Lecture note 4

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Lecture note 4

• Probabilities

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Topics to be covered

• Basic set notations.
• Basic properties of probabilities
• Bivariate Probabilities
• Conditional Probabilities
• Statistical Independence

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Sample Space

• Suppose that you roll a die once. There will be 6
  possible outcomes you may get either 1, 2, 3, 4,
  5 or 6. These possible outcomes of such a random
  experiment are called the basic outcomes. The set
  of all basic outcomes is called the sample space.
  The symbol S will be used to denote the sample
  space.

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Sample Space- An Example -

• What is the sample space for a roll of a single
  six-sided die?

S 1, 2, 3, 4, 5, 6
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An event

• A subset of a sample space is called an event. We
  usually use a capital letter to denote an event.
  Taking the rolling-of-a-die for example, A1, 2
  is an event.
• The meaning of an event is important. A1, 2
  means that this is the event that you get either
  1 or 2.
• B1,4,6 is the event that you get either 1, 4
  or 6.

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Some set notations-Intersection-

• Suppose you roll a die. Then the sample space is
  S1,2,3,4,5,6.
• Consider the following two events A1,3,5 and
  B1,3,6.
• Then , AnB is defined as an event that
  consists of the basic outcomes that are common to
  both A and B.
• AnB reads A intersection B.
• Exercise What is AnB?

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Some set notation-Union-

• Suppose you roll a die. Then the sample space is
  S1,2,3,4,5,6.
• Consider the following two events A1,3,5 and
  B1,3,6.
• A ? B is defined as an event that consists of
  the basic outcomes that are either in A or B.
• A ? B reads A union B.
• Exercise What is A ? B?

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Mutually Exclusive Events

• If the events A and B have no common basic
  outcomes, they are called mutually exclusive and
  their intersection A ? B is said to be the empty
  set indicating that A ? B cannot occur.
• More generally, the K events E1, E2, . . . , EK
  are said to be mutually exclusive if every pair
  of them is a pair of mutually exclusive events.

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Mutually exclusive events-Example 1-

• Suppose you roll a die. Then the sample space is
  S1, 2, 3, 4, 5, 6. Now, consider the following
  events. A1, 2, 3 and B4, 5, 6
• Then there is no common basic outcome in event A
  and B. Therefore, A and B are mutually exclusive
  events.

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Mutually exclusive events-Example 2-

• Consider you roll a die. Then the sample space is
  S1, 2, 3, 4, 5, 6. Now, consider the following
  3 events. E11, 2 , E23, 4 and E35,6
• Then there is no common basic outcome in any
  pair of events. Therefore, E1, E2 and E3 are
  mutually exclusive sets.

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Venn Diagrams

• Venn Diagrams are drawings, usually using
  geometric shapes, used to depict basic concepts
  in set theory and the outcomes of random
  experiments.

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Intersection of Events A and B
S
S
A
B
A
B
A?B
(a) A?B is the striped area
(b) A and B are Mutually Exclusive
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Union of events A and B
S
A
B
A ? B

• A ? B is the striped area.

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Collectively Exhaustive Events

• Given the K events E1, E2, . . ., EK in the
  sample space S. If E1 ? E2 ? . . . ?EK S,
  these events are said to be collectively
  exhaustive.

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Collectively Exhaustive Events-Example-

• Consider rolling a die. Then the sample space is
  S1, 2, 3, 4, 5, 6. Further consider the
  following 3 events. E11, 2, 3 , E22, 3, 4
  and E34,5,6
• Then E1, E2 and E3 are collectively exhaustive
  events since E1 ? E2 ? E3 1,2,3,4,5,6S

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Complement

• Let A be an event in the sample space S. The set
  of basic outcomes of a random experiment
  belonging to S but not to A is called the
  complement of A and is denoted by A.

