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Basics of Probability Theory
Dr. Gita A. Kumta
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Probability as a Numerical Measure
of the Likelihood of Occurrence
0 1.5
Increasing Likelihood of Occurrence
Probability:
The event
is very
unlikelyto occur.
The occurrence
of the event is
just as likely asit is unlikely.
The event
is almost
certainto occur.
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Probability
Experiment of chance: a phenomena whose outcome isuncertain.
Probabilities Chances
Probability Model
Sample Space
Events
Probability of Events
Sample Space: Set of all possible outcomesEvent: A set of outcomes (a subset of the sample space). An
eventEoccurs if any of its outcomes occurs.
Probability: The likelihood that an event will produce a
certain outcome.
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An Experiment and Its Sample Space
An experimentis any process that generateswell-defined outcomes.
The sample space for an experiment is the
set of all experimental outcomes.
An experimental outcome is also called a
sample point.
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A Counting Rule for
Multiple-Step Experiments
If an experiment consists of a sequence of ksteps
in which there are n1possible results for the first step,
n2possible results for the second step, and so on,
then the total number of experimental outcomes is
given by (n1)(n2) . . . (nk).
A helpful graphical representation of a multiple-step
experiment is a tree diagram.
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Govind Investments can be viewed as a
two-step experiment. It involves two stocks,each with a set of experimental outcomes.
Mukund Oil: n1= 4
Collins Mining: n2= 2
Total Number of
Experimental Outcomes: n1n2= (4)(2) = 8
A Counting Rule for
Multiple-Step Experiments
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A second useful counting rule enables us to count
the number of experimental outcomes when nobjects are to be selected from a set ofNobjects.
Counting Rule for Combinations
CN
n
N
n N nn
N
!
!( )!
Number of Combinations ofNObjects Taken nat a Time
where: N! =N(N- 1)(N- 2) . . . (2)(1)
n! = n(n- 1)(n- 2) . . . (2)(1)
0! = 1
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Number of Permutations ofN
Objects Takenn
at a Time
where: N! =N(N- 1)(N- 2) . . . (2)(1)
n! = n(n- 1)(n- 2) . . . (2)(1)
0! = 1
P nN
n
N
N nn
N
!
!
( )!
Counting Rule for Permutations
A third useful counting rule enables us to count the
number of experimental outcomes when nobjects are tobe selected from a set of Nobjects, where the order of
selection is important.
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Assigning Probabilities
Classical Method
Relative Frequency Method
Subjective Method
Assigning probabilities based on the assumption
of equally likely outcomes
Assigning probabilities based on experimentation
or historical data
Assigning probabilities based on judgment
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Classical Method
If an experiment has npossible outcomes, this method
would assign a probability of 1/nto each outcome.
Experiment: Rolling a dieSample Space: S= {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a
1/6 chance of occurring
Example
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Relative Frequency Method
Number of
Polishers Rented
Number
of Days
0
12
3
4
4
618
10
2
Lucas Tool Rental would like to assign probabilities to the number
of car polishers it rents each day. Office records show the
following frequencies of daily rentals for the last 40 days.
Example: Lucas Tool Rental
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Each probability assignment is given by dividing the
frequency (number of days) by the total frequency(total number of days).
Relative Frequency Method
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Probability
Number of
Polishers Rented
Number
of Days012
34
46
18
10240
.10
.15
.45
.25.051.00
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Subjective Method
When economic conditions and a companys circumstanceschange rapidly it might be inappropriate to assign probabilities
based solely on historical data.
We can use any data available as well as our experience and
intuition, but ultimately a probability value should express our
degree of belief that the experimental outcome will occur.
The best probability estimates often are obtained by combining
the estimates from the classical or relative frequency approachwith the subjective estimate.
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Events and Their Probabilities
The process of making an observation or recording a
measurement under a given set of conditions is a trialor
experiment.
Outcomes of an experiment are called events.An eventis
a collection of sample points.
The probability of any event is equal to the sum of the
probabilities of the sample points in the event.
