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Transcript of Basic concepts of probability. Numerical measure of likelihood that the event (Simple Event / Joint...
Basic Basic concepts concepts of of probabilityprobability
Numerical Numerical measure of likelihoodmeasure of likelihood that that
the event (Simple Event / Joint Event / the event (Simple Event / Joint Event /
Compound Event) will occur.Compound Event) will occur.
What is Probability?What is Probability?
Three concepts of probabilityThree concepts of probability
1.1. Classical probabilityClassical probability
2.2. Relative frequency probabilityRelative frequency probability
3.3. Subjective probability Subjective probability
(There are 2 ways to get one 6 and the other 4)
e.g. P( ) = 2/36
Computing ProbabilitiesComputing ProbabilitiesThe probability of an event E:The probability of an event E:
Each of the outcomes in the sample space is Each of the outcomes in the sample space is equally likely to occur equally likely to occur
number of event outcomes( )
total number of possible outcomes in the sample space
P E
X
T
Properties of ProbabilityProperties of Probability
1.1. P(A) lies between 0 & 1. i.e.P(A) lies between 0 & 1. i.e.1
.5 0
Certain
Impossible
If A any event in the sample space S then If A any event in the sample space S then
1)(0 AP
2.2. Let A be any event and Let A be any event and iodenote the basic outcomes. Then denote the basic outcomes. Then
)()( iOPAP
4.4. P(S)=1P(S)=1
3.3. Probability of sum of events is 1Probability of sum of events is 1
Compound ProbabilityCompound ProbabilityAddition RuleAddition Rule
1.1. P(A or B)P(A or B)= P(A = P(A B) B)
= = P(A) + P(B) P(A) + P(B) P(A P(A B) B)
2.2. For For Mutually ExclusiveMutually Exclusive Events: Events:
P(A or B) = P(A P(A or B) = P(A B) = P(A) + P(B) B) = P(A) + P(B)
3.3. Probability of ComplementProbability of Complement
P(A) + P(Ā) = 1. So, P(Ā) = 1 P(A) + P(Ā) = 1. So, P(Ā) = 1 P(A) P(A)
Conditional ProbabilityConditional Probability
Event Probability Event Probability GivenGiven that Another Event that Another Event OccurredOccurred
P(A | B) = P(A | B) = P(A and B)P(A and B) , P(B)>0 , P(B)>0 P(B) P(B)
)(
)()(
BP
BAPBAP
SimilarlySimilarly
)(
)()(
AP
BAPABP
, P (A)>0
Statistical IndependenceStatistical Independence
1.1. Event Occurrence Event Occurrence Does Does NotNot Affect Probability Affect Probability of Another Event.of Another Event.
e.g. Toss 1 Coin Twice, Throw 3 Dice e.g. Toss 1 Coin Twice, Throw 3 Dice
2.2. Tests For Independence Tests For Independence
P(A | B) = P(A), or P(B | A) = P(B), P(A | B) = P(A), or P(B | A) = P(B),
or P(A and B) = P(A)P(B)or P(A and B) = P(A)P(B)
Multiplication RuleMultiplication Rule1.1. Joint Probabilities for Joint Probabilities for IntersectionIntersection of Events (Joint of Events (Joint Events)Events)
2.2. P(A and B) = P(A P(A and B) = P(A B) B) P(A P(A B) = P(A)P(B|A) B) = P(A)P(B|A) = P(B)P(A|B) = P(B)P(A|B)
3.3. For For IndependentIndependent Events: Events:P(A and B) = P(AP(A and B) = P(AB) = P(A)P(B)B) = P(A)P(B)
A sample of executives were surveyed about A sample of executives were surveyed about
their loyalty to the company. One of the their loyalty to the company. One of the
questions was, “ If you were given an offer by questions was, “ If you were given an offer by
another company equal to or slightly better than another company equal to or slightly better than
your present position, would you remain with your present position, would you remain with
the company or take the other position?” the the company or take the other position?” the
responses of the 200 executives in the survey responses of the 200 executives in the survey
were cross-classified with their length of service were cross-classified with their length of service
with the company and their loyalty.with the company and their loyalty.
