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Casinos are big business. If you don’t think so, research the top ten casinos in the world and check out their buildings. They are huge. Some people all over the world like to gamble. They like the anticipation of winning, not to mention the financial benefit when they do. The question is, if casinos “give away” so much prize money, how can they stay in business? How can they afford to build and maintain those extravagant entertainment complexes? Does it come down to randomness and probability?

So now is your chance! Think about a casino game and look beyond the glimmer of the lights and free drinks! What is behind the gambling? Is it really a “numbers game?”

Conduct some research on a casino game or other game of chance. If you aren’t into that sort of thing, research a state or country lottery or games where you can’t control the results. If you need help selecting a game, reach out to your instructor. Remember to include a summary of your research

Summarize the articles that you read and then answer the following questions:

Do you think the organization makes a profit even if there is a “payoff” to some lucky customer?

Do you believe the game is a worthwhile investment?

Would you “put money down” if given a chance?

Chapter 5:
Randomness
and
Probability
1
5.1 Random Phenomena and Probability
With random phenomena, we can’t predict the individual
outcomes, but we can hope to understand
characteristics of their long-run behavior.
For any random phenomenon, each attempt, or trial,
generates an outcome.
We use the more general term event to refer to
outcomes or combinations of outcomes.
5-2
5.1 Random Phenomena and Probability
The sample space is the collection of all possible outcomes.
We denote the sample space S or sometimes Ω.
The probability of an event is its long-run relative frequency.
Independence means that the outcome of one trial doesn’t
influence or change the outcome of another.
5-3
5.1 Random Phenomena and Probability
The Law of Large Numbers (LLN) states that if the events
are independent, then as the number of trials increases, the
long-run relative frequency of an event gets closer and
closer to a single value.
Empirical probability is based on repeatedly observing the
event’s outcome.
5-4
5.1 Random Phenomena and Probability
Many people confuse the Law of Large numbers with the
so-called Law of Averages.
Many people believe that an outcome of a random event
that hasn’t occurred in many trials is “due” to occur.
The Law of Averages doesn’t exist.
5-5
5.2 The Nonexistent Law of Averages
Many people confuse the Law of Large numbers with the
so-called Law of Averages.
Many people believe that an outcome of a random event
that hasn’t occurred in many trials is “due” to occur.
The Law of Averages doesn’t exist.
5-6
5.3 Different Types of Probability
Model-Based (Theoretical) Probability
Example: Pew Research reports that of 10,190 randomly
generated working phone numbers, the initial results of
the calls were as follows:
Since the phone numbers were generated randomly, each
was equally likely, and the probability of a contact is
simply 7400 / 10,190 = 0.7262.
5-7
5.3 Different Types of Probability
Personal Probability
A subjective, or personal probability expresses your
uncertainty about the outcome.
Although personal probabilities may be based on
experience, they are not based either on long-run
relative frequencies or on equally likely events.
5-8
5.4 Probability Rules
Rule 1
If the probability of an event occurring is 0, the event can’t
occur.
If the probability is 1, the event always occurs.
For any event A,
0 ï‚£ P( A) ï‚£.1
5-9
5.4 Probability Rules
Rule 2: The Probability Assignment Rule
The probability of the set of all possible outcomes must
be 1.
P(S) = 1
where S represents the set of all possible outcomes and
is called the sample space.
5-10
5.4 Probability Rules
Rule 3: The Complement Rule
The probability of an event occurring is 1 minus the
probability that it doesn’t occur.
P( A) = 1 − P( AC )
where the set of outcomes that are not in event A is
called the “complement” of A, and is denoted AC.
5-11
5.4 Probability Rules
For Example: Lee’s Lights sell lighting fixtures. Lee
records the behavior of 1000 customers entering the store
during one week. Of those, 300 make purchases. What is
the probability that a customer doesn’t make a purchase?
If P(Purchase) = 0.30
then P(no purchase) = 1 – 0.30 = 0.70
5-12
5.4 Probability Rules
Rule 4: The Multiplication Rule
For two independent events A and B, the probability that
both A and B occur is the product of the probabilities of the
two events.
P( A and B) = P( A) ï‚´ P(B)
provided that A and B are independent.
