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Sets and Probability

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Title: Sets and Probability


1
Sets and Probability
  • Chapter 7

2
Ch. 7 Sets and Probabilities
  • 7.1 Sets
  • 7.2 Applications of Venn Diagrams
  • 7.3 Introduction to Probability
  • 7.4 Basic Concepts of Probability
  • 7.5 Conditional Probability Independent Events
  • 7.6 Bayes Theorem

3
7.1 Sets
  • Set A well defined collection of objects.
  • The set consisting of all students in the class.
  • The set consisting of all the coins produced by
    the U.S. government.
  • The set consisting of the numbers 3, 4, and 5
  • 3, 4, 5
  • 4 is an element of the set 3, 4, 5
  • 4?3, 4, 5
  • 8?3, 4, 5
  • n(A) the number of elements in finite set A
  • A 3, 4, 5, n(A) 3

4
7.1 Sets
  • Two sets are equal if they contain the same
    elements
  • 3, 4, 5 5, 4, 3 4, 3, 5 3, 3, 4, 5,
    5
  • Subsets
  • Set A is a subset of set B (A ? B) if every
    element of A is also an element of B.
  • Set A is a proper subset (A ? B) if A ? B and A ?
    B.
  • A (3, 4, 5, 6, B 2, 3, 4, 5, 6, 7, 8)
  • A ? B
  • The Universal Set (U)
  • The set that contains all objects under
    consideration.

5
Relationships Among Sets Venn Diagrams
  • (A ? B)

U
B
A
6
Relationships Among Sets Venn Diagrams
  • Complement of a set
  • Set A contains the elements of U that are not in
    A

U
A'
A
7
Relationships Among Sets Venn Diagrams
  • Intersection of Two Sets
  • A ? B

U
B
A
8
Relationships Among Sets Venn Diagrams
  • Union of Two Sets
  • A ? B

U
B
A
9
Relationships Among Sets Venn Diagrams
  • Disjoint Sets
  • A ? B 0

U
B
A
10
7.2 Applications of Venn Diagrams
  • One set leads to 2 regions

U
2
A
1
11
7.2 Applications of Venn Diagrams
  • One set leads to 2 regions
  • Region 1 A
  • Region 2 A

U
2
A
1
12
Applications of Venn Diagrams
  • Two sets leads to 3 regions

U
A
B
2
3
1
13
Applications of Venn Diagrams
  • Two sets leads to 3 regions
  • Or...

U
A
B
2
3
1
14
Applications of Venn Diagrams
  • Two sets leads to 3 regions
  • Or...

U
A
3
3
2
B
1
15
Applications of Venn Diagrams
  • Two sets leads to 4 regions

U
A
B
3
4
2
1
16
Applications of Venn Diagrams
  • 3 sets, 8 regions

U
A
3
B
2
8
4
5
7
1
6
C
17
Example
  • 140 people in a suburban shopping center were
    interviewed to discover some of their cooking
    habbits.
  • 58 use microwave ovens
  • 63 use electric ranges
  • 58 use gas ranges
  • 19 use microwave ovens and electric ranges
  • 17 use microwave ovens gas ranges
  • 4 use both gas and electric ranges
  • 1 uses all three
  • 2 use none of the three

18
Example
U
M
18
E
23
41
1
16
3
2
38
G
19
7.3 Introduction to Probability
  • Probability is a numerical measure of the
    likelihood that an event will occur.
  • Probability values are always assigned on a scale
    from 0 to 1.
  • A probability near 0 indicates an event is very
    unlikely to occur.
  • A probability near 1 indicates an event is almost
    certain to occur.
  • A probability of 0.5 indicates the occurrence of
    the event is just as likely as it is unlikely.

20
An Experiment and Its Sample Space
  • An experiment is an activity or occurrence with
    an observable result.
  • Each repetition of an experiment is called a
    trial.
  • The sample space for an experiment is the set of
    all experimental outcomes.
  • An event is a subset of the sample space.

21
An Experiment and Its Sample Space
  • An experiment consists of studying the number of
    boys and girls in families with exactly 3
    children.
  • Let b represent boy and g represent girl.
  • Event B the family has only boys. B b, b, b
  • Event H the family has exactly two girls

Sample Space
22
Set Operations for Events
  • Let E and F be events for a sample space S.
  • E ? F occurs when both E and F occur
  • E ? F occurs when E or F or both occur
  • E occurs when E does not occur.
  • Mutually Exclusive Events
  • E and F are mutually exclusive if E ? F Ø.

