bradley has invested in two stocks, markley oil
DESCRIPTION
Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. The number of experiment outcomes. Example: Bradley Investments. Investment Gain or Loss - PowerPoint PPT PresentationTRANSCRIPT
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
The number of experiment outcomes
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+5
+8
+8
+10
+8
+8
-20
-2
-2
-2
-2
0
Markley Oil(Stage 1)
Collins Mining(Stage 2)
ExperimentalOutcomes
Example: Bradley Investments
(10 + 8)1000 = $18,000
(10 – 2)1000 = $8,000
(5 + 8)1000 = $13,000
(5 – 2)1000 = $3,000
(0 + 8)1000 = $8,000
(0 – 2)1000 = –$2,000
(-20 + 8)1000 = –$12,000
(-20 – 2)1000 = –$22,000
8
The number of experiment outcomes
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(50)(50)(50)(50)(501 15,625,)(50) 000 01!
,00
Example: State lotteries
Politicians propose a new lottery. In this lottery there are 6 jars each filled with 50 ping pong balls numbered 1 to 50. How many distinct winning tickets could win this lottery if you have to pick the order in which they come out of the jars?
Since order matters and there is replacement the total number of experimental outcomes equals
The number of experiment outcomes
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1 11,441,304(50)(49) ,000(48)(47)( 15,890,7006 5 4 3 2 1
46
)(!
6 45)
Example: State lotteries
Politicians propose a new lottery. In this lottery there is one jar filled with 50 ping pong balls numbered 1 to 50. How many distinct winning tickets could win this lottery if the order of the balls does not have to be picked?
Since order does not matter and there is no replacement the total number of experimental
outcomes equals
!!
( )!Nn
NCNn n
-
The number of experiment outcomes
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Example: State lotteries
Since order matters and there is no replacement the total number of experimental outcomes equals
(50)(49)(48)(47)(461 11,441,)(45) 304 01!
,00
Politicians propose a new lottery. In this lottery there is one jar filled with 50 ping pong balls numbered 1 to 50. Six balls are selected one at a time. How many distinct winning tickets could win this lottery if the order of the balls must be chosen too?
1!!
( )!N
nNP
N n
-
The number of experiment outcomes
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Probability
0 1.5
Increasing Likelihood of Occurrence
Probability:
P(A U B) = P(A) + P(B) if A & B are ME
P(A ∩ B) = P(A) P(B) if A & B are Indep.
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If an experiment has n possible outcomes, the
classical method would assign a probability of 1/n
to each outcome.Experiment: Rolling a die
Sample Space: S = {1, 2, 3, 4, 5, 6}
Probabilities: Each sample point has a 1/6 chance of occurring
Example: Rolling a Die
Probability
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(50)(50)(50)(50)(501 15,625,)(50) 000 01!
,00
Example: State lottery 1
Since order matters and there is replacement the total number of experimental outcomes equals
1( ) 0.00000000006415,625,000,000
P win
Politicians propose a new lottery. In this lottery there are 6 jars each filled with 50 ping pong balls numbered 1 to 50. What is the probability of winning this lottery if you have to pick the order in which they come out of the jars?
Probability
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1 11,441,304(50)(49) ,000(48)(47)( 15,890,7006 5 4 3 2 1
46
)(!
6 45)
Since order does not matter and there is no replacement the total number of experimental
outcomes equals
1( ) 0.000000062915,890,700
P win
Example: State lottery 2Politicians propose a new lottery. In this lottery there is
one jar filled with 50 ping pong balls numbered 1 to 50. What is the probability of winning this lottery if the order of the balls does not have to be picked?
Probability
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Since order matters and there is no replacement the total number of experimental outcomes equals
1( ) 0.000000000087411,441,304,000
P win
(50)(49)(48)(47)(461 11,441,)(45) 304 01!
,00
Example: State lottery 3Politicians propose a new lottery. In this lottery there is
one jar filled with 50 ping pong balls numbered 1 to 50. Six balls are selected one at a time. What is the probability of winning this lottery if the order of the balls must be chosen too?
Probability
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Example: US population by age (The World Almanac 2004) is given below:
AGE Frequency RelativeLL UL (millions) Frequency0 19 80.5 0.29
20 24 19 0.0725 34 39.9 0.1435 44 45.2 0.1645 54 37.7 0.1355 64 24.3 0.0965 35 0.12Totals 281.6 1.00
Probability
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
Markley discovers 3 oil reserves under the ocean using its 3 R&D
vessels.
.10
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
Markley discovers 2 oil reserves under the ocean using its 3 R&D
vessels.