5-13
5.4 Probability Rules
For Example: If we can assume that customers behave
independently, what is the probability that the next two
customers entering Lee’s Lights both make purchases?
We can use the multiplication rule because the events are
independent:
P( A and B) = P( A) ï‚´ P(B)
P(first customer makes a purchase and second customer makes a purchase)
= P(purchase) x P(purchase)
= 0.30 x 0.30
= 0.09
5-14
5.4 Probability Rules
For Example: If we can assume that customers behave
independently, what is the probability that the next two
customers entering Lee’s Lights both make purchases?
We can use the multiplication rule because the events are
independent:
P( A and B) = P( A) ï‚´ P(B)
P(first customer makes a purchase and second customer makes a purchase)
= P(purchase) x P(purchase)
= 0.30 x 0.30
= 0.09
5-15
5.4 Probability Rules
For Example: Some customers prefer to see the
merchandise but then make their purchase online. Lee
determines that there’s an 8% chance of a customer
making a purchase in this way. We know that about 30%
of customers make purchases when they enter the store.
What is the probability that a customer who enters the
store makes no purchase at all?
P (purchase in the store or online)
= P (purchase in store) + P (purchase online)
= 0.30 + 0.09 = 0.39
P(no purchase) = 1 – 0.39 = 0.61
5-16
5.4 Probability Rules
Rule 6: The General Addition Rule
The General Addition Rule calculates the probability that
either of two events occurs. It does not require that the
events be disjoint.
P( A or B) = P( A) + P(B) − P( A and B)
5-17
5.4 Probability Rules
For Example: Lee notices that when two customers enter
the store together, their purchases are not disjoint. In fact,
there’s a 20% they’ll both make a purchase. When two
customers enter the store together, what is the probability
that at least one of them will make a purchase?
P(at least one purchases) =
P(A purchases or B purchases) =
P(A purchases) + P(B purchases)
– P(A and B both purchase) =
0.30 + 0.30 – 0.20 = 0.40
5-18
5.4 Probability Rules
Example: Car Inspections
You and a friend get your cars inspected. The event of your
car’s passing inspection is independent of your friend’s car. If
75% of cars pass inspection what is the probability that
Your car passes inspection?
Your car doesn’t pass inspection?
Both cars pass inspection?
At least one of two cars passes?
Neither car passes inspection?
5-19
5.4 Probability Rules
Example (continued): Car Inspections
You and a friend get your cars inspected. The event of your
car’s passing inspection is independent of your friend’s car.
If 75% of cars pass inspection what is the probability that
Your car passes inspection?
P(Pass) = 0.75
Your car doesn’t pass inspection?
P(PassC) = 1-0.75 = 0.25
Both cars pass inspection?
P(Pass)P(Pass) = (0.75)(0.75) = 0.5625
5-20
5.4 Probability Rules
Example (continued): Car Inspections
You and a friend get your cars inspected. The event of
your car’s passing inspection is independent of your
friend’s car. If 75% of cars pass inspection what is the
probability that
At least one of two cars passes?
1 – (0.25)2 = 0.9375 OR
0.75 + 0.75 – 0.5625 = 0.9375
Neither car passes inspection?
1 – 0.9375 = 0.0625
5-21
5.5 Joint Probability and
Contingency Tables
Events may be placed in a contingency table such as the
one in the example below.
As part of a Pick Your Prize Promotion, a store invited
customers to choose which of three prizes they’d like to win.
The responses could be placed in the following contingency
table:
5-22
5.5 Joint Probability and
Contingency Tables
Marginal probability depends only on totals found in
the margins of the table.
5-23
5.5 Joint Probability and
Contingency Tables
In the table below, the probability that a respondent
chosen at random is a woman is a marginal probability.
P(woman) = 251/478 = 0.525.
5-24
5.5 Joint Probability and
Contingency Tables
Joint probabilities give the probability of two events
occurring together.
P(woman and camera) = 91/478 = 0.190.
5-25
5.5 Joint Probability and
Contingency Tables
Each row or column shows a conditional distribution given
one event.
In the table above, the probability that a selected customer
wants a bike given that we have selected a woman is:
P(bike | woman) = 30/251 = 0.120.
5-26
5.6 Conditional Probability and the
General Multiplication Rule
In general, when we want the probability of an event from a
conditional distribution, we write P(B|A) and pronounce it
“the probability of B given A.”
