Introduction-to-Probability.pptx

To accompany Quantitative Analysis for
Management, 9e
by Render/Stair/Hanna
2-1 © 2006 by Prentice Hall, Inc.,
Upper Saddle River, NJ 07458
Introduction
 Life is uncertain!
 We must deal with risk!
 A probability is a numerical statement
about the likelihood that an event will
occur.

Management, 9e
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.

Management, 9e
Diversey Paint Example
Demand for white latex paint at Diversey
Paint and Supply has always been either 0,
1, 2, 3, or 4 gallons per day. Over the past
200 days, the frequencies of demand are
represented in the following table:
Qty Demanded No. of Days
0 40
1 80
2 50
3 20
4 10
Total 200

Management, 9e
Diversey Paint Example (continued)
Quantity Freq.
Demand (days)
0 40
1 80
2 50
3 20
4 10
Total days = 200
Probability
(Relative Freq)
(40/200) = 0.20
(80/200) = 0.40
(50/200) = 0.25
(20/200) = 0.10
(10/200) = 0.05
Total Prob =1.00
Probabilities of Demand
Note: 0  P(event)  1
and P(event) = 1


Management, 9e
Types of Probability
Objective probability is based on
logical observations:
Determined by:
 Relative frequency – Obtained using historical data
(Diversey Paint)
 Classical method – Known probability for each
outcome (tossing a coin)
occurrences
or
outcomes
of
number
Total
occurs
event
times
of
Number
)
( =
event
P

Management, 9e
Types of Probability
Subjective probability is based on
personal experiences.
Determined by:
 Judgment of experts
 Opinion polls
 Delphi method
 Others

Management, 9e
Mutually Exclusive Events
 Events are said to be mutually exclusive if
only one of the events can occur on any one
trial.
Example: a fair coin toss results in either a heads or a
tails.

Management, 9e
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: a collectively exhaustive list of possible
outcomes for a fair coin toss includes heads and
tails.

Management, 9e
Die Roll Example
Outcome
of Roll
1
2
3
4
5
6
Probability
1/6
1/6
1/6
1/6
1/6
1/6
Total = 1
This is a collectively
exhaustive list of potential
outcomes for a single die
roll.
The outcome is a mutually exclusive event because only one
event can occur (a 1, 2, 3, 4, 5, or 6) on any single roll.

Management, 9e
Twin Birth Example
A woman is pregnant with non- identical twins.
Following is a list of collectively exhaustive,
mutually exclusive possible outcomes:
Outcome Probability
of Birth
Boy/Boy ¼
Boy/Girl ¼
Girl/Girl ¼
Girl/Boy ¼
What is the probability that both babies will

Management, 9e
In-Class Practice
 Draw a spade and a club
 Draw a face card and a number card
 Draw an ace and a 3
 Draw a club and a nonclub
 Draw a 5 and a diamond
 Draw a red card and a diamond
Assuming a traditional 52-card deck, can you identify if
these outcomes are mutually exclusive and/or collectively
exhaustive ??

Management, 9e
Law of Addition:
Mutually Exclusive
P (event A or event B) =
P (event A) + P (event B)
or:
P (A or B) = P (A) + P (B)
Example:
P (spade or club) = P (spade) + P (club)
= 13/52 + 13/52
= 26/52 = 1/2 = 50%

Management, 9e
Law of Addition:
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)

Management, 9e
Venn Diagram
P(A) P(B)

Management, 9e
Venn Diagram
P(A or B)
+ -
=
P(A) P(B) P(A and B)
P(A or B)

Management, 9e
In-Class Example: Specialized
University
Specialized University offers four different graduate
degrees: business, education, accounting, and science.
Enrollment figures show 25% of their graduate students
are in each specialty. Although 50% of the students are
female, only 15% are female business majors. If a student
is randomly selected from the University’s registration
database:
 What is the probability the student is a business or
education major?
 What is the probability the student is a female or a business
major?

