uncertainty and probability using probabilities using decision trees probability revision

Uncertainty and probability

Using probabilities

Using decision trees

Probability revision

Today’s agenda

• Important terms

• Simple review (objective, subjective, marginal, joint, and conditional probabilities)

• Examples: outcomes, expected values, risk attitudes

• Examples: action choices, decision trees

Vocabulary

• A probability is a number between zero and one representing the likelihood of the occurrence of some event.

• Probability– objective vs. subjective– marginal vs. joint– joint vs. conditional– prior vs. posterior– likelihood vs. posterior

Vocabulary continued

• Outcomes or payoffs (mutually exclusive)

• Action choices

• States of nature

• Decision tree

• Expected value

• Risk

Probability

Imagine an urn containing 1500 red, pink, yellow, blueand white marbles.

Take one ball from the urn. What is:

P(black) =

P(~black) = ~ = NOT

0

1

Probabilities are all greater than or equal to zero and lessthan or equal to one.

Same urn:Suppose the number of balls is as follows:

Red 400Pink 100Yellow 400Blue 500White 100Total 1500

What is:

P(Red) =

P(Pink) =

P(Yellow) =

P(Blue) =

P(White) =

Total =

400/1500 = .267

100/1500 = .067

400/1500 = .267

500/1500 = .333

100/1500 = .067

1

Joint probabilities and independence

Define A as the event “draw a red or a pink marble.”

We know 500 marbles are either red or pink.

What are: P(A) =

P(~A) =

1500100400

(1 - P(A)) = .67

= .33

Joint probabilities and independence (we’re getting

there)

Define B as the event, “draw a pink or white marble.”

We know 200 marbles are pink or white.

What are: P(B) =

P(~B) =

.133

.867


Define A as the event “draw a red or a pink marble.”

Define B as the event “draw a pink or white marble.”

What is: P(A, B) = P(A B)

This is the joint probability of A and B.

What color is the marble? Pink

P(A, B) = P(pink) = 1500100

= .0667


Are A and B independent?

Note that P(A, B) P(A) * P(B) = .33 * .13 = .0429

Are A and B mutually exclusive?

What is the probability of A or B?

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

= .40


Suppose we draw one marble from the urn andreplace it. Then, we draw a second marble.

What is:P(Red, Red) = = .071

1500400

*1500400

Are (Red, Red) independent?

P(Red, Blue) = .088

Are (Red, Blue) independent?

Joint and marginal probabilities

100pink

100white

400red

900all others

A ~A

B

~B

200

1300

500 1000

What are:

P(B) =1500200

P(~B) = .867

P(A, B) =1500100

P(A or B) =

1500100

1500200

1500500

1500

= .133

Conditional probabilities

What is

P(A | B) = 200100

P(~A | ~B) =1300900

P(A | ~B) =1300400

P(~B | A) =500400

The probability that a particularevent will occur, given we alreadyknow that another event hasoccurred.

We have information to bringto bear on the base rate probability of the event

100pink

100white

400red

900all others

A ~A

B

~B

200

1300

500 1000 1500

Definition of independence

Events A and B are independent if P(B | A)

P(A)B)P(A,

= P(B)

100pink

100white

400red

900all others

A ~A

B

~B

200

1300

500 1000

Here P(B | A) =

500100

P(B)

1500

P(B) =1500200

P(B | A) =

Marginal and joint probability table:

.0667 .0667

.2666 .6000

.3333 .6667

.8666

.1334

1

A ~A

B

~B

The joint probabilities are in the box. The marginalsare outside.

P(B | A) =

20.3333.0667.

How do you compute conditionals from this?

Joint probability tables

.0667 .0667

.2666 .6000

.3333 .6667

.8666

.1334

1

A ~A

B

~B

What are:

P(A, B) =

P(~A, ~B) =

P(~A) = P(~B) = P(~B | A)

.0667

.6

.667 .867 = .8

Outcomes or payoffsExample: Win $1,000 if you draw a pink marble,

win $0 otherwise.

