probability for managers

7/30/2019 Probability for Managers

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The Chance factor in decision making


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Insurance premiums Banking industry

Betting on Cricket team

Weather forecasting Medical decisions

Getting a job

Lottery prize

Gambling

16-08-2013 2Probabilty: S C Agarkar, BRIMS


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Gamblers have been taking chances based ontheir judgements.

Seventeen century French mathematicianstried to give it a mathematical basis. Theywere Antoine Gombauld (1607-84), BlaisePascal (1623-62) and Pierre de Fermat(1601-65).

In 19th

century Pierrie Simon, Marquis deLaplace unified early ideas and compiled thefirst general theory of probability.



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The warning was issued by the SurgeonGeneral that Cigarette smoking is hazardousto your health. How might probability theoryhave played a part in this warning?

A well known manufacturer of childrenclothing decides to expand its product line byadding preteen clothing. In what way you

think the decision involves probability theory



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Event: It is one or more of the possible outcomesof doing something (head or tail during cointossing).

Experiment: It is an activity that produces an

event. Sample space: It is the set of all possible

outcomes of an experiment S= (head, tail).

Mutually exclusive events: If one and only one

event can take place (head or tail). Collectively exhaustive list: All possible events

can result in the experiment (HH, HT, TT, etc.).



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Give a collectively exhaustive list of all thepossible outcomes of tossing two dice.

Give the sample space of outcomes for thefollowing experiment in terms of their sexmake up: the birth of (a) twins, (b) triplets.

Give the probability of each of the followingtotals in the rolling of two dice: 1, 2, 5,

6,7,10,11Ans: 0, 1/36, 5/36, 6/36,4/36,3/36,2/36



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There are three basic ways of classifyingprobability. Classical Probability (based on theoretical

understanding of the situation)

Relative frequency probability (based on empiricalfindings of experiments)

Subjective probability (based on the experiencesand judgements)



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Classical probability is also called a prioryprobability because we can give answer inadvance (a priori) without conducting anexperiment. For example, coin tossing or dicerolling.

P(E)= Number of event outcomes/Total numberof possible outcomes Probability of getting head in coin tossing is

P(H)= =0.5. Probability of getting 5 in dice rolling is

P(5)= 1/6This is an ideal case and everyday cases would bemore complicated.



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This method uses the relative frequencies ofpast occurrences as probabilities. Look athow often an event has occurred in the pastan predict the occurrence of the event in the

future. Data on the number and nature of accidents

on the highway enables us to guess possibleaccidents and attempt to prevent them.

Data on the sale of previously launchedgadget enables one to predict the sale ofnewly launched gadget.



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The subjective approach of assigningprobabilities was introduced in 1926 by FrankRamsey in his book The Foundation ofMathematics and Other Logical Essays.

Subjective probability can be defined as theprobability assigned to an event by an individualbased on his/her beliefs and experiences.

Examples are: Choosing a right person for a job,

entering into a new business, launching a newproduct, giving a judgement on nuclear plant,etc.



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Determine the probabilities of the followingevents in drawing a card from a standard deck of52: (a) A queen, (b) A Club (c) An ace in red suit,(d) A red card, (e) A face card, (f) what type ofprobability estimates are these?

Classify the following probability estimates as totheir types (classical, relative frequency orsubjective)

(a) You will make B in this course is 0.75.(b) A family has two children

(c) My candidate will win(d) Student from this course will go abroad.(e) I will get selected for a football team.



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Marginal ProbabilityJoint Probability

Conditional Probability

16-08-2013Probabilty: S C Agarkar, BRIMS 12


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Marginal or unconditional probability is socalled when there is a possibility of only oneevent occurring.

It is given by the formula P(A)= 1/Totalnumber of cases. It is read as the probabilityof event A happening.

Examples are a person securing Gold Medal

in Olympics, a person winning the statelottery, a person becoming a governor of aReserve Bank, etc.



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The probability of two or more independentevents occurring together or in succession

P(AB) = P(A) * P(B)

The probability of heads on two successivetosses is the probability of heads on first tosstimes the probability of heads on secondtoss.

