03 probability distributions.ppt

7/27/2019 03 Probability Distributions.ppt

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Quantitative Methods

Varsha Varde


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Quantitative Methods

Quantifying Uncertainty:

Basic Concepts of Probability


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Uncertainty Each Statement Involves Uncertainty.

Chances = Odds = Likelihood = Probability Real Life is Usually Full of Uncertainty.

Almost Nothing is for Sure.

There are Chances of Something Happeningand Chances of Something Else Happening.

In Such Situations, You cant Prove Anything.

All You Can Do is to Assign a Probability toEach of the Different Possible Outcomes.


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Quotes from You and Me

After This MBA

Chances of your getting a handsome jobwould be 90% if you obtain an MBA.

I am 75% confident that collections will

jump this month. Odds are 80:20 for my promotion this time.

. Winning cricket match against Australia

is not impossible but has only 10% chance New machines churn out good product 97

out of 100 times.


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Probability Theory

How Do You Say 90% Chances, or80:20Odds, or75% Confidence?

Probability Theory Provides Tools toDecision Makers to Quantify Uncertainties.


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Assigning Probabilities

Classical Approach: Assumes equally likelyoutcomes (card games ,dice games, tossingcoins and the like)

Relative Frequency Approach: Uses relativefrequencies of past occurrences as probabilities(Decision problems in area of management.Delay in delivery of product)

Subjective Approach :Guess based on pastexperience or intuition.( At higher level ofmanagerial decisions for important ,specific andunique decisions


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Assume equally likely outcomes


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Use Relative Frequencies

Making use of relative frequencies

of past.

Suppose an organisation knows from pastdata that about 25 out of 300 employees

entering every year leave due to goodopportunities elsewhere

then the organisation can predict theprobability of employee turnover for this

reason

as 25/300=1/12=0.083


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Subjective Probability

Based on personal judgements

Uses individuals experience and familiaritywith facts

An expert analyst of share prices may give

his judgement as follows on price of ACCshares in next two months

20% probability of increase by Rs500or more

60% probability of increase by less thanRs500

20%probability of remaining unchanged


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Experiment Experiment: An experiment is some act, trial

or operation that results in a set of possibleoutcomes.

-The roll of two dice to note the sum of spots

-The toss of a coin to see the face that turnsup.

- polling

- inspecting an assembly line

- counting arrivals at emergency room

- following a diet


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Event Event: An event means any collection of

possible outcomes when an experiment isperformed. For example,

When an unbiased die is rolled we may

get either spot 1, spot 2, spot 3, spot 4,spot 5 or spot 6. Appearance of anyone of

the spots is an event.

Appearance of an even spot is also anevent.


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EVENT/OUTCOME -The roll of two dice (Appearance of

the sum of spots )-The toss of a coin( the face that turnsup)

- polling (Win or lose)- inspecting an assembly line(Numberof defectives)

- counting arrivals at emergencyroom(Number of arrivals in one hour)

- following a diet (weight loss or gain)


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Sample space

Sample space: the set of all samplepoints (simple events) for an experiment iscalled a sample space; or set of all

possible outcomes for an experiment

Venn diagram :It is a pictorialrepresentation of the sample space.It is

usually drawn as a rectangular figure

representing the sample space and circles

representing events in the sample space.

V Di


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Venn Diagram

For Roll of a die

A:Odd spots

B:Even Spots


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Equally Likely Events

Equiprobable or Equally Likely Events:

Events are said to be equiprobablewhen one does not occur more often

than the others.

When an unbiased die is thrown anyone of the six spots may appear.

When an unbiased coin is tossed either

a head or a tail appears


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Exhaustive Events

Exhaustive Events: Events are said to beexhaustive when they include all possible

cases or outcomes. For example, in

tossing of fair coin, the two eventsappearance of a head and appearanceof a tail are exhaustive events because

when a coin is tossed we would get eithera head or a tail.


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Independent Events

Independent Events: Two events A andB are said to be independent ifoccurrence of A does not affect and isnot affected by the occurrence of B.

When a coin is tossed twice the resultof the first toss does not affect and isnot affected by the result of the second

toss. Thus, the result of the first tossand the result of the second toss areindependent events.


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Dependent Events

Dependent Events: Two events A and B are

called dependent if the occurrence of Aaffects or is affected by the occurrence of B.

