8/14/2019 9185l_2013_Lecture2.pdf

1/62

Risk Analysis, Assessment & Management

9185 Chemical & Biochemical Engineering 2013

Housyn Mahmoud

Probability ReviewEvent Tree & Decision Tree Analysis

8/14/2019 9185l_2013_Lecture2.pdf

2/62

ObjectivesBy the end of this presentation you will

know:

Probability Requirements

Probability Approaches

Event Tree Analysis Decision Tree Analysis

8/14/2019 9185l_2013_Lecture2.pdf

3/62

Probabilities

http://www.google.ca/url?sa=i&rct=j&q=cards&source=images&cd=&cad=rja&docid=lmWNao-cJluzyM&tbnid=vf-JX3ATwW6E3M:&ved=0CAUQjRw&url=http://www.angelfire.com/anime3/db1234567/cards.html&ei=o16xUePmJ4bVyAGt9IHQAw&psig=AFQjCNEr2qflS6x61So-c4YLMAfuwOmqbQ&ust=1370664947570882http://www.google.ca/url?sa=i&rct=j&q=dice&source=images&cd=&cad=rja&docid=7c5Jqi8gsoU4wM&tbnid=z11KD5-xFg4N7M:&ved=0CAUQjRw&url=http://commons.wikimedia.org/wiki/File:Dice.png&ei=zVyxUe-vBOr1ygHw3YCwCw&bvm=bv.47534661,d.aWc&psig=AFQjCNEhjGvueoRYdDtvgQs8zKNIbBuonA&ust=1370664518279509

8/14/2019 9185l_2013_Lecture2.pdf

4/62

Random Experiment a random experimentis an action or

process that leads to one of several possible

outcomes. For example:

Experiment Outcomes

Flip a coin Heads, Tails

Exam Marks Numbers: 0, 1, 2, ..., 100

Assembly Time t > 0 seconds

Course Grades F, D, C, B, A, A+

8/14/2019 9185l_2013_Lecture2.pdf

5/62

ProbabilitiesList the outcomes of a random experimentList: Called the Sample Space

Outcomes: Called the Simple Events

This list must be exhaustive, i.e. ALL possible outcomes included.

Die roll {1,2,3,4,5} Die roll {1,2,3,4,5,6}

The list must be mutually exclusive, i.e. no two outcomes canoccur at the same time:

Die roll {odd number or even number}

Die roll{ number less than 4 or even number}

8/14/2019 9185l_2013_Lecture2.pdf

6/62

Sample Space A list of exhaustive [dont leave anything out]and mutually

exclusive outcomes [impossible for 2 different events tooccur in the same experiment]is called a sample spaceand is denoted by S.

The outcomes are denoted by O1, O2, , Ok

Using notation from set theory, we can represent thesample space and its outcomes as:

S = {O1, O2, , Ok}

8/14/2019 9185l_2013_Lecture2.pdf

7/62

Requirements of ProbabilitiesGiven a sample space S = {O1, O2, , Ok}, theprobabilities

assigned to the outcome must satisfy these requirements:

(1) The probability of any outcome is between 0 and 1

i.e. 0 P(Oi) 1 for each i, and

(2) The sum of the probabilities of all the outcomes equals 1i.e. P(O1) + P(O2) + + P(Ok) = 1

P(Oi) represents the probability of outcome i

8/14/2019 9185l_2013_Lecture2.pdf

8/62

Approaches to AssigningProbabilities

There are three ways to assign a probability, P(Oi), to anoutcome, Oi, namely:

Classical approach: make certain assumptions (such asequally likely, independence) about situation.

Relative frequency: assigning probabilities based onexperimentation or historical data.

Subjective approach: Assigning probabilities based on theassignors judgment. [Bayesian]

8/14/2019 9185l_2013_Lecture2.pdf

9/62

Classical ApproachIf an experiment has n possible outcomes[all equally

likely to occur], this method would assign a probabilityof 1/n to each outcome.

Experiment: Rolling a die

Sample Space: S = {1, 2, 3, 4, 5, 6}

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

occurring.What about randomly selecting a student and observing their

gender? S = {Male, Female}

Are these probabilities ?

8/14/2019 9185l_2013_Lecture2.pdf

10/62

Classical ApproachExperiment: Rolling 2 die [dice] and summing 2 numbers

on top.

Sample Space: S = {2, 3, , 12}Probability Examples:

P(2) = 1/36

P(7) = 6/36

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

What are the underlying,

unstated assumptions??

