amity mba chapter 5 probability

8/8/2019 Amity MBA Chapter 5 Probability

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Probability: Introduction



Sharad Kant Amity MBA QT 2

Introduction

• The study of Uncertainty – Changes “I’m not sure …”

• to “I’m positive we’ll succeed … with probability 0.8”

– Can’t predict “for sure” what will happen next• But can quantify the likelihood of what might happen• And can predict percentages well over the long run• Example: a 60% chance of rain• e.g., success/failure of a new business venture

– Investment Problem – Introducing a new product – Stocking Decision




Terminology

An Event• An event is one or more of the possible outcomes of

doing something.

– Examples of Event1. Toss of a coin, and getting a tail would be an event.2. From a pack of cards, drawing an ace of spade.3. One student is selected from a class of 60 students




Terminology (Contd.)

Random Experiment – A procedure that produces an outcome

• Not perfectly predictable in advance

– An activity that produces an event is referred in probabilitytheory as an Experiment

– Examples:1. In a coin toss experiment, the occurrence of Head is an event.2. In a card-drawing experiment, selection of a King is an event.




Terminology (Contd.)

Sample Space – A list of all possible outcomes is called sample space

• Each random experiment has a sample space

– Examples Sample Space• Toss of a coin {H, T}• Toss of two coins {HH, HT, TH, TT}• Toss of a Dice {1, 2, 3, 4, 5, 6}

• Toss of two Dice {11,12, … 21, 22, …, 61, ., 66}• Selection of a student

from a class of 54 {List of all the 54 students}




Terminology (contd.)

• Probability of an Event

– A number between 0 and 1• The likelihood of occurrence of an event

n e v e r h

a p p e n s

a l w a y s

h a p p e n

s





Contingency Table • It is way to present the sample space. It is a table of cross

classification.

Example: Contingency Table for Face-colour variable of aplaying card pack

Red Black Total

Ace 2 2 4

Non-Ace 24 24 48

Total 26 26 52





Mutually Exclusive Events• Events are said to be mutually exclusive if one and only one of them

can take place at a time.

• Example: – Toss of a coin: either head or tail may turn up but not both. Head

and tail are mutually exclusive events – Results of an examination: a student will either pass or fail but

not both. Pass and fail are mutually exclusive.





Collectively Exhaustive List• When a list of the possible events that can result from an

experiment includes all possible outcomes, the list is called acollectively exhaustive.

– Examples• Toss of a coin: {H, T}• Throw of a Dice: {1, 2, 3, 4, 5, 6}

• Result of a student {Pass, Fail}




Types of Probabilities

• Classical Probability – From mathematical theory – Make assumptions, draw conclusions

• Relative Frequency – From data – What percent of the time the event happened in the past

• Subjective Probability – Anyone’s opinion , perhaps even without data or theory

– Bayesian analysis uses subjective probability with data




Classical Probability

• If all N possible outcomes in the sample space areequally likely, then the probability of an event A is

Probability ( A) = (# of outcomes where event A occurs ) / N

Note: This probability is not a random number. The probability isbased on the entire sample space

Examples:

1. Suppose there are 35 defects in a production lot of 400.Choose item at random.Probability (defective) = 35/400 = 0.0875

2. Toss of a coin. Probability (heads) = 1/2




Relative Frequency• From data: Run random experiment N times and see

how often an event happened• (Relative Frequency of A) = (# of times A happened) / N

– Example:12 flights, 9 were on time.Relative frequency of the event “on time” is 9/12 = 0.75




Relative Frequency

Law of Large Numbers – If N is large, then the relative frequency will be close

to the probability of an event

• Probability is FIXED .• Relative frequency is RANDOM

Example:Toss coin 20 times.Probability of “heads” is 0.5Relative frequency is 12/20 = 0.6 , or 9/20 = 0.45




Relative Frequency (contd.)

