Probability Probability Concepts and Concepts and Applications Applications
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Probability Probability Concepts and Concepts and ApplicationsApplications
IntroductionIntroduction
• Life is uncertain!
• We must deal with risk!
• A probability is a numerical statement about the likelihood that an event will occur
Basic Statements About Basic Statements About ProbabilityProbability
1. The probability, P, of any
event or state of nature
occurring is greater than or
equal to 0 and less than or
equal to 1.
That is: 0 P(event) 1
2. The sum of the simple
probabilities for all possible
outcomes of an activity must
equal 1.
ExampleExample
• Demand for white latex paint at
Diversey Paint and Supply has
always been 0, 1, 2, 3, or 4
gallons per day. (There are no
other possible outcomes; when
one outcome occurs, no other
can.) Over the past 200 days,
the frequencies of demand are
represented in the following
table:
Example - continuedExample - continued
Quantity
Demanded
(Gallons)
0
1
2
3
4
Number of Days
40
80
50
20
10
Total 200
Frequencies of DemandFrequencies of Demand
Example - continuedExample - continued
Quant. Freq.
Demand (days)
0 40
1 80
2 50
3 20
4 10
Total days = 200
Probability
(40/200) = 0.20
(80/200) = 0.40
(50/200) = 0.25
(20/200) = 0.10
(10/200) = 0.05Total Prob = 1.00
Probabilities of DemandProbabilities of Demand
Types of ProbabilityTypes of Probability
Objective probability:
Determined by experiment or observation:• Probability of heads on coin flip• Probably of spades on drawing card
from deck
occurrencesor outcomes ofnumber Total
occursevent timesofNumber )( eventP
Types of ProbabilityTypes of Probability
Subjective probability:
Based upon judgement
Determined by:
• judgement of expert
• opinion polls
• Delphi method
• etc.
Mutually Exclusive EventsMutually Exclusive Events
• Events are said to be mutually
exclusive if only one of the
events can occur on any one
trial
Collectively Exhaustive Collectively Exhaustive EventsEvents
• Events are said to be
collectively exhaustive if the list
of outcomes includes every
possible outcome: heads and
tails as possible outcomes of
coin flip
ExampleExampleOutcome
of Roll
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
Total = 1
Rolling a die has six possible outcomes
ExampleExample
Outcome
of Roll = 5
Die 1 Die 2
1 4
2 3
3 2
4 1
Probability
1/36
1/36
1/36
1/36
Rolling two
dice results in
a total of five
spots
showing.
There are a
total of 36
possible
outcomes.
Probability : Probability : Mutually Exclusive Mutually Exclusive
P(event A or event B) =
P(event A) + P(event B)
or:
P(A or B) = P(A) + P(B)
i.e.,
P(spade or club) = P(spade) + P(club)
= 13/52 + 13/52
= 26/52 = 1/2 = 50%
Probability:Probability: Not Mutually Exclusive Not Mutually Exclusive
P(event A or event B) =
P(event A) + P(event B) -
P(event A and event B both occurring)
or
P(A or B) = P(A)+P(B) - P(A and B)
P(A and B)P(A and B)(Venn Diagram)(Venn Diagram)
P(A) P(B)
P(A and B
)
P(A or B)P(A or B)
+ -
=
P(A) P(B) P(A and B)
P(A or B)
Statistical DependenceStatistical Dependence
• Events are either
• statistically independent (the
occurrence of one event has no
effect on the probability of
occurrence of the other) or
• statistically dependent (the
occurrence of one event gives
information about the occurrence
of the other)
Probabilities - Probabilities - Independent EventsIndependent Events
• Marginal probability: the probability of an event occurring:
[P(A)]• Joint probability: the probability of
multiple, independent events, occurring at the same time
P(AB) = P(A)*P(B)
• Conditional probability (for
independent events):
• the probability of event B given that
event A has occurred P(B|A) = P(B)
• or, the probability of event A given that
event B has occurred P(A|B) = P(A)
Probability(A|B) Probability(A|B) Independent EventsIndependent Events
P(B
)
P(A
)
P(A|B)P(B|A)
Statistically Statistically Independent EventsIndependent Events
1. P(black ball drawn on first draw)
• P(B) = 0.30 (marginal probability)
2. P(two green balls drawn)
• P(GG) = P(G)*P(G) = 0.70*0.70 = 0.49 (joint probability for two independent events)
A bucket contains 3 black balls, and 7 green balls. We draw a ball from the bucket, replace it, and draw a second ball
Statistically Independent Statistically Independent Events - continuedEvents - continued
1. P(black ball drawn on second
draw, first draw was green)
• P(B|G) = P(B) = 0.30
(conditional probability)
2. P(green ball drawn on second
draw, first draw was green)
• P(G|G) = 0.70
(conditional probability)
Probabilities - Probabilities - Dependent EventsDependent Events
• Marginal probability: probability of
an event occurring P(A)
• Conditional probability (for
dependent events):
• the probability of event B given that
event A has occurred P(B|A) =
P(AB)/P(A)
• the probability of event A given that
event B has occurred P(A|B) =
P(AB)/P(B)
Probability(A|B)Probability(A|B)
/
P(A|B) = P(AB)/P(B)
P(AB) P(B)P(A)
Probability(B|A)Probability(B|A)
P(B|A) = P(AB)/P(A)
/
P(AB)P(B) P(A)
Statistically Dependent Statistically Dependent EventsEvents
Assume that we have an urn containing 10 balls of the following descriptions:
•4 are white (W) and lettered (L)
•2 are white (W) and numbered N
•3 are yellow (Y) and lettered (L)
•1 is yellow (Y) and numbered (N)
Then:
• P(WL) = 4/10 = 0.40
• P(WN) = 2/10 = 0.20
• P(W) = 6/10 = 0.60
• P(YL) = 3/10 = 0.3
• P(YN) = 1/10 = 0.1
• P(Y) = 4/10 = 0.4
Statistically Dependent Statistically Dependent Events - ContinuedEvents - Continued
Then:• P(L|Y) = P(YL)/P(Y)
= 0.3/0.4 = 0.75
• P(Y|L) = P(YL)/P(L)
= 0.3/0.7 = 0.43
• P(W|L) = P(WL)/P(L)
= 0.4/0.7 = 0.57
Joint Probabilities, Joint Probabilities, Dependent EventsDependent Events
Your stockbroker informs you that if
the stock market reaches the 10,500
point level by January, there is a
70% probability the Tubeless
Electronics will go up in value.