A probability that takes into account a given condition is
called a conditional probability.
P(A and B)
P(B | A) =
P(A)
5-27
5.6 Conditional Probability and the
General Multiplication Rule
Rule 7: The General Multiplication Rule
The General Multiplication Rule calculates the
probability that both of two events occurs. It does not
require that the events be independent.
P( A and B) = P( A) ï‚´ P(B | A)
5-28
5.6 Conditional Probability and the
General Multiplication Rule
What is the probability that a randomly selected customer
wants a bike if the customer selected is a woman?
P ( A and B)
P (B | A ) =
= P(bike|woman)
P( A)
P (bike and woman) 30 478
=
=
= 0.12
P(woman)
251 478
5-29
5.6 Conditional Probability and the
General Multiplication Rule
Events A and B are independent whenever P(B|A) = P(B).
This means knowing event A has occurred has no impact on
the probability of event B occurring.
5-30
5.6 Conditional Probability and the
General Multiplication Rule
Are Prize preference and Sex independent?
If so, P(bike | woman) will be the same as P(bike).
P(bike | woman)= 30/251 = 0.12
P(bike) = 90/478 = 0.265
0.12 ≠ 0.265
Since the two probabilities are not equal, Prize preference and
Sex and not independent.
5-31
5.6 Conditional Probability and the
General Multiplication Rule
Independent vs. Disjoint
For all practical purposes, disjoint events cannot be
independent.
Don’t make the mistake of treating disjoint events as if they
were independent and applying the Multiplication Rule for
independent events.
5-32
5.6 Conditional Probability and the
General Multiplication Rule
5-33
5.7 Constructing Contingency Tables
If you’re given probabilities without a contingency table,
you can often construct a simple table to correspond to the
probabilities and use this table to find other probabilities.
5-34
5.7 Constructing Contingency Tables
A survey classified homes into two price categories (Low
and High). It also noted whether the houses had at least 2
bathrooms or not (True or False). 56% of the houses had
at least 2 bathrooms, 62% of the houses were Low
priced, and 22% of the houses were both.
Translating the percentages to probabilities, we have:
5-35
5.7 Constructing Contingency Tables
The 0.56 and 0.62 are marginal probabilities, so they go in
the margins.
The 22% of houses that were both Low priced and had at
least 2 bathrooms is a joint probability, so it belongs in the
interior of the table.
5-36
5.7 Constructing Contingency Tables
Because the cells of the table show disjoint events, the
probabilities always add to the marginal totals going
across rows or down columns.
5-37
5.8 Probability Trees
Some business decisions involve more subtle evaluation
of probabilities.
Given the probabilities of various states of nature, we can
use a picture called a probability tree or tree diagram to
help think through the decision-making process.
5-38
5.8 Probability Trees
Some business decisions involve more subtle evaluation
of probabilities.
Given the probabilities of various states of nature, we can
use a picture called a probability tree or tree diagram to
help think through the decision-making process.
5-39
5.8 Probability Trees
Example:
Personal electronic devices, such as smart phones and
tablets, are getting more capable all the time. Microscopic
and even submicroscopic flaws that can cause intermittent
performance failures can develop during their manufacture.
Defects will always occur, so the quality engineer in charge
of the production process must monitor the number of
defects and take action if the process seems out of control.
5-40
5.8 Probability Trees
Example:
The probability tree on the next slide shows that a
manufacturing problem can occur with the motherboard, the
memory, or the case alignment for an electronic device.
The next set of branches denote the probability that a minor
adjustment will fix each type of problem.
5-41
5.8 Probability Trees
5-42
5.8 Probability Trees
Example:
At the end of each branch, joint probabilities can be
calculated by applying the multiplication rule for
independent events.
5-43
5.8 Probability Trees
5-44
5.9 Reversing the Conditioning: Bayes’
Rule
When you need to find reverse conditional probabilities,
we recommend drawing a tree and finding the
appropriate probabilities by using the definition of
conditional probability (see text).
5-45
5.9 Reversing the Conditioning: Bayes’
Rule
In general, we can write for n events Ai that are mutually
exclusive (each pair is disjoint) and exhaustive (their union is
the whole sample space):
5-46
WHAT CAN GO WRONG?
•
Beware of probabilities that don’t add up to 1.
