study of chances

8/3/2019 Study of Chances

1/33

1

Prof.S.Ganguly


2/33

THE STUDYOFCHANCES

What is the chance of all of you passing in theexam?

What is the chance of raining today.

What is the chance of the price of the equity sharesof company X to increase significantly.

What are the chances of a gambler winning a bet?

2

Prof.S.Ganguly


3/33

PROBABILITY

Probability is the chance that a particular event willoccur.

It is a term used wherever uncertainty of any eventis unavoidable, and an appropriate decision isimportant.

It is used to get deeper understanding of thedecision problems and base their decisions onrational considerations.

3

Prof.S.Ganguly


4/33

NEEDTOSTUDY PROBABILITY

Indispensible tool for decision making whichinvolves lot of uncertainty.

The basis for inferential statistics, Quality controland other management decisions.

To make decisions wherein the decision makersface a certain kind of risk while selecting aparticular course of action.

4

Prof.S.Ganguly


5/33

BASIC TERMINOLOGY

Experiments

An activity that results in one and only ne outcome outof a set of disjoint outcomes, where an outcome can-notbe predicted with certainty.

E.G. :- Throwing a die, coin,

Event

The results of the experiment.

E.g.: -Tossing two coins will give the following results

(H,H) (T,T) (H,T) (T,H) 4 EVENTS.

5

Prof.S.Ganguly


6/33

Sample space

The set of all possible outcomes of an event.

E.G.:- If we toss a fair coin, the sample space is

Sample space = [head, tail].

If we throw a fair die Sample space = [1,2,3,4,5,6]

6

Prof.S.Ganguly


7/33

APPROACHESTOPROBABILITY

The classical approach

The relative frequency approach

The subjective approach

7

Prof.S.Ga

nguly


8/33

THECLASSICALAPPROACH

Equally likely

We assume that is symmetry and homogeneity inthe occurrence of events.

Collectively exhaustive.The number of favorable plus unfavorable eventscan never exceed the total number of events.

Mutually Exclusive

One and only one event takes place at a time.

8

Prof.S.Ga

nguly


9/33

EXAMPLE

If two coins are tossed simultaneously, there areonly four possibilities.

(H,H) (T,T) (T,H) (H,T)

Hence each of them has a probability of .

The probability of getting at least one head is .(As the event (T,T) is excluded in this case.

9

Prof.S.Ga

nguly


10/33

THE RELATIVE FREQUENCY APPROACH

Here probability is defined as the proportions oftime an event occurs in the long run when theconditions are stable.

Observer relative frequency of an event in a verylarge number of events.

It uses statistical data. E.G calculation ifinsurance.

E.G.: - What if you toss the same fair coin 300 times,

what are the chances of getting a head every time.

10

Prof.S.Ga

nguly


11/33

EXPERIMENTALRESULTSFOR 300 TOSSES

0.2

0.4

0.5

0.6

0.8

50 100 150 200 250 300

Relative frequency H/n,Where H = total number of heads tossed,

N = total number of tosses

Number of Tosses

Re

lativeFrequency

11

Prof.S.Ga

nguly


12/33

P(A) = m/n.

Where A is the event of getting .

m = Number of times an event occurs.

n = Number of times the experiment is performed.

0 < P < A.

12

Prof.S.Ga

nguly


13/33

THE SUBJECTIVEAPPROACH

One can use whatever evidence that is availableand assign a probability of his own perception ofthe situation.

Doesnt involve any specificmathematicalcalculation. It gives a degree of freedom to thedecision maker.

Can be applied to an event that has not yetoccurred.

13

Prof.S.Ga

nguly


14/33

PROBABILITYRULES.(AXIOMS)

0 < P(A) < 1.

P(S) = 1.

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

(A and B are mutually exclusive) P(A) = 1 P(A).

Where A is the non-occurrence of the event.

14

Prof.S.Ga

nguly


15/33

EXAMPLE

Following are the grades as obtained by thenumber of students followed.

A 20.

B 25.

C 20.

D 35.

Find Probability of selecting a student who has

Either grade A or B

Either Grade C or D.

15

Prof.S.Ga

nguly


16/33

SOLUTION

There are four events, the probabilities of theseevents are.

P(grade A) = 20/100 = 0.2

P(grade B) = 25/100 = 0.25

P(grade C) = 20/100 = 0.2

P(grade D) = 35/100 = 0.35

Since A, B, C, D are mutually exclusive events,P(grade A or grade B) = P(A) + P(B) = 0.2 + 0.25 =

0.45.

P(grade C or grade D) = P(C) + P(D) = 0.2 + 0.35 =0.55.

16

Prof.S.Ga

nguly


17/33

MUTUALLY NON-EXCLUSIVEMETHOD.

200 university students

Full time Part Time Total

Boys 60 20 80

Girls 80 40 120

40140 6040

20

Full time, Part time boys Full time, All Girls

Q. Find the probability that the student selected is either Full time or a Girl.i.e. P(A or C)

20

17

Prof.S.Ga

nguly

80


18/33

SUMMARY

Hence for Mutually Exclusive events

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

And For Mutually non-exclusive events.

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

18

Prof.S.Ga

nguly


19/33

EXAMPLE 2

A Product Manager launches two new products.Probability of success of Product A is that of

product B is , and that of both is 1/8. What is the

probability that product A or B succeeds.

Solution

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

= + -1/8 = 5/8 = 0.625.