It will be .5*.5= .25



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Addition Rule for events that are mutuallyexcusive. It is used when we are interested inthe probability that one thing or the other willoccur assuming that each of them is mutually

exclusive.

Addition rule for events that are not mutuallyexclusive. It is used when we are interested inthe probability that one thing or the other willoccur assuming that each of them is notmutually exclusive.



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If A and B events are mutually exclusive then wecalculate the probability that one thing oranother will occur by addition formula:

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

It is read as probability of either A or B occurring isgiven by the addition of A occurring and Boccurring.

Suppose five candidates have been short listedfor a summer job. Chance of A getting a job is

1/5 and the chance of B getting the job is also1/5. Probability of A or B getting the job is1/5+1/5=2/5 that is 0.4



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As safety officer of an airline Debbie Best hasbeen asked to give a talk to the press concerningengine safety. As part of her talk she has decidedto include the probability of two-engine jethaving failure on flight. After consulting herrecord she find the following information aboutlast years operating record

29 reported failures of the right engine

33 reported failures of left engine

there was no crash attributedThere were 345,000 flights during the year

What probability should she report?



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When two events are not mutually exclusive thenthe addition rule needs to be modified. In thiscase we have to avoid double counting. Hencethe formula is modified as

P(A or B) = P(A) + P(B) P(AB) If we wish to determine the probability of

drawing either an ace or a heart from a deck ofcards then

P (Ace or Heart)= P(Ace) + P(Heart) P (A&H)

= 4/52 + 13/52 1/52= 16/25 or 4/13



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When two events happen, the outcome of thefirst event may or may not have an effect onthe outcome of the second event.

Events are statistically independent whenoccurrence of one event has no effect on theprobability of occurrence of any other event

Events are statistically dependent when

occurrence of one vent has direct effect onthe probability of occurrence of anotherevent.



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Let us examine the events that arestatistically independent. In this case theoccurrence of one event has no effect on theprobability of the occurrence on any other

event. There are three types of probabilities under

statistical independence: Marginal Probability

Joint Probability

Conditional Probability.



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A marginal or unconditional probability is thesimple probability of occurrence of an event.

In a fair coin toss P(H)= 0.5 and P(T)= 0.5.This is true for every toss.

Even if the coin is unfair or biased theoutcome of any particular toss is completelyunrelated to the outcome of tosses made

earlier.



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The probability of two or more independentevents occurring together or in succession isthe product of their marginal probabilities.

Multiplication rule for joint probabilities

under statistical independence is expressedas

P(AB) = P(A) * P(B)P(AB): Probability of events A and B occurring

together or in successionP(A): Marginal probability of event A occurringP(B): Marginal probability of event A occurring



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Probability of getting two heads on twosuccessive tosses is 0.5 * 0.5 = 0.25.

Probability of getting three heads on threesuccessive tosses is 0.5*0.5*0.5= 0.125

Probability of tails, tails and heads in thatorder on three successive tosses is0.5*0.5*0.5= 0.125.

Probability of at least two heads on threetosses is 0.125+0.125= 0.5



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Conditional probability is the probability thata second event (B) will occur if the first event(A) has already happened.

For statistical independent event the

conditional probability of event B given that Ahas occurred is simply the probability ofevent B: P(A\B) = P(B).

The probability that second toss would result

into head even if the result of the first tosswas head is 0.5 not affected by previousresult.



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What is the probability that a couples secondchild will be(a) A boy, given that their first child was a girl

(b) A girl given that their first child was a girl

What is the probability that in selecting twocards one at a time from the deck withreplacement the second card is(a) A spade, given that the first card was a heart

(b) Black given that the first card was red

(c) A queen, given that the first card was a queen



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Statistical dependence exists when theprobability of some event is dependent uponor affected by the occurrence of some otherevent.

Conditional and joint probabilities understatistical dependence are more involved thanmarginal probabilities.



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The general formula for conditional probabilityunder statistical dependence is P(B\A) =P(BA)P(A) if the occurrence of B is dependent onthe occurrence of A.

It is P(A\B) = P(AB)P(B) if the occurrence of A isdependent on the occurrence of B.

