Transcript + P(B)

José Rizal University
Graduate School
MPA Program
Reported by Group I
Probability Concepts
1
Chance, Consequence &
Strategy:
Likelihood or Probability
Since there is little in life that occurs with
absolute certainty, probability theory has
found application in virtually every field
of human endeavor.
2
Why Probability Theory?
• As we observe the universe about us, wonderful
Craftsmanship can be seen.
• As we examine the elements of this creation we
discover that there is incredible order, but also
variation therein.
• Probability theory seeks to describe the variation
or randomness within order so that underlying
order may be better understood.
• Once understood, strategies can be more
effectively formulated and their risks evaluated.
3
Definitions





Probability: A measure of the likelihood that
an event in the future will happen; it can only
assume a value between 0 and 1, inclusive.
Experiment: The observation of some activity
or the act of taking some measurement.
Outcome: A particular result of an experiment.
Event: A collection of one or more outcomes
of an experiment.
Sample space: It is the set of all possible
simple outcomes, responses, or measurements
of an experiment.
4
Mutually Exclusive Events
• Mutually Exclusive Events: The
occurrence of any one event means
that none of the others can occur at
the same time.
Mutually exclusive: Rolling a 2
precludes rolling a 1, 3, 4, 5, 6 on
the same roll.
5
Collectively Exhaustive Events
• Collectively exhaustive: At least one of
the events must occur when an
experiment is conducted.
• In the following EXAMPLE 1, the four
possible outcomes are collectively
exhaustive. In other words, the sum of
probabilities = 1 (.25 + .25 + .25 + .25).
6
EXAMPLE 1
• Consider the experiment of tossing two
coins once.
• The sample space S = {HH, HT, TH, TT}
• Consider the event of one head.
• Probability of one head = 2/4 = 1/2.
7
A
B
Mutually
Exclusive &
Exhaustive Events
A and B are mutually exclusive.
• A collection of events is exhaustive if,
taken in totality, they account for all
possible results or outcomes.
8
There are three approaches of classifying probability:
classical, relative frequency, and subjective.
The Classical
probability is
often called a
priori
probability
which applies
when there are
n equally likely
outcomes.
The Relative
Frequency
approach applies
when the number
of times the event
happens is divided
by the number of
observations.
The Subjective
probability is
based on
personal belief
or feelings
with whatever
the evidence is
available.
9
A priori (Classical) Probability
In some experiments, such as the tossing of a coin or
rolling of a die, it is reasonable to believe that each of
the possible outcomes has the same chance of
happening.
If so, the probability of the various events can be
determined before the fact, or a priori (prior to any
statistical experiments).
P E  
number of simple outcomes favorable to the event f

total number of simple outcomes
n
Ex.: If A = {a die lands on a number less than 5}, then P (A)
= 4/6 since there are four ways of being successful (1 or 2 or 3
or 4) out of the six possible outcomes.
10
Relative Frequency Probability
• The probability of an event happening in
the long run is determined by observing
what fraction of the time like events
happened in the past:
P E  
number of favorable outcomes f

total number of trials
n
11
EXAMPLE 2
• Throughout her career Professor Jones
has awarded 186 A’s out of the 1200
students she has taught. What is the
probability that a student in her section
this semester will receive an A?
• By applying the relative frequency
concept, the probability of an P(A)=
186/1200=.155
12
Subjective Probability
• The likelihood (probability) of a particular
event happening that is assigned by an
individual based on whatever information
is available.
Examples:
estimating the probability the
Washington Redskins will win the
Super Bowl this year.
estimating the probability
mortgage rates for home loans
will top 8 percent.
13
Basic Rules of Probability
•
•
If events are mutually exclusive, then the
occurrence of any one of the events precludes
any of the other events from occurring.
Rules of addition: If two events A and B are
mutually exclusive, the special rule of addition
states that the probability of A or B occurring
equals the sum of their respective probabilities:
P(A U B) = P(A) + P(B)
A Venn diagram
illustration appears as:
A
B
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EXAMPLE 3
If A is the event that a
flight arrives early, then
P(A) = 100/1000 = .10.
If B is the event that a flight
arrives late, then P(B) =
75/1000 = .075.
The probability that a flight is either early or late is:
P(A U B) = P(A) + P(B) = .10 + .075 =.175.
15
Basic Rules of Probability (continued)
P(A U B) for Events— Not Mutually Exclusive
When events A and B are not mutually
exclusive, we must adjust the addition formula
to compute the probability that at least one of
them occurs.
P(A U B) = P(A) + P(B) – P(A ∩ B)
Another Venn diagram
illustration appears as:
16
EXAMPLE 4
In a sample of 500 students, 320 said they had a
stereo, 175 said they had a TV, and 100 said they had
both. 5 said they had neither.
