Transcript Basic Probability And Probability Distributions

Business Statistics
Basic Probability And
Probability Distributions
Chapter Topics
• Basic
Probability Concepts:
Sample Spaces and Events, Simple
Probability, and Joint Probability,
• Conditional Probability
• Bayes’ Theorem
• The Probability Distribution for a
Discrete Random Variable
Chapter Topics
• Binomial
and Poisson Distributions
• Covariance and its Applications in
Finance
• The Normal Distribution
• Assessing the Normality Assumption
Sample Spaces
Collection of all Possible Outcomes
e.g. All 6 faces of a die:
e.g. All 52 cards of a bridge deck:
Events
• Simple Event: Outcome from a Sample Space
with 1 Characteristic
e.g.
A Red Card from a deck of cards.
• Joint Event: Involves 2 Outcomes
Simultaneously
e.g.
An Ace which is also a Red Card from a
deck of cards.
Visualizing Events
• Contingency Tables
Ace
• Tree Diagrams
Black
Red
2
2
Total
4
Not Ace
24
24
48
Total
26
26
52
Simple Events
The Event of a Happy Face
There are 5 happy faces in this collection of 18 objects
Joint Events
The Event of a Happy Face AND Light Colored
3 Happy Faces which are light in color
Special Events
Null event
Club & diamond on
1 card draw
Complement of event
For event A,
'
All events not In A: A
Null Event

Dependent or
Independent Events
The Event of a Happy Face GIVEN it is Light Colored
E = Happy FaceLight Color
3 Items: 3 Happy Faces Given they are Light Colored
Contingency Table
Red Ace
A Deck of 52 Cards
Ace
Not an
Ace
Total
Red
2
24
26
Black
2
24
26
Total
4
48
52
Sample Space
Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed | Total
MALE
900
200
400
|
FEMALE
300
100
600
|
Total
1200
300
1000
|
Sample Space
1500
1000
2500
Tree Diagram
Event Possibilities
Full
Deck
of Cards
Red
Cards
Ace
Not an Ace
Ace
Black
Cards
Not an Ace
Probability
•Probability is the numerical
measure of the likelihood
that the event will occur.
•Value is between 0 and 1.
•Sum of the probabilities of
all mutually exclusive and
collective exhaustive events
is 1.
1
Certain
.5
0
Impossible
Computing Probability
• The Probability of an Event, E:
P(E) =
=
Number of Event Outcomes
Total Number of Possible Outcomes in the Sample Space
X
T
e.g. P(
) = 2/36
(There are 2 ways to get one 6 and the other 4)
• Each of the Outcome in the Sample Space
equally likely to occur.
Computing
Joint Probability
The Probability of a Joint Event, A and B:
P(A and B)
=
Number of Event Outcomes from both A and B
Total Number of Possible Outcomes in Sample Space
e.g. P(Red Card and Ace)
2 Red Aces
1

=
52 Total Number of Cards 26
Joint Probability Using
Contingency Table
Event
B1
Event
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
Joint Probability
P(B1)
P(B2)
1
Marginal (Simple) Probability
Computing
Compound Probability
The Probability of a Compound Event, A or B:
Numer of Event Outcomes from Either A or B
P  A or B 
Total Outcomes in the Sample Space
e.g.
P(Red Card or Ace)
4 Aces + 26 Red Cards  2 Red Aces 28 7



52 Total Number of Cards
52 13
Contingency Table
2500 Employees of Company ABC
Agree
Neutral
Opposed | Total
MALE
900
200
400
|
FEMALE
300
100
600
|
Total
1200
300
1000
|
Sample Space
1500
1000
2500
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
600/1000 = 0.60
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
600/1000 = 0.60
4. Either a female or opposed to the
proposal
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
600/1000 = 0.60
4. Either a female or opposed to the
proposal ……….. 1000/2500 + 1000/2500 - 600/2500 =
1400/2500 = 0.56
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
600/1000 = 0.60
4. Either a female or opposed to the
proposal ……….. 1000/2500 + 1000/2500 - 600/2500 =
1400/2500 = 0.56
5. Are Gender and Opinion (statistically) independent?
The pervious table refers to 2500 employees of ABC Company, classified by gender
and by opinion on a company proposal to emphasize fringe benefits rather than wage
increases in an impending contract discussion
Calculate the probability that an employee selected (at random) from this group will be:
1. A female opposed to the proposal
600/2500 = 0.24
2. Neutral
300/2500 = 0.12
3. Opposed to the proposal, GIVEN that
the employee selected is a female
600/1000 = 0.60
4. Either a female or opposed to the
proposal ……….. 1000/2500 + 1000/2500 - 600/2500 =
1400/2500 = 0.56
5. Are Gender and Opinion (statistically) independent?
For Opinion and Gender to be independent, the joint probability of each pair of
A events (GENDER) and B events (OPINION) should equal the product of the
respective unconditional probabilities….clearly this does not hold…..check, e.g.,
the prob. Of MALE and IN FAVOR against the prob. of MALE times the prob. of
IN FAVOR …they are not equal….900/2500 does not equal 1500/2500 * 1200/2500
Compound Probability
Addition Rule
P(A1 or B1 ) = P(A1) +P(B1) - P(A1 and B1)
Event
Event
B1
B2
Total
A1
P(A1 and B1) P(A1 and B2) P(A1)
A2
P(A2 and B1) P(A2 and B2) P(A2)
Total
P(B1)
P(B2)
1
For Mutually Exclusive Events: P(A or B) = P(A) + P(B)
Computing
Conditional Probability
The Probability of Event A given that Event B has
occurred:
P  A and B
P(A B) =
P  B
e.g.
2 Red Aces 1
P(Red Card given that it is an Ace) =

