a2 Probability

02-Aug-13

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Probability

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Probability Models

� Concerned with the study of random (or chance) phenomena

� Random experiment: An experiment that can result in different

outcomes, even though it is repeated in the same manner every

time

� Managerial Implication: Due to this random nature additional

capacities for any setup are required

� Although random, certain statistical regularities are to be captured

to form the model

� Model: An abstraction of real world problem

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Sample Space

� A set of all possible outcomes of a random experiment

� Denoted by ‘S’

� Examples:

• Tossing of a coin: S = {‘Head’, ‘Tail’}

• Throwing of a dice: S = {1, 2, 3, 4, 5, 6}

• Marks of a student in a 20 mark paper with only integer negative marking

possible: S = {-20, -19, …, -1, 0, 1, …, 19, 20}

• Number of people arriving at a bank in a day: S = {0, 1, 2, …}

• Inspection of parts till one defective part is found: S = {d, gd, ggd, gggd, …}

• Temperature of a place with a knowledge that it ranges between 10 degrees

and 50 degrees: S = {any value between 10 to 50}

• Speed of a train at a given time, with no other additional information: S =

{any value between 0 to infinity}

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Sample Space (cont…)

� Discrete sample space: One that contains either finite or

countable infinite set of outcomes

• Out of the previous examples, which ones are discrete sample spaces???

� Continuous sample space: One that contains an interval of real

numbers. The interval can be either finite or infinite

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Events

� A collection of certain sample points

� A subset of the sample space

� Denoted by ‘E’

� Examples:

• Getting an odd number in dice throwing experiment

S = {1, 2, 3, 4, 5, 6}; E = {1, 3, 5}

• Getting a defective part within the first three inspected parts

S = {d, gd, ggd, gggd, ggggd, …}; E = {d, gd, ggd}

• Event of a component not failing before time t:

S = [0, ∞); E = [t, ∞)

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Events (cont…)

� Complement of an event E1: E1’ = S – E1

� Intersection of two events: E3 = E1 E2

� Union of two events: E4 = E1 U E2

� Two events A and B are mutually exclusive if: A B = Ø

� ‘Union’ and ‘Intersection’ can be extended to more than two

events

� Two events A and B are mutually exclusive and exhaustive if:

A B = Ø and A U B = S

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U

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Graphical Representation of Events

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Venn Diagram

A B

S

Disjoint events ‘A’ and ‘B’ A B

A

S

B

U

A U B

A

S

B

C

BS

Mutually exclusive and exhaustiveevents: A, B, C, and D

A

D

Graphical Representation of Events (cont…)

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Tree Diagram

Defective Not-Defective



Head Tail

Head Tail Head Tail

Probability

� Probability of an event represents the ‘relative likelihood’ that

performance of the experiment will result in occurrence of that

event

� P(A) = Probability of the event A in the sample space

� Examples:

• A = Getting an odd number in dice throwing experiment

P(A) = 0.5

• B = Getting two heads in two successive tosses of a coin

P(B) = 0.25

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Concept of ‘Sample’ and ‘Population’

� A population is the entire collection units in which we are interested

� A sample is a subset of a population

� The objective of ‘inferential statistics’ is to make an inference about a

population of interest based on the information obtained from the

sample from that population

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Probability in the Context of Sample and Population

� Proportion � Sample

� Probability � Population

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The Approaches for Determining Probability

� Classical Approach: probability is based on the idea that certain

occurrences are equally likely (e.g. throwing of dice)

� Relative Frequency Approach: probability based on long term

relative frequency --- requires a large amount of historical data

� Subjective Probability: probability based on one’s judgement (e.g.

how may days will it take for a consignment to reach from Indore

to Delhi?)

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02-Aug-13

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The Three Axioms of Probability

� For any event A, 0 <= P(A) <= 1

� P(S) = 1

� If A and B are mutually exclusive events then,

P(A U B) = P(A) + P(B)

� Some subsequent results:

• P(E1’) = 1 – P(E1)

• If event E1 is contained in event E2, then

P(E1) <= P(E2)

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Addition Rules

� If A and B are any two events then,

P(A U B) = P(A) + P(B) – P(A B)

� If A, B and C are any three events then,

P(A U B U C) = P(A) + P(B) + P(C) – P(A B) – P(A C) – P(B C) + P(A B C)

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U

U UUUU

Conditional Probability

� The conditional probability of A given B is:

P(A|B) = P(A B) / P(B)

if P(B) ≠ 0;

else it is undefined

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Multiplication Rule

� The conditional probability rule can be rewritten to get the

multiplication rule:

P(A B) = P(A|B) x P(B) = P(B|A) x P(A)

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Example

� A consulting firm is bidding for two jobs, one with each of two

large multinational corporations. The company executives

estimate that the probability of obtaining the consulting job with

firm A, event A, is 0.45. The executives also feel that if the

company should get the job with firm A, then there is a 0.90

probability that firm B will also give the company the consulting

job. What are the company’s chances of getting both jobs?