EXERCISE
LoyaltyLoyalty
Length of serviceLength of service
Less Less than 1 than 1 yearyear
BB11
1 – 5 1 – 5 yearyearss
BB22
6 – 6 – 10 10 yearsyears
BB33
More More than 10 than 10 yearsyears
BB44
Would remain, AWould remain, A1 1
Would not remain, Would not remain, AA22
1010
25253030
151555
10107575
3030
a.a.What is the probability of randomly selecting an What is the probability of randomly selecting an executive who is loyal to the company (would executive who is loyal to the company (would remain) and who has more than 10 years of remain) and who has more than 10 years of service?service?
b.b.What is the probability of randomly selecting an What is the probability of randomly selecting an executive who would remain with the company or executive who would remain with the company or has less than 1 year of experience?has less than 1 year of experience?
c.c.What is the probability of randomly selecting What is the probability of randomly selecting
an executive with more than 10 years of an executive with more than 10 years of
service?service?
d.d.What is the probability of randomly selecting What is the probability of randomly selecting
an executive who would not remain with the an executive who would not remain with the
company, given that he or she has more than company, given that he or she has more than
10 years of service?10 years of service?
e.e.What is the probability of randomly selecting What is the probability of randomly selecting
an executive with more than 10 years of an executive with more than 10 years of
service or one who would not remain with the service or one who would not remain with the
company?company?
Joint and marginal probabilitiesJoint and marginal probabilities
In the context of bivariate probabilities, In the context of bivariate probabilities,
the intersection probabilities the intersection probabilities
are called joint probabilities and the are called joint probabilities and the
probabilities of individual events, P(Aprobabilities of individual events, P(Aii) )
or P(Bor P(Bii) are called marginal ) are called marginal
probabilities.probabilities.
)( ii BAP
P(A1 and B2) P(A1)
TotalEvent(A)
Joint Probability Using Contingency TableJoint Probability Using Contingency Table
P(A2 and B1)
P(A1 and B1)
Event (B)
Total 1
Joint ProbabilityJoint ProbabilityJoint ProbabilityJoint Probability Marginal (Simple) ProbabilityMarginal (Simple) Probability
A1
A2
B1 B2
P(B1) P(B2)
P(A2 and B2) P(A2)
A survey carried out for a super market shows the A survey carried out for a super market shows the
number of customers according to 2 characteristicsnumber of customers according to 2 characteristics
Frequency Frequency of visitof visit
Purchase of generic productsPurchase of generic products
Often Often Sometimes Sometimes Never Never
Frequent Frequent 1212 4848 1919
Infrequent Infrequent 77 66 88
• What is the probability that a customer is both a What is the probability that a customer is both a
frequent shopper and often purchases generic frequent shopper and often purchases generic
products?products?• What is the probability that a customer who never What is the probability that a customer who never
buys generic products visits the store frequently?buys generic products visits the store frequently?• Are the events “ never buys generic products” and Are the events “ never buys generic products” and
“visits the store frequently” independent?“visits the store frequently” independent?
• What is the probability that a customer who What is the probability that a customer who
infrequently visits the store often buys generic infrequently visits the store often buys generic
products?products?
• What is the probability that a customer frequently What is the probability that a customer frequently
visits the store?visits the store?
• What is the probability that the customer often What is the probability that the customer often
buys generic products?buys generic products?
• What is the probability that a customer either What is the probability that a customer either
frequently visits the store or never buys generic frequently visits the store or never buys generic
products or do both?products or do both?
Consumers are surveyed on the relative Consumers are surveyed on the relative
number of visits to a circuit city store and if number of visits to a circuit city store and if
the store was conveniently located. the store was conveniently located.