4 Aces
2
Conditional Probability
Using Contingency Table
Conditional Event: Draw 1 Card. Note Kind & Color
Color
Type
Red
Black
Total
Ace
2
2
4
Non-Ace
24
24
48
Total
26
26
52
P(Ace | Red) =
Revised
Sample
Space
P(Ace AND Red)
2 / 52
2


P(Red)
26 / 52
26
Conditional Probability and
Statistical Independence
Conditional Probability:
P ( A and B )
P(AB) =
P( B )
Multiplication Rule:
P(A and B) = P(A B) • P(B)
= P(B A) • P(A)
Conditional Probability and
Statistical Independence (continued)
Events are Independent:
P(A  B) = P(A)
Or, P(B  A) = P(B)
Or, P(A and B) = P(A) • P(B)
Events A and B are Independent when the
probability of one event, A is not affected by
another event, B.
Bayes’ Theorem
P(Bi A) =
P( A Bi )  P( Bi )
P( A B1 )  P( B1 )      P( A Bk )  P( Bk )
P( Bi and A)

P( A)
Same
Event
Adding up
the parts
of A in all
the B’s
A manufacturer of VCRs purchases a particular microchip, called the LS-24, from three suppliers: Hall
Electronics, Spec Sales, and Crown Components. Thirty percent of the LS-24 chips are purchased from
Hall, 20% from Spec, and the remaining 50% from Crown. The manufacturer has extensive past records
for the three suppliers and knows that there is a 3% chance that the chips from Hall are defective, a 5%
chance that chips from Spec are defective and a 4% chance that chips from Crown are defective. When
LS-24 chips arrive at the manufacturer, they are placed directly into a bin and not inspected or otherwise
identified as to supplier. A worker selects a chip at random.
What is the probability that the chip is defective?
A worker selects a chip at random for installation into a VCR and finds it is defective. What is the
probability that the chip was supplied by Spec Sales?
Bayes’ Theorem: Contingency Table
What are the chances of repaying a loan,
given a college education?
Loan Status
Education Repay Default Prob.
College
No College
Prob.
P(RepayCollege) =
.2
?
.05
?
.25
?
?
?
1
P(College and Repay)
P(College and Repay) + P(College and Default)
= .20
.25 = .80
Discrete Random Variable
•
Random Variable: outcomes of an
experiment expressed numerically
e.g.
Throw a die twice: Count the number of times 4
comes up (0, 1, or 2 times)
Discrete Random Variable
•Discrete Random Variable:
• Obtained by Counting (0, 1, 2, 3, etc.)
• Usually finite by number of different
values
e.g.
Toss a coin 5 times. Count the number of
tails. (0, 1, 2, 3, 4, or 5 times)
Discrete Probability
Distribution Example
Event: Toss 2 Coins.
Count # Tails.
Probability distribution
Values
probability
T
T
T
T
0
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Discrete
Probability Distribution
•
List of all possible [ xi, p(xi) ] pairs
Xi = value of random variable
P(xi) = probability associated with value
•
Mutually exclusive (nothing in common)
•
Collectively exhaustive (nothing left out)
0  p(xi)  1
 P(xi) = 1
Discrete Random Variable
Summary Measures
Expected value (The mean)
Weighted average of the probability distribution
 = E(X) = xi p(xi)
E.G. Toss 2 coins, count tails, compute expected value:
= 0  .25 + 1 .50 + 2  .25 = 1
Number of Tails
Discrete Random Variable
Summary Measures
Variance
Weighted average squared deviation about mean
 = E [ (xi -  )2]= (xi -  )2p(xi)
E.G. Toss 2 coins, count tails, compute variance:
 = (0 - 1)2(.25) + (1 - 1)2(.50) + (2 - 1)2(.25)
= .50
Important Discrete Probability
Distribution Models
Discrete Probability
Distributions
Binomial
Poisson
Binomial Distributions
•
‘N’ identical trials