For example, there are four kings in a pack of52 cards. The event of drawing a king at the

first draw and the event of drawing anotherking at the second draw when the first drawnking is not replaced, are two dependentevents. In the first event there are four kings

in a pack of 52 cards and in the second eventthere are only three kings left in the pack ofremaining 51 cards

M t ll E l i E t


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Mutually Exclusive Events Events are termed mutually exclusive if they cannot occur together

so that in any one trial of an experiment at most one of the eventswould occur.

Mutually Exclusive Events: throwing even and throwing odd with one die, drawing the spade, drawing a diamond and drawing a club

while drawing one card from a deck.

purchase of a machine out of 3 brands available

Not mutually exclusive drawing a spade and drawing a queen even number and at least 3 with one die Selection of a candidate with post graduate qualification and over 3

years experience

A particular easy way to obtain two mutually exclusive events is toconsider an event and its negative(Complement). Such as evenand not even, spade, not spade or in general A and not A.

o a on


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o a on.

Sample space : S

Sample point: E1, E2, . . . etc. Event:A,B,C,D,E etc. (any capital letter).

Venn diagram:

Example. S = {E1, E2, . . ., E6}.

That is S = {1, 2, 3, 4, 5, 6}. We may think

ofS as representation of possibleoutcomes of a throw of a die.

Venn Diagram


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Venn Diagram

A:Candidates over 3 years experience

B:Candidates with post graduate qualification

S

AB

22

A B

ore e n ons


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ore e n ons Union, Intersection and Complementation

GivenA and B two events in a sample space S.

1. The union ofA and B,AUB, is the event containing allsample points in eitherA or B or both. Sometimes weuseA or B for union.

2. The intersection ofA and B,AB, is the eventcontaining all sample points that are both inA and B.

Sometimes we useAB orA and B for intersection.3. The complementofA, the event containing all sample

points that are not in A. Sometimes we use not A or Acfor complement.

Mutually Exclusive Events (Disjoint Events)4 Two events are said to be mutually exclusive (or disjoint)

if their intersection is empty. (i.e.A B = ).


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Example

Suppose S = {E1, E2, . . ., E6}. Let

A = {E1, E3, E5}; B = {E1, E2, E3}. Then

(i)A U B = {E1, E2, E3, E5}.

(ii)A B = {E1, E3}. (iii)= {E2, E4, E6}; Bc ={E4, E5, E6}; (iv)A and B are not mutually exclusive

(why?) (v) Give two events in S that are mutually

exclusive.

Probability of an event


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Probability of an event

Relative Frequency Definition If an

experiment is repeated a large number, n,of times and the eventA is observed nAtimes, the probability ofA is

P(A) = nA / n

Interpretation n = # of trials of an experiment nA= frequency of the eventA

nA/n = relative frequency ofA P(A) = nA /n , ifn is large enough.

B i F l f P b bilit


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Basic Formula of Probability Probability of an Event A:

No. of Outcomes Favourable to Event A= ----------------------------------------------------Total Number of All Possible Outcomes

Probability is a Ratio. (A Distribution Ratio)It varies from 0 to 1.

Often, It is Expressed in Percentage

Terms Ranging from 0% to 100%. It is denoted as P(A) and termed as

marginal or unconditional probability


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Rules of Probability:

Multiplication Rule It is for Probability of Simultaneous

Occurrence of Two Events

If A and B are two independent events,P(A & B) = P(A) x P(B)

Example: Experiment: Toss Two Coins

A: Getting Head on Coin No. 1

B: Getting Head on Coin No. 2

P(A)= , P(B)= , P(A&B)= =0.25


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General Multiplication Rule If A and B are two dependent events,

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

P(B|A) The conditional probabilityof the event B

given that event Ahas occurred Example: Draw Two Cards from a Deck

A: First Card a King

B: Second Card also a King

P(A)=4/52=1/13, P(B|A)=3/51

P(A & B)=1/13 x 3/51=3/204=0.015=1.5%


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Addition Rule It is for Probability of Occurrence of Either of the

Two Events

If A and B are two mutually exclusive events,

P(A or B) = P(A) + P(B) Example: Experiment: Roll a Die

A: Getting the No. 5 B: Getting the No. 6

P(A)=1/6, P(B)=1/6, P(A or B)=1/3=0.33=33% Note: Two Events are Mutually Exclusive if They

Cannot Occur Together


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General Addition Rule If A and B are any two events,

P(A or B) = P(A) + P(B) P(A & B)

Example: Toss Two Coins

A: Getting Head on Coin No. 1

B: Getting Head on Coin No. 2

P(A)= , P(B)= , P(A & B)= So, P(A or B)= + - = =0.75=75%


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Exercise

If 80% Company guests visit the HO,

70% visit the Plant, and 60% visit both,

what is the chance that a guest will visit

HO or Plant or both?