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Example

� In a standard government run lottery in Europe and North

America, you choose 6 out of 49 numbers (1 through 49). You win

the biggest prize if these 6 are drawn. (The prize money is divided

between all those who choose the lucky numbers. If no one wins,

then most of the prize money is put back into next week's

lottery). You are offered two tickets A = (1,2,3,4,5,6) or B =

(39,36,32,21,14, and 3). Do you prefer A, B, or are you

indifferent between the two?

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Independence

� Two events A and B are independent if any one of the following

equivalent statements are true:

P(A|B) = P(A)

P(B|A) = P(B)

P(A B) = P(A) x P(B)

� If a list of events is mutually independent, the probability of their

intersection is the product of their probabilities

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More about Statistical Independence

� Assume that biologically, the probability of giving birth to a girl is

0.5 and the probability of giving birth to a boy is 0.5

� A couple has two girls. What is the chance of delivering a baby girl

for the third time?

Does this imply statistical independence?

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In many practical business scenarios, the assumption of statistical

independence may not be valid.

Loan default example:

A bank has assigned probabilities of default to three different loans:

P(Loan 1 defaults) = p1



If the three loans are from three different companies, but in the same

business, example suppliers of auto parts.

Loan repayment problem with one of the customers is very likely an

indication of similar problems with remaining two customers.

Is the assumption of independence valid in such a case? 21


Dependence may be built in purposely -

Advertising campaign example: The aim of any advertising

campaign is to attract potential customers and boost up the sales

Suppose, a company has launched a new advertisement for a product

Event A = Potential customer sees an advertisement for a product

Event B = Potential customer buys the product

You can estimate P(B) from the historical data

What can you say about the estimate for P(B|A)?

Is the independence of events desirable in such a case?

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Consider the statement:

If two events are mutually exclusive then they are

independent.

The above statement is FALSE

In fact, the exact reverse is true, that is,

“If two events are mutually exclusive then they can never be

independent.”

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Conditional Independence:

Two events A and B are said to be conditionally independent,

given another event C with P(C) > 0, if

P(A ∩ B |C) = P(A|C) x P(B |C)

Independence does not imply conditional independence, and

vice versa.

Hint example:

A = {1st toss result is a tail},

B = {2nd toss result is a tail},

C = {the two tosses have different results}.

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Independence of three events:

� If we have a collection of three events, A1, A2, and A3,

independence amounts to satisfying all the four conditions:

� P(A1 ∩ A2) = P(A1) x P(A2),

� P(A1 ∩ A3) = P(A1) x P(A3),

� P(A2 ∩ A3) = P(A2) x P(A3),

� P(A1 ∩ A2 ∩ A3) = P(A1) x P(A2) x P(A3).

only then the three events are said to be independent events.

� First three conditions simply assert that any two events are

independent, a property known as pairwise independence.

� The fourth condition is also important and does not automatically

follow from the first three.

� Conversely, the fourth condition does not imply the first three25


Independence of three events:

Hint examples:

� A1 = {1st toss result is a tail}, A2 = {2nd toss result is a tail},

A3 = {the two tosses have different results}

Exercise: Verify that the first three conditions (pairwise

independence) are satisfied, but the fourth is not satisfied.

� A1 = {Roll of first dice is 1, 2, or 3},

A2 = {Roll of first dice is 3, 4, or 5},

A3 = {Sum of rolls of first dice and second dice is 9}

Exercise: Verify that the fourth condition is satisfied, but the

pairwise independence is not satisfied.26

Joint and Marginal Probabilities

Example: Does education really give you a better income?

Annual household income and education in US 2000 census

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Education Poor

(<$25K)

Lower

($25K to

$50K)

Middle

($50K to

$75K)

Upper

(>$75K)

Row total

No HS diploma 10.51 4.26 1.32 0.68 16.77

High school grad. 10.93 10.57 5.84 3.87 31.21

Some college 6.36 8.36 5.84 5.23 25.79

Bachelor’s deg. 2.23 4.16 4.15 6.57 17.11

Master’s deg. 0.62 1.26 1.36 2.84 6.08

Professional deg. 0.16 0.22 0.29 0.99 1.66

Doctorate 0.09 0.21 0.29 0.80 1.39

Column total 30.9 29.04 19.09 20.98 100

All cells show percentages

Sample size is 98 million households

Source: Statistics for Business Decision Making and Analysis by Stine and Foster

Law of Total Probability

� For any two events A and B:

