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Probability Concepts and
Applications
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
Frequencies of Demand
Quantity Demanded (Gallons)
0
1
Number of Days
40
80
2
50
3
20
4
10
Total 200
Example 2.1 - continued
Probabilities of Demand
Quant.
Freq.
Demand
(days)
0
40
(40/200) = 0.20
1
80
(80/200) = 0.40
2
50
(50/200) = 0.25
3
20
(20/200) = 0.10
4
10
Total days = 200
Probability
(10/200) = 0.05
Total Prob
= 1.00
Types of Probability
Objective probability:
P ( event
) =
Total
Number
of times
event
number
of outcomes
occurs
or occurrences
Determined by experiment or observation:
– Probability of heads on coin flip
– Probably of spades on drawing card from deck
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
Rolling a die
has six possible
outcomes
Outcome
Probability
of Roll
1
1/6
2
1/6
3
1/6
4
1/6
5
1/6
6
1/6
Total = 1
Example 2a
Rolling two dice
results in a total of
five spots
showing. There
are a total of 36
possible outcomes.
Outcome
Probability
of Roll = 5
Die 1 Die 2
1
4
1/36
2
3
1/36
3
2
1/36
4
1
1/36
Example 3
Draw
Mutually Collectively
Exclusive Exhaustive
Draw a spade or a club
Yes
No
Draw a face card or a
Yes
Yes
Draw an ace or a 3
Yes
No
Draw a club or a nonclub
Yes
Yes
Draw a 5 or a diamond
No
No
No
No
number card
Draw a red card or a
diamond
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 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(B|A)
P(A|B)
Statistically 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
1. P(black ball drawn on first draw)
2. P(two green balls drawn)
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)
P(AB)
P(A|B) = P(AB)/P(B)
P(B)
Probability(B|A)
/
P(B)
P(AB)
P(B|A) = P(AB)/P(A)
P(A)
Statistically Dependent Events
Then:
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)
• 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
Let M represent the event of
the stock market reaching
the 10,500 point level, and T
represent the event that
Tubeless goes up.
Then:
P(MT) =P(T|M)P(M)
= (0.70)(0.40)
= 0.28
Revising Probabilities: Bayes’ Theorem
Bayes’ theorem can be used to calculate
revised or posterior probabilities
Prior
Probabilities
New Information
Bayes’
Process
Posterior
Probabilities
General Form of Bayes’ Theorem
P ( AB)
P( A | B) =
P( B)
P( A | B) =
or
P ( B | A) P ( A)
P ( B | A) P ( A)  P ( B | A ) P ( A )
where A = complement of the event A
For example, if the event A is " fair" die,
then the event A is " unfair" die.
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:
P(Fair | 3) =
P(Fair and 3)
0.083
=
= 0.22
P(3)
0.383
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
P(3,3 and Fair)
P(Fair | 3,3) =
P(3,3)
0.013
=
= 0.067
0.193
P(3,3 and Loaded)
P(Loaded | 3,3) =
P(3,3)
0.18
=
= 0.933
0.193
Further Probability Revisions - continued
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
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
Stock 50
Xmas trees
Inspect 600
items
Send out
5,000 sales
letters
Build an
apartment
building
Test the
lifetime of a
light bulb
(minutes)
Outcome
Number of
trees sold
Number
acceptable
Number of
people e
responding
%completed
after 4
months
Time bulb
lasts - up to
80,000
minutes
Random Variable
X = number of
trees sold
Y = number
acceptable
Z = number of
people responding
R = %completed
after 4 months
S = time bulb
burns
Range of
Random
Variable
0,1,2,, 50
0,1,2,…,
600
0,1,2,…,
5,000
0  R  100
0  S  80,000
Random Variables (Non-numeric)
Experiment
Students
respond to a
questionnaire
One machine is
inspected
Consumers
respond to how
they like a
product
Outcome
Strongly agree (SA)
Agree (A)
Neutral (N)
Disagree (D)
Strongly Disagree (SD)
Defective
Not defective
Good
Average
Poor
Random
Variable
X = 5 if SA
4 if A
3 if N
2 if D
1 if SD
Y = 0 if defective
1 if not defective
Z = 3 if good
2 if average
1 if poor
Range of
Random
Variable
1,2,3,4,5
0,1
1,2,3
Probability Distributions
Table 2.4
Outcome
X
SA
5
Number
Responding
10
P(X)
A
4
20
0.20
N
3
30
0.30
D
2
30
0.30
SD
1
10
0.10
0.10
D
0.30
0.25
Figure 2.5
Probability Function
0.20
0.15
0.10
0.05
0.00
1
2
3
4
5
Expected Value of a Discrete Probability Distribution
n
E ( X ) =  X iP ( X
i =1
E( X ) =
i
)
5
X
i =1
i
P( X i )
= X 1 P( X 1 )  X 2 P( X 2 )  X 3 P( X 3 )
 X 4 P( X 4 )  X 5 P( X 5 )
= (5)( 0.1)  ( 4)( 0.2)  (3)( 0.3)
 ( 2)( 0.3)  (1)( 0.1)
= 2.9
Variance of a Discrete Probability Distribution

2
=
n
 X
i =1

2
i
 EX

2
P X i

= 5  2.9 
0.1  4  2.9 2 0.2 
2
 3  2.9  0.3  (2 - 2.9) 2 (0.3)
2
 (1  2.9) 2 (0.1)
= 0.44 - 0.242  0.003  0.243  0.361
= 1.29
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
n = number of trials
r = number of successes
p = probability of success
q = probability of failure
Probability of r successes in n trials
n!
r
nr
=
p q
r!(n - r)!
Binomial Distribution
 = np
  = np( 1  p )
Binomial Distribution
N = 5, p = 0.50
0.35
0.30
P(r)
0.25
0.20
0.15
0.10
0.05
0.00
1
2
3
4
5
(r) Number of Succes s es
6
Probability Distribution Continuous Random Variable
Normal Distribution
Probability density function - f(X)
5
5.05
5.1
f (X ) =
5.15
1

2
5.2
5.25
5.3
1 / 2 ( X   ) 2
e

2
5.35
5.4
Normal Distribution for Different Values of 
=40
=50
40
50
=60
0
30
60
70
Normal Distribution for Different Values of 
=1
=0.1
=0.3
0
=0.2
0.5
1
1.5
2
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
=100
Z =
=15
x

55
70
85
100
115
130
145
-3
-2
-1
0
1
2
3
Haynes Construction Company Example Fig. 2.12
Haynes Construction Company Example
Fig. 2.13
Haynes Construction Company Example Fig. 2.14