Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson3-1 Lesson 3: A Survey of Probability Concepts

Lesson3-1 Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

Lesson 3:

A Survey of Probability Concepts


Outline

Learning Exercises

Definitions

Basic rules of Probability

Independence

Tree Diagram

Bayes’ Theorem


Learning exercise 1: University Demographics

Current enrollments by college and by sex appear in the following table.

College

Ag-For

Arts-Sci

Bus-Econ

Educ Engr

Law Undecl

Totals

Female

500 1500 400 1000

200 100 800 4500

Male 900 1200 500 500 1300

200 900 5500

Totals 1400 2700 900 1500

1500

300 1700 10000 If we select a student at random, what is the probability

that the student is : A female or male, i.e., P(Female or Male). Not from Agricultural and Forestry, i.e., P(not-Ag-For) A female given that the student is known to be from

BusEcon, i.e., P(Female |BusEcon). A female and from BusEcon, i.e., P(Female and BusEcon). From BusEcon, i.e., P(BusEcon).


Learning exercise 1: University Demographics

College

Ag-For

Arts-Sci

Bus-Econ

Educ

Engr

Law Undecl

Totals

Female

500 1500 400 1000

200 100 800 4500

Male 900 1200 500 500 1300

200 900 5500

Totals 1400 2700 900 1500

1500

300 1700 10000P(Female or Male)=(4500 + 5500)/10000 = 1

P(not-Ag-For)=(10000 – 1400) /10000 = 0.86

P(Female | BusEcon)= 400 /900 = 0.44

P(Female and BusEcon)= 400 /10000 = 0.04

P(BusEcon)= 900 /10000 = 0.09

P(Female and BusEcon) = P(BusEcon) P(Female | BusEcon)


Learning exercise 2: Predicting Sex of Babies

Many couples take advantage of ultrasound exams to determine the sex of their baby before it is born. Some couples prefer not to know beforehand. In any case, ultrasound examination is not always accurate. About 1 in 5 predictions are wrong. In one medical group, the proportion of girls correctly

identified is 9 out of 10, i.e., applying the test to 100 baby girls, 90 of the tests will indicate girls.

and the number of boys correctly identified is 3 out of 4.

i.e., applying the test to 100 baby boys, 75 of the tests will indicate boys.

The proportion of girls born is 48 out of 100.

What is the probability that a baby predicted to be a girl actually turns out to be a girl? Formally, find P(girl | test says girl).



P(girl | test says girl) In one medical group, the proportion of girls correctly

identified is 9 out of 10 and the number of boys correctly identified is 3 out of 4. The proportion of girls born is 48 out of 100.

Think about the next 1000 births handled by this medical group. 480 = 1000*0.48 are girls 520 = 1000*0.52 are boys Of the girls, 432 (=480*0.9) tests indicate that they are girls. Of the boys, 130 (=520*0.25) tests indicate that they are

girls. In total, 562 (=432+130) tests indicate girls. Out of these

562 babies, 432 are girls. Thus P(girl | test says girl ) = 432/562 = 0.769



Test says girl

Test says boy

Totals

Girl (4) 432 (7) 48 (2) 480

Boy (5) 130 (8) 390 (3) 520

Totals (6) 562 (9) 438 (1) 1000

With the information given, we can fill in the following table in sequence from (1) to (9), with the initial assumption of 1000 babies in total.

For the question at hand, i.e., P(girl | test says girl ), we only need to fill in the cells from (1) to (6).

P(girl | test says girl ) = 432/562 = 0.769



480 = 1000*0.48 are girls 520 = 1000*0.52 are boys Of the girls, 432 (=480*0.9) tests indicate that they are girls.

Of the boys, 130 (=520*0.25) tests indicate that they are girls.

In total, 562 tests indicate girls.

Out of these 562 babies, 432 are girls. Thus P(girls | test syas girls ) = 432/562 = 0.769

1000*P(girls)

1000*P(boys)

1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]

1000*P(boys)*P(test says girls | boys)

1000*P(girls)*P(test says girls|girls)

1000*P(girls)*P(test says girls|girls)

1000*[P(girls)*P(test says girls|girls) + P(boys)*P(test says girls|boys)]

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data

A fair die is rolled once. Peter is concerned with whether the resulted

number is even, i.e., 2, 4, 6. Paul is concerned with whether the resulted number

is less than or equal to 3, i.e., 1, 2, 3. Mary is concerned with whether the resulted number

is 6. Sonia is concerned with whether the resulted

number is odd, i.e., 1, 3, 5.