Outcomes: $1,000 or $0

Probabilities: P(Pink) = 1/15P(~Pink) = 14/15

The expected value of this gamble:

E(gamble) = (1/15)*($1,000) + (14/15)*($0) = $66.67

Events: A pink marble or a marble ofanother color

Example

We expect to sell 10,000 computers if the market isgood and to sell 1,000 computers if the market isbad. The marketing department’s best estimate ofthe likelihood of a good market is .5.

Outcomes: 10,000 computers sold or 1,000 computerssold.

Probabilities: P(good market) = .5P(bad market) = .5

Are these objective or subjective probabilities?

What are expected computer sales? 5,500

More expected valuesWith discrete outcomes, an expected value is theprobability-weighted sum of the outcomes for thedecision of interest.

Here are some two-outcome lotteries. Compute theexpected values.

L1: Win $1,000 with probability .5 or lose $500 withprobability .5.

L2: Win $2,000 with probability .5 or lose $1,000with probability .5.

E(L1) = $250

E(L2) = $500

More lotteriesHere are some two-outcome lotteries. Compute theexpected values.

L3: Win $1,500 with probability 1/3 or lose $750 with probability 2/3.

L4: Win $750 with probability 2/3 or lose $750with probability 1/3.

L5: Win $300 with probability .5 or lose $200 withprobability .5.

L6: Win $10,000 with probability 9/10 or lose$85,000 with probability 1/10.

E(L3) = $0

E(L4) = $250

E(L5) = $50

E(L6) = $500

L3: Win $1,500 with probability 1/3 or lose $750 with probability 2/3.

L4: Win $750 with probability 2/3 or lose $750with probability 1/3.

L5: Win $300 with probability .5 or lose $200 withprobability .5.

L6: Win $10,000 with probability 9/10 or lose$85,000 with probability 1/10.

E(L3) = $0

E(L4) = $250

E(L5) = $50

E(L6) = $500

L1: Win $1,000 with probability .5 or lose $500 withprobability .5.

L2: Win $2,000 with probability .5 or lose $1,000with probability .5.

E(L1) = $250

E(L2) = $500

Action choices

If I build a large hotel (cost = $5,000,000) and tourismis high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000.

If I build a small hotel (cost = $2,000,000) and tourismis high, I will make $5,000,000, but if tourism is low, Iwill make $2,000,000.

I can also choose to do nothing.

Action choicesIf I build a large hotel (cost = $5,000,000) and tourism is high (P(high) = 2/3), I will make $15,000,000 in revenue, but if it is low, I will make $2,000,000.

If I build a small hotel (cost = $2,000,000) and tourism is high, I will make $5,000,000, but if tourism is low, I will make $2,000,000.

I can also choose to do nothing.

Action choices: Do nothing, build large, build small

Outcomes: $15,000,000; $2,000,000; $5,000,000, $0

States of nature: high tourism, low tourism

Probabilities: P(high) = 2/3; P(low) = 1/3

Decision treesSuppose I need to decide whether to invest $10,000 inthe market or leave it in the bank to earn interest. If Iinvest, there is a 50% chance that the market willincrease 20% over the coming year and a 50% chancethat the market will be stagnant (no change). If Ileave the money in the bank, there is an 80% chancethat interest rates will increase to 10% and a 20%chance that interest rates will remain at 5%.

What should I do? Use a decision tree.

A decision by an individual is required

Nature makes these decisions

I decide StockMarket

Bank

Nature decides$2,000 orEV = $1,000

$0 or EV = $0

$1,000 orEV = $800

$500 or EV = $100

.5

.5

.8

.2

(Good)

(Stagnant)

(Increase)

(Same)

$900

$1,000

Homework assignment

Problems 5-15 and 16

Consider two urns: Urn 1 Urn 2

Red ballsBlack balls

73

46

P(R1) = Probability of red on first drawP(R2) = Probability of red on second drawP(B1) = Probability of black on first drawP(B2) = Probability of black on second draw

a(1) Take one ball from urn 1, replace it, and take a second ball. What is the probability of two reds beingdrawn? P(R1, R2) = .7 x .7 = .49

Homework

a(2) What is the probability of a red on the seconddraw if a red is drawn on the first draw?

