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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

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Uncertainty and probability l.jpg

Uncertainty and probability

Using probabilities

Using decision trees

Probability revision


Today s agenda l.jpg
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 l.jpg
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 l.jpg
Vocabulary continued

  • Outcomes or payoffs (mutually exclusive)

  • Action choices

  • States of nature

  • Decision tree

  • Expected value

  • Risk


Probability l.jpg
Probability

Imagine an urn containing 1500 red, pink, yellow, blue

and white marbles.

Take one ball from the urn. What is:

P(black) =

0

~ = NOT

P(~black) =

1

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


Same urn l.jpg
Same urn:

Suppose the number of balls is as follows:

Red 400

Pink 100

Yellow 400

Blue 500

White 100

Total 1500

What is:

P(Red) =

400/1500 = .267

P(Pink) =

100/1500 = .067

P(Yellow) =

400/1500 = .267

P(Blue) =

500/1500 = .333

P(White) =

100/1500 = .067

Total =

1


Joint probabilities and independence l.jpg
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) =

= .33

(1 - P(A)) = .67

P(~A) =


Joint probabilities and independence we re getting there l.jpg
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) =

.133

P(~B) =

.867


Joint probabilities and independence9 l.jpg
Joint probabilities and independence

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) =

= .0667


Joint probabilities and independence10 l.jpg
Joint probabilities and independence

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


Joint probabilities and independence11 l.jpg
Joint probabilities and independence

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

What is:

P(Red, Red) =

= .071

P(Red, Blue) = .088

Are (Red, Red) independent?

Are (Red, Blue) independent?


Joint and marginal probabilities l.jpg
Joint and marginal probabilities

What are:

= .133

P(B) =

P(~B) =

.867

1500

P(A, B) =

P(A or B) =


Conditional probabilities l.jpg
Conditional probabilities

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

What is

P(A | B) =

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

P(~A | ~B) =

1500

P(A | ~B) =

P(~B | A) =


Definition of independence l.jpg
Definition of independence

P(B | A) =

Events A and B are independent if

P(B | A)

= P(B)

Here P(B | A) =

P(B)

P(B) =

1500


Marginal and joint probability table l.jpg
Marginal and joint probability table:

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

How do you compute conditionals from this?

A

~A

P(B | A) =

B

~B


Joint probability tables l.jpg
Joint probability tables

A

~A

What are:

B

P(A, B) =

.0667

~B

P(~A, ~B) =

.6

P(~A) =

.667

P(~B) =

.867

= .8

P(~B | A)


Outcomes or payoffs l.jpg
Outcomes or payoffs

Example: Win $1,000 if you draw a pink marble,

win $0 otherwise.

Outcomes: $1,000 or $0

Events: A pink marble or a marble of another color

Probabilities:

P(Pink) = 1/15

P(~Pink) = 14/15

The expected value of this gamble:

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


Example l.jpg
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 computers

sold.

Probabilities: P(good market) = .5

P(bad market) = .5

Are these objective or subjective probabilities?

What are expected computer sales?

5,500


More expected values l.jpg
More expected values

With 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 with probability .5.

E(L1) = $250

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

E(L2) = $500


More lotteries l.jpg
More lotteries

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

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

E(L3) = $0

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

E(L4) = $250

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

E(L5) = $50

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

E(L6) = $500


Slide21 l.jpg

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

E(L1) = $250

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

E(L2) = $500

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

E(L3) = $0

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

E(L4) = $250

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

E(L5) = $50

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

E(L6) = $500


Action choices l.jpg
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, I

will make $2,000,000.

I can also choose to do nothing.


Action choices23 l.jpg
Action choices

If 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 trees l.jpg
Decision trees

Suppose I need to decide whether to invest $10,000 in

the market or leave it in the bank to earn interest. If I

invest, there is a 50% chance that the market will

increase 20% over the coming year and a 50% chancethat the market will be stagnant (no change). If I

leave the money in the bank, there is an 80% chance

that 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.


Slide25 l.jpg

A decision by an individual is required

Nature makes these decisions

$2,000 or

EV = $1,000

Nature decides

.5

(Good)

(Stagnant)

.5

I decide

StockMarket

$0 or EV = $0

$1,000

$1,000 orEV = $800

.8

Bank

(Increase)

(Same)

.2

$500 or EV = $100

$900


Homework assignment l.jpg
Homework assignment

Problems 5-15 and 16

Urn 1

Urn 2

Consider two urns:

Red balls

7

4

Black balls

3

6

P(R1) = Probability of red on first draw

P(R2) = Probability of red on second draw

P(B1) = Probability of black on first draw

P(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 l.jpg
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) =

=

a(3) What is the probability of a red on the second

draw if a black is drawn on the first draw?

P(R2 | B1) =


Homework28 l.jpg
Homework

b(1) Take a ball from urn 1; replace it. Take a ball

from urn 2 if the first ball was black; otherwise, draw

a 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 second

draw if a red is drawn on the first draw?

P(R2 | R1) =


Homework29 l.jpg
Homework

b(3) What is the probability of a red on the second

draw if a black is drawn on the first draw?