•
Don’t add probabilities of events if they’re not
disjoint.
• Don’t multiply probabilities of events if they’re not
independent.
•
Don’t confuse disjoint and independent.
5-47
FROM LEARNING TO EARNING
Apply the facts about probability to determine whether
an assignment of probabilities is legitimate.
• Probability is long-run relative frequency.
• Individual probabilities must be between 0 and 1.
• The sum of probabilities assigned to all outcomes
must be 1.
Understand the Law of Large Numbers and that the
common understanding of the “Law of Averages” is
false.
5-48
FROM LEARNING TO EARNING
Know the rules of probability and how to apply them.
•
•
•
•
•
The Complement Rule says that P(A) = 1 – P(Ac).
The Multiplication Rule for independent events say
that
P(A and B) = P(A)*P(B), provided A and B are
independent.
The General Multiplication Rule says that
P(A and B) = P(A)*P(B|A).
The Addition Rule for disjoint events says that
P(A or B) = P(A) + P(B), provided events A and B
are disjoint.
The General Addition Rule says that
P(A or B) = P(A) + P(B) – P(A and B).
5-49
FROM LEARNING TO EARNING
Know how to construct and read a contingency table.
Know how to define and use independence.
• Events A and B are independent if P(B|
A) = P(B).
Know how to construct tree diagrams and use them to
calculate probabilities.
Know how to use Bayes’ Rule to compute conditional
probabilities.
5-50
Chapter 6:
Random
Variables
and
Probability
Models
1
6.1 Expected Value of a Random
Variable
A variable whose value is based on the outcome of a
random event is called a random variable.
If we can list all possible outcomes, the random variable is
called a discrete random variable.
If a random variable can take on any value between two
values, it is called a continuous random variable.
6-2
6.1 Expected Value of a Random
Variable
For both discrete and continuous random variables, the set
of all the possible values and their associated probabilities is
called the probability model.
When the probability model is known, then the expected
value can be calculated:
E ( X ) =  x  P ( x ) (discrete random variable)
E ( X ) can be written as  but never as x or y
6-3
6.1 Expected Value of a Random
Variable
The probability model for a particular life insurance policy
is shown. Find the expected annual payout on a policy.
6-4
6.1 Expected Value of a Random
Variable
The probability model for a particular life insurance policy
is shown. Find the expected annual payout on a policy.
 1 
 2 
 997 
 = E ( X ) = $100, 000 
 + $50, 000 
 + $0 

1000
1000
1000






= $200
We expect that the insurance company will pay out $200
per policy per year.
6-5
6.1 Expected Value of a Random
Variable
Standard Deviation of a discrete random variable:
 = Var ( X ) =  ( x −  )  P ( x )
2
2
 = SD ( X ) = Var ( X )
6-6
6.2 Standard Deviation of a Random
Variable
The probability model for a particular life insurance policy is
shown. Find the standard deviation of the annual payout.
6-7
6.2 Standard Deviation of a Random
Variable
Example: Book Store Purchases
Suppose the probabilities of a customer purchasing 0, 1, or
2 books at a book store are 0.2, 0.4 and 0.4 respectively.
What is the expected number of books a customer will
purchase?
What is the standard deviation of the book purchases?
6-8
6.2 Standard Deviation of a Random
Variable
Example: Book Store Purchases
 = 0 ( 0.2 ) + 1( 0.4 ) + 2 ( 0.4 )
= 1.2
 2 = ( x −  )  P ( x)
2
= (0 – 1.2) 2 (0.2) + (1 – 1.2) 2 (0.4) + (2 – 1.2) 2 (0.4)
= 0.288 + 0.016 + 0.256 = 0.56
 = 0.56 = 0.748
6-9
6.3 Properties of Expected Values and
Variances
Adding a constant c to X:
Multiplying X by a constant a:
6-10
6.3 Properties of Expected Values and
Variances
Addition Rule for Expected Values of Random Variables
Addition Rule for Variances of (independent) Random
Variables
6-11
6.3 Properties of Expected Values and
Variances
The expected annual payout per insurance policy is
$200 and the variance is $14,960,000.
If the payout amounts are doubled, what are the new
expected value and variance?