P(R2 | R1) = )P(R

)R ,P(R1

21

= )P(R.7

.7x2 7.

7.

a(3) What is the probability of a red on the seconddraw if a black is drawn on the first draw?

P(R2 | B1) = )P(B

)R ,P(B1

21

)P(R.3

.7x2 7.

3.

Homeworkb(1) Take a ball from urn 1; replace it. Take a ballfrom urn 2 if the first ball was black; otherwise, drawa ball from urn 1.

What is the probability of two reds being drawn?

P(R1, R2) = .7 x .7 = .49

b(2) What is the probability of a red on the seconddraw if a red is drawn on the first draw?

P(R2 | R1) = )P(R)R ,P(R

1

21

)P(R.7

.7x2 7.

7.

Homework

b(3) What is the probability of a red on the seconddraw if a black is drawn on the first draw?

P(R2 | B1) = )P(B

)R ,P(B1

21

4.104

What is the unconditional probability of red on thesecond draw?

P(R2) = P(B1, R2) + P(R1, R2) = (.3 x .4) + (.7 x .7) = .61

Homework5-16. Draw a tree diagram for Problem 5-15a

Draw 1

Draw 2

.7

.3

.7

.3

.7

.3

Red

Black

Red

Black

Red

Black

P(R1, R2) = .49

P(R1, B2) = .21

P(B1, R2) = .21

P(B1, B2) = .09

Homework5-17. Draw a tree diagram for Problem 5-15b

Draw 1

Draw 2

.7

.3

.7

.3

.4

.6

Red

Black

Red

Black

Red

Black

P(R1, R2) = .49

P(R1, B2) = .21

P(B1, R2) = .12

P(B1, B2) = .18

Homework5-29. This is the survey problem involving home-ownership and income levels. The results can besummarized by the table below

Homeownership

Income

> $25,000

<= $25,000

Yes No

.6

.2

.1

.1 .3

.7

.8 .2 1

SurveyHomeownership

Income

> $25,000

<= $25,000

Yes No

.6

.2

.1

.1 .3

.7

.8 .2 1

A. Suppose a reader of this magazine is selected atrandom and you are told that the person is a home-owner. What is the probability that the person has income in excess of $25,000?

P(>$25,000 | homeowner) = 75.8.6.

SurveyHomeownership

Income

> $25,000

<= $25,000

Yes No

.6

.2

.1

.1 .3

.7

.8 .2 1

b. Are home ownership and income (measured only asabove or below $25,000) independent factors for thisgroup?

If yes, then P(>$25 | home) = P(>$25)But, P(>$25) = .7 and P(>$25 | home) = .75

They are NOT independent.

Homework

5-38. The president of a large electric utility has todecide whether to purchase one large generator (BigJim) or four smaller generators (Little Arnies) to attain a given amount of electric generating capacity.On any given summer day, the probability of a generator being in service is 0.95 (the generators areequally reliable). Equivalently, there is a 0.05 probability of a failure.

Homework5-38.

a. What is the probability of Big Jim’s being out ofservice on a given day?

Let P(out) = the probability of any generator being out of service = .05

If P(BJout) = the probability of Big Jim’s being out of service.

Then P(out) = P(BJout) = .05

Homework5-38.

b. What is the probability of either zero or one of thefour Arnies being out? (At least three will be running.)