• Suppose event “Yes” has probability 0.25• In N = 5 runs of random experiment

– Event happens: No, Yes, No, No, Yes – Relative frequency of Yes is 2/5 = 0.4

• Graph of relative frequencies for n = 1 to 5

0.5

1 2 3 4 50.0

Relative frequency

Number n of times random experiment was run




Relative Frequency (contd.)• As N gets larger

– Relative frequency gets closer to probability

• Graph of relative frequencies for N = 1 to 200 – Relative frequency approaches the probability

0

0.5

0 50 100 150 200Number n of times random experiment was run

R e l a t i v e f r e q u e n

c y

Probability = 0.25




Subjective Probability

•Anyone’s opinion – What do you think the chances are that the Indian economy will

have steady expansion in the near future? – An economist’s answer

• Bayesian analysis – Combines subjective probability with data to get results

• But subjective opinions (prior beliefs) can still play a backgroundrole, even when they are not introduced as numbers into acalculation, when they influence the choice of data and themethodology (model) used




Exercises

1. From a pack of cards find the probabilities of the followingevents

a) Drawing a card of spadeb) Drawing a Kingc) Drawing a 9 of heart

d) Drawing a Red Queen

2. Random experiment of throwing two dice find theprobabilities of the following events:

a) Probability of total of both dice appearing (1)

b) Probability of total of both dice appearing (2)c) Probability of total of both dice appearing (4)d) Probability of total of both dice appearing (6)e) Probability of total of both dice appearing (7)




Complement of an Event

• The event “ not A ” happens whenever A does not• Venn diagram: A (in circle), “ not A” (shaded)

• Prob( not A ) = 1 – Prob( A) – If Prob (Success) = 0.7 , then Prob (Failure) = 1– 0.7 = 0.3

A

not A

A




Union of Two Events• Union happens whenever either (or both) happen• Venn diagram: Union “ A or B” (shaded) AυB

– e.g., A = “get Amity job offer”, B = “get Professor job offer”• Did the union happen? Congratulations! You have a job

A B




Intersection of Two Events• Intersection happens whenever both events happen• Venn diagram: Intersection “ A and B” (shaded) A∩B

– e.g., A = “sign contract”, B = “get financing”• Did the intersection happen? Great! Project has been

launched!

A B




Relationship Between and and or

• Prob( A or B) = Prob ( A) + Prob ( B) – Prob ( A and B )

= + –

• Prob ( A and B) = Prob ( A) + Prob ( B) – Prob ( A or B)• Example: Customer purchases at appliance store

• Prob( Washer ) = 0.20• Prob( Dryer ) = 0.25

• Prob( Washer and Dryer ) = 0.15 – Then we must have

• Prob( Washer or Dryer ) = 0.20 + 0.25 – 0.15 = 0.30




Conditional Probability

• Examples

– Prob ( Success given Good results in test market ) is Higher thanProb ( Success ) evaluated before marketing study

– Prob ( Get job given Poor interview ) is Lower than Prob ( Get job given Good interview )




Conditional Probability (contd.)

• Given the extra information that B happens for sure, how must youchange the probability for A to correctly reflect this new knowledge?

– This is a (conditional) probability about A – The event B gives information

• Unconditional

– The probability of A• Conditional

– A new universe, since B must happen

Prob ( A given B) =Prob ( A and B)

Prob ( B)

A B

A and B B




Conditional Probability (contd.)• Example: appliance store purchases

P ( Washer ) = 0.20P ( Dryer ) = 0.25P ( Washer and Dryer ) = 0.15

– Conditional probability of buying a Dryer given that they bought aWasher P ( Dryer given Washer )

= P ( Washer and Dryer ) / P ( Washer ) = 0.15 /0.20 = 0.7575% of those buying a washer also bought a dryer

– Conditional probability of Washer given Dryer = P ( Washer and Dryer ) / P ( Dryer ) = 0.15 /0.25 = 0.6060% of those buying a dryer also bought a washer

Watch the denominator!




Independent Events

• Two events are Independent if information about one does not change thelikelihood of the other – Three equivalent ways to check independence

Prob ( A given B) = Prob ( A)Prob ( B given A) = Prob ( B)Prob ( A and B) = Prob ( A) × Prob ( B)

• Two events are Dependent if not independent – Prob( Washer and Dryer ) = 0.15 – Prob ( Washer ) × Prob ( Dryer ) = 0.20 × 0.25 = 0.05 – Washer and Dryer are not independent

• They are dependent

I f i n d e p e n d

e n t,

a l l t h r e e a r e

t r u e.