A fair die is rolled twice. John is concerned with whether the resulted number

of first roll is even, i.e., 2, 4, 6. Sarah is concerned with whether the resulted

number of second roll is even, i.e., 2, 4, 6.

Example 1(to be used to illustrate the definitions)


Definitions: Experiment and outcome

A random experiment is the observation of some activity or the act of taking some measurement. The experiment is rolling the one die in the first

example, and rolling one die twice in the second example.

An outcome is the particular result of an experiment. The possible outcomes are the numbers 1, 2, 3, 4,

5, and 6 in the first example. The possible outcomes are number pairs (1,1),

(1,2), …, (6,6), in the second example.

Sample Space – the collection of all possible outcomes of a random experiment.


Definition: Event

An event is the collection of one or more outcomes of an experiment. For Peter: the occurrence of an even number, i.e.,

2, 4, 6. For Paul: the occurrence of a number less than or

equal to 3, i.e., 1, 2, 3. For Mary: the occurrence of a number 6. For Sonia: the occurrence of an odd number, i.e.,

1, 3, 5. For John: the occurrence of (2,1), (2,2), (2,3),…,

(2,6), (4,1),…,(4,6), (6,1),…,(6,6) [John does not care about the result of the second roll].


Definition: Intersection of Events

Intersection of Events – If A and B are two events in a sample space S, then the intersection, A ∩ B, is the set of all outcomes in S that belong to both A and B

A BAB

S

The intersection of Peter’s event and Paul’s event contains 2. The intersection of Peter’s event and Mary’s event contains 6. The intersection of Paul’s event and Mary’s event contains nothing. The intersection of Peter’s event and Sonia’s event contains nothing.


Definition: Mutually Exclusive events

A and B are Mutually Exclusive Events if they have no basic outcomes in common i.e., the set A ∩ B is empty

A B

S


Example 2: Intersections and Mutually Exclusive events

Peter’s event and Paul’s event are not mutually exclusive – both contains 2. The intersection of Peter’s event and Paul’s event contains 2.

Peter’s event and Mary’s event are not mutually exclusive – both contains 6. The intersection of Peter’s event and Mary’s event contains 6.

Paul’s event and Mary’s event are mutually exclusive – no common numbers. The intersection of Paul’s event and Mary’s event contains

nothing.

Peter’s event and Sonia’s event are mutually exclusive – no common numbers. The intersection of Peter’s event and Sonia’s event contains

nothing.


Definition: Union

Union of Events – If A and B are two events in a sample space S, then the union, A U B, is the set of all outcomes in S that belong to either A or B

A B

The entire shaded area represents A U B

S


Definition: Exhaustive events

Events are collectively exhaustive if at least one of the events must occur when an experiment is conducted. Peter’s event (even numbers) and Sonia’s event

(odd numbers) are collectively exhaustive. Peter’s event (even numbers) and Mary’s event

(number 6) are not collectively exhaustive.

Events E1, E2, … Ek are Collectively Exhaustive events if

E1 U E2 U . . . U Ek = S

i.e., the events completely cover the sample space


Complement

The Complement of an event A is the set of all basic outcomes in the sample space that do not belong to A. The complement is denoted A

AS

A


Example 3

Let the Sample Space be the collection of all possible outcomes of rolling one die:

S = [1, 2, 3, 4, 5, 6]

Let A be the event “Number rolled is even”

Let B be the event “Number rolled is at least 4”

Then the two events contains

A = [2, 4, 6] B = [4, 5, 6]


Example 3

S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]

A

BA

BA

AA

Complements:

BA

B 5] 3, [1, 3] 2, [1,

Intersections:

Unions:

6] [4, [5]

6] 5, 4, [2,

S 6] 5, 4, 3, 2, [1,


Mutually exclusive: Are A and B mutually exclusive?

No. The outcomes 4 and 6 are common to both

Collectively exhaustive: Are A and B collectively exhaustive?

No. A U B does not contain 1 and 3

Example 3

S = [1, 2, 3, 4, 5, 6] A = [2, 4, 6] B = [4, 5, 6]


Probability

Probability – the chance that an uncertain event will occur (always between 0 and 1)

0 ≤ P(A) ≤ 1 For any event A

Certain

Impossible

.5

1

0


Assessing Probability

There are three approaches to assessing the probability of an uncertain event:

1. classical probability

spacesampletheinoutcomesofnumbertotal

eventthesatisfythatoutcomesofnumber

N

NAeventofyprobabilit A

(Assumes all outcomes in the sample space are equally likely to occur.)