Since the probability of a failure (f) for one Arnie is .05,

P(f = 0 | n=4, p=.05) = 8145.)1()!0(!0

!4 040

ppn

P(f = 1 | n=4, p=.05) = 1715141111

4.)(

)!(!!

ppn

P(f 1|n=4, p=.05) = .8145 + .1715 = .9860

We want P(f 1|n=4, p=.05)

Homework5-38.

c. If five Little Arnies are purchased, what is the probability of at least four operating?

P(f 1 | n=5, p=.05) =

9774.2036.7738.

)1()!15(!1

!5)1(

)!05(!0!5 151050

pppp

Homework5-38.

d. If six Little Arnies are purchased, what is the probability of at least four operating?

P(f 2 | n=6, p=.05) =

9977.0305.2321.7351.

)1()!26(!2

!6)1(

)!16(!1!6

)1()!06(!0

!6

262161

060

pppp

pp

Homework5-40. Newspaper articles frequently cite the fact that inany one year, a small percentage (say, 10%) of alldrivers are responsible for all automobile accidents.The conclusion is often reached that if only we couldsingle out these accident-prone drivers and eitherretrain them or remove them from the roads, we coulddrastically reduce auto accidents. You are told that of100,000 drivers who were involved in one or moreaccidents in one year, 11,000 of them were involved inone or more accidents in the next year.

A. Given the above information, complete the entriesin the joint probability table in Table 5-21.

Accidents: Joint probability table

Accident No accident

Accident

No accident

Year 2

Year 1

Marginalprobability of

event insecond year


event infirst year

0.10 0.90 1

0.10

0.90

P(A1, A2) =

A1 = accident in year 1, A2 = accident in year 2

Given: P(A2 | A1) = 11,000/100,000 = .11

.11

P(A2 | A1)=

)),

1

21

P(AAP(A

Therefore,P(A1, A2) =P(A2 | A1)xP(A1) =

.11 x .10 = .011

.011.089

.089 .811

Accidents

B. Do you think searching for accident-pronedrivers is an effective way to reduce auto accidents?Why?

If the information in the problem is representative,then searching for accident-prone drivers will notbe very helpful, since having had an accident in Year 1 has only a minor effect on the probability ofan accident in Year 2.

Uncertainty continued . . .

Probability revisions

Continue decision trees

Today’s agenda

• Finish the homework problems

• Work through a decision tree example that– Uses no information– Uses perfect information– Uses imperfect information

• Briefly discuss Freemark Abbey Winery

• Group problem solving

Homework5-42. A safety commissioner for a certain city performed a study of the pedestrian fatalities atintersections. He noted that only 6 of the 19 fatalitieswere pedestrians who were crossing the intersectionagainst the light (i.e., in disregard of the propersignal), whereas the remaining 13 were crossing withthe light. He was puzzled because the figures seemedto show that it was roughly twice as safe for a pedestrian to cross against the light as with it. Canyou explain this apparent contradiction to thecommissioner?

Homework5-42.

The commissioner is looking at the wrong conditionalfrequencies (probabilities).

P(?) = 6/19 It’s a conditional probability

It is not the probability of being killed if you crossagainst the light. Further, 13/19 is not the probabilityof being killed if you cross with the light.

P(crossing against the light | killed at intersection) = 6/19

Homework5-42.

The relevant probabilities are: P(killed | crossed with light) andP(killed | crossed against light)

The deaths must be considered relative to the numberof pedestrians who cross with and against the light.

As an extreme possibility, it may be that the only sixpersons who crossed against the light were killed, afatality rate of 100%; whereas, 1 million crossed with the light, a fatality rate of .000013. It isn’t likely that this is the case, but the commissioner’s data do not rule it out.

Homework5-43.

Probability revision

Suppose a new test is available to test for drugaddiction. The test is 95 percent accurate “each way”;that is, if the person is an addict, there is a 95 percentchance the test will indicate “yes”; if the person is notan addict, then 95 percent of the time the test willindicate “no.”

Suppose it is known that the incidence of drug addictionin urban populations is about 1 out of 1,000. Given a positive (yes) test result, what are the chances that theperson being tested is addicted?

