isnot

affectedby the occurrence of the other event.

Two events A and B are said to be independent if

P(A|B) = P(A)

or

P(B|A) = P(B)

Independence


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6.27

Independence

For example, we saw that

P(B1 | A1) = .275

The marginal probability for B1 is: P(B1) = 0.17

Since P(B1|A1) P(B1), B1 and A1 arenot independentevents.

Stated another way, they aredependent. That is, theprobability of one event (B1) is affectedby the occurrence of

the other event (A1).

Union


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6.28

Union

We stated earlier that the union of two events is denoted as:A or B. We can use this concept to answer questions like:

Determine the probability that a fund outperforms the market

or the manager graduated from a top-20 MBA program.

Union


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6.29

Union

Determine the probability that a fund outperforms (B1

)

or the manager graduated from a top-20 MBA program (A1).

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

A1 or B1 occurs whenever:

A1 and B1 occurs, A1 and B2 occurs, orA2 and B1 occurs

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

Union


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6.30

Union

Determine the probability that a fund outperforms (B1

)

or the manager graduated from a top-20 MBA program (A1).

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

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

B1

A1

Alternatively


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6.31

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

B1

A1

Alternatively

Take 100% and subtract off when doesnt A1 or B1 occur?

i.e. at A2 and B2

P(A1

or B1

) = 1 P(A2

and B2

) = 1 .54 = .46

Probability Rules and Trees


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6.32

Probability Rules and Trees

We introduce three rules that enable us to calculate theprobability of more complex events from the probability of

simpler events

The Complement Rule,

The Multiplication Rule, and

The Addition Rule

Complement Rule


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6.33

Complement Rule

As we saw earlier with the complement event, thecomplement rule gives us the probability of an event NOT

occurring. That is:

P(AC) = 1 P(A)

For example, in the simple roll of a die, the probability of the

number 1 being rolled is 1/6. The probability that some

number other than 1 will be rolled is 1 1/6 = 5/6.

Multiplication Rule


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6.34

Multiplication Rule

Themultiplication rule is used to calculate thejointprobability of two events. It is based on the formula for

conditional probability defined earlier:

If we multiply both sides of the equation by P(B) we have:

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

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

If A and B are independent events, then P(A and B) = P(A)P(B)

Example of Multiplication I


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6.35


A graduate statistics course has seven male and three femalestudents. The professor wants to select two students at

random to help her conduct a research project. What is the

probability that the two students chosen are female?

Let A represent the event that the first student is female

P(A) = 3/10 = .30

What about the second student?



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6.36


A graduate statistics course has seven male and three femalestudents. The professor wants to select two students atrandom to help her conduct a research project. What is the

probability that the two students chosen are female?

Let B represent the event that the second student is female

P(B | A) = 2/9 = .22

That is, the probability of choosing a female studentgiventhat the first student chosen is 2 (females) / 9 (remainingstudents) = 2/9



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6.37


A graduate statistics course has seven male and three femalestudents. The professor wants to select two students atrandom to help her conduct a research project. What is theprobability that the two students chosen are female?

Thus, we want to answer the question: what is P(A and B) ?

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

There is a 6.7% chance that the professor will choose twofemale students from her grad class of 10.

Example of Multiplication II


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6.38


Refer to Example 6.5. The professor who teaches the courseis suffering from the flu and will be unavailable for twoclasses. The professors replacement will teach the next twoclasses. His style is to select one student at random and pickon him or her to answer questions during that class. What isthe probability that the two students chosen are female?

Let A represent the event that the first student is female

P(A) = 3/10 = .30

What about the second student?



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6.39

p p

Let B represent the event that the second student is female

P(B | A) = 3/10 = .30

That is, the probability of choosing a female studentgiven

that the first student chosen is unchanged since the studentselected in the first class can be chosen in the second class.

P(A and B) = P(A)P(B|A) = (3/10)(3/10) = 9/100 = .090

What is the different

between

Multiplication

Example I and II?

Addition Rule


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6.40

Recall: theaddition rule is used to compute the probabilityof event Aor Borboth A and B occurring; i.e. the union of

A and B.