Counting the Possible Outcomes

Use the Combinations formula to determine the number of combinations of n things taken k at a time

where n! = n(n-1)(n-2)…(1) 0! = 1 by definition

k)!(nk!

n! k)C(n,


Example 4: Combination

Suppose there are 2 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. [1], [2] C(2,1) = 2!/(1!(2-1)!)=2

k)!(nk!

n! k)C(n,



Suppose there are 3 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. [1], [2], [3] C(3,1) = 3!/(1!(3-1)!)=3*2*1/[1*(2*1)]=3

Suppose there are 3 stocks in our portfolio. We would like to select 2 stocks and sell them. What are all the possible combinations. [1,2], [1,3], [2,3] C(3,2) = 3!/(2!(3-2)!)=3*2*1/[(2*1)*1]=3

k)!(nk!

n! k)C(n,



Suppose there are 4 stocks in our portfolio. We would like to select 1 stocks and sell them. What are all the possible combinations. [1], [2], [3], [4] C(4,1) = 4!/(1!(4-1)!)=4*3*2*1/[1*(3*2*1)]=4

Suppose there are 4 stocks in our portfolio. We would like to select 2 stocks and sell them. What are all the possible combinations. [1,2], [1,3], [1,4], [2,3], [2,4], [3,4] C(4,2) = 4!/(2!(4-2)!)=4*3*2*1/[(2*1)*(2*1)]=6

k)!(nk!

n! k)C(n,


Assessing Probability

Three approaches (continued)

2. relative frequency probability

the limit of the proportion of times that an event A occurs in a large number of trials, n

3. subjective probability

populationtheineventsofnumbertotal

Aeventsatisfythatpopulationtheineventsofnumber

n

nAeventofyprobabilit A

an individual opinion or belief about the probability of occurrence


Probability Postulates

1. If A is any event in the sample space S, then

2. Let A be an event in S, and let Oi denote the basic outcomes (mutually exclusive). Then

(the notation means that the summation is over all the basic outcomes in A)

3. P(S) = 1

1P(A)0

)P(OP(A)A

i


Probability Rules

The Complement rule:

The Addition rule: The probability of the union of two events is

1)AP(P(A)i.e., P(A)1)AP(

B)P(AP(B)P(A)B)P(A


A Probability Table

B

A

A

B

)BP(A

)BAP( B)AP(

P(A)B)P(A

)AP(

)BP(P(B) 1.0P(S)

Probabilities and joint probabilities for two events A and B are summarized in this table:


Example 5: Addition Rule

Consider a standard deck of 52 cards, with four suits:

♥ ♣ ♦ ♠Let event A = card is an Ace

Let event B = card is from a red suit

B)P(AP(B)P(A)B)P(A


P(Red U Ace) = P(Red) + P(Ace) - P(Red ∩ Ace)

= 26/52 + 4/52 - 2/52 = 28/52Don’t count the two red aces twice!

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

Example 5: Addition Rule B)P(AP(B)P(A)B)P(A


A conditional probability is the probability of one event, given that another event has occurred:

P(B)

B)P(AB)|P(A

P(A)

B)P(AA)|P(B

The conditional probability of A given that B has occurred

The conditional probability of B given that A has occurred

Conditional Probability


Example 6: Conditional Probability

What is the probability that a car has a CD player, given that it has AC ?

i.e., we want to find P(CD | AC)

Of the cars on a used car lot, 70% have air conditioning (AC) and 40% have a CD player (CD). 20% of the cars have both.

P(B)

B)P(AB)|P(A



No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .6 1.0


P(B)

B)P(AB)|P(A



No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .6 1.0

Given AC, we only consider the top row (70% of the cars). Of these, 20% have a CD player. 20% of 70% is 28.57%.

.2857.7

.2

P(AC)

AC)P(CDAC)|P(CD

P(B)

B)P(AB)|P(A


Multiplication Rule

Multiplication rule for two events A and B:

also

P(B)B)|P(AB)P(A

P(A)A)|P(BB)P(A

Examples:1. P(test says girl and girl) = P(girls) * P(test says girls |

girls)2. P(test says boy and boy) = P(boys) * P(test says boys |

boys)


Example 7: Multiplication Rule

P(Red ∩ Ace) = P(Red| Ace)P(Ace)

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52

2

52

4

4

2

52

2

cards of number total

ace and red are that cards of number

P(B)B)|P(AB)P(A


Example 8: IndependenceShould I go to a party without my girlfriend?