What is the probability that he will visit

neither the HO nor the Plant, but meet

Company Executives at the Taj?


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Solution

P(A)=0.8 P(B)=0.7 P(A&B)=0.6

Prob that a guest will visit HO or Plant orboth = P(A&B)=0.8 + 0.7 0.6=0.9 = 90%

Prob that he will visit neither the HO northe Plant, but meet Company Executives

at the Taj = 1 - Prob that a guest will visit

HO or Plant or both = 1 0.9 = 0.1 = 10%

C t l D fi iti f P b bilit


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Conceptual Definition of Probability

Consider a random experiment whose

sample space is S with sample points E1,E2, . . . ,.

For each event Eiof the sample space S

let P(Ei) be the probability of Ei(i) 0 P(Ei) 1 for all i

(ii) P(S) = 1

(iii)P(Ei) = 1,where the summation isover all sample points in S.


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Example

Definition The probability of any eventA isequal to the sum of the probabilities of the

sample points inA.

Example. Let S = {E1, . . ., E10}.

Ei E1 E2 E3 E4 E5 E6 E7 E8 E9 E10 P(Ei) 1/20 1/20 1/20 1/20 1/20 1/20 1/5 1/5 1/5 1/10

Question: Calculate P(A) whereA = {Ei, i6}.

P(A) = P(E6) + P(E7) + P(E8) + P(E9) + P(E10)= 1/20 + 1/5 + 1/5 + 1/5 + 1/10 = 0.75

Steps in calculating probabilities of events


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Steps in calculating probabilities of events

1. Define the experiment

2. List all simple events

3. Assign probabilities to simple events4. Determine the simple events that constitute the

given event

5. Add up the simple events probabilities to obtain

the probability of the given event Example Calculate the probability of observing

one Hin a toss of two fair coins.

Solution.

S = {HH,HT,TH, TT} A = {HT, TH} P(A) = 0.5


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Example.Example. Toss a fair coin 3 times.

(i) List all the sample points in the samplespace

Solution: S = {HHH, TTT} (Complete

this)

(ii) Find the probability of observing exactlytwo heads and at most one head.

Probability Laws


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Probability Laws Complementation law: P(A) = 1 - P()

Additive law: P(A U B) = P(A) + P(B) - P(A B) Moreover, ifA and B are mutually exclusive,

then P(AB) = 0 and

P(A U B) = P(A) + P(B) Multiplicative law (Product rule) P(A B) = P(A|B)P(B)

= P(B|A)P(A)

Moreover, ifA and B are independent P(AB) = P(A)P(B)

E l


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Example

Let S = {E1, E2, . . ., E6};A = {E1, E3, E5}; B =

{E1, E2, E3}; C= {E2, E4, E6};D ={E6}. Supposethat all elementary events are equally likely.

(i) What does it mean that all elementary eventsare equally likely?

(ii) Use the complementation rule to find P(Ac). (iii) Find P(A|B) and P(B|A)

(iv) Find P(D) and P(D|C) (v) AreA and B independent? Are Cand D

independent?

(vi) Find P(A B) and P(A UB).


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Law of total probability

Let A, Acbe complementary eventsand let B denote an arbitrary event.

Then

P(B) = P(B A) + P(B Ac

) ,or

P(B) = P(B/A)P(A) + P(B/Ac)P(Ac).


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Bayes Law Let A,Acbe complementary events and let Bdenote an arbitrary

event. Then

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

P(B/A)P(A)

P(A|B) =- ---------------------------------P(B/A)P(A) + P(B/Ac)P(Ac)

Remarks.

(i) The events of interest here are A, Ac,

(ii) P(A) and P(Ac) are calledpriorprobabilities,

(iii) P(A|B) and P(Ac|B) are calledposterior(revised) probabilities.

(iv) Bayes Law is important in several fields of applications.