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Venn Diagram for the Complement of Event A
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Unions, Intersections, and Complements(Example
4.3)
A die is rolled. Let A be the event Number
rolled is even and B be the event Number rolled
is at least 4. Then A 2, 4, 6 and B
4, 5, 6
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Classical Probability

• The classical definition of probability is the
  proportion of times that an event will occur,
  assuming that all outcomes in a sample space are
  equally likely to occur. The probability of an
  event is determined by counting the number of
  outcomes in the sample space that satisfy the
  event and dividing by the number of outcomes in
  the sample space.

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Classical Probability

• The probability of an event A is
• where NA is the number of outcomes that satisfy
  the condition of event A and N is the total
  number of outcomes in the sample space.

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Classic Probability-Example-

• A die is rolled. Let A be the event Number
  rolled is even
• Then A2,4,6. Therefore,
• NA3, and N6. Therefore,
• P(A)3/60.5

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Probability Postulates

• Let S denote the sample space of a random
  experiment, Oi, the basic outcomes, and A, an
  event. For each event A of the sample space S,
  we assume that a number P(A) is defined and we
  have the postulates
• If A is any event in the sample space S
• Let A be an event in S, and let Oi denote the
  basic outcomes. Then
• 
  
  
  
  where the notation implies that the summation
  extends over all the basic outcomes in A.
• 3. P(S) 1

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Probability Postulate -Example for Postulate 2-

• Consider a roll of a die. Let A be the event
  Number rolled is even. Then, A2, 4, 6 and
  P(A)0.5.
• The notation in the postulate 2 means,
  O12, O24 and O36, and

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Bivariate Probabilities

• Bivariate Probabilities is the intersection
  probabilities of two distinct sets of events.

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Bivariate Probabilities-Example-

• Consider that you are an advisor for a particular
  TV show. You want to know both the income and
  other characteristics of the audience of the
  show. You can consider the following 2 distinct
  sets of events about the potential audiences.
• Next Slide

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Bivariate Probabilities-Example- Contd

• The first set of events is the following.
• A1Regular watcher
• A2Occasional watcher
• A3Never Watch
• The second set of the events is
• B1High income
• B2Middle income
• B3Low income

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Bivariate Probabilities, -Example- contd

• Then, the Bivariate Probabilities of the two sets
  of events, A1, A2, A3 and B1, B2, B3 can be
  represented by the following table.

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Joint probabilities

• In the context of bivariate probabilities, the
  intersection probabilities P(Ai ? Bj) are called
  joint probabilities.

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Joint Probabilities for the Television Viewing
and Income Example
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Bivariate Probabilities-TV viewer example, contd-

• Often we also want to know the probability that a
  person is a frequent watcher of the program
  P(A1), or the probability that a person has high
  income P(B1).
• Such probabilities are called the marginal
  probabilities.

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Marginal Probabilities

• In the context of bivariate probabilities, the
  probabilities for individual events P(Ai) and
  P(Bj) are called marginal probabilities.
• They can be computed by summing the corresponding
  row or column.

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Exercise Compute the following marginal
probabilities
0.21
0.27
0.52
0.27
0.41
1
0.32
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Bivariate Probabilities and tree diagram

• We have represeted the Bivariate Probabilities
  using a table.
• Often it is represented by a tree diagram.
• Example is in the next slide

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Tree Diagrams
P(A1 ? B1) .04

P(A1 ? B2) .13
P(A1 ? B3) .04
P(A1) .21
P(A2 ? B1) .10
P(A2) .27
P(A2 ? B2) .11
P(S) 1
P(A2 ? B3) .06
P(A3) .52
P(A3? B1) .13
P(A3 ? B2) .17
P(A3 ? B3) .22
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Probability Rules

• Conditional Probability
• Let A and B be two events. The conditional
  probability of event A, given that event B has
  occurred, is denoted by the symbol P(AB) and is
  found to be
• provided that P(B gt 0).

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Probability Rules

• Conditional Probability
• Let A and B be two events. The conditional
  probability of event B, given that event A has
  occurred, is denoted by the symbol P(BA) and is
  found to be
• provided that P(A gt 0).

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Conditional Probability-Exercise-

• Continue using the TV viewer example. Suppose
  that a person is in high income range. Given
  this information, what is the probability that
  this person is a occasional viewer of the
  program?