23
Basic Probability Principle
  • For sample spaces with equally likely outcomes,
    the probability of an event is defined as follows.

BASIC PROBABILITY PRINCIPLE Let S be a sample
space of equally likely outcomes, and let Event E
be a subset of S. Then the probability that E
occurs is
24
Basic Probability Principle
If a single playing card is drawn at random from
a standard 52-card deck, find the probability of
each event
  • Drawing an ace

25
Basic Probability Principle
If a single playing card is drawn at random from
a standard 52-card deck, find the probability of
each event
  • Drawing a spade
  • Drawing a spade or a heart

26
7.4 Basic Concepts of Probability
  • Union (Addition) Rule for Probability
  • Union Rule for Mutually Exclusive Events
  • Complement Rule
  • Odds

27
Union of Two Events
  • The union of events A and B is the event
    containing all sample points that are in A or B
    or both.
  • The union is denoted by A ??B?
  • The union of A and B is illustrated below.
  • P(A ??B) The probability of the occurrence of
    Event A or Event B.

Sample Space S
28
Addition Rule
  • The addition rule provides a way to compute the
    probability of event A, or B, or both A and B
    occurring.
  • The rule is written as
  • P(A ??B) P(A) P(B) - P(A ? B?

Event A
29
Addition Rule
  • The addition rule provides a way to compute the
    probability of event A, or B, or both A and B
    occurring.
  • The rule is written as
  • P(A ??B) P(A) P(B) - P(A ? B?

Event B
30
Addition Rule
  • The addition rule provides a way to compute the
    probability of event A, or B, or both A and B
    occurring.
  • The rule is written as
  • P(A ??B) P(A) P(B) - P(A ? B?

Event A
Event B
31
Union of Two Events
  • ADDITION RULE FOR PROBABILITY
  • For any events E and F from sample space S

Consider an assembly plant with 50 employees.
Each worker is expected to complete work
assignments on time and in such a way that the
assembled product will pass a final inspection.
On occasion, some of the workers fail to meet the
performance standards by completing work late
and/or assembling a defective product. Recently,
the manager found that 5 workers had completed
work late, 6 had assembled a defective product,
and 2 had both completed work late and assembled
a defective product.
32
Union of Two Events
  • Let
  • L the event that the work is completed late
  • D the event that the assembled product is
    defective

The manager decided to assign a poor performance
rating to any employee whose work was either late
or defective. What is the probability an employee
was given a poor performance rating?
33
Addition Rule forMutually Exclusive Events
  • Two events are said to be mutually exclusive if
    the events have no sample points in common. That
    is, two events are mutually exclusive if, when
    one event occurs, the other cannot occur.
  • Addition Rule for Mutually Exclusive Events
  • P(A ??B) P(A) P(B)

Event A
Event B
34
Roll the Dice
  • If you roll 2 dice, whats the probability of
    rolling a 7 or 11?

35
Die 1
Die 2
P(7) 6/36 .167
P(11) 2/36 .056
36
Roll the Dice
  • If you roll 2 dice, whats the probability of
    rolling a 7 or 11?

37
Complement of an Event
  • The complement of event A is defined to be the
    event consisting of all sample points that are
    not in A.
  • The complement of A is denoted by A.

Event A
A
COMPLEMENT RULE P(E) 1 P(E) and
P(E) 1 P(E)
38
7.5 Conditional Probability
  • The probability of an event given that another
    event has occurred is called a conditional
    probability.
  • The conditional probability of A given B is
    denoted by P(AB).

39
Conditional Probability
40
Joint Probability Table
P(A ??M)
P(M)
41
Joint Probability Table
Joint probabilities
42
Joint Probability Table
Marginal probabilities
43
Joint Probability Table
44
Conditional Probability
  • The probability of an event given that another
    event has occurred is called a conditional
    probability.
  • The conditional probability of A given B is
    denoted by P(AB).
  • A conditional probability is computed as follows
  • If P(AB) 0, then event A and event B are
    mutually exclusive.