19

Prof.S.Ga

nguly


20/33

PROBABILITYUNDERCONDITIONALSTATISTICALINDEPENDENCE

When a statistically independent event occurs, itdoesnt have effect on the happenings of another

event. Three types.

Marginal.

Joint.

Conditional.

20

Prof.S.Ga

nguly


21/33

MARGINAL PROBABILITY

The tossing of a coin. Here the probability of gettinga head and tail, both are 0.5. These are known asMarginal probabilitiesas a toss of a fair coin isstatically independent.

21

Prof.S.Ga

nguly


22/33

JOINT PROBABILITY

The product of two or more events occurringtogether is the product of their marginalprobabilities.

P(AB) = P(A) x P(B).

P(AB) - Probability of the events A and B occurring together orin succession.

P(A) - Probability of the event A.

P(B) - Probability of the event B.

E.G.: - Two coins tossed simultaneously. Probabilityof getting two successive heads.

P(H1H2) = P(H1) x P(H2). = 0.5 x 0.5 = 0.25.

22

Prof.S.Ga

nguly


23/33

CONDITIONAL PROBABILITIES

Conditional Probability means the probability ofevent A, given that the event B has occurred.P(A/B).

E.G.: - What is the probability that the second toss of a

fair coin will result in tail, given that a tail resulted on thefirst toss.

Solution

P(T2/T1) = 0.5, as once it is clear that the first toss resulted intail, now the second toss in independent. So its probability is

just the Marginal one. Hence

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

23

Prof.S.Ga

nguly


24/33

PROBABILITYUNDERCONDITIONSOFSTATISTICALDEPENDENCE

Conditional: -

E.G.: - A bag contains ten balls of different colours, suchthat.

2 balls are red and dotted.

1 ball is green and dotted. 4 balls are red and striped.

3 balls are green and striped.

What is the probability of drawing a dotted ball, giventhat it is green.

24

Prof.S.Ga

nguly


25/33

SOLUTION.

Color and pattern of ten balls

Event Probability of Event

1 0.1

2 0.1 Red and Dotted

3 0.1 Green and Dotted

4 0.1

5 0.1

6 0.1

7 0.1 Red and Striped8 0.1

9 0.1

10 0.1 Green and Striped25

Prof.S.Ga

nguly


26/33

Total probability of green balls = 0.4

Hence the probability of a ball been green anddotted i.e.

P(D/G) = P(DG)/P(G) = 0.1/0.4 = .

We can also find the probability of a green stripedball: -

P(S/G) = P(SG)/P(G) = 0.3/0.4 = .

26

Prof.S.Ga

nguly


27/33

JOINT PROBABILITIES

Joint Probability of events A and B is equal to theprobability of event A, given that event B hasalready occurred, multiplied by the probability ofevent B.

As P(A/B) = P(AB)/P(B), hence

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

27

Prof.S.Ga

nguly


28/33

E.g.: - Find the probability of a red and striped ball.

Solution

P(SR) = P(S/R) x P(R) = 2/3 x 6/10 = 0.4

Similarly P(DR) = P(D/R) x P(R) = 1/3 x 6/10 = 0.2

P(DG) = P(D/G) x P(G) = x 4/10 = 0.1

P(SG) = P(S/G) x P(G) = x 4/10 = 0.3

28

Prof.S.Ga

nguly


29/33

MARGINAL PROBABILITIES.

The Marginal probabilities of the green ball can bedetermined by adding the events in which the greenball is contained.

P(G) = P(GD) + P(GS) = 0.4

Similarly

P(R) = P(RS) + P(RD) = 0.6

P(D) = P(GD) + (RD) = 0.3

P(S) = P(RS) + (GS) = 0.7

29

Prof.S.Ga

nguly


30/33

BAYES THEOREM

When the estimate of Probabilities is based onlimited information, and after some time, someadditional information is available, it becomesnecessary to revise the prior estimate. These new

probabilities are known as the revised r posteriorprobabilities, and is calculated by the Bayes

Theorem.

Makes it unnecessary to collect huge data over a

long period in order to make good decisions.

30

Prof.S.Ga

nguly


31/33

EXAMPLE

Two machines I and II are used to produce shoes.E1 is the event that a shoe is produced by MachineI, and E2 is that by machine II. Machine I prodeces60% of the shoes and Machine produces 40%. It is

also estimated that I produces 10% defective and IIproduces 20% defectives.

What is the probability that a non-defective shoewas produced by machine I?

31

Prof.S.Ga

nguly


32/33

By formula for conditional probability

P(E1/A) = P(E1A)/P(A).-----------------------(i)

But from the formula on total probabilities

P(A) = P(AE1) + P(AE2)

= {P(A/E1) x P(E1)} + {P(A/E2) x P(E2)}

= P(A/Ei) x P(Ei).

Substituting in result (i), we have

P(E1/A) = P(E1A) ------Bayes theorem.

P(A/Ei) x P(Ei)

32

Prof.S.Ga

nguly


33/33

COMPUTATIONOF POSTERIORPROBABILITIES

Event Prior P(Ei) ConditionalP(A/Ei)

Joint P(EiA) PosteriorP(Ei/A)

(1) (2) (3) (4) (5) =(4)/P(A)

Machine I 0.6 0.9 0.54 0.54/0.86 =

0.63

Machine II 0.4 0.8 0.32 0.32/0.86 =0.37

Total 1.0 P(A) = 0.86 1.00

33

Prof.S.Ga

nguly

study of chances

Documents