Homework: Probability revision5-43.

We were given P(“yes” | addicted) = .95and P(“no” | not addicted) = .95

We want P(addicted | “yes”) and P(not addicted | “no”)

the prior P(addicted) = .001

and

This is a different conditional probability called a“revised” or “posterior” probability.

Test for drug addiction

We know:

P(addicted | yes) = P(yes)yes) ,P(addicted

and

P(yes | addicted) = )P(addicted

yes) ,P(addicted= .95

P(addicted) = .001We can solve for the joint.


P(addicted, yes).00095 .001 * .95

)P(addicted * addicted) | P(yes

What is P(yes)? It consists of two joint probabilities.

The test can say “yes” and the subject is addictedORThe test can say “yes” and the subject is not addicted

P(addicted, yes) + P(not addicted, yes) = P(yes)


P(not addicted, yes).04995 .999 * .05

addicted)P(not * addicted)not | P(yes

Therefore, P(yes) = .00095 + .04995 = .0509

Given a positive test result, the probability that aperson chosen at random from an urban populationis a drug addict is

P(addict | yes) = .00095/.0509 = .0187

This is a posterior probability.

Drug test: joint probability table



1

Event

Report

Yes

No

addiction

.001 .999

.0509

.9491

.00095 .04995

.94905.00005

Addict No addictreport

content

.95

.95

Another revision example

Priors: P(disease) = .01P(~disease) = .99

Test accuracy: P(positive | disease) = .97P(positive | ~disease) = .05P(negative | disease) = .03P(negative | ~disease) = .95

Note that: false positives > false negatives

Disease detection continuedPriors: P(disease) = .01

P(~disease) = .99

Test accuracy: P(positive | disease) = .97P(positive | ~disease) = .05

P(negative | disease) = .03P(negative | ~disease) = .95

What is the probability that an individual chosen at random who tests positive has the disease?

P(positive, disease) = .97 * .01 = .0097

P(positive) = (.97 * .01) + (.05 * .99) = .0592

P(disease | positive) = .0097/.0592 = .1639

Disease detection continued

Suppose the tested individual was not chosenfrom the population at random, but insteadwas selected from a subset of the populationwith a greater chance of getting the disease?

Prior: Suppose P(disease) = .2

Then,P(disease | positive) = 829

058972972

.))(.(.))(.(.

))(.(.

Homework5-45 Revision

This is a classical probability problem. Tryout your intuition before solving it systematically.

Assume there are three boxes and each box has two drawers. There is either a gold or silver coin in eachdrawer. One box has two gold, one box two silver, and one box one gold and one silver coin. A box ischosen at random and one of the two drawers is opened.A gold coin is observed. What is the probability ofopening the second drawer in the same box and observing a gold coin?

Coin and box problemHere is a helpful visualization:

We know we chose a box with a gold coin.

We want P(gold2 | gold1) =

Gold

Gold

Gold

Silver

Silver

Silver

)

),

1

21

P(goldgoldP(gold

32

2131

Silver

Silver

Buying informationAs manager of a post office, you are trying to decidewhether to rearrange a production line and facilitiesin order to save labor and related costs. Assume thatthe only alternatives are to “do nothing” or “rearrange.”Assume also that the choice criterion is that the expectedsavings from rearrangement must equal or exceed$11,000.Operating costs if you do nothing will be $200,000

If you rearrange successfully, operating costs will be$100,000.

If you rearrange unsuccessfully, operating costs willbe $260,000.

Post Office ExampleBuying information

Operating costs if you do nothing will be $200,000

If you rearrange successfully (P(success) = .6), operating costs will be $100,000.

If you rearrange unsuccessfully (P(fail) = .4), operating costs will be $260,000.

What is the expected value of each action choice?

Rearrange: .6 x $100,000 + .4 x $260,000 = $164,000

Do nothing: $200,000 What would you choose?