8/14/2019 9185l_2013_Lecture2.pdf

11/62

Relative Frequency ApproachBB Computer Shop tracks the number of desktop computer

systems it sells over a month (30 days):

For example,

10 days out of 30

2 desktops were sold.

From this we can construct

theestimated probabilities of an event

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

DesktopsSold # of Days

0 1

1 2

2 103 12

4 5

8/14/2019 9185l_2013_Lecture2.pdf

12/62

Relative Frequency Approach

There is a 40% chance BB will sell 3 desktops on anygiven day [Based on estimates obtained from sample of30 days]

Desktops Sold [X] # of Days Desktops Sold

0 1 1/30 = .03= P(X=0)

1 2 2/30 = .07= P(X=1)2 10 10/30 = .33= P(X=2)

3 12 12/30 = .40= P(X=3)

4 5 5/30 =.17 = P(X=4)

= 1.00

8/14/2019 9185l_2013_Lecture2.pdf

13/62

Subjective ApproachIn the subjective approach we define probability as the

degree of belief that we hold in the occurrence of an event

P(you drop this course)

P(NASA successfully land a man on the moon)

P(girl says yes when you ask her to marry you)

8/14/2019 9185l_2013_Lecture2.pdf

14/62

Events & ProbabilitiesAn individual outcome of a sample space is called a simple

event [cannot break it down into several other events],

An eventis a collection or set of one or more simple events

in a sample space.

Roll of a die: S = {1, 2, 3, 4, 5, 6}Simple event:the number 3 will be rolled

Event:an even number (one of 2, 4, or 6) will be rolled

8/14/2019 9185l_2013_Lecture2.pdf

15/62

Events & ProbabilitiesTheprobability of an eventis the sumof the probabilities

of the simple events that constitute the event.

E.g. (assuming a fair die) S = {1, 2, 3, 4, 5, 6} and

P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = 1/6

Then:

P(EVEN) = P(2) + P(4) + P(6) = 1/6 + 1/6 + 1/6 = 3/6 = 1/2

8/14/2019 9185l_2013_Lecture2.pdf

16/62

Interpreting Probability

One way to interpret probability is this:

If a random experiment is repeated an infinitenumber oftimes, the relative frequency for any given outcome is the

probability of this outcome.

For example, the probability of heads in flip of a balanced coinis .5, determined using the classical approach. The probability

is interpreted as being the long-term relative frequency of

heads if the coin is flipped an infinite number of times.

8/14/2019 9185l_2013_Lecture2.pdf

17/62

Joint, Marginal, ConditionalProbability

We study methods to determine probabilities of events that

result from combiningother events in various ways.

There are several types of combinations and relationships

between events:

Complement of an event [everything other than that event]

Intersection of two events [event A andevent B] or [A*B] Union of two events [event A orevent B] or [A+B]

8/14/2019 9185l_2013_Lecture2.pdf

18/62

Example 1

Why are some mutual fund managers more successful than others?

One possible factor is where the manager earned his or her MBA.

The following table compares mutual fund performance against the

ranking of the school where the fund manager earned their MBA:Where do we get these probabilities from? [population or sample?]

Mutual fund

outperforms the market

Mutual fund doesnt

outperform the market

Top 20 MBA program .11 .29

Not top 20 MBA program .06 .54

E.g. This is the probability that a mutualfund outperformsANDthe manager was

in a top-20 MBA program; its ajointprobability [intersection].

8/14/2019 9185l_2013_Lecture2.pdf

19/62

Example 1

Alternatively, we could introduce shorthand notation to represent theevents:

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(A2and B1) = .06

= the probability a fund outperforms the market

andthe manager isnt from a top-20 school.

8/14/2019 9185l_2013_Lecture2.pdf

20/62

Marginal Probabilities

Marginal probabilitiesare computed by adding across rows and down

columns; that is they are calculated in the marginsof 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)

8/14/2019 9185l_2013_Lecture2.pdf

21/62

Conditional Probability

Conditional probabilityis used to determine how two

events are related; that is, we can determine the probability

of one event giventhe occurrence of another related event.Experiment: random select one student in class.

P(randomly selected student is male) =

P(randomly selected student is male/student is on 3rdrow) =

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

the probability of A givenB and is calculated as:

8/14/2019 9185l_2013_Lecture2.pdf

22/62

Conditional Probability

Again, the probability of an event giventhat another eventhas occurred is called a conditional probability

P( A and B) = P(A)*P(B/A) = P(B)*P(A/B) both are true

Keep this in mind!