The probability of the my going to party is 0.7 (i.e., I go to 70 out of 100 parties on average).

If I tend to go to whichever party my girlfriend (Venus) goes, my party behavior depends on Venus’s. That is, my probability of going to a party conditional on Venus’s presence is larger than 0.7 (extreme case being 1.0).

If I tends to avoid going to whichever party Venus goes, my party behavior also depends on Venus. That is, my probability of going to a party conditional on Venus’s presence is less than 0.7 (extreme case being 0.0).

If in making the party decision, I never consider whether Venus is going to a party, my party behavior does not depends on Venus’s. That means, the probability of going to a party conditional on Venus’s presence is 0.7.


Example 8: IndependenceShould I go to a party without my girlfriend?

Define events: A: a young man goes to a party B: his girlfriend goes to the same party.

Assume P(A) =0.7 His party behavior does not depend on his girlfriend’s

only if P(A|B) =P(A) = 0.7. And, event A is said to be independent of event B.

P(the young man and his girlfriends shows up in a party) = P(A & B) = P(B)*P(A|B). If he always goes to whichever party his girlfriend

goes, P(A|B) = 1. Hence, P(A & B) = P(B)*P(A|B) = P(B).

If he always avoid to whichever party his girlfriend goes, P(A|B) = 0. Hence, P(A & B) = P(B)*P(A|B) = 0.

If in making the party decision, he never considers whether his girlfriend is going to a party, P(A|B) = 0.7. Hence, P(A & B) = P(B)*P(A|B) = P(B)*P(A) = P(B)*0.7.


Statistical Independence

Two events are statistically independent if and only if:

Events A and B are independent when the probability of one event is not affected by the other event

If A and B are independent, thenP(A)B)|P(A

P(B)P(A)B)P(A

P(B)A)|P(B

if P(B)>0

if P(A)>0


Example 9: Statistical Independence

No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .6 1.0


Are the events AC and CD statistically independent?


Example 9: Statistical Independence

No CDCD Total

AC .2 .5 .7

No AC .2 .1 .3

Total .4 .6 1.0

P(AC ∩ CD) = 0.2

P(AC) = 0.7

P(CD) = 0.4P(AC)P(CD) = (0.7)(0.4) = 0.28

P(AC ∩ CD) = 0.2 ≠ P(AC)P(CD) = 0.28

So the two events are not statistically independent


Bivariate Probabilities

B1 B2 . . . Bk

A1 P(A1B1) P(A1B2) . . . P(A1Bk)

A2 P(A2B1) P(A2B2) . . . P(A2Bk)

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

Ah P(AhB1) P(AhB2) . . . P(AhBk)

Outcomes for bivariate events:


Joint and Marginal Probabilities

The probability of a joint event, A ∩ B:

Computing a marginal probability:

Where B1, B2, …, Bk are k mutually exclusive and collectively exhaustive events

outcomeselementaryofnumbertotal

BandAsatisfyingoutcomesofnumberB)P(A

)BP(A...)BP(A)BP(AP(A) k21


Example 10: Marginal Probability

P(Ace)

BlackColor

Type Red Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52

52

4

52

2

52

2Black)P(AceRed)P(Ace


Tree Diagrams

A tree diagram is useful for portraying conditional and joint probabilities. It is particularly useful for analyzing business decisions involving several stages.


EXAMPLE 11: Tree Diagram

R1

B1

R2

B2

R2

B2

7/12

5/12

6/11

5/11

7/11

4/11

In a bag containing 7 red chips and 5 blue chips you select 2 chips one after the other without replacement. Construct a tree diagram showing this information.


EXAMPLE 11: Tree Diagram

R1

B1

R2

B2

R2

B2

7/12

5/12

6/11

5/11

7/11

4/11

The tree diagram is very illustrative about the relation between joint probability and conditional probabilityLet A (B) be the event of a red chip in the first (second) draw.