E ( 2 X ) = 2 E ( X ) = 2  200 = $400
Var ( 2 X ) = 22 Var ( X ) = 4 14,960,000 = 59,840,000
6-12
6.3 Properties of Expected Values and
Variances
Compare this to the expected value and variance on two
independent policies at the original payout amount.
E ( X + Y ) = E ( X ) + E (Y ) = 2  200 = $400
Var ( X + Y ) = Var ( X ) + Var (Y ) = 2 14,960,000 = 29,920,00
Note: The expected values are the same but the variances are different.
6-13
6.3 Properties of Expected Values and
Variances
The association of two random variables can be measured
using the covariance of X and Y:
Cov ( X , Y ) = E ( ( X −  )(Y − ) )
Then, the covariance gives us the extra information needed
to find the variance of the sum or difference of two random
variables when they are not independent:
Var ( X  Y ) = Var ( X ) + Var (Y )  2Cov ( X , Y )
6-14
6.3 Properties of Expected Values and
Variances
Covariance doesn’t have to be between –1 and +1, which
makes it harder to interpret. To fix this “problem”, we can
divide the covariance by each of the standard deviations to
get the correlation:
Corr ( X , Y ) =
Cov ( X , Y )
 XY
6-15
6.4 Bernoulli Trials
Bernoulli trials have the following characteristics:
• There are only two outcomes per trial, called success
and failure.
• The probability of success, called p, is the same on
every trial (the probability of failure, 1 – p, is often called
q).
• The trials are independent.
6-16
6.4 Bernoulli Trials
Examples of Bernoulli trials include tossing a coin, collecting
yes / no responses from a survey, or shooting free throws in
basketball.
When using Bernoulli trials to develop probability models, we
require that trial are independent.
We can check the 10% Condition: As long as the number of
trials or sample size is less than 10% of the population size,
we can proceed with confidence that the trials are
independent.
6-17
6.5 Discrete Probability Models
The Uniform Model
If X is a random variable with possible outcomes 1, 2, …, n
and P ( X = i ) = 1 / n for each i, then we say X has a discrete
Uniform distribution U[1, …, n].
When tossing a fair die, each number is equally likely to
occur. So tossing a fair die is described by the Uniform
model
U[1, 2, 3, 4, 5, 6], with P ( X = i ) = 1/ 6.
6-18
6.5 Discrete Probability Models
The Geometric Model
Predicting the number of Bernoulli trials required to achieve
the first success.
6-19
6.5 Discrete Probability Models
The Binomial Model
Predicting the number of successes in a fixed number of
Bernoulli trials.
6-20
6.5 Discrete Probability Models
The Binomial Model
Predicting the number of successes in a fixed number of
Bernoulli trials.
6-21
6.5 Discrete Probability Models
Example: Probability in customer acquisition
A venture capital firm has a list of potential investors who
have previously invested in new technologies. On average,
these investors invest about 5% of the time. A new client
of the firm is interested in finding investors for a mobile
phone application that enables financial transactions, an
application that is finding increasing acceptance in much of
the developing world. An analyst at the firm is about to start
calling potential investors.
6-22
6.5 Discrete Probability Models
Example: Closing Sales
A salesman normally closes a sale on 80% of his
presentations. Assuming the presentations are
independent:
What model should be used to determine the probability
that he closes his first presentation on the fourth attempt?
What is the probability he closes his first presentation on
the fourth attempt?
6-23
6.5 Discrete Probability Models
Example (continued): Closing Sales
What model should be used to determine the probability that
he closes his first presentation on the fourth attempt?
Geometric
What is the probability he closes his first presentation on the
fourth attempt?
P ( X = 4) = (0.20) (0.80) = 0.0064
3
6-24
6.5 Discrete Probability Models
Example: Professional Tennis
A tennis player makes a successful first serve 67% of the
time. Of the first 6 serves of the next match:
What model should be used to determine the probability that
all 6 first serves will be in bounds?
What is the probability that all 6 first serves will be in bounds?
How many first serves can be expected to be in bounds?
6-25
6.5 Discrete Probability Models
Example (continued): Professional Tennis
What model should be used to determine the probability
that all 6 first serves will be in bounds? Binomial
What is the probability that all 6 first serves will be
inbounds?
 6
P( X = 6) =   (0.67)6 (0.33)0 = 0.0905
 6
How many first serves can be expected to be inbounds?