8/14/2019 9185l_2013_Lecture2.pdf

23/62

Example 2Conditional Probability

Whats the probability that a fund will outperform the market

giventhat the manager graduated from a top-20 MBA

program?

Recall:

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

A2= Fund manager did not graduate from a top-20 MBA program

B1= Fund outperforms the market

B2= Fund does not outperform the market

Thus, we want to know what is P(B1|A1)?

8/14/2019 9185l_2013_Lecture2.pdf

24/62

Conditional Probability

We want to calculate P(B1| A1)

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

8/14/2019 9185l_2013_Lecture2.pdf

25/62

Independence

One of the objectives of calculating conditional probability is to

determine whether two events are related.

In particular, we would like to know whether they are

independent, that is, if the probability of one event is not

affectedby the occurrence of the other event.

Two events A and B are said to be independentif

P(A|B) = P(A)

and

P(B|A) = P(B)

P(you have a flat tire going home/radio quits working)

8/14/2019 9185l_2013_Lecture2.pdf

26/62

Independence

For example, we saw that

P(B1| A1) = .275

The marginal probability for B1is: P(B1) = 0.17

Since P(B1|A1) P(B1), B1and A1are not independent

events.

Stated another way, they are dependent. That is, the

probability of one event (B1) is affectedby the occurrence of

the other event (A1

).

8/14/2019 9185l_2013_Lecture2.pdf

27/62

Union

Determine the probability that a fund outperforms (B1)

orthe 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

A1or B1occurs whenever:

A1and B1occurs,A1and B2occurs, orA2and B1occurs

P(A1or B1) =.11+.06+.29 =.46

8/14/2019 9185l_2013_Lecture2.pdf

28/62

Probability Rules and Trees

We introduce three rules that enable us to calculate the

probability of more complex events from the probability of

simpler events The Complement Rule:May be easier to calculate the probability

of the complement of an event and then subtract it from 1.0 to getthe probability of the event.

P(at least one head when you flip coin 100 times)

= 1P(0 heads when you flip coin 100 times)

The Multiplication Rule: P(A*B) way I write it

The Addition Rule: P(A+B) way I write it

8/14/2019 9185l_2013_Lecture2.pdf

29/62

Example 3

A graduate Risk Assessment course has seven maleand three

Female students. The professor wants to select two students at

random to help her conduct a research project. What is theprobability that the two students chosen are female?

P(F1 * F2) = ???

Let F1represent the event that the first student is female

P(F1) = 3/10 = .30

What about the second student?

P(F2/F1) = 2/9 = .22

P(F1 * F2) =P(F1) * P(F2/F1) = (.30)*(.22) = 0.066

NOTE: 2 events are NOTindependent.

8/14/2019 9185l_2013_Lecture2.pdf

30/62

Example 4

The professor in Example 3 is unavailable. Her replacement

will teach two classes. His style is to select one student at

random and pick on him or her in the class. What is theprobability that the two students chosen are female?

Both classes have 3 female and 7 male students.

P(F1 * F2) =P(F1) * P(F2/F1) = P(F1) * P(F2)= (3/10) * (3/10) = 9/100 = 0.09

NOTE: 2 eventsAREindependent.

8/14/2019 9185l_2013_Lecture2.pdf

31/62

Addition Rule

Additionruleprovides a way to compute the probability of

eventA orBorboth A and Boccurring; i.e. the union of A

and B.

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

Why do we subtract the joint probability P(A and B) from thesum of the probabilities of A and B?

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

8/14/2019 9185l_2013_Lecture2.pdf

32/62

Addition Rule

P(A1) = .11 + .29 = .40

P(B1) = .11 + .06 = .17

By adding P(A) plus P(B) we add P(A and B) twice.To correct we subtract P(A and B) from P(A) + P(B)

B1 B2 P(Ai)

A1 .11 .29 .40

A2 .06 .54 .60

P(Bj) .17 .83 1.00

P(A1or B1) =P(A) + P(B)P(A and B) =.40+.17- .11 =.46

B1

A1

8/14/2019 9185l_2013_Lecture2.pdf

33/62

Addition Rule for MutuallyExcusive Events

If and A and B are mutually exclusive the occurrence of one

event makes the other one impossible. This means that

P(A and B) = P(A * B) = 0

The addition rule for mutually exclusive events is

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

Only if A and B are Mutually Exclusive.