P(A) = P(A B) + P(A Bc)

= P(B) x P(A|B) + P(Bc) x P(A|Bc)

� In general,

Given k mutually exclusive and exhaustive events S1, S2, … Sk and any

other event A, the probability of event A can be expressed as:

P(A) = P(A S1) + P(A S2) + … + P(A Sk)

= P(S1) x P(A|S1) + P(S2) x P(A|S2) + … + P(sk) x P(A|Sk)

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U U

U U U

An Example towards Bayes’ Rule

A petroleum company uses a specialized equipment for sensing

the presence of oil wells in a region, which can then be drilled to

extract the oil. The machine can accurately sense the presence of

oil for 95% of the times. However, it also gives a ‘false-positive’

result (that is incorrectly showing the presence of oil when in fact

there is none) for 25% of the times.

It is estimated that 10% of the region under consideration actually

has oil underneath. For a certain survey, if the machine has given

a positive result for a certain area under consideration within the

region, what is the probability that the oil is actually present in

that area?

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Bayes’ Rule

P(Si) is called as the prior probability of Si because it does not

take into account any other behaviour and is based on historical

data

In practice, Bayes’ Rule can be applied to re-evaluate (or update)

the prior belief of probabilities of S1, S2, …, Sk, when new

information is obtained (as a result of event A happening)

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02-Aug-13

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Bayes’ Rule – Some More Examples

ABC Plastics specializes in manufacturing of plastic bottles. For this

purpose, it has two production facilities in two different cities – say City 1

and City 2. The plant in City 1 contributes to 60% of the total annual

output, and that in City 2 the remaining 40%. The central quality

assurance reported that 5% of the bottles produced in City 1 plant were

defective. Likewise, 10% of the bottles from the City 2 plant were

defective. The Controller of Operations for ABC Plastics wants to apportion

the cost of poor quality fairly between the two plants. How can that be

done?

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A startup software development firm has prepared a customized software

application for a large business organization. The sales manager of the

startup firm has a hunch that the probability of selling the application is

60%.

Usually in this kind of process, after making an initial presentation to the

potential client by the vendor, the client organization most often asks the

vendor to make a repeat presentation to the higher authorities from the

client side. Historically suppose for this particular client, 70% of the

successful vendors had been asked to do a second presentation. However,

50% of the unsuccessful vendors were also asked to do the second

presentation.

Suppose the startup firm in question has been asked to do a second

presentation. What is the revised estimate of the probability of selling the

application using the additional information in the previous paragraph?

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You are a physician. You think it is quite likely that one of your patients has strep

throat, but you are not sure. You take some swabs from the throat and send them

to a lab for testing.

The test is (like nearly all lab tests) not perfect. If the patient has strep throat,

then 70% of the time the lab says YES. but 30% of the time it says NO. If the

patient does not have strep throat, then 90% of the time the lab says NO. But

10% of the time it says YES.

You send five successive swabs to the lab, from the same patient. You get back

these results, in order; YNYNY. What do you conclude among the following?

1. These results are worthless

2. It is likely that the patient does not have the strep throat.

3. It is slightly more likely than not, that patient does have the strep throat.

4. It is very much more likely than not, that patient does have the strep throat.

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Additional Examples

� Samples of emission from three suppliers are classified for

conformance to air-quality specifications. The results from 100

samples are summarized as follows:

Confirms Not confirms

Supplier 1 22 8

Supplier 2 25 5

Supplier 3 30 10

Let ‘A’ denote the event that the sample is from supplier 2 and let B

denote the event that the sample confirms the specifications. If a

sample is selected at random, determine the following probabilities:

(a) P(A), (b) P(B), (c) P(A’),

(d) P(A B), (e) P(A U B), (f) P(A’ U B)

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Additional Examples

� A lot contains 15 castings from a local supplier and 25 castings

from a supplier from the next state. Two castings are selected

randomly without replacement from the lot of 40. Let event A be

that the first casting is selected from the local supplier and the

event ‘B’ be that the second casting is selected from the local

supplier. Determine the following:

• P(A)

• P(B|A)

• P(A B)

• P(A U B)

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Additional Examples

� Software to detect fraud in customer phone cards tracks the

number of metropolitan areas where the calls originate each day.

It is found that 1% of legitimate callers originate calls from more

than one metropolitan areas in a single day. As against these,

30% of the fraudulent users originate calls from more than one

metropolitan areas in a single day. The proportion of fraudulent

users is 0.01%.

If a same user originates calls from more than one metropolitan

area in a single day, what is the probability that the user is

fraudulent?

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a2 Probability