E(X) = np = 6(0.67) = 4.02
6-26
6.5 Discrete Probability Models
Example (continued): Professional Tennis
What model should be used to determine the probability that
all 6 first serves will be in bounds? Binomial
What is the probability that all 6 first serves will be inbounds?
 6
P( X = 6) =   (0.67)6 (0.33)0 = 0.0905
 6
How many first serves can be expected to be inbounds?
E(X) = np = 6(0.67) = 4.02
6-27
6.5 Discrete Probability Models
Example (continued): Satisfaction Survey
What probability model would be used to model the selection
of a single number? Uniform, all numbers are equally likely.
What is the probability the number selected will be
an even number? 0.5
What is the probability the number selected will
end in 000?
0.001
6-28
6.5 Discrete Probability Models
Example: Probability in customer acquisition
A venture capital firm has a list of potential investors who
have previously invested in new technologies. On average,
these investors invest about 5% of the time. A new client
of the firm is interested in finding investors for a mobile phone
application that enables financial transactions, an application
that is finding increasing acceptance in much of the
developing world. An analyst at the firm is about to start
calling potential investors.
6-29
6.5 Discrete Probability Models
Example: (Continued) Probability in customer acquisition
What is the probability that the first person she calls will want
to invest?
How many investors will she have to call, on average, to find
someone interested?
6-30
6.5 Discrete Probability Models
Example: (Continued) Probability in customer acquisition
What is the probability that the first person she calls will want
to invest?
Each investor has a 5% or 1/20 chance of wanting to invest,
so the chance that the first person she calls is interested is
1/20.
How many investors will she have to call, on average, to find
someone interested?
This uses a Geometric model. Let X = number of people she
calls until the first interested person.
E( X ) = 1 p – 1/ (1/ 20) = 20 people.
6-31
6.5 Discrete Probability Models
Example: (Continued) Probability in customer acquisition
If she calls 10 investors, what is the probability that exactly 2
of them will be interested?
Using the Binomial model, let Y = number of people
interested in 10 calls, then
10 2
10 ´ 9
8
P(Y = 2) = ( ) p (1- p) =
(1/ 20)2 (19 / 20)8 = 0.0746
2
2
What assumptions are you making to answer these
questions?
We are assuming that the trials are independent and that the
probability of being interested in investing is the same for all
potential investors.
6-32
WHAT CAN GO WRONG?
• Probability distributions are still just models.
• If the model is wrong, so is everything else.
• Watch out for variables that aren’t independent.
• Don’t write independent instances of a random
variable with notation that looks like they are the
same variables.
6-33
WHAT CAN GO WRONG?
• Don’t forget: Variances of independent random
variables add. Standard deviations don’t.
• Don’t forget: Variances of independent random
variables add, even when you’re looking at the
difference between them.
• Be sure you have Bernoulli trials.
6-34
FROM LEARNING TO EARNING
Apply the facts about probability to determine whether
an assignment of probabilities is legitimate.
• Probability is long-run relative frequency.
• Individual probabilities must be between 0 and 1.
• The sum of probabilities assigned to all outcomes
must be 1.
Understand the Law of Large Numbers and that the
common understanding of the “Law of Averages” is
false.
6-35
FROM LEARNING TO EARNING
Understand how probability models relate values to
probabilities.
• For discrete random variables, probability models
assign a probability to each possible outcome.
Know how to find the mean, or expected value, of a
discrete probability model and the standard
deviation:
E ( X ) =  x  P ( x)
 =
 (x −  )  P (x )
2
6-36
FROM LEARNING TO EARNING
Foresee the consequences of shifting and scaling
random variables:
E(X ± c) = E(X) ± c
E(aX) = aE(X)
Var(X ± c) = Var(X)
Var(aX) = a2Var(X)
SD(X ± c) = SD(X)
SD(aX) = |a|SD(X)
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FROM LEARNING TO EARNING
Understand that when adding or subtracting random
variables the expected values add or subtract well:
E(X ± Y) = E(X) ± E(Y).
However, when adding or subtracting independent
random variables, the variances add:
Var(X ±Y) = Var(X) + Var(Y)
Be able to explain the properties and parameters of the
Uniform, the Binomial, the Geometric, and the Poisson
distributions.
6-38

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