.10
.25
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
Markley discovers 1 oil reserves under the ocean using its 3 R&D
vessels.
.10
.25
.40
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
Markley discovers 0 oil reserves under the ocean using its 3 R&D
vessels.
.10
.25
.40
.25
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
.10
.25
.40
.25
.80
The FED keeps interest rates set a
0.25%
Probability
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Bradley has invested in two stocks, Markley Oil
and Collins Mining. Bradley has determined that the
possible outcomes of these investments three months
from now are as follows. Investment Gain or Loss in 3 Months (in $000)
Markley Oil Collins Mining 10 5 0-20
8-2
Example: Bradley Investments
The FED raises interest rates to
2.50%
.10
.25
.40
.25
.80
.20
Probability
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+5
+8
+8
+10
+8
+8
-20
-2
-2
-2
-2
0
Markley Oil(Stage 1)
Collins Mining(Stage 2)
ExperimentalOutcomes
Example: Bradley Investments
$18,000
$8,000
$13,000
$3,000
$8,000
–$2,000
–$12,000
–$22,000
Probability(.10)(.80) =.08(.10)(.20) =.02(.25)(.80) =.20(.25)(.20) =.05(.40)(.80) =.32(.40)(.20) =.08(.25)(.80) =.20(.25)(.20) =.05
1.00
Probability
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Example: US population by ageLet A be the event “55 years of age or older.” Compute the probability of Ac.
Complement of an Event
( ) 0.09 0.12 0.21P A
( ) 0.29 0.07 0.14 0.16 0.13 0.79cP A
( ) 1 ( ) 1 0.21 0.79cP A P A - -
AGE RelativeLL UL Frequency0 19 0.29
20 24 0.0725 34 0.1435 44 0.1645 54 0.1355 64 0.0965 0.12
Totals 1.00
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( ) 0.09P A B
Example: US population by ageLet A be the event: “55 years of age or older.” Let B be the event:“64 years of age or younger.” Compute the probability of A and B.
( ) 0.29 0.07 0.14 0.16 0.13 0.880.09P B
0.09( ) 0.12 0.21P A
Intersection of Two Events
AGE RelativeLL UL Frequency0 19 0.29
20 24 0.0725 34 0.1435 44 0.1645 54 0.1355 64 0.0965 0.12
Totals 1.00
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( ) 0CP A
Example: US population by age
( ) 0.29 0.07 0.36P C
( ) 0.09 0.12 0.21P A
Intersection of Mutually Exclusive Events
Let A be the event: “55 years of age or older.” Let C be the event:“24 years of age or younger.” Compute the probability of A and C.
0.36
AGE RelativeLL UL Frequency0 19 0.29
20 24 0.0725 34 0.1435 44 0.1645 54 0.1355 64 0.0965 0.12
Totals 1.00
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( ) ( ) ( )0.21 0.8 98
( )0.0
1
P AP A B P A P BB - -
Example: US population by age
( ) 0.88P B
( ) ( ) 0.21 0.88 1.09P A P B
Union of Two Events
( ) 0.21P A
Let A be the event: “55 years of age or older.” Let B be the event:“64 years of age or younger.” Compute the probability of A or B.
AGE RelativeLL UL Frequency0 19 0.29
20 24 0.0725 34 0.1435 44 0.1645 54 0.1355 64 0.0965 0.12
Totals 1.00
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( ) ( ) ( )0.21 0.360
0.
( )
57
P AP A C P A P C C - -
Example: US population by age
Union of Mutually Exclusive Events
Let A be the event “55 years of age or older.” Let C be the event“24 years of age or younger.” Compute the probability of A or C.
( ) 0.29 0.07 0.36P C
( ) 0.09 0.12 0.21P A
AGE RelativeLL UL Frequency0 19 0.29
20 24 0.0725 34 0.1435 44 0.1645 54 0.1355 64 0.0965 0.12
Totals 1.00
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The conditional probability of A given B is denoted by P(A|B).
Conditional Probability
( )( | ) ( )P A BP A B
P B
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700( ) 0.7143980
P W
Conditional ProbabilityExample: Consider the hiring of black and white workers at
BigMart
White (W)
Black (B) Total
Hired (H) 130 30 160
Not Hired (Hc) 570 250 820
Total 700 280 980
( ) 0.1327( | ) 0.1858( ) 0.7143
P H WP H WP W
130( ) 0.1327980
P H W
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130( | ) 0.1858700
P H W
700( ) 0.7143980
P W
Conditional ProbabilityExample: Consider the hiring of black and white workers at
BigMart
White (W)
Black (B) Total
Hired (H) 130 30 160
Not Hired (Hc) 570 250 820
Total 700 280 980
130( ) 0.1327980
P H W
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( )( | ) ( )P A BP A B
P B
( )( ) ( | ) ( )( )
P A BP B P A B P BP B
( ) ( | ) ( )P B P A B P A B ( ) ( ) ( | )P A B P B P A B
Multiplication Law For Dependent Events
The multiplication law provides a way to compute the probability of the intersection of two events.