We denote events by capital letters A, B, C,
The probability of an event A, denoted by P(A), in general,
is the chance A will happen.
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Probability
Consider a deck of playing cards
Sample Space: Set of 52 cards
Event: R: The card is red. F:The card is a face card
.
A:The card is a heart. B:The card is a 3.
Probability: P(R) = 26/52 P(F) = 12/52
P(A) = 13/52 P(B) = 3/52
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Events and variables
Can be described as random or deterministic:
The outcome of a random event cannot be predicted:
The sum of two numbers on two rolled dice.
The time of emission of the i
th
particle fromradioactive material.
The measured length of a table to the nearest cm. Motion of macroscopic objects (projectiles, planets, space
craft) as predicted by classical mechanics.
The outcome of a deterministic event can be predicted:
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Random variables
Can be described as discrete or continuous:
A discrete variable has a countable number of values.Number of customers who enter a store before one purchases
a product.
The values of a continuous variable can not be listed:
Distance between two oxygen molecules in a room.
Random Variable Possible Values
Gender Male, Female
Class Fresh, Soph, Jr, Sr
Height (inches) # in interval {30,90}
College Arts, Education, Engineering, etc.
Shoe Size 3, 3.5 18
Consider data collected for undergraduate students:
Is the height a discrete or continuous variable?
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Tree Diagram
A tree diagram is a way of describing all the
possible outcomes from a series of events
A tree diagram is a way of calculating theprobability of all the possible outcomes from a
series of events
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Examplea fair coin is flipped twice
H
H
H
T
T
T
HH
HT
TH
TT
2nd1st
Possible
Outcomes
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Outcome Table
If you flip a coin twice, you can model also model
the results with an outcome table
Flip 1 Flip 2 SimpleEvent
H H HH
H T HTT H TH
T T TT
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Tree DiagramsFor flipping a coin
Probability of two or more events
1st Throw 2ndThrow
THHHHH TTTT 1/21/21/21/21/21/21/2
OUTCOMES
H,H
H,T
T,H
T,T
P(Outcome)
P(H,H) =1/4=1/2x1/2
P(H,T) =1/4=1/2x1/2
P(T,H) =1/4=1/2x1/2
P(T,T) =1/4=1/2x1/2
Total P(all outcomes) = 1
Total=4 (2x2)
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Rules for calculating probability
The addition rule
for pairwise mutually exclusive events
P(A1+ A2+ ...+An)= P(A1)+P(A2)+ ...+P(An)
for two non-mutually exclusive events A and B
P(A+B) = P(A) + P(B)P(AB).
Multiplicative rule
P(AB) = P(A) P(B|A) = P(B) P(A|B).
Formula of total probabilityP(B)= P(A1)P(B|A1)+P(A2)P(B|A2)+ ...+P(An)P(B|An).
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Some Basic Relationships of Probability
There are some basic probability relationships that can be used to
compute the probability of an event without knowledge of all thesample point probabilities.
Complement of an Event
Intersection of Two Events
Mutually Exclusive Events
Union of Two Events
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The complement ofAis denoted byAc.
The complement of eventA is defined to be the eventconsisting of all sample points that are not inA.
Complement of an Event
EventA AcSampleSpace S
VennDiagram
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The union of eventsAand Bis denoted byA B
The union of eventsAand Bis the event containingall sample points that are inA orB or both.
Union of Two Events
SampleSpace SEventA Event B
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The intersection of eventsAand Bis denoted byA
The intersection of eventsAand Bis the set of allsample points that are in bothA and B.
SampleSpace SEventA Event B
Intersection of Two Events
Intersection ofAandB
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The addition law provides a way to compute theprobability of eventA,or B,or bothAand B occurring.
Addition Law
The law is written as:
P(A B) = P(A) + P(B) P(AB
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Mutually Exclusive Events
Two events are said to be mutually exclusive if theevents have no sample points in common.
Two events are mutually exclusive if, when one eventoccurs, the other cannot occur.