U s e a n y o n

e t o c h e c k.

N o t e q u a

l




Mutually Exclusive Events

• Two events are Mutually Exclusive if they cannot both happen, thatis, if

Prob ( A and B) = 0

• No overlapin Venn diagram

• Examples – Profit and Loss (for a selected business division) – Green and Purple (for a manufactured product)

• Mutually exclusive events are dependent events

A B




Additive Law of Probability• The probability that at least one of the several mutually exclusive

events A1, A2, … , An will occur is the sum of the probability of theoccurrences of the individual events.

• P(A1υA2 υ … υ An) = P(A1) + P(A2)+ … + P(An)

• Example• In a Horse race, the probabilities of winning horses are as follows:

Horse A winning = P(A) = 1/4Horse B winning = P(B) = 1/5Horse C winning = P(C) = 1/6Horse D winning = P(D) = 1/7

– What is the probability of winning of at least one of them in therace?




Additive Law of Probability

A.notB notA.B

AB

A B

P(A) = P(AB) + P(A.notB) P(B) = P(AB) + P(notA.B)

P(A + B) = P(A) + P(B) – P(AB)




Additive Law of ProbabilityExercise

80 % of all the tourist who come to India will visit Delhi, 70%of them will visit Mumbai and 60% of them will visit bothDelhi and Mumbai.

What is the Probability that a tourist coming to India will visitDelhi or Mumbai or both.

What is the probability that he will visit neither city?




Additive Law of ProbabilityExercise

Solution80 % of all the tourist who come to India will visit Delhi, 70% of them

will visit Mumbai and 60% of them will visit both Delhi and Mumbai.

P(D) = 0.8 P(M) = 0.7 P(DM) = 0.6

Probability that a tourist will visit Delhi or Mumbai or both= P(D + M)= P(D) + P(M) – P(DM)

= 0.8 + 0.7 – 0.6= 0.9

Probability that he will visit neither city = 1 – P(D + M) = 0.1




Permutations and Combinations

• n! = Factorial n• 0! = Factorial zero = 1• Combinations: Selecting r items out of n when the order

in which these are selected in not important

• Permutations: Selecting r items out of n when the order in which these are selected makes the difference

[ ]1.2.3.4)...2)(1(−−=

nnn

[ ])!(!!

r nr

nCr

n

−

=

))1()......(2)(1()!(

! −−−−=

−

= r nnnnr n

n P r

n




Probability Trees

A method for solving probability problems• Given probabilities for some events (perhaps union, intersection, or

conditional) – Find probabilities for other events – Record the basic information on the tree

• Usually three probability numbers are given – Perhaps two probability numbers if events are independent

• The tree helps guide your calculations – Each column of circled probabilities adds up to 1 – Circled prob times conditional prob gives next probability

– For each group of branches• Conditional probabilities add up to 1• Circled probabilities at end add up to probability at start




Probability Tree (contd.)• Shows probabilities and conditional probabilities

P( A and B)

P( A and “not B”)

P(“not A” and B)

P(“ not A” and “not B”)

P( A)

P(not A) Ye s

Y e s

N o

P ( “ n o t B ” g i v e n A )

P( B gi ve n A )

P( B gi ve n “ not A ” )

P ( “ n o t B ” g i v e n “ n o t A” )

N o

Event B

Y e s

N o

Event A




Example: Appliance Purchases• First, record the basic information

• Prob( Washer ) = 0.20 , Prob( Dryer ) = 0.25• Prob( Washer and Dryer ) = 0.15

0.15

0.20 Y e s

Y e s

Y e s

N o

N o

N o

Dryer ?Washer ?

T h e s e a d

d

u p t o P ( D r y

e r )

= 0. 2 5




Example (contd.)

• Next, subtract: 1 – 0.20 = 0.80, 0.25 – 0.15 = 0.10

0.15

0.10

0.20

0.80

Y e s

Y e s

Y e s

N o

N o

N o

Dryer ?Washer ?

T h e s e a d d

u p t o P ( D r y e r

)

= 0. 2 5




Example (contd.)