It is derived by manipulating the conditional probability:
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White (W)
Black (B) Total
Hired (H) 130 30 160
Not Hired (Hc) 570 250 820
Total 700 280 980
Example: Consider the hiring of black and white workers at BigMart
700 130( ) ( ) ( | ) 0.1327980 700
P H W P W P H W
( ) 0.1327P H W
Multiplication Law For Dependent Events
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Independent Events
If the probability of event A is not changed by the existence of event B, we would say that events A and B are independent.
Two events A and B are independent if:
P(A|B) = P(A) P(B|A) = P(B)or
( ) ( ) ( )P A B P B P A
This changes the multiplication law:
( ) ( ) ( | )P A B P B P A B
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Example: Consider the promotion status of economists at some economic research think tank:
Men (M)
Women (W) Total
Promoted (P) 100 20 120
Not Promoted (N) 500 100 600
Total 600 120 720
100( | ) 0.1667600
P P M 120( ) 0.1667720
P P
Independent Events
“Getting the promotion” and “being male” are independent events
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160( ) 0.1633980
P H
Example: Consider the hiring of black and white workers at BigMart
White (W)
Black (B) Total
Hired (H) 130 30 160
Not Hired (Hc) 570 250 820
Total 700 280 980
30( | ) 0.1071280
P H B
Independent Events
“Getting hired” and “being black” are not independent eventsdata_simpson.xls
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Bayes’ Theorem
Tells us about how the probability of something changeswhen we learn information.
For example, we know from a drug lab’s claim:
P(testing positive | employee is a druggie)
Since we fire druggies, we want to know:
P(employee is a druggie | testing positive)
To compute the latter we need additional information (e.g.,the false positive rate, prevalence of drug use among our employees)
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We want to ensure that our employees are not taking drugs because this is a safety risk.
We contract with a laboratory that claims their drug test is 94% accurate but there is a 5% chance of a false positive.
Suppose we have 10,000 employees and that 1% of them are druggies. If an employee is found to be a druggie, we fire them for safety reasons.
P = test is PositiveN = test is NegativeD = employee is a DruggieDc = employee is NOT a Druggie
Bayes’ Theorem
Example: Cocaine drug testing
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From the drug lab we know:P(P | D) = 0.94
Example: Cocaine drug testing
Bayes’ Theorem
(accuracy of the test)
P(N | D) = 0.06
P(P | Dc) = 0.05 (false positive rate)
P(N | Dc) = 0.95
We believe:P(D) = 0.01 (prevalence)
P(Dc) = 0.99
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P(N|D) = .06P(D) = .01
P(Dc) = .99P(P|Dc) = .05
P(N|Dc) = .95
P(P|D) = .94 P(P D) = .0094
P(P Dc) = .0495
P(N Dc) = .9405
P(N D) = .0006
Prevalence Drug Lab ExperimentalOutcomes
P(P) = .0589
P(N) = .9411
Bayes’ Theorem
Example: Cocaine drug testing
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Bayes’ Theorem
( )( )
( | )( | ) ( ) (| ) | )( c cP P D
P PP D P
P D PDP PD
DP
D
To find the (posterior) probability that an employee is a druggiegiven he tested positive, we apply Bayes’ theorem.
The drug lab’s accuracy claim
The proportion of our employees that are not
druggies.
The drug lab’s false positive rate.
Prevalence of drug use in our
company.
Example: Cocaine drug testing
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Q1 What is the probability a worker is a druggie given he tested positive for cocaine use?
(.01)(.94)(.01)(.94) (.99)(.05)
= .1596
Bayes’ Theorem
( ) ( | )( | ) ( ) ( | ) ( ) ( | )c cP D P P DP D P
P D P P D P D P P D
Example: Cocaine drug testing
.0094.0094 .0495
.0094
.0589
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Q2 What is the probability a worker isn’t a druggie given he tested positive for cocaine use?
Q3 What is the probability a worker is a druggie given he did not test positive for cocaine use?
Q4 What is the probability a worker is NOT a druggie given he did not test positive for cocaine use?
Bayes’ Theorem
Example: Cocaine drug testing