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

A B A B= +

P(A or B) = P(A) + P(B) P(A and B)If A and B are

mutually exclusive,

then this term

goes to zero

Example of Addition


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6.41

p

In a large city, two newspapers are published, the Sun andthe Post. The circulation departments report that 22% of the

citys households have a subscription to the Sun and 35%

subscribe to the Post. A survey reveals that 6% of allhouseholds subscribe to both newspapers. What proportion

of the citys households subscribe to either newspaper?

That is, what is the probability of selecting a household at

random that subscribes to the Sun or the Post or both?

i.e. what is P(Sun or Post) ?

Example of Addition


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6.42

p

In a large city, two newspapers are published, the Sun andthe Post. The circulation departments report that 22% of thecitys households have a subscription to the Sun and 35%subscribe to the Post. A survey reveals that 6% of allhouseholds subscribe to both newspapers. What proportionof the citys households subscribe to either newspaper?

P(Sun or Post) = P(Sun) + P(Post) P(Sun and Post)

= .22 + .35.06 = .51

There is a 51% probability that a randomly selectedhousehold subscribes to one or the other or both papers

Probability Trees


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6.43

y

An effective and simpler method of applying the probabilityrules is the probability tree, wherein the events in an

experiment are represented by lines. The resulting figure

resembles a tree, hence the name. We will illustrate theprobability tree with several examples, including two that we

addressed using the probability rules alone.

Probability Trees


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6.44

First selection Second selection

P(F)=3/10

P(M)=7/10 P

(F|M)=3/9

P(F|F)=2/9

P(M|M)=6/9

P(M|F)=7/9

This is P(F), the probability ofselecting a female student first

This is P(F|F), the probability ofselecting a female studentsecond, given that a female wasalready chosen first

Probability Trees


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6.45

At the ends of the branches, we calculatejointprobabilities as the product of the individual probabilities

on the preceding branches.


P(F)=3/10

P(M)=7/10 P

(F|M)=3/9

P(F|F)=2/9

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 probabilities

Example of Probability Trees


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6.46

Suppose we have our grad class of 10 students again, butmake the student sampling independent, that is with

replacement a student could be picked first and picked

again in the second round. Our tree and joint probabilitiesnow look like: FF

MF

MM

FMP(F)

=3/10

P(M)=7/10 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)

Probability Trees


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6.47

3/9 + 6/9= 9/9 = 1

2/9 + 7/9

= 9/9 = 1

3/10 + 7/10

= 10/10 = 1

The probabilities associated with any set of branches fromone node must add up to 1.00


P(F)=3/10

P(M)=7/10

P(F|M)=3/9

P(F|F)=2/9

P(M|M)=6/9

P(M|F)=7/9

Handy way to check

your work !

Probability Trees


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6.48

Note: there is no requirement that the branches splits bebinary, nor that the tree only goes two levels deep, or that

there be the same number of splits at each sub node

Probability Trees


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6.49

Law school grads must pass a bar exam. Suppose pass ratefor first-time test takers is 72%. They can re-write if they fail

and 88% pass their second attempt. What is the probability

that a randomly grad passes the bar?P(Pass) = .72

P(Fail and Pass)=

(.28)(.88)=.2464

P(Fail and Fail) =

(.28)(.12) = .0336

First exam

P(Pass)=

.72

P(Fail)=.28

Second exam

P(Pass|Fail)=.88

P(Fail|Fail)=.12

Probability Trees


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6.50

What is the probability that a randomly grad passes the bar?There is almost a 97% chance they will pass the bar

P(Pass) = P(Pass 1st) + P(Fail 1st and Pass 2nd) =

= 0.7200 + 0.2464 = .9664P(Pass) = .72

P(Fail and Pass)=

(.28)(.88)=.2464

P(Fail and Fail) =

(.28)(.12) = .0336

First exam

P(Pass)=

.72

P(Fail)=.28

Second exam

P(Pass|Fail)=.88

P(Fail|Fail)=.12

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BM6043_Probability