Visits Visits
Convenient Convenient
Yes Yes NoNo
Often Often
OccasionalOccasional
NeverNever
6060
2525
55
2020
3535
5050
EXERCISE
a.a.What is this table called?What is this table called?
b.b.Are visits and convenience Are visits and convenience
independent? Interpret your independent? Interpret your
conclusion.conclusion.
c.c.Draw a tree diagram and Draw a tree diagram and
determine the joint determine the joint
probabilities.probabilities.
Tree diagramA tree diagram is a diagram that branches out and ends in leaves that correspond to the final variety.
Often
Occasional
Never P(Y
|O)
P(N|O)
P(Y|O)
P(N|O)
P(Y|n)
P(N|n)
.410
.308
.282
.308
.103
.128
.179
.026
.256
.751
.416
.092
.042
.581
.908
Bayes’ TheoremBayes’ Theorem
)|()()|()(
)|()()|(
2211 ABPAPABPAP
ABPAPBAP ii
i
Prior probabilityPrior probability
The initial The initial probability based on probability based on the present level of the present level of information.information.
Posterior ProbabilityPosterior Probability
A revised probability A revised probability based on additional based on additional information.information.
Example Bayes’ theoremExample Bayes’ theorem
Suppose 5% of a population have a disease. There is Suppose 5% of a population have a disease. There is
a diagnostic technique to detect the disease , but it a diagnostic technique to detect the disease , but it
is not very accurate. Historical evidence shows that is not very accurate. Historical evidence shows that
if a person has actually if a person has actually has the diseasehas the disease, the , the
probability that the test will indicate the probability that the test will indicate the presence of presence of
the diseasethe disease is 0.9. Further the probability for a is 0.9. Further the probability for a
person who actually person who actually does not have the diseasedoes not have the disease but but
the test will indicate the the test will indicate the disease is presentdisease is present is 0.15. If is 0.15. If
a person is randomly selected and the test is a person is randomly selected and the test is
performed. The results indicate the performed. The results indicate the disease is disease is
presentpresent. What is the probability the person actually . What is the probability the person actually
has the diseasehas the disease ? ?
Solution Solution
Let,Let,
AA11 refer to the event ‘ refer to the event ‘ has the diseasehas the disease’. ’.
A2A2 refer to the event ‘ refer to the event ‘does not have the diseasedoes not have the disease’. ’.
BB denote the event “ denote the event “ test shows the disease is presenttest shows the disease is present”.”.
So P ( ASo P ( A11 ) = 0.05, ) = 0.05,
P (A2 ) = 0.95, P (A2 ) = 0.95,
P ( B I AP ( B I A11)= 0.90)= 0.90
and P( B I Aand P( B I A22) = 0.15) = 0.15
So P ( ASo P ( A11 I B ) = 0.24 I B ) = 0.24
Example Bayes’ theoremExample Bayes’ theorem
EventPrior prob.
P (Ai)
Conditional prob.
P ( B I Ai)
Joint prob.
P (Ai and B)
Posterior prob.
P (Ai I B)
Disease A1 .05 .90 .0450 .24
No Disease A2 .95 .15 .1425 .76
The credit department of Lion’s Department The credit department of Lion’s Department Store in Anaheim, California, reported that Store in Anaheim, California, reported that 30 percent of their sales are cash, 30 percent 30 percent of their sales are cash, 30 percent are paid for by check at the time of the are paid for by check at the time of the purchase, and 40 percent are charged. 20% purchase, and 40 percent are charged. 20% of the cash purchasers, 90% of the checks, of the cash purchasers, 90% of the checks, and 60% of the charges are for more than and 60% of the charges are for more than $50. Ms. Tina Stevens just purchased a new $50. Ms. Tina Stevens just purchased a new dress that cost $120. what is the probability dress that cost $120. what is the probability that she paid cash? that she paid cash?