Your own feeling is that there is
only a 40% chance of the market
reaching 10,500 by January.
What is the probability that both the
stock market will reach 10,500
points, and the price of Tubeless
will go up in value?
Joint Probabilities, Dependent Joint Probabilities, Dependent Events - continuedEvents - continued
Then:
P(MT) =P(T|M)P(M)
= (0.70)
(0.40)
= 0.28
Let M represent
the event of the
stock market
reaching the
10,500 point
level, and T
represent the
event that
Tubeless goes
up.
Revising Probabilities: Revising Probabilities: Bayes’ TheoremBayes’ Theorem
Bayes’ theorem can be used to calculate revised or posterior probabilities
Prior Probabilities
Bayes’ Process
Posterior Probabilities
New Information
General Form of General Form of Bayes’ TheoremBayes’ Theorem
Aevent theof complementA where
)()|()()|(
)()|()|(
)(
)()|(
APABPAPABP
APABPBAP
orBP
ABPBAP
die.unfair"" is Aevent then the
die,fair"" isA event theifexample,For
Posterior ProbabilitiesPosterior ProbabilitiesA cup contains two dice identical in
appearance. One, however, is fair (unbiased), the other is loaded (biased). The probability of rolling a 3 on the fair die is 1/6 or 0.166. The probability of tossing the same number on the loaded die is 0.60.
We have no idea which die is which, but we select one by chance, and toss it. The result is a 3.
What is the probability that the die rolled was fair?
Posterior Probabilities Posterior Probabilities ContinuedContinued
• We know that:P(fair) = 0.50 P(loaded) = 0.50
• And:
P(3|fair) = 0.166 P(3|loaded) = 0.60
• Then:P(3 and fair) = P(3|fair)P(fair)
= (0.166)(0.50)
= 0.083
P(3 and loaded) = P(3|loaded)P(loaded)
= (0.60)(0.50)
= 0.300
Posterior Probabilities Posterior Probabilities ContinuedContinued
• A 3 can occur in combination with
the state “fair die” or in combination
with the state ”loaded die.” The sum
of their probabilities gives the
unconditional or marginal probability
of a 3 on a toss:
P(3) = 0.083 + 0.0300 = 0.383.
• Then, the probability that the die
rolled was the fair one is given by:
0.22 0.383
0.083
P(3)
3) andP(Fair 3)|P(Fair
Further Probability Further Probability RevisionsRevisions
• To obtain further information as
to whether the die just rolled is
fair or loaded, let’s roll it again.
• Again we get a 3.
Given that we have now rolled
two 3s, what is the probability
that the die rolled is fair?
Further Probability Further Probability Revisions - continuedRevisions - continued
P(fair) = 0.50, P(loaded) = 0.50 as before
P(3,3|fair) = (0.166)(0.166) = 0.027
P(3,3|loaded) = (0.60)(0.60) = 0.36
P(3,3 and fair) = P(3,3|fair)P(fair)
= (0.027)(0.05)
= 0.013
P(3,3 and loaded) = P(3,3|loaded)P(loaded)
= (0.36)(0.5)
= 0.18
P(3,3) = 0.013 + 0.18 = 0.193
Further Probability Further Probability Revisions - continuedRevisions - continued
933.00.193
0.18
P(3,3)
Loaded) and P(3,3 3,3)|P(Loaded
067.00.193
0.013
P(3,3)
Fair) and P(3,33,3)|P(Fair
To give the final comparison:
P(fair|3) = 0.22
P(loaded|3) = 0.78
P(fair|3,3) = 0.067
P(loaded|3,3) = 0.933
Further Probability Further Probability Revisions - continuedRevisions - continued
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