SampleSpace S
EventA
EventB
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Mutually Exclusive Events
If eventsAand Bare mutually exclusive, P(AB= 0.
The addition law for mutually exclusive events is:
P(A
B
) =P
(A
) +P
(B
)
theres no need toinclude P(AB
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The probability of an event given that another eventhas occurred is called a conditional probability.
A conditional probability is computed as follows :
The conditional probability ofAgiven Bis denotedby P(A|B).
Conditional Probability
( )
( | ) ( )
P A BP A B
P B
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Multiplication Law
The multiplication law provides a way to compute theprobability of the intersection of two events.
The law is written as:
P(A B) = P(B)P(A|B)
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Independent Events
If the probability of eventA
is not changed by theexistence of event B, we would say that eventsAand Bare independent.
Two eventsAand Bare independent if:
P(A|B) = P(A) P(B|A) = P(B)or
Multiplication Law
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The multiplication law also can be used as a test to seeif two events are independent.
The law is written as:
P(A B) = P(A)P(B)
Multiplication Law
for Independent Events
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Bayes Theorem
NewInformation
Applicationof BayesTheorem
PosteriorProbabilities
PriorProbabilities
Often we begin probability analysis with initial or
prior probabilities.
Then, from a sample, special report, or a producttest we obtain some additional information.
Given this information, we calculate revised orposterior probabilities.
Bayes theoremprovides the means for revising theprior probabilities.
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A proposed shopping center will provide strong
competition for suburban businesses like
L. S. Clothiers. If the shopping center is built, the owner of L.S. Clothiers feels it would be best to relocate to the center.
Bayes Theorem
Example: L. S. Clothiers
The shopping center cannot be built unless a zoning changeis approved by the town council. The planning board mustfirst make a recommendation, for or against the zoning
change, to the council.
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Prior ProbabilitiesLet:
Bayes Theorem
A1= town council approves the zoning change
A2= town council disapproves the change
P(A1) = .7, P(A2) = .3
Using subjective judgment:
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New Information
The planning board has recommended
against the zoning change. LetBdenote the
event of a negative recommendation by the
planning board.
Given thatBhas occurred, should L. S.
Clothiers revise the probabilities that the
town council will approve or disapprove the
zoning change?
Bayes Theorem
B Th
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Conditional ProbabilitiesPast history with the planning board and the towncouncil indicates the following:
Bayes Theorem
P(B|A1) = .2 P(B|A2) = .9
P(BC|A1) = .8 P(BC|A2) = .1
Hence:
B Th
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P(Bc|A1) = .8
P(A1) = .7
P(A2) = .3
P(B|A2) = .9
P(Bc|A2) = .1
P(B|A1) = .2 P(A1B) = .14
P(A2B) = .27
P(A2Bc) = .03
P(A1Bc) = .56
Bayes Theorem
Tree Diagram
Town Council Planning Board ExperimentalOutcomes
B Th
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Bayes Theorem
1 1 2 2
( ) ( | )( | )
( ) ( | ) ( ) ( | ) ... ( ) ( | )
i i
i
n n
P A P B AP A B
P A P B A P A P B A P A P B A
To find the posterior probability that eventA
i will occur giventhat eventB has occurred, we apply Bayes theorem.
Bayes theorem is applicable when the events for
which we want to compute posterior probabilities
are mutually exclusive and their union is the entire
sample space.
B Th
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Posterior Probabilities
Given the planning boards recommendation not to
approve the zoning change, we revise the prior
probabilities as follows:
1 11
1 1 2 2
( ) ( | )( | )
( ) ( | ) ( ) ( | )
P A P B AP A B
P A P B A P A P B A
(. )(. )
(. )(. ) (. )(. )
7 2
7 2 3 9
Bayes Theorem
= .34
B Th
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Conclusion
The planning boards recommendation is good newsfor L. S. Clothiers. The posterior probability of the
town council approving the zoning change is .34compared to a prior probability of .70.
Bayes Theorem