Exercise
Assignment #1
Lind, Marchal and Wathen. Statistical Techniques in Business and Economics.
Page # 174
Problem # 67, 70, 71, 74, and 79
Random variableRandom variable
A variable that can take numerical A variable that can take numerical
values determined by the outcomes of values determined by the outcomes of
a random experiment.a random experiment.
Discrete r. v
A random variable that can A random variable that can
assume only certain clearly assume only certain clearly
separated values i.e. countable separated values i.e. countable
number of values.number of values.
e.g. number of runs of a cricketer in a match.e.g. number of runs of a cricketer in a match.
e.g. number of heads after tossing a coin 4 times.e.g. number of heads after tossing a coin 4 times.
Continuous r. vContinuous r. vA random variable that can A random variable that can
assume one of an infinitely assume one of an infinitely
large number of values, large number of values,
within certain limitations.within certain limitations.
e.g. height, weight, temperature etc.e.g. height, weight, temperature etc.
What is probability distribution?What is probability distribution?
A listing of all the outcomes of A listing of all the outcomes of
an experiment and the an experiment and the
probability associated with probability associated with
each outcome. each outcome.
What is probability distribution What is probability distribution function?function?
Let X is a discrete random variable Let X is a discrete random variable
and P(x) is called the probability and P(x) is called the probability
distribution function if and only ifdistribution function if and only if
1. 1.
2.2.
0)( xP
1)( xP
Mathematical Expectation Mathematical Expectation
Sum of the product of the values of Sum of the product of the values of
a discrete r.v. with the a discrete r.v. with the
corresponding probabilities.corresponding probabilities.
The supervisor of traffic signals must decide whether to The supervisor of traffic signals must decide whether to
install a traffic light at the reportedly dangerous install a traffic light at the reportedly dangerous
intersection of 2 roads. The policy is to install a traffic intersection of 2 roads. The policy is to install a traffic
light at an intersection at which the monthly expected light at an intersection at which the monthly expected
number of accidents is higher than 7. Should he number of accidents is higher than 7. Should he
recommend that traffic light be installed at this recommend that traffic light be installed at this
intersection?intersection?
Year Year
Number of accidentsNumber of accidents
J J F F M M AA
M M J J J J A A S S O O N N D D
20072007 1010 88 1010 66 99 1212 22 1010 1010 00 77 1010
20082008 1212 99 77 88 44 33 77 1414 88 88 88 44
Exercise
TheThe supervisorsupervisor
has has collectedcollecteddata on data on
Accidents Accidents at the intersection:at the intersection:
What is the average number of accidents? What is the average number of accidents? Standard deviation?Standard deviation?
Two types of losses are incurred in Two types of losses are incurred in business : business :
i)i) Obsolescence losses, Obsolescence losses, caused by caused by stocking too much stocking too much
ii)ii) Opportunity losses , Opportunity losses , caused by caused by
being out of stockbeing out of stock
Neither of the losses occurs when Neither of the losses occurs when the demand and supply are same.the demand and supply are same.
Combining probabilities and Monetary values Combining probabilities and Monetary values
Example Example
Levin and Rubin. Levin and Rubin. Statistics for Statistics for ManagementManagement. . Page # 233-235Page # 233-235
A fruit and vegetable wholesaler sells strawberries. A fruit and vegetable wholesaler sells strawberries.
This product has a very limited useful life. If This product has a very limited useful life. If not soldnot sold on on
the day of delivery it is the day of delivery it is worthlessworthless. One case of . One case of
strawberries strawberries cost $20cost $20, and the wholesaler , and the wholesaler receives $50receives $50
for it. The wholesaler can not specify the number of for it. The wholesaler can not specify the number of
cases customers will call for on any one day, but his cases customers will call for on any one day, but his
analysis of past records produced the following analysis of past records produced the following
information information
Sales Sales during during
100 100 daysdays
Daily Daily salessales
Number of Number of days solddays sold
1010
1111
1212
1313
1515
2020
4040
2525
i.i. Calculate probability of each Calculate probability of each number being sold.number being sold.