Stereo
320
Stereo
320
Both
100
Both
00
TV
175
TV
175
17
EXAMPLE 4 continued
• If a student is selected at random, what
is the probability that the student has
only a stereo, only a TV, and both a
stereo and TV?
• P(S) = 320/500 = .64.
• P(T) = 175/500 = .35.
• P(S and T) = 100/500 = .20.
18
EXAMPLE 4 continued
• If a student is selected at random, what
is the probability that the student has
either a stereo or a TV in his or her
room?
• P(S or T) = P(S) + P(T) - P(S and T)
= .64 +.35 -.20 = .79.
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The Complement Rule
• The complement rule is used to
determine the probability of an event
occurring by subtracting the probability
of the event not occurring from 1. If P(A)
is the probability of event A and P(~A) is
the complement of A,
P(A) + P(~A) = 1 or P(A) = 1-P(~A).
20
The Complement Rule continued
• A Venn diagram illustrating the
complement rule would appear as:
A
~A
21
Independence - Key Concept
• If two events, A & B, are independent then
the occurrence of one of the two events does
not change the LIKELIHOOD or probability
that the other of the two events will occur.
• Occurrence of one of the two events does
alter the MANNER in which the other of the
two events may occur.
22
Dependence - Key Concept
• If two events A & B are dependent, then
occurrence of one of the two events will alter the
likelihood and the manner in which the other of
the two events may occur.
• In the case of mutually exclusive events,
occurrence of one of the two events will
preclude occurrence of the other event.
• Mutually exclusive events are always dependent.
23
Probability Under INDEPENDENCE
• The marginal probability of the event A is given by: P(A)
• Joint probability: It follows that if A and B are independent then P(AB)
= P(A)*P(B) This is the multiplication rule for independent events.
• Conditional probability: Two events A and B are said to be independent
if and only if: P(A|B) = P(A)
and
P(B|A) = P(B)
－ Probability Under DEPENDENCE
 The marginal probability of the event A is given by: P(A)
 The joint probability of AB is P(A and B) = P(A|B) x P(B) where P(A
and B) is the probability of the union of events A and B.
 The conditional probability of the event A given that the event B has
occurred is: P(A|B) = P(AB)/P(B)
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Joint Probability under statistical independence
•
Joint Probability is a probability that measures the
likelihood that two or more events will happen
concurrently. That is:
The idea of independence can be extended to more than
two events. For example, A, B and C are independent if:
a) A and B are independent; A and C are independent and
B and C are independent (pairwise independence);
b)
25
EXAMPLE 5
Suppose that a man and a woman each have a pack of 52
playing cards. Each draws a card from his/her pack. Find
the probability that they each draw the ace of clubs.
We define the events:
A = probability that man draws ace of clubs = 1/52
B = probability that woman draws ace of clubs = 1/52
Clearly events A and B are independent so:
That is, there is a very small chance that the man
and the woman will both draw the ace of clubs.
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Conditional Probability
under statistical independence
Conditional probability is the probability that a second
event (B) will occur if a first event (A) has already
happened. Note: The probability of the event A given that the event
B has occurred is denoted by P(A|B).
Symbolically, we say events A and B are independent
when P(B|A) = P(B).
Example: To toss a fair coin, the probabilities of heads
on 1st toss and 2nd toss are .5 each always, so we write:
P(H2|H1) = P(H2) = .5.
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Conditional Probability
under statistical dependence
The conditional probability of an event B in
relationship to an event A is the probability that
event B occurs given that event A has already
occurred. The notation for conditional
probability is P(B|A), read as the probability of
B given A. The formula for conditional
probability is:
28
EXAMPLE 6
Problem:
A math teacher gave her class two tests. 25% of the class
passed both tests and 42% of the class passed the first test.
What percent of those who passed the first test also passed
the second test?
Analysis:
This problem describes a conditional probability since it
asks us to find the probability that the second test was
passed given that the first test was passed. In the last
lesson, the notation for conditional probability was used in
the statement of Multiplication Rule 2.
Solution:
P(Second | First)
=
0.25
P(First & Second)
P(First)
=
0.42
=
0.60
=
60%
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EXAMPLE 7
•
The Dean of the School of Business at Miami
collected the following information about
undergraduate students in her college:
MAJOR
Male
Female
Total
Accounting
170
110
280
Finance
120
100
220
Marketing
160
70
230
Management
150
120
270
Total
600
400
1000
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EXAMPLE 7
continued
• If a student is selected at random, what is the
probability that the student is a female accounting
major? P(AF) = 110/1000.
• Given that the student is a female, what is the
probability that she is an accounting major?