P(R2 | B1) =

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


Homework30 l.jpg
Homework

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

P(R1, R2) = .49

Red

Draw 2

.7

Red

.3

Draw 1

.7

Black

P(R1, B2) = .21

P(B1, R2) = .21

Red

.3

.7

Black

.3

P(B1, B2) = .09

Black


Homework31 l.jpg
Homework

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

P(R1, R2) = .49

Red

Draw 2

.7

Red

.3

Draw 1

.7

Black

P(R1, B2) = .21

P(B1, R2) = .12

Red

.3

.4

Black

.6

P(B1, B2) = .18

Black


Homework32 l.jpg
Homework

5-29. This is the survey problem involving home-

ownership and income levels. The results can be

summarized by the table below


Survey l.jpg
Survey

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) =


Survey34 l.jpg
Survey

b. Are home ownership and income (measured only as

above or below $25,000) independent factors for this

group?

They are NOT independent.

If yes, then P(>$25 | home) = P(>$25)

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


Homework35 l.jpg
Homework

5-38. The president of a large electric utility has to

decide whether to purchase one large generator (Big

Jim) 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 are

equally reliable). Equivalently, there is a 0.05

probability of a failure.


Homework36 l.jpg
Homework

5-38.

a. What is the probability of Big Jim’s being out of

service 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


Homework37 l.jpg
Homework

5-38.

b. What is the probability of either zero or one of the

four Arnies being out? (At least three will be running.)

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

We want P(f  1|n=4, p=.05)

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

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

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


Homework38 l.jpg
Homework

5-38.

c. If five Little Arnies are purchased, what is the

probability of at least four operating?

P(f  1 | n=5, p=.05) =


Homework39 l.jpg
Homework

5-38.

d. If six Little Arnies are purchased, what is the

probability of at least four operating?

P(f  2 | n=6, p=.05) =


Homework40 l.jpg
Homework

5-40. Newspaper articles frequently cite the fact that in

any one year, a small percentage (say, 10%) of all

drivers are responsible for all automobile accidents.

The conclusion is often reached that if only we could

single out these accident-prone drivers and either

retrain them or remove them from the roads, we could

drastically reduce auto accidents. You are told that of

100,000 drivers who were involved in one or more

accidents in one year, 11,000 of them were involved in

one or more accidents in the next year.

A. Given the above information, complete the entries

in the joint probability table in Table 5-21.


Accidents joint probability table l.jpg
Accidents: Joint probability table

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

11,000/100,000 = .11

Given: P(A2 | A1) =

P(A2 | A1)=

P(A1, A2) =

.089

.011

Therefore,

.11

P(A1, A2) =

P(A2 | A1)xP(A1) =

.089

.811

.11 x .10 = .011


Accidents l.jpg
Accidents

B. Do you think searching for accident-prone

drivers is an effective way to reduce auto accidents?

Why?

If the information in the problem is representative,

then searching for accident-prone drivers will not

be very helpful, since having had an accident in

Year 1 has only a minor effect on the probability of

an accident in Year 2.


Uncertainty continued l.jpg

Uncertainty continued . . .

Probability revisions

Continue decision trees


Today s agenda44 l.jpg
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


Homework45 l.jpg
Homework

5-42. A safety commissioner for a certain city

performed a study of the pedestrian fatalities at

intersections. He noted that only 6 of the 19 fatalities

were pedestrians who were crossing the intersection

against the light (i.e., in disregard of the proper

signal), whereas the remaining 13 were crossing with

the light. He was puzzled because the figures seemed

to show that it was roughly twice as safe for a

pedestrian to cross against the light as with it. Can

you explain this apparent contradiction to the

commissioner?


Homework46 l.jpg

5-42.

Homework

The commissioner is looking at the wrong conditional

frequencies (probabilities).

P(?) = 6/19

It’s a conditional probability

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

It is not the probability of being killed if you cross

against the light. Further, 13/19 is not the probability

of being killed if you cross with the light.


Homework47 l.jpg

5-42.

Homework

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

P(killed | crossed against light)

The deaths must be considered relative to the number

of pedestrians who cross with and against the light.

As an extreme possibility, it may be that the only six

persons who crossed against the light were killed, a

fatality 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.


Homework48 l.jpg

5-43.

Homework

Probability revision

Suppose a new test is available to test for drug

addiction. The test is 95 percent accurate “each way”;

that is, if the person is an addict, there is a 95 percent

chance the test will indicate “yes”; if the person is not

an addict, then 95 percent of the time the test will

indicate “no.”

Suppose it is known that the incidence of drug addiction

in urban populations is about 1 out of 1,000. Given a

positive (yes) test result, what are the chances that the

person being tested is addicted?


Homework probability revision l.jpg
Homework: Probability revision

5-43.

We were given P(“yes” | addicted) = .95

and P(“no” | not addicted) = .95

and

the prior P(addicted) = .001

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

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


Test for drug addiction l.jpg
Test for drug addiction

We know:

P(addicted | yes) =

and

P(yes | addicted) =

= .95

P(addicted) = .001

We can solve for the joint.


Test for drug addiction51 l.jpg
Test for drug addiction

P(addicted, yes)

What is P(yes)?