Management, 9e
Specialized University Solution
The probability that the student is a business or education
major is mutually exclusive event. Thus:
P(Bus or Edu) = P(Bus) + P(Edu)
= .25 + .25
= .50 or 50%
The probability that the student is a female or a business
major is not mutually exclusive because the student could be
a female business major. Thus:
P(Fem or Bus) = P(Fem) + P(Bus)
– P(Fem and Bus)
= .50 + .25 - .15
= .60 or 60%

Management, 9e
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).

Management, 9e
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
(a) Chicago Cubs win the National League
pennant
(b) Chicago Cubs win the World Series

Management, 9e
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:

Management, 9e
Venn Diagram: P(A|B)
P(B
)
P(A|B)
P(B|A)

Management, 9e
Independent Events
Example
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.

Management, 9e
Independent Events Example 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)

Management, 9e
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)
 Joint probability: The probability of
multiple events occurring at the same
time: P(AB) = P(B|A)*P(A)

Management, 9e
/
P(B|A) = P(AB)/P(A)
P(AB) P(A)
P(B)
P(A) P(B)
Venn Diagram: P(B|A)

Management, 9e
/
P(A|B) = P(AB)/P(B)
P(AB) P(B)
P(A)
P(A) P(B)
Venn Diagram: P(A|B)

Management, 9e
Dependent Events Example
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

Management, 9e
Dependent Events Example
(continued)
Then:
 P(Y) = .4
- marginal probability
 P(L|Y) = P(YL)/P(Y)
= 0.3/0.4 = 0.75
- conditional probability
 P(W|L) = P(WL)/P(L)
= 0.4/0.7 = 0.57 - conditional
probability

Management, 9e
Dependent Events: Joint
Probability Example
Your stockbroker informs you that if the stock
market reaches the 10,500 point level by January,
there is a 70% probability that 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?

Management, 9e
Dependent Events: Joint Probabilities
Solution
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.

Management, 9e
Revising Probabilities: Bayes’
Theorem
Bayes’ theorem can be used to calculate
revised or posterior probabilities.
Prior
Probabilities
Bayes’
Process
Posterior
Probabilities
New Information

Management, 9e
General Form of Bayes’
Theorem
A
event
the
of
complement
A
where
)
(
)
|
(
)
(
)
|
(
)
(
)
|
(
)
|
(
)
(
)
(
)
|
(
=

=
=
A
P
A
B
P
A
P
A
B
P
A
P
A
B
P
B
A
P
or
B
P
AB
P
B
A
P
die.
unfair"
"
is
A
event
then the
die,
fair"
"
is
A
event
the
if
example,
For

Management, 9e
Posterior Probabilities
Example
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

Management, 9e
Posterior Probabilities Example
(continued)
 We know that:
P(fair) = 0.50 P(loaded) = 0.50
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)
- joint probability

Management, 9e
Posterior Probabilities
Example 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 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)
and
P(Fair
3)
|
P(Fair =
=
=
- conditional probability

Management, 9e
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?

Management, 9e
continued
 We know from before that:
P(fair) = 0.50, P(loaded) = 0.50
 Then:
P(3,3|fair) = P(3|fair)*P(3|fair)
= (0.166)(0.166) = 0.027
P(3,3|loaded) = P(3|loaded)*P(3|loaded)
= (0.60)(0.60) = 0.36
 So:
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
Thus, the probability of getting two 3s is a marginal
probability obtained from the sum of the probability

Management, 9e
continued
933
.
0
0.193
0.18
P(3,3)
Loaded)
and
P(3,3
3,3)
|
P(Loaded
067
.
0
0.193
0.013
P(3,3)
Fair)
and
P(3,3
3,3)
|
P(Fair
=
=
=
=
=
=
 Using the probabilities from the
previous slide:

Management, 9e
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
continued

Introduction-to-Probability.pptx

More Related Content

Similar to Introduction-to-Probability.pptx

Recently uploaded

Introduction-to-Probability.pptx