Probability Concepts and Probability Concepts and ApplicationsApplications
Chapter Outline
2.1 Introduction
2.2 Fundamental Concepts
2.3 Mutually Exclusive and Collectively Exhaustive Events
2.4 Statistically Independent Events
2.5 Statistically Dependent Events
2.6 Revising Probabilities with Bayes’ Theorem
Chapter Outline - continued
2.7 Further Probability Revisions
2.8 Random Variables
2.9 Probability Distributions
2.10 The Binomial Distribution
2.11 The Normal Distribution
2.12 The Exponential Distribution
2.13 The Poisson Distribution
Introduction
• 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 Probability
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.
Example 2.1
• 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 2.1 - 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 2.1 - 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 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 Probability
Subjective probability:
Based upon judgement
Determined by:
– judgement of expert
– opinion polls
– Delphi method
– etc.
Mutually Exclusive Events
• Events are said to be mutually
exclusive if only one of the events
can occur on any one trial
Collectively Exhaustive Events
• 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
Example 2
Outcome
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
Example 2a
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.
Example 3
Draw a spade or a club
Draw a face card or a
number card
Draw an ace or a 3
Draw a club or a nonclub
Draw a 5 or a diamond
Draw a red card or a
diamond
Yes No
Yes Yes
Yes No
Yes Yes
No No
No No
Draw Mutually Collectively Exclusive Exhaustive
Probability : 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: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)(Venn Diagram)
P(A) P(B)
P(A and B
)
P(A or B)
+ -
=
P(A) P(B) P(A and B)
P(A or B)
Statistical 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)
Which Are Independent?
• (a) Your education
(b) Your income level
• (a) Draw a Jack of Hearts from a full 52 card deck
(b) Draw a Jack of Clubs from a full 52 card deck
Probabilities - Independent 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) Independent Events
P(B
)
P(A
)
P(A|B)P(B|A)
Statistically Independent Events
1. P(black ball drawn on first draw)
2. P(two green balls drawn)
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 Events - continued
1. P(black ball drawn on second draw, first draw was
green)
2. P(green ball drawn on second draw, first draw was
green)
Probabilities - Dependent 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)
/
P(A|B) = P(AB)/P(B)
P(AB) P(B)P(A)
Probability(B|A)
P(B|A) = P(AB)/P(A)
/
P(AB)P(B) P(A)
Statistically Dependent Events
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 Events - 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, Dependent 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 Events - 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: Bayes’ Theorem
Bayes’ theorem can be used to calculate revised or posterior probabilities
Prior Probabilities
Bayes’ Process
Posterior Probabilities
New Information
General Form of Bayes’ Theorem
Aevent theof complementA where
)()|()()|(
)()|()|(
)(
)()|(
APABPAPABP
APABPBAP
orBP
ABPBAP
die.unfair"" is Aevent then the
die,fair"" isA event theifexample,For
Posterior Probabilities
A 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 Continued
• 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 Continued
• 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 Revisions
• 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 Revisions - 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 Revisions - 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 Revisions - continued
Random Variables
• Discrete random variable - can assume only a
finite or limited set of values- i.e., the number of
automobiles sold in a year
• Continuous random variable - can assume any
one of an infinite set of values - i.e., temperature,
product lifetime
Random Variables (Numeric)
Experiment Outcome Random Variable Range of Random Variable
Stock 50 Xmas trees
Number of trees sold
X = number of trees sold
0,1,2,, 50
Inspect 600 items
Number acceptable
Y = number acceptable
0,1,2,…, 600
Send out 5,000 sales letters
Number of people e responding
Z = number of people responding
0,1,2,…, 5,000
Build an apartment building
%completed after 4 months
R = %completed after 4 months
0R100
Test the lifetime of a light bulb (minutes)
Time bulb lasts - up to 80,000 minutes
S = time bulb burns
0S80,000
Random Variables (Non-numeric)
Experiment Outcome Random
Variable
Range of
Random
Variable
Students respond to a questionnaire
Strongly agree (SA)Agree (A)Neutral (N)Disagree (D)Strongly Disagree (SD)
X = 5 if SA4 if A3 if N2 if D1 if SD
1,2,3,4,5
One machine is inspected
DefectiveNot defective
Y = 0 if defective1 if not defective
0,1
Consumers respond to how they like a product
GoodAveragePoor
Z = 3 if good2 if average1 if poor
1,2,3
Probability Distributions
Table 2.4
Outcome X Number Responding
P(X)
SA 5 10 0.10
A 4 20 0.20
N 3 30 0.30
D 2 30 0.30
SD 1 10 0.10
D
0.00
0.05
0.10
0.15
0.20
0.25
0.30
1 2 3 4 5
Figure 2.5Probability Function
Expected Value of a Discrete Probability Distribution
n
iii )X(PX)X(E
2.9
)1.0)(1()3.0)(2(
)3.0)(3()2.0)(4()1.0)(5(
)()(
)()()(
)()(
5544
332211
5
1
XPXXPX
XPXXPXXPX
XPXXE ii
i
Variance of a Discrete Probability Distribution
i
n
ii XPXEX
1
22
29.1
0.3610.2430.0030.242- 0.44
)1.0()9.21(
)3.0(2.9)-(2 3.09.23
2.09.241.09.25
2
22
222
Binomial Distribution
Assumptions:
1. Trials follow Bernoulli process – two possible outcomes
2. Probabilities stay the same from one trial to the next
3. Trials are statistically independent
4. Number of trials is a positive integer
Binomial Distribution
rnr qp r)!-(nr!
n!
Probability of r successes in n trials
n = number of trials
r = number of successes
p = probability of success
q = probability of failure
Binomial Distribution
)p(np
np
0.00
0.05
0.10
0.15
0.20
0.25
0.30
0.35
1 2 3 4 5 6
(r) Number of Successes
P(r
)
N = 5, p = 0.50
Binomial Distribution
Probability Distribution Continuous Random Variable
Probability density function - f(X)
5 5.05 5.1 5.15 5.2 5.25 5.3 5.35 5.4
Normal Distribution
2
2)(2/1
2
1)(
X
eXf
Normal Distribution for Different Values of
0
30 40 50 60 70
=50 =60=40
0 0.5 1 1.5 2
Normal Distribution for Different Values of
=0.1
=0.2=0.3
= 1
Three Common Areas Under the Curve
• Three Normal distributions with different areas
Three Common Areas Under the Curve
Three Normal
distributions
with different
areas
The Relationship Between Z and X
55 70 85 100 115 130 145
-3 -2 -1 0 1 2 3
x
Z=100
=15
Haynes Construction Company Example Fig. 2.12
Haynes Construction Company ExampleFig. 2.13
Haynes Construction Company Example Fig. 2.14