P(B|A) = 6/11

P(A) = 7/12

P(A and B) = P(A)*P(B|A)= 6/11 * 7/12

7/12

6/11


Example 12: Using a Tree Diagram

Has AC

Does not have AC

Has CD (.2/.7)

Does not have CD(.5/.7)

Has CD (.2/.3)

Does not have CD(.1/.3)

P(AC)= .7

P(AC)= .3

P(AC ∩ CD) = .2

P(AC ∩ CD) = .5

P(AC ∩ CD) = .1

P(AC ∩ CD) = .2

AllCars



Odds

The odds in favor of a particular event are given by the ratio of the probability of the event divided by the probability of its complement

The odds in favor of A are

)AP(

P(A)

P(A)1-

P(A) odds


Example 13: Odds

Calculate the probability of winning if the odds of winning are 3 to 1:

Now multiply both sides by 1 – P(A) and solve for P(A):

3 x (1- P(A)) = P(A) 3 – 3P(A) = P(A) 3 = 4P(A) P(A) = 0.75

P(A)1-

P(A)

1

3 odds


Overinvolvement Ratio

Does A1 depends on B1 and B2 differently?

The probability of event A1 conditional on event B1

divided by the probability of A1 conditional on activity B2 is defined as the overinvolvement ratio:

An overinvolvement ratio greater than 1 implies that event A1 increases the conditional odds ratio in favor of B1:

)B|P(A

)B|P(A

21

11

)P(B

)P(B

)A|P(B

)A|P(B

2

1

12

11


Example 14: Overinvolvement ratio

If the probability to get lung cancer is 0.5% for smokers and 0.1% for non-smokers, what is the overinvolvement ratio?

Prob(lung cancer | smokers) = 0.005

Prob(lung cancer | non-smokers) = 0.001

Overinvolvement ratio = P(L|S)/P(L|N) = 5


Bayes’ Theorem

where:Ei = ith event of k mutually exclusive and

collectively exhaustive events

A = new event that might impact P(Ei)

))P(EE|P(A...))P(EE|P(A))P(EE|P(A

))P(EE|P(A

P(A)

))P(EE|P(AA)|P(E

kk2211

ii

iii


Bayes’ Theorem

Bayes’ Theorem can be derived based on simple manipulation of the general multiplication rule.

P(A1|B)

= P(A1 & B) /P(B)

= [P(A1) P(B|A1)] / P(B)

= [P(A1) P(B|A1)] / [P(A1 & B) + P(A2 & B)]

= [P(A1) P(B|A1) ]/ [P(A1) P(B|A1) + P(A2) P(B|A2)]

)A|)P(BP(A)A|)P(BP(A)A|)P(BP(A

B)|P(A2211

111


Example 15: Bayes’ Theorem

A drilling company has estimated a 40% chance of striking oil for their new well.

A detailed test has been scheduled for more information. Historically, 60% of successful wells have had detailed tests, and 20% of unsuccessful wells have had detailed tests.

Given that this well has been scheduled for a detailed test, what is the probability that the well will be successful?



Let S = successful well

U = unsuccessful well

P(S) = .4 , P(U) = .6 (prior probabilities)

Define the detailed test event as D

Conditional probabilities:

P(D|S) = .6 P(D|U) = .2

Goal is to find P(S|D)


667.12.24.

24.

)6)(.2(.)4)(.6(.

)4)(.6(.

U)P(U)|P(DS)P(S)|P(D

S)P(S)|P(DD)|P(S


Apply Bayes’ Theorem:

So the revised probability of success (from the original estimate of .4), given that this well has been scheduled for a detailed test, is .667



Duff Cola Company recently received several complaints that their bottles are under-filled. A complaint was received today but the production manager is unable to identify which of the two Springfield plants (A or B) filled this bottle. The following table summarizes the Duff production experience.

What is the probability that the under-filled bottle came from plant A?

% of Total Production % of under-filled bottles

A 55 3

B 45 4



.4783.45(.04).55(.03)

.55(.03)

B)|P(B)P(UA)|P(A)P(UA)|P(A)P(U

P(A/U)

The likelihood the bottle was filled in Plant A is reduced from .55 to .4783. Without the information about U, the manager will say the under-filled bottle is likely from plant A. With the additional information about U, the manager will say the under-filled bottle is likely from plant B.

What is the probability that the under-filled bottle came from plant A?

% of Total Production % of under-filled bottles

A 55 3

B 45 4


- END -

Lesson 3: Lesson 3: A Survey of Probability Concepts

Documents

Ka-fu Wong © 2007 ECON1003: Analysis of Economic Data Lesson3-1 Lesson 3: A Survey of Probability Concepts