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Bayesian Approach

English mathematician Thomas Bayes

(1702-61) set out his theory of probability

It is being revived now 250 years later

Step ahead from Subjective Prob Method

A: Digestive disorder, B: Drinking Coke

Bayes Rule:P(B|A) P(A) 0.65x0.3P(A|B) = ------------------- = ----------- = 0.53

P(B) 0.37


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P(Ai) P(B/Ai) P(AiB) P(Ai/B)

Prior Probabilities

Conditional

Probabilities

JointProbabilitie

s

Posterior

Probabilities

P(A1)=0.30 P(B/A1)=0.65 P(A1B)=0.195

P(A1/B)=.195/.37

=.527

P(A2)=0.70 P(B/A2)=0.25 P(A2B)=0.175

P(A2/B)=.175/.37

=.473

Example .


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p

A laboratory blood test is 95 percent effective indetecting a certain disease when it is, in fact, present.However, the test also yields a false positive results for1 percent of healthy persons tested. (That is, if a healthyperson is tested, then, with probability 0.01, the testresult will imply he or she has the disease.) If 0.5 percentof the population actually has the disease, what is theprobability a person has the disease given that the test

result is positive? Solution Let D be the event that the tested person hasthe disease and Ethe event that the test result ispositive. The desired probability P(D|E) is obtained by

P(D/E) =P(D E)/P(E)

=P(E/D)P(D)/P(E/D)P(D) + P(E/Dc)P(Dc) =(.95)(.005)/(.95)(.005) + (.01)(.995) =95/294 0 .323. Thus only 32 percent of those persons whose test

results are positive actually have the disease.


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General BayesTheorom A1,A2,..Ak are k mutually exclusive and

exhaustive events with known prior probabilitiesP(A1),P(A2),.P(Ak)

B is an event that follows or is caused by priorevents A1,A2, Ak with

Conditional probabilitiesP(B/A1),P(B/A2),P(B/Ak) which are known

Bayes formula allows us to calculate posterior

(revised) probabilitiesP(A1/B),P(A2/B),.P(Ak/B) P(Ai/B)=P(Ai)P(B/Ai)/{P(A1)P(B/A1)++P(Ak)P(B/Ak)}

Counting Sample Points


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Counting Sample Points

Is it always necessary to list all sample points in S?

Coin Tosses Coins sample-points Coins sample-points 1 2 2 4 3 8 4 16 5 32 6 64

10 1024 20 1,048,576 30 109 40 10 12

50 1015 60 1019

Note that 230109 = one billion, 240 1012 = one

thousand billion, 250

1015

=one trillion. RECALL: P(A) = nA/n , so for some applications we needto find n, nA where n and nA are the number of points inS andA respectively.

B i i i l f ti l


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Basic principle of counting: mn rule

Suppose that two experiments are to beperformed. Then if experiment 1 can result

in any one ofm possible outcomes and if,

for each outcome of experiment 1, thereare n possible outcomes of experiment 2,

then together there are mn possible

outcomes of the two experiments.


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.(i) Toss two coins: mn = 22 = 4

(ii) Throw two dice: mn = 6 6 = 36

(iii) A small community consists of 10 men, eachof whom has 3 sons. If one man and one of his

sons are to be chosen as father and son of the

year, how many different choices are possible?

Solution: Let the choice of the man as the

outcome of the first experiment and the

subsequent choice of one of his sons as the

outcome of the second experiment, we see,fromthe basic principle, that there are 10 3 = 30

possible choices.

G li d b i i i l f i


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Generalized basic principle of counting

Ifrexperiments that are to be performed are

such that the first one may result in any ofn1 possible outcomes, and if for each ofthese n1 possible outcomes there are n2possible outcomes of the second

experiment, and if for each of the possibleoutcomes of the first two experimentsthere are n3 possible outcomes of the thirdexperiment, and so on,. . ., then there area total ofn1 x n2 xnrpossible outcomesof the rexperiments.

E l


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Examples (i) There are 5 routes available betweenA and

B; 4 between B and C; and 7 between Cand D.What is the total number of available routesbetweenA and D?

Solution: The total number of available routes ismnt= 5.4.7 = 140.

(ii) A college planning committee consists of 3freshmen, 4 parttimers, 5 juniors and 2 seniors.

A subcommittee of 4, consisting of 1 individualfrom each class, is to be chosen. How manydifferent subcommittees are possible?

