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8/3/2019 Study of Chances
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Prof.S.Ganguly
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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?
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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.
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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.
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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.
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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]
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APPROACHESTOPROBABILITY
The classical approach
The relative frequency approach
The subjective approach
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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.
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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.
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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.
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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
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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.
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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.
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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.
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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.
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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.
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MUTUALLY NON-EXCLUSIVEMETHOD.
200 university students
Full time Part Time Total
Boys 60 20 80
Girls 80 40 120
40140 6040
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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
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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)
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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.
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PROBABILITYUNDERCONDITIONALSTATISTICALINDEPENDENCE
When a statistically independent event occurs, itdoesnt have effect on the happenings of another
event. Three types.
Marginal.
Joint.
Conditional.
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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.
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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.
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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).
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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.
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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
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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 = .
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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).
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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
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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
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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.
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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?
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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)
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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
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