• Now subtract: 0.20 – 0.15 = 0.05, 0.80 – 0.10 = 0.70

0.15

0.05

0.10

0.70

0.20

0.80

Y e s

Y e s

Y e s

N o

N o

N o

Dryer ?Washer ?




Example (completed tree)• Now divide to find conditional probabilities

0.15 /0.20 = 0.75, 0.05/ 0.20 = 0.250.10/0.80 = 0.125, 0.70/0.80 = 0.875

0.15

0.05

0.10

0.70

0.20

0.80

Y e s

Y e s

Y e s

N o

0 .2 5

N o

0. 7 5

0. 1 2 5

0 .8 7 5

N o

Dryer ?Washer ?




Example (finding probabilities)• Finding probabilities from the completed tree

P( Washer ) = 0.20

P( Dryer ) = 0.15+0.10 = 0.25

P( Washer and Dryer ) = 0.15

P( Washer or Dryer ) =0.15 +0.05+0.10 = 0.30

P( Washer and not Dryer ) = 0.05

P( Dryer given Washer ) = 0.75

P( Dryer given not Washer ) = 0.125

P( Washer given Dryer ) = 0.15 /0.25 = 0.60(using the conditional probability formula)

0.15

0.05

0.10

0.70

0.20

0.80

Y e s

Y e s

Y e s

N o

0 .2 5

N o

0. 7 5

0. 1 2 5

0 .8 7 5

N o

Dryer ?Washer ?

E i




Exercise

• A business firm has invited applications for a managerialpost. The probability that an applicant has apostgraduate qualification is 0.3 and that he hasadequate work experience is 0.7, and that he has both

the postgraduate qualification and work experience is0.4. Assuming that 50 persons have applied for thismanagerial post in the company, find out how manyapplicants would have either a postgraduate degree or adequate work experience.




Exercise - Solution

• Event A: – Applicant has a postgraduate degree: P(A) = 0.3

• Event B: – Applicant has an adequate work experience: P(B) = 0.7

• A and B are not mutually exclusive: P(A and B) = 0.4

• P(A or B) = P(A) + P(B) – P(A and B) = 0.3 + 0.7 – 0.4 = 0.6

• No. of applicants that have either a postgraduate degree or adequatework experience = 0.6 x 50 = 30




Probabilities under conditions of statistical Dependence

• Conditional Probability for StatisticallyDependent Events

P(B/A) = P(BA) / P(A)




Example / Exercise• A box containing 10 balls as described below:

2 balls – Red and Dotted1 ball – Green and Dotted4 balls – Red and Striped

3 balls – Green and Stripped

1. What is the probability of drawing any particular ball from this box?2. We draw a ball from the box and find it is red, what is probability

that it is striped?3. What is the probability of getting dotted ball given that it is green?




Example - Solution

1. P(A) = 1/10

2. P(S/R) = P(SR) / P(R) = (4/10) / (6/10) = 4/6 = 2/3

3. P(D/G) = P(DG) / P(G) = (1/10) / (4/10) = 1/4




Joint Probability for StatisticallyDependent Events

Joint probability of events Band A happening together or in succession

Probability of event Bgiven that event A hashappened

P(BA) = P(B / A) x P(A)

Probability that event A will happen




Marginal Probabilities under StatisticalDependence

• P(R) = P(SR) + P(DR) = 4/10 + 2/10 = 6/10

• P(G) = P(SG) + P(DG) = 3/10 + 1/10 = 4/10

• P(S) = P(SG) + P(SR) = 3/10 + 4/10 = 7/10

• P(D) = P(DR) + P(DG) = 2/10 + 1/10 = 3/10




Probabilities under StatisticalDependence

Green Red Total

Dotted 1 2 3

Stripped 3 4 7

Total 4 6 10

Note: Distinguish between conditional probability and joint probability bycareful use of terms given that and both … and:

P(A/B) is “the probability that A will occur given that B has occurred” and

P(AB) is “the probability that both A and B will occur”.