P(A|F) = P(AF)/P(F) = (110/1000)/(400/1000)
= .275.
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Joint Probability
under statistical dependence
• The general rule of multiplication is used to
find the joint probability that two events will
occur, as it states: for two events A and B, the
joint probability that both events will happen is
found by multiplying the probability that event
A will happen by the conditional probability of
B given that A has occurred.
P(AB) = P(A) × P(B|A) or
P(AB) = P(B) × P(A|B)
Exercise: to use multiplication choice of Conditional Probability formula
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Objective Assessment:
A priori & A posteriori Probability
 A priori means “before the fact” and hence
probability assessments of this sort typically rely
on a study of traits of the phenomenon under
consideration.
 Based on Theory.
A posteriori means “after the fact”.
This approach to likelihood assessment is
also called the “relative frequency”
approach.
Based on repeated observation.
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Bayes’ Theorem
Bayes' Theorem is a result that allows new information of posterior probability
to be used to update or to revise the prior conditional probability of an event.
Bayes’ Theorem is given by the following formula. This is Bayes' Theorem
in its simplest form.
Using the Law of Total Probability:
where:
P(A) = probability that event A occurs
P(B) = probability that event B occurs
P(A') = probability that event A does not occur
P(A|B) = probability that event A occurs given that event B has occurred already
P(B|A) = probability that event B occurs given that event A has occurred already
P(B|A') = probability that event B occurs given that event A has not occurred already
34
Taipei 101
The tallest building in the world
Thank you!
35
Exercise 2-2
Of the repair jobs that Bennie’s Machine Shop receives, 20
percent are welding jobs and 80 percent are machining jobs.
a.
What is the probability that the next three jobs to come in will
be welding jobs?
b. What is the probability that two of the next three jobs to come
in will be the machining jobs?
Solution:
a.
P(A) = 0.2 Thus P(A)xP(A)xP(A) = 0.2 x 0.2 x 0.2 = 0.008
b. P(B) = 0.8 Thus P(B)xP(B)xP(B) = 0.8 x 0.8 x 0.8 = 0.512
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Exercise 2-3
A building contractor knows that the probability of rain in his
region during any given day in August is 30%. He has 2 days’
work remaining on a job and there are 4 days during which the
work can be done without incurring penalty costs. If he gets no
work done on rainy days, what is the probability that he will pay a
penalty? Assume that rain on any given day is independent of the
fact that it did or did not rain on any other day.
Solution:
(0.3x0.3x0.3x0.3) + (0.3x0.3x0.3x0.7) = 0.027
Because only any 3 and more rainy days occur, then he will pay
a penalty.
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Exercise 2-11
As a means of increasing worker morale and performance, the
management of an assembly shop was considering a plan
providing job enrichment (such as self-supervision, team working,
and the individual selection of working hours). Two groups of 100
workers were selected at random. Then first group worked under
existing conditions; the second was placed in a separate room and
was allowed to function under the proposed job enrichment plan.
After a year, management compared the performance ratings of
all workers and found that in the group operating under existing
conditions, for 30 employees the performance ratings improved,
for 60 employees the ratings remained the same, and for 10
employees the ratings dropped. Within the group working under
the job enrichment plan, 40 performance rating improved, 55
remained the same, and 5 dropped. How should management
assess the effects of the job enrichment program on worker morale
and performance?
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Exercise 2-11
Condition Improved
Same
Dropped
Group I 30/100 =
(existing) 0.3
60/100 =
0.6
10/100 =
0.1
Group II 40/100 =
0.4
(new)
55/100 =
0.55
5/100 =
0.05
Solution: We take improved and same performance rating to
compare, then we find
Group I: P(A U B) = P(A) + P(B) = 0.3 + 0.6 = 0.9
Group II: P(A U B) = P(A) + P(B) = 0.4 + 0.55 = 0.95
Since 0.95 > 0.9, so we consider Group II has better
performance than Group I, it means the job enrichment
program has positive effects on worker morale and performance.
39
Exercise 2-19
International Foods, Inc., manufacturer of Zesty, a special Italian
tomato sauce, is interested in discovering whether consumers can
tell the difference between a competitor’s more expensive sauce
and its own. Fifteen persons are given a sample of Zesty and then
a sample of the competitor’s sauce. Then they are each asked to
identify the competitor’s brand. It is customary to hypothesize
that there is no difference in the tastes; if this is so, the chances of
a person identifying the competitor’s brand are 0.5. What is the
probability that 12 of 15 persons in the test will identify the
competitor’s brand if the hypothesis is really true? Use the
binomial tables to compute your answer.
Solution:
n = 15 p=0.5 r=12 from Appendix Table II, we find the P(12
or more) = 0.0176
So P(12) = 0.0176-0.0037 = 0.0139
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