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Answer

• This problem can be formulated in the following
  way.
• The event the person has high income is B1,
  and the event the person is an occasional viewer
  of the problem is A2.
• The probability that that the person is an
  occasional viewer given the information that the
  person has high income person can be written as
• P(A2B1)P(A2 ? B1)/P(B1)0.1/0.270.37

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Probability Rules

• The Multiplication Rule of Probabilities
• Let A and B be two events. The probability of
  their intersection can be derived from the
  conditional probability as
• Also,

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Multiplication rules of the probability-Example-

• When we describe a situation that involves
  sequential decision making, multiplication rules
  become convenient.
• Consider the following investment problem.
  See next page.

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Multiplication rules of the probability-Example,
contd-

• A company is considering to invest in a project.
  Before investing in a full scale project, the
  company will undertake a test marketing. The
  probability that test marketing turns out to be
  successful is 0.6.
• Continues to the next slide

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• If the test marketing is successful, you may go
  ahead with the full scale investment. Given the
  successful test marketing result, there will be
  0.7 probability that the full scale investment
  will generate \100 million, and 0.3 probability
  that full scale project will generate \70
  million.
• Continue to the next slide

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• If the test marketing turns out to be
  unsuccessful, you can still go ahead with the
  full scale project. However, given unsuccessful
  test marketing, there will be only 0.15
  probability that the full scale project generate
  \70 million, and 0.85 probability that full scale
  project generate only 40 million.

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• We would like to represent this investment
  problem using a tree diagram. First define the
  following
• A1Test marketing successful
• A2Test marketing fail
• B1Full scale project generate \100 million
• B2Full scale project generate \70 million
• B3Full scale project generate \40 million

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Sequential decision making
Test marketing Successful
\100 million
P(B1A1)0.7

P(B2A1)0.3
P(A1) 0.6
Test Marketing
70 million
P(B2A2)0.15
P(A2) 0.4
Test marketing unsuccessful
P(B3A2)0.85
40 million
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Exercise

• Using the tree diagram in the previous slide find
  the following probabilities.
• 1. P(A1 ?B1)
• 2. P(A1 ?B2)
• 3. P(A2 ?B2)
• 4. P(A2 ?B3)
• 5. P(B1)
• 6. P(B2)
• 7. P(B3)

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Statistical Independence

• Let A and B be two events. These events are said
  to be statistically independent if and only if
• If A and B are statistically independent, from
  the multiplication rule, it also follows that
• More generally, the events E1, E2, . . ., Ek are
  mutually statistically independent if and only if

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Statistical Independence-Example 1-

• Consider tossing coins twice. Define the
  following events.
• A1Get the heads for the first toss
• A2Get the tails for the first toss
• B1Get the heads for the second toss
• B2Get the tails for the second toss
• The Bivariate Probabilities are given in the
  table on the next slide.

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Statistical Independence -Example 1, contd-
Exercise Show that A1 and B1 are statistically
independent.
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Statistical Independence-Example 2-

• A survey carried out for a supermarket clasified
  the cutomers according to whether their visits to
  the store are frequent or infreqent, and to
  whether they often, sometimes, or never purchases
  generic products. The table in the next slide
  gives the proportions of people surveyed in each
  of the six joint classifications.

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Exercise Check to see whether the event Visit
frequently and the event never purchase are
statistically independent.
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Probability Rules

• Let A be an event and A its complement. The the
  complement rule is

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Probability Rules

• The Addition Rule of Probabilities
• Let A and B be two events. The probability of
  their union is

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Probability RulesVenn Diagram for Addition
Rule(Figure 4.8)
P(A?B)
A
B

P(A)
P(B)
P(A?B)
A
B
A
B
A
B

-
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Set operation theorem

• De Morgans law

An application of De Morgans law can be found in
Exercise 4.61 (d) of the text book
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Corresponding chapters in the textbook

• 4.1
• 4.2
• 4.3
• 4.4