8/14/2019 9185l_2013_Lecture2.pdf

34/62

Example 5In a large city, two newspapers are published, the Sun and

the Post. The circulation departments report that 22%of the

citys households have a subscription to the Sunand 35%

subscribe to the Post. A survey reveals that 6%of all

households 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?

P(Sunor Post)= P(Sun) + P(Post)P(Sunand Post)

= .22 + .35 .06= .51

8/14/2019 9185l_2013_Lecture2.pdf

35/62

Probability Trees [DecisionTrees]

Aprobability treeis a simple and effective method of applying the

probability rules by representing events in an experiment by lines. The

resulting figure resembles a tree.

First selection Second selectionThis is P(F), the probability ofselecting a female student first

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

second, given that a female wasalready chosen first

8/14/2019 9185l_2013_Lecture2.pdf

36/62

Probability TreesAt the ends of the branches, we calculatejoint probabilities asThe productof the individual probabilities on the preceding branches.

First selection Second selectionP(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

Sample Space:[F1*F2, F1*M2, M1*F2, M1*M2]

8/14/2019 9185l_2013_Lecture2.pdf

37/62

Probability Trees

Note: there is no requirement that the branches splits bebinary, nor that the tree only goes two levels deep, or thatthere be the same number of splits at each sub node

8/14/2019 9185l_2013_Lecture2.pdf

38/62

Example 6

Law school grads must pass a bar exam. Suppose pass rate

for first-time test takers is 72%.They can re-write if they

fail and 88% pass their second attempt

[P(pass take2/fail take 1)].

What is the probability that a randomly grad passes the bar?

[sample space?]

8/14/2019 9185l_2013_Lecture2.pdf

39/62

Example 6

P(Pass) = .72

P(Fail and Pass)=

(.28)(.88)=.2464

P(Fail and Fail) =

(.28)(.12) = .0336

First exam Second exam

8/14/2019 9185l_2013_Lecture2.pdf

40/62

Bayes Law

Bayes Law is named for Thomas Bayes, an eighteenthcentury mathematician.

In its most basic form, if we knowP(B | A),

we can apply Bayes Law to determine P(A | B)

P(B|A) P(A|B)for example

8/14/2019 9185l_2013_Lecture2.pdf

41/62

Breaking News: New test for early detection ofcancer has been developed.

LetC = event that patient has cancer

Cc = event that patient does not have cancer

+ = event that the test indicates a patient has cancer- = event that the test indicates that patient does not have cancer

Clinical trials indicate that the test is accurate 95% of the time

in detecting cancer for those patients who actually havecancer: P(+/C) = .95

but unfortunately will give a + 8% of the time for those

patients who are known not to have cancer: P(+/ Cc) = .08

8/14/2019 9185l_2013_Lecture2.pdf

42/62

Breaking News: New test for early detection ofcancer has been developed.

It has also been estimated that approximately 10% of the

population have cancer and dont know it yet:P(C) = .10

You take the test and receive a + test results.

Should you be worried? P(C/+) = ?????

8/14/2019 9185l_2013_Lecture2.pdf

43/62

What we know.P(+/C) = .95 P(+/ Cc) = .08 P(C) = .10From these probabilities we can find

P(-/C) = .05 P(-/ Cc) = .92 P(Cc) = .90

Have Cancer: C Do Not Have Cancer: CC

+

-

True State of Nature

TestR

esults

8/14/2019 9185l_2013_Lecture2.pdf

44/62

Bayesian Terminology

The probabilities P(A) and P(AC) are calledprior Probabilities

because they are determinedpriorto the decision about

taking the preparatory course.

The conditional probability P(A | B) is called aposterior

probability(or revised probability), because the priorrobability

is revised afterthe decision about taking the preparatory

course.

8/14/2019 9185l_2013_Lecture2.pdf

45/62

Students WorkBayes Problem

The Rapid Test is used to determine whether someone hasHIV [H]. The false positive and false negative rates are0.05P(+/ Hc)and 0.09P(-/H)respectively.

The doctor just received a positive test results on one oftheir patients [assumed to be in a low risk group for HIV].The low risk group is known to have a 6% P(H)probability of having HIV. What is the probability that thispatient actually has HIV [after they tested positive]. Feel

free to use a table to work this problemP(H) = 0.06 **** P(Hc) = ?

P(+/ Hc) = 0.05 **** P(-/ Hc) = ?