Marginal probability P(A) is the “probability that A will occur, whether or not Bhappens”




Bayes’ Theorem




Bayes’ TheoremPosterior Probabilities definition:

• Certain probabilities change after the people involved got additionalinformation. The new probabilities are known as revised, or posterior, probabilities. Because probabilities can be revised asmore information is gained, probability theory is of great value in

managerial decision-making.• Example: In a competitive Examination

– P (selection) = 0.05 (1/20) – The candidate is called for an interview – P ( selection / interview call) = 0.3 (3/10)




Bayes’ Theorem

•Bayes’ Theorem deals with the posterior probability assessment of an event.

Suppose B1 and B2 are mutually exclusive and collectivelyexhaustive events with prior probabilities P(B1) and P(B2) which areknown.

Suppose A is an event having conditional probability P(A / B1), P(A /B2) which are also known.

Then, Posterior probability of events B1 given that the event A hasalready happened is given by

P(A / B1) x P(B1)P(B1 / A) = --------------------------------------------------------

P(A / B1) x P(B1) + P(A / B2) x P(B2)




Bayes’ TheoremExample

• An item is manufactured by three machines M1, M2, and M3. Out of thetotal items manufactured during specified production period, 50% are

manufactured on M1, 30% on M2, and 20% on M3.

• It is also known that 2% of the item produced by M1 and M2 are defective,

while 3% of those manufactured on M3 are defective.

• All the items are put into one bin.

• From the bin, one item is drawn at random and is found to be defective.

What is the probability that it was made on M1, M2, or M3.

Bayes’ Theorem




Bayes TheoremExample - Solution

A = Item is defectiveB1 = Item is manufactured on M1

B2 = Item is manufactured on M2 Mutually ExclusiveB3 = Item is manufactured on M3

Prior Probabilities of manufacturing on M1 = P(B1) = 0.5Prior Probabilities of manufacturing on M2 = P(B2) = 0.3Prior Probabilities of manufacturing on M3 = P(B3) = 0.2

Probability of Defective item on M1 = P(A/B1) = 0.02Probability of Defective item on M2 = P(A/B2) = 0.02Probability of Defective item on M3 = P(A/B3) = 0.03

To calculate Posterior Probability [knowing that the event A has already happened (itemis defective)] that it is manufactured on M1 is given by

P(A / B1) x P(B1) (0.02)(0.5)P(B1 / A) = ------------------------------------------------- = ------------------------------------------------

P(A / B1) x P(B1) + P(A / B2) x P(B2) (0.02)(0.5) + (0.02)(0.3) + (0.03)(0.2)

= [0.01] / [ 0.01 + 0.006 + 0.006] = 0.01 / 0.022 = 0.454

Bayes’ Theorem




Bayes TheoremExample - Solution

Prior Probabilities of manufacturing on M1 = P(B1) = 0.5Prior Probabilities of manufacturing on M2 = P(B2) = 0.3

Prior Probabilities of manufacturing on M3 = P(B3) = 0.2

Probability of Defective item on M1 = P(A/B1) = 0.02Probability of Defective item on M2 = P(A/B2) = 0.02Probability of Defective item on M3 = P(A/B3) = 0.03

To calculate Posterior Probability [knowing that the event A has already happened (itemis defective)] that it is manufactured on M1 is given by

P(A / B1) x P(B1) (0.02)(0.5)P(B1 / A) = ------------------------------------------------- = ------------------------------------------------

P(A / B1) x P(B1) + P(A / B2) x P(B2) (0.02)(0.5) + (0.02)(0.3) + (0.03)(0.2)

= [0.01] / [ 0.01 + 0.006 + 0.006] = 0.01 / 0.022 = 0.454P(B2/A) = (0.02)(0.3)/ 0.022 = 0.006 / 0.022 = 0.273P(B3/A) = (0.03)(0.2)/ 0.022 = 0.006 / 0.022 = 0.273




Bayes’ TheoremExercise

• When a machine is set correctly, it produces 25% defectives,

otherwise it produces 60% defectives.

• From the past experience, the manufacturer knows that the chances

that the machine is set correctly or wrongly is 50 : 50.

• The machine was set before the commencement of production. One

piece was inspected and found to be defective.

• What is the probability of machine set being correct?



Sh d K t A it MBA QT 55

Probability IntroductionThe End

amity mba chapter 5 probability

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