•
E.G. 15 tosses of a coin, 10 light bulbs
taken from a warehouse
2 mutually exclusive outcomes on each
trial

E.G. Heads or tails in each toss of a coin,
defective or not defective light bulbs
Binomial Distributions
• Constant Probability for each Trial
• e.g. Probability of getting a tail is the same
each time we toss the coin and each light bulb
has the same probability of being defective
• 2 Sampling Methods:
• Infinite Population Without Replacement
• Finite Population With Replacement
• Trials are Independent:
• The Outcome of One Trial Does Not Affect the
Outcome of Another
Binomial Probability
Distribution Function
P(X) 
n!
X
nX
p (1  p )
X ! (n  X)!
P(X) = probability that X successes given a knowledge of n
and p
X = number of ‘successes’ in
sample, (X = 0, 1, 2, ..., n)
p = probability of each ‘success’
n = sample size
Tails in 2 Tosses of Coin
X
0
P(X)
1/4 = .25
1
2/4 = .50
2
1/4 = .25
Binomial Distribution
Characteristics
P(X)
Mean
  E ( X )  np
e.g.  = 5 (.1) = .5
.6
.4
.2
0
n = 5 p = 0.1
X
0
1
2
3
4
5
Standard Deviation
 
np (  p )
e.g.  = 5(.5)(1 - .5)
= 1.118
P(X)
.6
.4
.2
0
n = 5 p = 0.5
X
0
1
2
3
4
5
Poisson Distribution
Poisson process:
• Discrete events in an ‘interval’


The probability of one success in
an interval is stable
The probability of more than one
success in this interval is 0
• Probability of success is
Independent from interval to
Interval
E.G. # Customers arriving in 15 min
# Defects per case of light
bulbs
P( X  x | 
-
e 
x!
x
Poisson Distribution
Function
 X
P (X )  e 
X!
P(X ) = probability of X successes given 
 = expected (mean) number of ‘successes’
e = 2.71828 (base of natural logs)
X = number of ‘successes’ per unit
e.g. Find the probability of 4
customers arriving in 3
minutes when the mean is 3.6.
-3.6
P(X) = e
4
3.6 = .1912
4!
Poisson Distribution
Characteristics
Mean
  E (X )  
N
  Xi P( Xi )
= 0.5
P(X)
.6
.4
.2
0
X
0
i 1
1
 

3
4
5
= 6
P(X)
Standard Deviation
2
.6
.4
.2
0
X
0
2
4
6
8
10
Covariance
 XY   X i  E( X )  Yi  E(Y )  P( X iYi )
N
i 1
X = discrete random variable X
Xi = value of the ith outcome of X
P(xiyi) = probability of occurrence of the
ith outcome of X and ith outcome of Y
Y = discrete random variable Y
Yi = value of the ith outcome of Y
I = 1, 2, …, N
Computing the Mean for
Investment Returns
Return per $1,000 for two types of investments
P(XiYi)
Investment
Economic condition Dow Jones fund X Growth Stock Y
.2
Recession
-$100
-$200
.5
Stable Economy
+ 100
+ 50
.3
Expanding Economy
+ 250
+ 350
E(X) = X = (-100)(.2) + (100)(.5) + (250)(.3) = $105
E(Y) = Y = (-200)(.2) + (50)(.5) + (350)(.3) = $90
Computing the Variance for
Investment Returns
P(XiYi)
Investment
Economic condition Dow Jones fund X Growth Stock Y
.2
Recession
-$100
-$200
.5
Stable Economy
+ 100
+ 50
.3
Expanding Economy
+ 250
+ 350
Var(X) =  2 = (.2)(-100 -105)2 + (.5)(100 - 105)2 + (.3)(250 - 105)2
X
= 14,725,
Var(Y) = 
2
Y=
X = 121.35
(.2)(-200 - 90)2 + (.5)(50 - 90)2 + (.3)(350 - 90)2
= 37,900,
Y = 194.68
Computing the Covariance for
Investment Returns
P(XiYi)
Investment
Economic condition Dow Jones fund X Growth Stock Y
.2
Recession
-$100
-$200
.5
Stable Economy
+ 100
+ 50
.3
Expanding Economy
+ 250
+ 350
XY = (.2)(-100 - 105)(-200 - 90) + (.5)(100 - 105)(50 - 90)
+ (.3)(250 -105)(350 - 90) = 23,300
The Covariance of 23,000 indicates that the two investments are
positively related and will vary together in the same direction.
The Normal Distribution
• ‘Bell Shaped’
• Symmetrical
f(X)
• Mean, Median and
Mode are Equal
• ‘Middle Spread’
Equals 1.33 
• Random Variable has
Infinite Range