ii.ii. Find the conditional loss table.Find the conditional loss table.
iii.iii.Calculate expected lossesCalculate expected losses
iv.iv.Comment on the wholesaler actionComment on the wholesaler action
Mario, owner of Mario’s Pizza emporium, has a Mario, owner of Mario’s Pizza emporium, has a difficult decision on his hands. He has found that he difficult decision on his hands. He has found that he always sells between one and four of his famous always sells between one and four of his famous ‘everything but’ pizzas per night. These pizzas take ‘everything but’ pizzas per night. These pizzas take so long to prepare, however, that Mario prepares all so long to prepare, however, that Mario prepares all of them in advance and stores them in the of them in advance and stores them in the refrigerator. Because the ingredients go bad within refrigerator. Because the ingredients go bad within one day, Mario always throws out any unsold pizzas one day, Mario always throws out any unsold pizzas at the end of each evening. The cost of preparing at the end of each evening. The cost of preparing each pizza is $7, and Mario sells each one for $12. each pizza is $7, and Mario sells each one for $12. He has the following informationHe has the following information
Number of pizzas demanded Number of pizzas demanded 11 22 3344ProbabilityProbability .4.4 .3.3 .2.2 .1.1
1.1. Find the conditional loss table.Find the conditional loss table.2.2. How many “everything but” pizzas should Mario How many “everything but” pizzas should Mario
stock each night in order to minimize expected stock each night in order to minimize expected loss?loss?
Probability Probability DistributionDistribution
Binomial Binomial DistributionDistribution
We call a distribution a We call a distribution a binomial distribution binomial distribution if all of if all of
the following are true the following are true
1.1. There are a There are a fixed number of trialsfixed number of trials, n, which are all , n, which are all
independent.independent.
2.2. There must be There must be exactly two mutually exclusive exactly two mutually exclusive
outcomes in a trialoutcomes in a trial, such as True or False, yes or , such as True or False, yes or
no, success or failure.no, success or failure.
3.3. The The probability of successprobability of success is the is the samesame for each for each
trial. trial.
What is binomial distribution?What is binomial distribution?
The binomial distribution is used to obtain the The binomial distribution is used to obtain the
probability of observing probability of observing xx successes in successes in nn trials, with trials, with
the probability of success on a single trial denoted the probability of success on a single trial denoted
by by pp. .
The formula for the binomial probability mass The formula for the binomial probability mass
function is function is
where where
Probability mass functionProbability mass function
PP((SS) = ) = pp. . PP((FF) = ) = qq = 1- = 1-pp. . nn indicates the fixed number of trials. indicates the fixed number of trials. xx indicates the number of successes (any indicates the number of successes (any
whole number [0,whole number [0,nn]). ]). pp indicates the probability of success for any indicates the probability of success for any
one trial. one trial. qq indicates the probability of failure (not indicates the probability of failure (not
success) for any one trial. success) for any one trial. PP((xx) indicate the probability of getting exactly ) indicate the probability of getting exactly
xx successes in successes in nn trials. trials.
Some features of binomial distributionSome features of binomial distribution
Some statistics
Average = Mean = np
Standard Deviation = )1( pnp
The Bernoulli distribution is a special The Bernoulli distribution is a special
case of the binomial distribution, case of the binomial distribution,
where n=1. where n=1.
Bernoulli distributionBernoulli distribution
Suppose that each time you take a free throw shot, you have a 25% chance of making it. If you take 15 shots, a. what is the probability of making exactly 5 of them.
SolutionSolution
We haveWe have n = 15, r = 5 p = .25 q = .75n = 15, r = 5 p = .25 q = .75
P(5) = 0.165 P(5) = 0.165
There is a 16.5 percent chance of making exactly 5 There is a 16.5 percent chance of making exactly 5 shots.shots.