P(-/H) = 0.09 **** P(+/H) = ?

8/14/2019 9185l_2013_Lecture2.pdf

46/62

9185 Chemical & Biochemical Engineering 2013

Event Tree & Decision Tree Analysis

8/14/2019 9185l_2013_Lecture2.pdf

47/62

Event Tree

An event tree is a graphical representationof a series of possible events in an accident

sequence. Using this approach assumes that as each event occurs,

there are only two outcomes, failure or success. A successends the accident sequence and the postulated outcome

is either the accident sequence terminated successfully orwas mitigated successfully.

8/14/2019 9185l_2013_Lecture2.pdf

48/62

Example Event Tree

A fire starts in a plant. This is the initiating event.

Then the fire suppression system is challenged. If

the system actuates, the fire is extinguished orsuppressed and the event sequence ends. If the fire

suppression system fails then the fire is not

extinguished or suppressed and the accidentsequence progresses.

8/14/2019 9185l_2013_Lecture2.pdf

49/62

Accident Sequence

8/14/2019 9185l_2013_Lecture2.pdf

50/62

Accident Sequence Event Tree

8/14/2019 9185l_2013_Lecture2.pdf

51/62

ProbabilityPath Line

h

8/14/2019 9185l_2013_Lecture2.pdf

52/62

Event Sequence withProbability

8/14/2019 9185l_2013_Lecture2.pdf

53/62

Results

The probability derived for a fire in which the fire suppressionsystem actuates and the consequence is minimal damage isapproximately 1/1000 or 1 103.

The probability derived for a fire in which the fire suppressionsystem fails to actuate but a fire alarm signal is successfullytransmitted to the local fire department and there is onlymoderate damage is 1/100,000 or 1 105.

Finally, the probability that a fire occurs and both the firesuppression system and the fire alarm system fail and severedamage occurs is 5/10,000,000 or 5 108.

8/14/2019 9185l_2013_Lecture2.pdf

54/62

Decision Trees

Decision analysis and their associated tree analysis techniquesassume that a decision is being made under risk.

The style presented here is one that is commonly used inindustry. Influence diagrams are a close cousin to this style ofdecision tree.

Decision tree format

8/14/2019 9185l_2013_Lecture2.pdf

55/62

Example:

Decision Tree Analysis

In this scenario, you are evaluating four systems.

Table :The information about the systems.

8/14/2019 9185l_2013_Lecture2.pdf

56/62

Decision Tree 2.

8/14/2019 9185l_2013_Lecture2.pdf

57/62

Final outcomes

System A

0.3 100,000,000 = 30,000,000

0.5 25,000,000 = 12,500,000

0.2 ()4,000,000 = ()800,000 Average benefit = $41,700,000

Cost is 1,000,000

Net benefit = Benefit Cost = 41,700,000 1,000,000 = $40,700,000

8/14/2019 9185l_2013_Lecture2.pdf

58/62

Final outcomes

System B

0.3 80,000,000 = 24,000,000

0.5 35,000,000 = 17,500,000

0.2 ()1,000,000 = ()200,000 Average benefit = $41,300,000

Cost is 750,000

Net benefit = Benefit Cost = 41,300,000 750,000 =

$40,550,000

8/14/2019 9185l_2013_Lecture2.pdf

59/62

Final outcomes

System C

0.3 125,000,000 = 37,500,000

0.5 45,000,000 = 22,500,000

0.2 0 = 0 Average benefit = $60,000,000

Cost is 1,500,000

Net benefit = Benefit Cost = 60,000,000 1,500,000 =

$58,500,000

8/14/2019 9185l_2013_Lecture2.pdf

60/62

Final outcomes

System D

0.3 95,000,000 = 28,500,000

0.5 65,000,000 = 32,500,000

0.2 ()80,000,000 = ()1,600,000 Average benefit = $45,000,000

Cost is 450,000

Net benefit = Benefit Cost = 45,000,000 450,000 =$44,550,000

System C: provides the best return on the investment.Another benefit of System C is that there is no potential for a negative outcome.

8/14/2019 9185l_2013_Lecture2.pdf

61/62

Summary

Event and decision trees are highly effectivetools that have broad application in providing

visual and analytical means to betterunderstand both simple as well as complexevents.

One can use the tools to model events in

failure, as well as success space to helpbetter understand them.

8/14/2019 9185l_2013_Lecture2.pdf

62/62

Questions?