Mean
Median
Mode
X
The Mathematical Model
f X  

1
2
2
e
1
2
 X  

2
f(X) =
frequency of random variable X

=
3.14159;

=
population standard deviation
X
=
value of random variable (- < X < )

=
population mean
e = 2.71828
Many Normal Distributions
There are
an Infinite
Number
Varying the Parameters  and , we obtain
Different Normal Distributions.
Normal Distribution:
Finding Probabilities
Probability is the
area under the
curve!
P (c  X  d )
f(X)
c
d
X
?
Which Table?
Each distribution
has its own table?
Infinitely Many Normal Distributions Means
Infinitely Many Tables to Look Up!
Solution (I): The Standardized
Normal Distribution
Standardized Normal Distribution
Table (Portion)
 Z = 0 and Z = 1
Z
.00
.01
.0478
.02
Shaded Area
Exaggerated
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
Z = 0.12
0.3 .0179 .0217 .0255 Probabilities
Only One Table is Needed
Solution (II): The Cumulative
Standardized Normal Distribution
Cumulative Standardized Normal
Distribution Table (Portion)
  0 and    1
Z .00
.01
.02
0.0 .5000 .5040 .5080
.5478
Shaded Area
Exaggerated
0.1 .5398 .5438 .5478
0.2 .5793 .5832 .5871
0.3 .5179 .5217 .5255
Z = 0.12
Probabilities
Only One Table is Needed
Standardizing Example
Normal
Distribution
6 .2  5
X


Z

 0 . 12

10
Standardized
Normal Distribution
 = 10
Z = 1
 = 5 6.2 X
 = 0 .12
Shaded Area Exaggerated
Z
Example:
P(2.9 < X < 7.1) = .1664
z
Normal
Distribution
z 
x

x  

2.9  5

 .21
10

7 .1  5
Standardized
 . 21
10
Normal Distribution
 = 10
Z = 1
.1664
.0832 .0832
2.9 5 7.1 X
-.21 0 .21
Shaded Area Exaggerated
Z
Example: P(X  8) = .3821
z
x

Normal
Distribution
85

 ..30
10
Standardized
Normal Distribution
 = 10
 =1
.5000
.1179
 =5
8
X
.3821
 = 0 .30 Z
Shaded Area Exaggerated
Finding Z Values
for Known Probabilities
What Is Z Given
Probability = 0.1217?
.1217
 =1
Standardized Normal
Probability Table (Portion)
Z
.00
.01
0.2
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
 = 0 .31 Z
Shaded Area
Exaggerated
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
Recovering X Values
for Known Probabilities
Normal Distribution
Standardized Normal Distribution
 = 10
 =1
.1217
 =5
?
X
.1217
 = 0 .31
X  Z= 5 + (0.31)(10) = 8.1
Shaded Area Exaggerated
Z
Assessing Normality
Compare Data Characteristics to
Properties of Normal Distribution
• Put Data into Ordered Array
• Find Corresponding Standard Normal
Quantile Values
• Plot Pairs of Points
• Assess by Line Shape
Assessing Normality
Normal Probability Plot for
Normal Distribution
90
X 60
Z
30
-2 -1 0 1 2
Look for Straight Line!
Normal Probability Plots
Left-Skewed
Right-Skewed
90
90
X 60
X 60
Z
30
-2 -1 0 1 2
-2 -1 0 1 2
Rectangular
U-Shaped
90
90
X 60
X 60
Z
30
-2 -1 0 1 2
Z
30
Z
30
-2 -1 0 1 2
Chapter Summary
• Discussed
Basic Probability Concepts:
Sample Spaces and Events, Simple
Probability, and Joint Probability
• Defined Conditional Probability
• Discussed Bayes’ Theorem
• Addressed the Probability of a Discrete
Random Variable
Summary
• Discussed Binomial and Poisson
Distributions
• Addressed Covariance and its
Applications in Finance
• Covered Normal Distribution
• Discussed Assessing the Normality
Assumption