Probability

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Probability

• Chapter 6

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6.2 Assigning probabilities to Events

• Random experiment
• a random experiment is a process or course of
  action, whose outcome is uncertain.
• Examples
• Experiment Outcomes
• Flip a coin Heads and Tails
• Record a statistics test marks Numbers between 0
  and 100
• Measure the time to assemble numbers from zero
  and abovea computer

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6.2 Assigning probabilities to Events

• Performing the same random experiment repeatedly,
  may result in different outcomes, therefore, the
  best we can do is consider the probability of
  occurrence of a certain outcome.
• To determine the probabilities we need to define
  and list the possible outcomes first.

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Sample Space

• Determining the outcomes.
• Build an exhaustive list of all possible
  outcomes.
• Make sure the listed outcomes are mutually
  exclusive.
• A list of outcomes that meets the two conditions
  above is called a sample space.

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Sample Space S O1, O2,,Ok
Sample Space a sample space of a random
experiment is a list of all possible outcomes of
the experiment. The outcomes must be mutually
exclusive and exhaustive.
Simple Events The individual outcomes are called
simple events. Simple events cannot be further
decomposed into constituent outcomes.
Event An event is any collection of one or more
simple events
Our objective is to determine P(A), the
probability that event A will occur.
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Assigning Probabilities

• Given a sample space SO1,O2,,Ok, the
  following characteristics for the probability
  P(Oi) of the simple event Oi must hold
• Probability of an event The probability P(A) of
  event A is the sum of the probabilities assigned
  to the simple events contained in A.

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Approaches to Assigning Probabilities and
Interpretation of Probability

• Approaches
• The classical approach
• The relative frequency approach
• The subjective approach
• Interpretation
• If a random experiment is repeated an infinite
  number of times, the relative frequency for any
  given outcome is the probability of this outcome.

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6.3 Joint, Marginal, and Conditional Probability

• We study methods to determine probabilities of
  events that result from combining other events in
  various ways.
• There are several types of combinations and
  relationships between events
• Intersection of events
• Union of events
• Dependent and independent events
• Complement event

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Intersection

• The intersection of event A and B is the event
  that occurs when both A and B occur.
• The intersection of events A and B is denoted by
  (A and B).
• The joint probability of A and B is the
  probability of the intersection of A and B, which
  is denoted by P(A and B)

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Intersection

• Example 6.1
• A potential investor examined the relationship
  between the performance of mutual funds and the
  school the fund manager earned his/her MBA.
• The following table describes the joint
  probabilities.

Mutual fund outperform the market Mutual fund doesnt outperform the market
Top 20 MBA program .11 .29
Not top 20 MBA program .06 .54
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Intersection

• Example 6.1 continued
• The joint probability of mutual fund
  outperform and from a top 20 .11
• The joint probability ofmutual fund
  outperform and not from a top 20 .06

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2)
Top 20 MBA program (A1) .11 .29
Not top 20 MBA program (A2) .06 .54
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Intersection
Intersection

• Example 6.1 continued
• The joint probability of mutual fund
  outperform and from a top 20 .11
• The joint probability ofmutual fund
  outperform and not from a top 20 .06

P(A1 and B1)
P(A2 and B1)
Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2)
Top 20 MBA program (A1) .11 .29
Not top 20 MBA program (A2) .06 .54
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Marginal Probability

• These probabilities are computed by adding across
  rows and down columns

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2) Marginal Prob. P(Ai)
Top 20 MBA program (A1)
Not top 20 MBA program (A2)
Marginal Probability P(Bj)
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Marginal Probability

• These probabilities are computed by adding across
  rows and down columns

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2) Marginal Prob. P(Ai)
Top 20 MBA program (A1) .11 .29 .40
Not top 20 MBA program (A2) .06 .54 .60
Marginal Probability P(Bj)




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Marginal Probability

• These probabilities are computed by adding across
  rows and down columns

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2) Marginal Prob. P(Ai)
Top 20 MBA program (A1) .40
Not top 20 MBA program (A2) .60
Marginal Probability P(Bj)
P(A1 and B1) P(A2 and B1 P(B1)
P(A1 and B2) P(A2 and B2 P(B2)
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Marginal Probability

• These probabilities are computed by adding across
  rows and down columns

Mutual fund outperforms the market (B1) Mutual funddoesnt outperform the market (B2) Marginal Prob. P(Ai)
Top 20 MBA program (A1) .11 .29 .40
Not top 20 MBA program (A2) .06 .54 .60
Marginal Probability P(Bj) .17 .83
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Conditional Probability

• Example 6.2 (Example 6.1 continued)
• Find the conditional probability that a randomly
  selected fund is managed by a Top 20 MBA Program
  graduate, given that it did not outperform the
  market.
• Solution
• P(A1B2) P(A1 and B2) .29 .3949
  P(B2) .83

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Conditional Probability

• Example 6.2
• Find the conditional probability that a randomly
  selected fund is managed by a Top 20 MBA Program
  graduate, given that it did not outperform the
  market.
• Solution
• P(A1B2)
• P(A1 and B2) P(B2).29/.83 .3949

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2) Marginal Prob. P(Ai)
Top 20 MBA program (A1) .11 .29 .40
Not top 20 MBA program (A2) .06 .54 .60
Marginal Probability P(Bj) .17 .83
.29
.83
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Conditional Probability

• Before the new information becomes available we
  have P(A1) 0.40
• After the new information becomes available P(A1)
  changes to P(A1 given B2) .3494
• Since the the occurrence of B2 has changed the
  probability of A1, the two event are related and
  are called dependent events.