45
Example Conditional Probability
  • Conditional Probability
  • Officer promoted given the officer is a man
  • Officer promoted given the officer is a woman

46
Intersection of Two Events
  • The intersection of events A and B is the set of
    all sample points that are in both A and B.
  • The intersection is denoted by A ????
  • The intersection of A and B is the area of
    overlap in the illustration below.
  • P(A ???) The probability of the occurrence of
    Event A and Event B.

Sample Space S
Intersection
Event A
Event B
47
Product Rule
  • The product rule provides a way to compute the
    probability of an intersection of two events.
  • The law is written as

Event A
Event B
48
Product Rule for Independent Events
  • Events A and B are independent if P(AB) P(A).
  • Multiplication Law for Independent Events
  • P(A ? B) P(A)P(B)
  • The product rule also can be used as a test to
    see if two events are independent.

49
You are given the following information on Events
A, B,
C, and D.
È
P(A) .4
P(A
D) .6
ô
P(B) .2
P(A
B) .3
Ç
P(C) .1
P(A
C) .04
Ç
P(A
D) .03
b. Compute P(A ? B)
50
7.5 Bayes Theorem
  • The probability of an event A B is generally
    different from the probability of B A.
  • However, there is a definite relationship between
    the two.
  • Bayes' theorem is the statement of that
    relationship.
  • Medical researchers know that the probability of
    getting lung cancer if a person smokes is .34.
  • The probability that a nonsmoker will get lung
    cancer is .03.
  • With Bayes theorem, we can calculate the
    probability that a person with lung cancer is (or
    was) a smoker.

51
Bayes Theorem
  • Prior Probabilities
  • Let

S Person is a smoker N Person is a
non-smoker
According to the Center for Disease Control and
Prevention, approximately 22 of the population
18 years or older smokes tobacco products
regularly.
P(S) .22, P(N) .78
52
Bayes Theorem
  • Conditional Probabilities
  • Let

C Person has (or will have) lung cancer H
Person will not have lung cancer
Based upon medical research
P(CS) .34
P(CN) .03
P(HS) .66
P(HN) .97
Hence
53
Bayes Theorem
  • We can illustrate the different possible outcomes
    with a tree diagram (2-step experiment).

Step 2 Health
Step 1 Smoker or non-smoker
Experimental Outcomes
C
(S, C)
S
(S, H)
H
C
(N, C)
N
(N, H)
H
54
Bayes Theorem
  • Now we can fill in the probabilities

Step 2 Health
Step 1 Smoker or non-smoker
Experimental Outcomes
P(CS) .34
P(S ? C) P(S)P(CS) .07
P(S) .22
P(S ? H) .15
P(HS) .66
P(CN) .03
P(N ? C) .02
P(N) .78
P(N ? H) .76
P(HN) .97
55
Bayes Theorem
  • Now suppose we want to determine the probability
    that a person who has been diagnosed with lung
    cancer is a smoker. In other words,
  • From the law of conditional probabilities, we
    know that
  • From the probability tree, we know that
  • Event C can occur in only two ways (S ? C) and
    (N ? C)

Posterior probability
Equation 1
Equation 2
56
Bayes Theorem
Bayes Theorem (2 events)
57
Bayes Theorem
  • To find the posterior probability that event
    Ai will
  • occur given that event B has occurred, we
    apply
  • Bayes theorem.
  • Bayes theorem is applicable when the events
    for
  • which we want to compute posterior
    probabilities
  • are mutually exclusive and their union is
    the entire
  • sample space.

58
Bayes Theorem, example
  • A local bank reviewed its credit card policy with
    the intention of recalling some of its credit
    cards. In the past, approximately 5 of
    cardholders defaulted, leaving the bank unable to
    collect the outstanding balance. Hence,
    management established a prior probability of .05
    that any particular cardholder will default. The
    bank also found that the probability of missing a
    monthly payment is .20 for customers who do not
    default. Of course, the probability of missing a
    monthly payment for those who default is 1.
  • Given that a customer missed a monthly payment,
    compute the posterior probability that the
    customer will default.

59
Bayes Theorem
  • M missed payment
  • D1 customer defaults
  • D2 customer does not default
  • P(D1) .05 P(D2) .95 P(MD2) .2
    P(MD1) 1

60
End of Chapter 4
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