Example
b. What is the probability of making fewer than 3
shots?
Solution
The possible outcomes that will make this happen
are 2 shots, 1 shot, and 0 shots. Since these are
mutually exclusive, we can add these probabilities.
P(2)+P(1)+P(0)
= .156 + .067 + .013 = 0.236
There is a 24 percent chance of sinking fewer than
3 shots.
Shape of binomial distribution for different Shape of binomial distribution for different values of values of nn and and r.r.
•When p is small (<0.5) , the binomial distribution is When p is small (<0.5) , the binomial distribution is skewed to the right.skewed to the right.
•As p increases and approaches to 0.5, the As p increases and approaches to 0.5, the skewness is less noticeable.skewness is less noticeable.
•When p = 0.5, the binomial distribution is When p = 0.5, the binomial distribution is symmetrical.symmetrical.
•When p is larger than 0.5, the distribution is When p is larger than 0.5, the distribution is skewed to the left.skewed to the left.
•When n increases binomial distribution When n increases binomial distribution approaches to symmetrical.approaches to symmetrical.
Problems with binomial distributionProblems with binomial distribution
The requirement that the probability of the outcome The requirement that the probability of the outcome must be fixed overtime is very difficult to meet in must be fixed overtime is very difficult to meet in practice. practice. In many industrial processes, however, it is In many industrial processes, however, it is extremely difficult to guarantee that this is indeed the extremely difficult to guarantee that this is indeed the case.case.
Again, the outcome of one trial may affect in any way Again, the outcome of one trial may affect in any way the outcome of any other trial. the outcome of any other trial. Consider an Consider an interviewing process in which high-potential interviewing process in which high-potential candidates are being screened for top positions. If the candidates are being screened for top positions. If the interviewer has talked to five unacceptable candidates interviewer has talked to five unacceptable candidates in a row, he may not view the sixth with complete in a row, he may not view the sixth with complete partiality. Therefore, independence is violated.partiality. Therefore, independence is violated.
Hypergeometric DistributionHypergeometric Distribution
1.1. An outcome on each trial of an experiment is classified An outcome on each trial of an experiment is classified
into into one of two mutually exclusiveone of two mutually exclusive categories-a categories-a
success or a failure success or a failure
2.2. The random variable is the number of successes in a The random variable is the number of successes in a
fixed number of trialsfixed number of trials
3.3. The trials are The trials are not independentnot independent. .
4.4. We assume that we sample from a finite population We assume that we sample from a finite population
without replacement. So, the without replacement. So, the probability of a success probability of a success
change for each trialchange for each trial. .
P.M.F of the Hypergeometric Distribution
If If XX is the number of is the number of SS’s in a completely random ’s in a completely random sample of size sample of size nn drawn from a population consisting drawn from a population consisting of of MM SS’s and (’s and (NN––MM)) FF’s, then the probability ’s, then the probability distribution is hypergeometric distribution and is distribution is hypergeometric distribution and is given bygiven by : :
n
N
xn
MN
x
M
NMnxhxXP ),,;()(
Mean and VarianceThe mean and variance of a hypergeometric rv The mean and variance of a hypergeometric rv XX
having pmf having pmf hh((x;n,M,Nx;n,M,N) are:) are:
N
M
N
Mn
N
nNXV
N
MnXE 1
1)(,)(
ExampleIn a lot of 20 units out of the production line, 2 units In a lot of 20 units out of the production line, 2 units
are known to be defective. If the inspector picks a are known to be defective. If the inspector picks a
sample of 3 units at random, what is the distribution sample of 3 units at random, what is the distribution
of the number of defectives in the sample?of the number of defectives in the sample?
What is the probability of having 0, 1, 2 and 3 What is the probability of having 0, 1, 2 and 3
defectives in the sample?defectives in the sample?What is the probability that none of the chosen What is the probability that none of the chosen
sample are defective?sample are defective?