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Independence

• Independent events
• Two events A and B are said to be independent if
  P(AB) P(A) or P(BA) P(B)
• That is, the probability of one event is not
  affected by the occurrence of the other event.

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Dependent and independent events
Dependent and Independent Events

• Example 6.3 (Example 6.1 continued)
• We have already seen the dependency between A1
  and B2.
• Let us check A2 and B2.
• P(B2) .83
• P(B2A2)P(B2 and A2)/P(A2) .54/.60 .90
• Conclusion A2 and B2 are dependent.

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Union

• The union event of A and B is the event that
  occurs when either A or B or both occur.
• It is denoted A or B.
• Example 6.4 (Example 6.1 continued)
  Calculating P(A or B))
• Determine the probability that a randomly
  selected fund outperforms the market or the
  manager graduated from a top 20 MBA Program.

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Union
Union of events

• Solution

Mutual fund outperforms the market (B1) Mutual fund doesnt outperform the market (B2)
Top 20 MBA program (A1) .11 .29
Not top 20 MBA program (A2) .06 .54
A1 or B1 occurs whenever either A1 and B1
occurs,
Comment P(A1 or B1) 1 P(A2 and B2) 1
.54 .46
A1 and B2 occurs,
A2 and B1 occurs.
P(A1 or B1) P(A1 and B1) P(A1 and B2) P(A2
and B1) .11 .29 .06 .46
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6.4 Probability Rules and Trees

• We present more methods to determine the
  probability of the intersection and the union of
  two events.
• Three rules assist us in determining the
  probability of complex events from the
  probability of simpler events.

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Complement Rule
Complement event

• The complement of event A (denoted by AC) is the
  event that occurs when event A does not occur.
• The probability of the complement event is
  calculated by

A and AC consist of all the simple events in the
sample space. Therefore,P(A) P(AC) 1
P(AC) 1 - P(A)
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Multiplication Rule

• For any two events A and B
• When A and B are independent

P(A and B) P(A)P(BA) P(B)P(AB)
P(A and B) P(A)P(B)
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Multiplication Rule

• Example 6.5
• What is the probability that two female students
  will be selected at random to participate in a
  certain research project, from a class of seven
  males and three female students?
• Solution
• Define the eventsA the first student selected
  is a femaleB the second student selected is a
  female
• P(A and B) P(A)P(BA) (3/10)(2/9) 6/90
  .067

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Multiplication Rule

• Example 6.6
• What is the probability that a female student
  will be selected at random from a class of seven
  males and three female students, in each of the
  next two class meetings?
• Solution
• Define the eventsA the first student selected
  is a femaleB the second student selected is a
  female
• P(A and B) P(A)P(B) (3/10)(3/10) 9/100
  .09

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Addition Rule

• For any two events A and B

P(A or B) P(A) P(B) - P(A and B)
P(A) 6/13
A

P(B) 5/13
_
B
P(A and B) 3/13
P(A or B) 8/13
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Addition Rule

• When A and B are mutually exclusive,

P(A or B) P(A) P(B)
B
A
B
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Addition Rule

• Example 6.7
• The circulation departments of two newspapers in
  a large city report that 22 of the citys
  households subscribe to the Sun, 35 subscribe to
  the Post, and 6 subscribe to both.
• What proportion of the citys household subscribe
  to either newspaper?

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Addition Rule
The Addition and Multiplication Rules

• Solution
• Define the following events
• A the household subscribe to the Sun
• B the household subscribe to the Post
• Calculate the probabilityP(A or B) P(A) P(B)
  P(A and B) .22.35 - .06 .51

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Probability Trees

• This is a useful device to calculate
  probabilities when using the probability rules.

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Probability Trees
Dependent events

• Example 6.5 revisited (dependent events).
• Find the probability of selecting two female
  students (without replacement), if there are 3
  female students in a class of 10.

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Probability Trees
Independent events

• Example 6.6 revisited (independent events)
• Find the probability of selecting two female
  students (with replacement), if there are 3
  female students in a class of 10.

FF
FM
MF
MM
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Probability Trees

• Example 6.8 (conditional probabilities)
• The pass rate of first-time takers for the bar
  exam at a certain jurisdiction is 72.
• Of those who fail, 88 pass their second attempt.
• Find the probability that a randomly selected law
  school graduate becomes a lawyer (candidates
  cannot take the exam more than twice).

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Probability Trees

• Solution

P(Pass) P(Pass on first exam) P(Fail on first
and Pass on second) .9664
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Bayes Law

• Conditional probability is used to find the
  probability of an event given that one of its
  possible causes has occurred.
• We use Bayes law to find the probability of the
  possible cause given that an event has occurred.

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Bayes Law

• Example 6.9
• Medical tests can produce false-positive or
  false-negative results.
• A particular test is found to perform as follows
• Correctly diagnose Positive 94 of the time.
• Correctly diagnose Negative 98 of the time.
• It is known that 4 of men in the general
  population suffer from the illness.
• What is the probability that a man is suffering
  from the illness, if the test result were
  positive?

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Bayes Law

• Solution
• Define the following events
• D Has a disease
• DC Does not have the disease
• PT Positive test results
• NT Negative test results
• Build a probability tree

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Bayes Law

• Solution Continued
• The probabilities provided are
• P(D) .04 P(DC) .96
• P(PTD) .94 P(NTD) .06
• P(PTDC) .02 P(NTDC) .98
• The probability to be determined is

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Bayes Law
D
PT
D
D
PTD
PT
D
PTD
PTD
D
PTD
D
PT
PTD
PT
PT
PT
PT
PTD
P(D and PT) .0376
PTD
P(DC and PT) .0192
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Bayes Law
Prior probabilities
Likelihood probabilities
Posterior probabilities