If there are ten balls (3 coloured and dotted, 1coloured and stripped, 2 gray and dotted and 4

gray and striped) in a box. To calculate theprobability of dotted given coloured we will have

P(D\C) = P(DC)P(C) = 0.30.4



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The formula for conditional probability underconditions of statistical dependence is

P(B\A)=P(BA)P(A)

By cross multiplication we get the following

formula for joint probabilities

P(BA) = P(B\A) P(A)

Continuing with example of coloured gray,

dotted and striped balls we will haveP(CD)= P(C\D) P(D)= 0.60.5 = 0.3



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Marginal probabilities under statisticaldependence are computed by summing upthe probabilities of all the joint events inwhich the simple event occurs.

P(C)= P(CD) + P(CS)

P(G)= P(GD + P(GS)

P(C)= 0.3 + 0.1= 0.4

P(G)= 0.2 + 0.4= 0.6P(D)= 0.3 + 0.2= 0.5

P(S)= 0.1 + 0.4 = 0.5



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Type ofProbability Symbol Under statisticalindependence Under statisticaldependenceMarginal P(A) P(A) Sum of probabilities of

the joint events in

which A occurs

Joint P(AB)P(BA)

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

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

Conditional P(B\A)P(A\B)

P(B)P(A)

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



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Given that P(A)=1/6, P(B)=1/3, P(C)= 4/9,P(A&C)= 1/12 and P(B\C)=1/4, find theprobabilities P(A\C), P(C\A), P(B&C), P(C\B).

Myers clothiers knows that 1 out of 10

families in its trading area qualifies for theircharge accounts and that 1 out of 15 familiesin this area has applied for an account, Frompast records, 90 percent of creditapplications are accepted. What is theprobability that an area family will apply for aMyers charge card and be accepted.



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A dice is rolled twice and the sum of the numbersappearing on them is observed to be 7. What isthe conditional probability that the number 2 hasappeared at least once?

The required probability is P(A\B)=P(AB)/P(B)

A= getting number 2 at least onceP(A)= 11/36: (2,1); (2,2); (2,3), etc.

B= getting 7 as sum of the numbers on two diceP(B)= 6/36: (2,5); (5,2); (1,6); (6,1); (3,4); (4,3) P(AB)= 2/36: (2,5); (5,2) P(A/B)= P(AB) P(B) = 2/36 6/36

=1/3



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We estimate certain probability based on thedata and experiences.

It needs to be modified as more data is madeavailable or collected.

The new probabilities thus estimated arecalled revised or posterior probabilities.

Posterior probabilities play a major role in

decision making in business world.



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The concept of posterior probabilities isattributed to Reverend Thomas Bayes (1702-1761).

He wanted to prove the existence of God with

the help of mathematics. He developed a lotof mathematics related to probability but didnot publish the work.

His work was brought to light after his death.Bayes theorem is one of them.



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The basic formula for conditional probabilityunder dependence is P(B/A)= P(BA)P(A). It iscalled Bayes theorem.

Bayes theorem offers a powerful statistical

method of evaluating new information andrevising prior estimates.

If correctly used it makes it unnecessary togather masses of data over long periods oftime in order to make decisions based onprobabilities.



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Assume that we have equal number of twodeformed dice in a bowl. On half of them ace(one dot) comes up 40% of the time. Therefore,P(ace) for type 1 is .4. On the other half acecomes 70% of the time. Hence P(ace) for type 2 is.7.

One dice is drawn and rolled once shows up ace(one dot). Which one it is? What is the probabilitythat it is type 1 dice?

Since the number of type 1 and type 2 dice areequal we might think that the probability ofgetting type 1 dice id 0.5.



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Elementaryevent Probability ofelementaryeventP(ace/ele event) P(ace, ele event)

Type 1 0.5 0.4 0.4*0.5=0.20

Type 2 0.5 0.7 0.7*0.5=0.35


The joint probability of ace and type 1 diceis .4*.5=0.2,the joint probability of ace and type 2 dice

is .7*.5=05.