Solution: It follows from the generalized principleof counting that there are 3452 = 120 possiblesubcommittees.

E l


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Examples

(iii) How many different 7-place license platesare possible if the first 3 places are to be

occupied by letters and the final 4 by numbers? Solution: It follows from the generalized principle

of counting that there are 26 26 26 10 10 10 10 = 175, 760, 000 possible license plates.

(iv) In (iii), how many license plates would bepossible if repetition among letters or numberswere prohibited?

Solution: In this case there would be 26 25 24 10 9 8 7 = 78, 624, 000 possible licenseplates.

Permutations: (Ordered arrangements)


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Permutations: (Ordered arrangements) Permutations: (Ordered arrangements) The number of

ways of ordering n distinct objects taken rat a time

(order is important) is given by n! /(n - r)! = n(n - 1)(n - 2) (n - r+ 1) Examples

(i) In how many ways can you arrange the letters a, b

and c. List all arrangements. Answer: There are 3! = 6 arrangements or permutations. (ii) A box contains 10 balls. Balls are selected without

replacement one at a time. In how many different wayscan you select 3 balls?

Solution: Note that n = 10, r= 3. Number of differentways is

= 10! /7! = 10 9 8= 720,

C bi ti


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Combinations

Combinations Forr n, we define nCr =n! / (n - r)! r!

and say that n and r represents the number ofpossible combinations ofn objects taken rat a time(with no regard to order).

Examples

(i) A committee of 3 is to be formed from a group of 20people. How many different committees are possible?

Solution: There are 20C3 = 20! /3!17! = 20.19.18/3.2.1 =1140 possible committees.

(ii) From a group of 5 men and 7 women, how manydifferent committees consisting of 2 men and 3 womencan be formed?

Solution: 5C2 x 27C3 = 350 possible committees.

R d S li


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Random Sampling

Definition.A sample of size n is said to be a randomsample if the n elements are selected in such a way that

every possible combination ofn elements has an equalprobability of being selected .In this case the samplingprocess is called simple random sampling.

Remarks. (i) Ifn is large, we say the random sampleprovides an honest representation of the population.

(ii) For finite populations the number of possible samplesof size n is NCn

For instance the number of possible samples when N=28 and n = 4 is 28C4=20475

Tables of random numbers may be used to selectrandom samples.

F Di t ib ti


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Frequency Distribution:Number of Sales Orders Booked by 50 Sales Execs April 2006

Number of Orders Number of SEs

00 04 14

05 - 09 19

10 14 0715 19 04

20 24 02

25 29 01

30 34 02

35 39 00

40 44 01

TOTAL 50


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Probability Distribution

Number of Orders Number of SEs Probability

00 04 14 0.28

05 - 09 19 0.38

10 14 07 0.1415 19 04 0.08

20 24 02 0.04

25 29 01 0.02

30 34 02 0.04

35 39 00 0.00

40 44 01 0.02

TOTAL 50 1.00


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Standard Discrete Prob Distns

Binomial Distribution: When a Situationcan have Only Two Possible Outcomes

e.g. PASS or FAIL, ACCEPT or REJECT.

This distribution gives probability of anoutcome (say, ACCEPT) occurring exactly

m times out of n trials of the situation, i.e.

probability of 10 ACCEPTANCES out of15 items tested.


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Standard Discrete Prob Distns

Poisson Distribution: When a Situationcan have Only Two Possible Outcomes, &

When the Total Number of Observations is

Large (>20), Unknown or Innumerable. This distribution gives the probability of an

outcome (say, ACCEPT) occurring m

times, i.e. probability of say 150ACCEPTANCES.


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Standard Continuous Prob Distn

Normal Distribution: Useful & Important

Several Variables Follow Normal Distn ora Pattern Nearing It. (Weights, Heights)

Skewed Distns Assume This Shape AfterGetting Rid of Outliers

For Large No. of Observations, DiscreteDistributions Tend to Follow Normal Distn

It is Amenable to Mathematical Processes


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Features of Normal Distribution

Symmetrical and Bell Shaped

Mean at the Centre of the Distribution

Mean = mode = Median

Probabilities Cluster Around the Middleand Taper Off Gradually on Both Sides

Very Few Values Beyond Three Times theStandard Deviation from the Mean


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03 probability distributions.ppt

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