Post Office Decision Tree

You decide

Do nothing

Rearrange Succeed

Fail

.6

.4

$100,000

$260,000

$200,000

$164,000

You can hire a consultant, Joan Zenoff, to study thesituation. She would then render a flawless predictionof whether the rearrangement would succeed or fail.Compute the maximum amount you would be willing to pay for the errorless prediction.

Don’t buy info.$164,000

Buy

Positive

Negative

rearrange

do nothing

rearrange

do nothing

$100,000

$200,000

$260,000

$200,000$200,000

$100,000

.6

.4

$140,000

$140,000

How much would you pay for Joan’s report?

Compute the expected value of perfect information= EVPI

EVPI = The expected value of the decision with thereport ($140,000) - The expected value of the decisionwithout the report ($164,000)

EVPI = $140,000 - $164,000 = -$24,000

The report saves us $24,000 in expected value

We would pay up to $24,000

Suppose now that Joan’s reports are not flawless.Suppose you have been provided the followingposterior probabilities:

EVENTSPROBABILITY OFEVENT IF success failure Optimistic report .818 .182 Pessimistic report .333 .667

This means that:

P(success | optimistic) = .818, not 1P(failure | optimistic) = .182, not 0

What would you now be willing to pay for Joan’sreport?


EVII = E(decision with imperfect information)- E(decision with no information)

Recall that E(decision with no information) = $164,000

Required:

1. Compute the expected cost assuming an optimistic report.

optimistic

Wedecide

rearrange

do nothing$200,000

success

failure

$100,000

$260,000?

.818

.182

$129,120

$129,120


2. Compute the expected costs assuming a pessimisticreport.

pessimistic

Wedecide

rearrange

do nothing$200,000

success

failure

$100,000

$260,000?

.333

.667

$200,000

$206,720


3. We were given the probability of an optimisticreport (P(optimistic) = .55) and the probability ofa pessimistic report (P(pessimistic) = .45).

Compute the expected value of imperfect information.

First we need to finish the decision tree.What is the first decision we must show on the tree?

Wedecide

Buy information

Don’t buy info $164,000

optimistic

pessimistic

$129,120

$200,000

.55

.45

$161,016

$161,016

E(decision with imperfect information) = $161,016

E(decision with no information) = $164,000

EVII = $164,000 - $161,016 = $2,984

We were not given likelihoods. We do not know theprobability that Joan will render an optimistic report given the rearrangement is a success.

What is that probability?

P(optimistic | success) = P(success)success) ic,P(optimist

P(success | optimistic) =ic)P(optimist

success) ic,P(optimist= .818

P(success) = .6

P(optimistic) = .55

P(optimistic, success) = .55 x .818 = .45

P(optimistic | success) = .45/.6 = .75

Also: P(pessimistic | failure) = (.667 x .45)/.4 = .75

Normally we would be given likelihoods and priorsand we would expect to compute:

1. The posterior probability of the outcome givena particular kind of information, and

2. The marginal probability of receiving that particular kind of information

Therefore, given the following information aboutthe accuracy of Joan Zenoff’s forecasts, completea joint probability table and compute the necessaryposterior (revised) probabilities.

P(optimistic | success) = .75P(pessimistic | failure) = .75



1

Event

Reportreport

contentSuccess Failure

Optimistic

Pessimistic

success

.6 .4

.75

.75

P(optimistic | success) =

P(success)success) ic,P(optimist

P(opt, success) = .75 x .6 = .45

.45

.15

P(pess, failure)= .75 x .4= .30

.30

.10 .55

.45



1

Event

Reportreport

contentSuccess Failure

Optimistic

Pessimistic

success

.6 .4

.75

.75

.45

.15 .30

.10 .55

.45

P(success | opt) = .45/.55 P(failure | opt) = .1/.55

P(failure | pess) = .3/.45 P(success | pess) = .15/.45

Which ones go on the decision tree?

uncertainty and probability using probabilities using decision trees probability revision

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