# prob by suchenfz

VIEWS: 8 PAGES: 75

• pg 1
```									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, blue
and white marbles.
Take one ball from the urn. What is:
P(black) = 0

P(~black) = 1            ~ = NOT

Probabilities are all greater than or equal to zero and less
than or equal to one.
Same urn:
Suppose the number of balls is as follows:
Red                400
Pink               100
Yellow             400 What is:
Blue               500
White              100 P(Red) =         400/1500 = .267
Total             1500 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

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

P(~A) = (1 - P(A)) = .67
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
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
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
independence
Suppose we draw one marble from the urn and
replace 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

What are:

P(B) =      = .133

P(~B) = .867
1500

P(A, B) =    P(A or B) =
Conditional probabilities
The probability that a particular
What is
event will occur, given we already
know that another event has
P(A | B) =
occurred.

We have information to bring            P(~A | ~B) =
to bear on the base rate  1500
probability of the event

P(A | ~B) =               P(~B | A) =
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:
The joint probabilities are in the box. The marginals
are outside. How do you compute conditionals from this?
A          ~A

P(B | A) =         B

~B
Joint probability tables
A         ~A
What are:
B
P(A, B) = .0667
~B
P(~A, ~B) = .6

P(~A) = .667   P(~B) = .867   P(~B | A) = .8
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
We expect to sell 10,000 computers if the market is
good and to sell 1,000 computers if the market is
bad. The marketing department’s best estimate of
the likelihood of a good market is .5.

Outcomes: 10,000 computers sold or 1,000 computers
sold.
Probabilities:     P(good market) = .5
Are these objective or subjective probabilities?
What are expected computer sales?          5,500
More expected values
With discrete outcomes, an expected value is the
probability-weighted sum of the outcomes for the
decision of interest.

Here are some two-outcome lotteries. Compute the
expected 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
Here are some two-outcome lotteries. Compute the
expected 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(L ) = \$250
4
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
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

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
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
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% chance
that 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.
A decision by an individual is required

Nature makes these decisions

\$2,000 or
Nature decides                .5 EV = \$1,000
(Good)

(Stagnant)
I decide                                       .5
Stock                                   \$0 or EV = \$0
Market          \$1,000
\$1,000 or
.8    EV = \$800
Bank                     (Increase)

(Same)
.2 \$500 or
\$900
EV = \$100
Homework assignment
Problems 5-15 and 16
Consider two urns:         Urn 1           Urn 2
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 being
drawn?          P(R1, R2) = .7 x .7 = .49
Homework
a(2) What is the probability of a red on the second
draw 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) =
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) =
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 the
second draw?

P(R2) = P(B1, R2) + P(R1, R2) = (.3 x .4) + (.7 x .7) = .61
Homework
5-16. Draw a tree diagram for Problem 5-15a

Red      P(R1, R2) = .49
Draw 2
.7
Red                .3
Draw 1
.7             Black    P(R1, B2) = .21

.3            Red      P(B1, R2) = .21
Black             .7

.3
Black    P(B1, B2) = .09
Homework
5-17. Draw a tree diagram for Problem 5-15b

Red     P(R1, R2) = .49
Draw 2
.7
Red
Draw 1                        .3
.7            Black    P(R1, B2) = .21

.3             Red     P(B1, R2) = .12
.4
Black
.6
Black    P(B1, B2) = .18
Homework
5-29. This is the survey problem involving home-
ownership and income levels. The results can be
summarized by the table below
Survey

A. Suppose a reader of this magazine is selected at
random 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) =
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
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.
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
5-38.               Homework
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
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) =
5-38.               Homework
d. If six Little Arnies are purchased, what is the
probability of at least four operating?

P(f  2 | n=6, p=.05) =
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
A1 = accident in year 1, A2 = accident in year 2
Given: P(A2 | A1) = 11,000/100,000 = .11

P(A2 | A1)=

P(A1, A2) =
.011
.089
Therefore,                          .11
P(A1, A2) =
P(A2 | A1)x                    .089       .811
P(A1) =
.11 x .10 = .011
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
Year 1 has only a minor effect on the probability of
an 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
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?
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.
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.
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
Homework: Probability revision
5-43.
We were given P(“yes” | addicted) = .95
and P(“no” | not addicted) = .95        and

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

We know:

and

P(yes | addicted) =                 = .95

We can solve for the joint.

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

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

Given a positive test result, the probability that a
person chosen at random from an urban population
P(addict | yes) = .00095/.0509 = .0187
This is a posterior probability.
Drug test: joint probability table

.95

.95
Bayes Theorem and the
Multiplication Rule

P(Ai | B) =

and
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
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 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) =
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

We know we chose a box with a gold coin.

We want P(gold2 | gold1) =
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
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
Do nothing
\$200,000
You decide

\$100,000
.6
Rearrange                    Succeed

Fail
\$164,000                .4
\$260,000
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
\$100,000
\$140,000           \$100,000
Positive               do nothing
.6                   \$200,000
\$260,000
Negative
\$140,000
do nothing
\$200,000                     \$200,000
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
\$140,000                                           \$100,000
\$100,000
Positive                do nothing
.6                    \$200,000
\$260,000
Negative
\$140,000
do nothing
\$200,000                      \$200,000
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 the
report (\$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
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
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
Required:

1. Compute the expected cost assuming an optimistic
report.                                success
.818 \$100,000
We        rearrange
decide                     .182
failure
optimistic                ?
\$129,120                 \$260,000

\$129,120                do nothing
\$200,000
2. Compute the expected costs assuming a pessimistic
report.                                   success
.333 \$100,000
rearrange
We
.667
decide                          failure
\$206,720
?                    \$260,000
pessimistic

\$200,000               do nothing
\$200,000
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 the first finish thewe must show on the tree?
What is need to decision decision tree.
optimistic

We
decide                                     .45
pessimistic
\$161,016                     \$200,000

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 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
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.
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
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?

```
To top