It consists of two joint probabilities.

The test can say “yes” and the subject is addicted

OR

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

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


Test for drug addiction52 l.jpg
Test for drug addiction

P(not addicted, yes)

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

Given a positive test result, the probability that a

person chosen at random from an urban population

is a drug addict is

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

This is a posterior probability.




Another revision example l.jpg
Another revision example

Priors: P(disease) = .01

P(~disease) = .99

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

P(positive | ~disease) = .05

P(negative | disease) = .03

P(negative | ~disease) = .95

Note that: false positives > false negatives


Disease detection continued l.jpg
Disease detection continued

Priors: P(disease) = .01

P(~disease) = .99

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

P(positive | ~disease) = .05

P(negative | disease) = .03

P(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 continued57 l.jpg
Disease detection continued

Suppose the tested individual was not chosen

from the population at random, but instead

was selected from a subset of the population

with a greater chance of getting the disease?

Prior: Suppose P(disease) = .2

Then,

P(disease | positive) =


Homework58 l.jpg

5-45 Revision

Homework

This is a classical probability problem. Try

out 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 each

drawer. One box has two gold, one box two silver,

and one box one gold and one silver coin. A box is

chosen at random and one of the two drawers is opened.

A gold coin is observed. What is the probability of

opening the second drawer in the same box and

observing a gold coin?


Coin and box problem l.jpg
Coin and box problem

Here is a helpful visualization:

We know we chose a box with a gold coin.

We want P(gold2 | gold1) =


Buying information l.jpg
Buying information

As manager of a post office, you are trying to decide

whether to rearrange a production line and facilities

in order to save labor and related costs. Assume that

the only alternatives are to “do nothing” or “rearrange.”

Assume also that the choice criterion is that the expected

savings 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 will

be $260,000.


Post office example l.jpg

Buying information

Post Office Example

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 l.jpg
Post Office Decision Tree

Do nothing

$200,000

You decide

$100,000

.6

Rearrange

Succeed

Fail

$164,000

.4

$260,000


Slide63 l.jpg

You can hire a consultant, Joan Zenoff, to study the

situation. She would then render a flawless prediction

of whether the rearrangement would succeed or fail.

Compute the maximum amount you would be willing to pay for the errorless prediction.

$164,000

rearrange

Don’t buy info.

$100,000

$100,000

$140,000

Positive

do nothing

.6

$200,000

rearrange

Buy

.4

$260,000

Negative

$140,000

do nothing

$200,000

$200,000


Slide64 l.jpg

You can hire a consultant, Joan Zenoff, to study the

situation. She would then render a flawless prediction

of whether the rearrangement would succeed or fail.

Compute the maximum amount you would be willing to pay for the errorless prediction.

$164,000

rearrange

Don’t buy info.

$100,000

$140,000

$100,000

Positive

do nothing

.6

$200,000

rearrange

Buy

.4

$260,000

Negative

$140,000

do nothing

$200,000

$200,000


Slide65 l.jpg

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 decision

without 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


Slide66 l.jpg

Suppose now that Joan’s reports are not flawless.

Suppose you have been provided the following

posterior probabilities:

This means that:

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

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


Slide67 l.jpg

What would you now be willing to pay for Joan’s

report?

EVII = E(decision with imperfect information)

- E(decision with no information)

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


Slide68 l.jpg

Required:

1. Compute the expected cost assuming an optimistic

report.

success

.818

$100,000

rearrange

Wedecide

.182

failure

optimistic

?

$129,120

$260,000

do nothing

$129,120

$200,000


Slide69 l.jpg

2. Compute the expected costs assuming a pessimistic

report.

success

$100,000

.333

rearrange

Wedecide

.667

failure

$206,720

$260,000

?

pessimistic

do nothing

$200,000

$200,000


Slide70 l.jpg

3. We were given the probability of an optimistic

report (P(optimistic) = .55) and the probability of

a 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?

optimistic

$129,120

.55

Buy information

We

decide

.45

pessimistic

$161,016

$200,000

Don’t buy info

$161,016

$164,000


Slide71 l.jpg

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

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

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


Slide72 l.jpg

We were not given likelihoods. We do not know the

probability that Joan will render an optimistic report

given the rearrangement is a success.

What is that probability?

P(optimistic | success) =

P(success) = .6

P(success | optimistic) =

= .818

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


Slide73 l.jpg

Normally we would be given likelihoods and priors

and we would expect to compute:

1. The posterior probability of the outcome given

a particular kind of information, and

2. The marginal probability of receiving that

particular kind of information

Therefore, given the following information about

the accuracy of Joan Zenoff’s forecasts, complete

a joint probability table and compute the necessary

posterior (revised) probabilities.


Slide74 l.jpg

P(optimistic | success) =

P(optimistic | success) = .75

P(pessimistic | failure) = .75

P(opt, success)

= .75 x .6

= .45

.45

.10

.55

P(pess, failure)

= .75 x .4

= .30

.15

.30

.45


Slide75 l.jpg

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

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

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

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

.45

.10

.55

.15

.30

.45

Which ones go on the decision tree?


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