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Probability Concepts

José Rizal University MPA Program Graduate School Reported by Group I

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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.

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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.

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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.

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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.

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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).

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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.

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• A collection of events is exhaustive if, taken in totality, they account for all possible results or outcomes.

A

B A and B are mutually exclusive.

Mutually Exclusive & Exhaustive

Events

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There are three approaches of classifying probability:classical, relative frequency, and subjective.

The ClassicalClassical probability is

often called a priori

probability which applies when there are n equally likely

outcomes.

The Relative Relative FrequencyFrequency

approach applies when the number of times the event happens is divided by the number of

observations.

The Subjective Subjective probability is

based on personal belief

or feelings with whatever the evidence is

available.

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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).

outcomes simple ofnumber total

event the to favorable outcomes simple ofnumber

n

fEP

outcomes simple ofnumber totalevent the to favorable outcomes simple ofnumber

n

fEP

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.

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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:

trials ofnumber totaloutcomes favorable ofnumber

n

fEP

trials ofnumber totaloutcomes favorable ofnumber

n

fEP

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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

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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.

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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:

AA BB

P(A U B) = P(A) + P(B)P(A U B) = P(A) + P(B)

A Venn diagram illustration appears as:

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EXAMPLE 3

The probability that a flight is either early or late is:

P(A U B) = P(A) + P(B) = .10 + .075 =.175.

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.

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P(A U B) for Events— Not Mutually ExclusiveWhen 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) P(A U B) = P(A) + P(B) – P(A ∩ B)

Another Venn diagram illustration appears as:

Basic Rules of Probability (continued)

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EXAMPLE 4

Stereo 320

Both 00

TV175StereoStereo

320320

BothBoth 100100

TVTV175175

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.

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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.

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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).

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The Complement Rule continued

• A Venn diagram illustrating the complement rule would appear as:

A ~A

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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.

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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.

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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)

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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:

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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.

EXAMPLE 6

Solution: 

P(Second | First)

  =  

P(First & Second)   =  

0.25

  =   0.60   =   60% P(First) 0.42

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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)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’ TheoremBayes' 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

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Taipei 101Taipei 101The tallest building in the world

Thank you!

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Exercise 2-2Of 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 2-3A 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 2-11As 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

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.

Condition Improved Same Dropped

Group I

(existing)

30/100 = 0.3

60/100 = 0.6

10/100 = 0.1

Group II

(new)

40/100 = 0.4

55/100 = 0.55

5/100 = 0.05

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Exercise 2-19International 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