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P(B\A)= P(BA)/P(A) P(type 1\ace)= P(type 1, ace)/P(ace) P (type 1\ace)= .20/.55= .364 Thus the probability that we have drawn type 1

dice is 0.364. P(type2\ace) = P(type 2, ace)/P(ace)

= .35/.55 = .636The probability that we have drawn type 2 dice is.636.

Our new posterior estimate is that there is a higherprobability that the dice in our hand is type 2 thantype 1.



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Assume that the same dice is rolled secondtime and again comes up one dot.

The joint probability of two one dots on twosuccessive rolls for type 1 is .4*.4= .16. It is

.7*.7= .49 for the second dice.Joint probabilities of two aces on two

successive rolls in type 1 is .16*.5= .080

Joint probabilities of two aces on twosuccessive rolls in type 2 is .49*.5= .245



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P(type1\2 aces)= P(type 1, 2 aces)/P(aces)=.80/.325= .246

P(type2\2 aces)= P(type 2, 2 aces)/ P(2 aces).245/.325= .754

Probability that it is type 1 is .246 and theprobabilities that it is type 2 is .754

We have thus changed the original probability

from .5 to .246 for type 1 and .754 for type 2based on the information.



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A manufacturing firm produces pipes in two plants I and II withdaily production of 1,500 and 2,000 pipes respectively. Thefraction of defective pipes produced by two plants I and II are0.006 and 0.008 respectively. If a pipe is selected at randomfrom days production is found to be defective, what is thechance that it has come from plant I, plant II?

P(E1) probability that a pipe is made in plant 1 is

1500/1500+2000 = 3/7 P(E2) probability that a pipe is made in plant II is

2000/1500+2000=4/7 P(A) is the probability that a defective pipe is drawn P(E1A)= 3/7*0.006 and P(E2A)=4/7*0.008 P(E1\A)= P(E1A) P(E1A)+P(E2A) = 3/7*0.0063/7*0.006+4/7*0.008

=9/25Similarly P(E2\A) = 16/25



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There was a fire accident in a building. Theinvestigation undertaken later by an expertshowed that

(a) Probability of short circuit is 0.8

(b) Probability of LPG cylinder explosion is 0.2

(c) Chance of fire accident is 30% for shortcircuit and it is 95% for LPG explosion.

What do you think is the most likely cause offire?



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Let A be the event that there was fire accident Let E1 be the event that there could have been

short circuit, Let E2 be the vent that there could have been

an LPG explosion. Given P(E1) = 0.8 and P(E2)=0.2 Also P(A/E1)= 0.3 and P(A/E2)= 0.95 P(E1) * (P(A/E1) = 0.8 * 0.3 = 0.24----- I

P(E2) * P (A/E2)= 0.2 * 0.95 = 0.19 ----- IISince I > II, it is more likely that the fire took

place because of short circuit.



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The probability that a person stopping at gasstation will ask to have his tyres checked is 0.12,the probability that he will ask to have his oilchecked is 0.29 and the probability that he willask to have them both checked is 0.07

(a) what is the probability that a person stoppingat gas station will have either his oil or tyrechecked?

(b) that a person who has checked his tyres will

also check his oil? (c) that a person who has checked his oil will aso

get the tyres checked?



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P(A)=0.12, P(B)= 0.29 P(AB)= 0.07 P(AorB)= P(A)+P(B)-P (AB)

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

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


Start


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Areevents

mutuallyexclusive

?

Addition rule:P(A Or B) = P(A) + P(B) P(AB)

Addition rule:P(A Or B) = P(A) + P(B)

Are eventsstatically

independent?

The marginal probability of event Aoccurring is P(A)

The joint probability of two eventsoccurring together or in succession is

P(BA) = P(B\A) * P(A)

The joint probability of two eventsoccurring together or in succession is

P(AB) =P(A)*P(B)

The conditional probability of oneevent occurring given that another has

already occurred is P(B\A) =P(B)

The conditional probability of oneevent occurring given that another hasalready occurred is P(B\A) = P(BA)/P(A)

Determine posterior probabilities withBayes Theorem

The marginal probability of event Aoccurring is the sum of the

probabilities of all joint events inwhich A occurs

This is known asBayes Theorem

No

No

Yes

Yes

probability for managers

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