Lecture Notes in Statistics 121 Chapter 1 (Daquis)

8/9/2019 Lecture Notes in Statistics 121 Chapter 1 (Daquis)

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LECTURE NOTES

Statistics 121

Probability Theory I

John Carlo P. Daquis

Assistant Professor 1UP School of Statistics


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Whenever I look through my notes on

probability distributions, my eyes will

look for her.

For I am always fascinated by the

normal distribution

the elegance she has formed

from irrationalities,

the majestic curve that flows

from a point in eternity

towards the other unfathomable

infinity,

the power emanating from her

that challenges the impossible.

She has given me the power that love

could only give:

Now I can defy probabilities.

MAJC


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SCHOOL OF STATISTICS

University of the Philippines Diliman

Statistics 121 (3 Units)

PROBABILITY THEORY 1

COURSE SYLLABUS

First Semester, School Year 2014 2015

Instructor: John Carlo P. Daquis

Instructor Information

OFFICE : School of Statistics Faculty Room 26

OFFICE HOURS : 1:00 3:00 Tuesday to Friday

OFFICE PHONE : (+632) 928 0881

OFFICE WEBSITE : stat.upd.edu.ph

E-MAIL ADRESS : [email protected]

[email protected]

CLASS HOURS : 8:30 10:00 (WFR)

10:00 11:30 (WFU)

SCHEDULE OF : 8:30 11:30 (TTh)OTHER CLASSES

Student Information Card

On a 3x5 index card, provide the following:

FRONT, left side: FRONT, upper right:

Last Name, Given Name M.I. a recognizable 1x1 photo

Mobile Number nickname below photo

Email Address

Person to contact in case of emergency BACK:class schedule

Contact number of the person


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Course Objectives

By the end of the course, a student enrolled in this class must be able

to have a heartfelt and working knowledge about all the lessons taught in

class. The student must be able to acquire an advanced skill in logicalreasoning. In particular, one must be able to:

Define probability and other concepts such as conditionality,

independence, random variables and probability distributions;

Master the basic properties of the probability function;

Compute for the probability of an event;

Use calculus skill in obtaining the distribution function from a density

or mass function and vice versa;

Evaluate expectations and moments; Be well aware of some special univariate distributions and its basic

properties; and

Derive the distribution of a function of a random variable.

Course Prerequisites

Course Prerequisites:

Statistics 117/Mathematics for Statistics (or equiv.)

-

For proving methods, cardinality and basic combinatorics, settheory and evaluating sums

Mathematics 53/Elementary Analysis 1

- For evaluating limits, derivatives and integrals

Course Co-requisite:

Mathematics 54/Elementary Analysis 2

Course Requirements

Requirement Breakdown

3 Long Exams 60%

1 Final Exam 15%

Problem Sets 07%Peer Eval. of PS 03%


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Quizzes/Assignments 10%

Attendance 05%

Long Exams

Materials needed:2 blue books, ballpens, pencils, calculators, tables (if

necessary), a fully-functioning brain

Duration: 3 hours

Schedule: to be announced, but will be on Saturday or Monday

Coverage:Our course will span 5 chapters (please see table of contents),

long exam 1 will cover chapters 1 and 2, long exam2 will cover chapters 3

and 4 and long exam 3, the last chapter.

Final Exam

Materials needed: ballpens, pencils, a fully-functioning brain

Duration:2 hours

Coverage:Though the coverage is all the five chapters, the final exam is an

evaluation on how well you know the concepts. Thus, computation and

proving is very minimal.

Exemptions: None.

Problem Sets

Number of members: 3 cooperative members. You must be, since 3% of

your overall grade will be based on how you contributed to your group.

Perks: There will be a special discussion session right before an exam. A

sign-up sheet for volunteers will be posted at the door of my room.

Boardwork volunteers who satisfactorily get a correct answer will be

rewarded a bonus 5% in the long exam. Volunteers whose solutions are wrong

will still get bonus points, though it is reduced to 4%. The class will assist

him/her in getting the right answer.

Attendance, Assignments and Quizzes

Attendance: Always checked. Again, being focused and attentive is

imperative. You might miss important details.


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Assignments/Quizzes:The schedule of giving of assignments or quizzes is

unstructured. By default, assignments are done by student and to be

submitted the next meeting while quizzes are announced.

Grading System

95 100 1.00 72


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Chapter 1

Probability


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1.1 Introduction: Modeling Reality

1.1.1 Theories and Models

Theories are ideas or concepts that are used to explain a phenomenonhappening in the real world. Thus, theories are just approximations

of reality but are not necessarily true. For example, it was then

believed that the universe is always expanding while maintaining a

constant average density. A universe following this Steady State

Theory has no beginning or end. As of this moment, it has been

superseded by the Big Bang Theory, which says that the universe

expanded from a singularity or a zone with infinite density. The Big

Bang is now the prevailing theory of the universes development, yet

it does not mean it is true, entirely or partially. A scientific process is

done to disprove or refine a theory.

Models on the other hand is a theoretical construct representing a

process or describing a phenomenon using a set of variables and

quantitative relationships between them. Models, though they are

theoretical approximations of reality are of big help in understanding

what is happening in our world. Thus, theres this famous quote

essentially, all models are wrong, but some are useful by statistician

George Box (1979).

1.1.2 Deterministic and Probabilistic Models

Suppose we wish to measure the area of a rectangular lot. Denoting

the area by R, the formula we will be using is R=lwwhere l is the

length of the lot and wis the width of a lot. This is what we call a

deterministic model. The deterministic comes from the part that

once the length and width are known, the area is assumed to be

known. The formula models reality because even though the lot is not

perfectly a rectangle, it has the quantitative relationship that helps usapproximate measurements of the area. Deterministic models describe

a phenomenon which will always produce the same outcome without

any room for variation.


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Now consider an experiment of tossing a balanced coin. There are two

possible outcomes: heads or tails. No matter what, one cannot

perfectly predict what exactly the next outcome will be. However, we

can say that the chance that the next outcome will come up heads is

0.5. The model described is probabilistic in nature. It describes thefact that, in the long run the chance that a head will show up in a

single toss is 0.5 but cannot assume with certainty what the next value

will be. Probabilistic models describe different outcomes not by fixed

values, but rather by taking into account the presence of randomness

in a phenomenon.

1.1.3 Applications of Probability

Probability models are used to describe a phenomenon.

Example: Population models often take into account the birth and

death rates as well as the population size at a particular

given time. One probabilistic model shows that it is very

likely for a population to go extinct if the birth rate is

equal to the death rate. The population will likely

survive if the death rate is lower than the birthrate.

Probability is a useful tool in decision-making.

Example: Weather reports nowadays also include a chance of

raining the next day. This would assist viewers in

making decisions like whether or not to bring umbrellas

and cancelling or pushing through an appointment

tomorrow.

Probability theory is the foundation of statistical inference.

Example: A car manufacturer claims that their new car model ismore efficient than the leading model in the market. To

validate this claim, the manufacturer conducted an

experiment and the resulting sample mean mileage,

is indeed lower than the leading model claim by

2kms/liter. Is the difference significant to support the


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manufacturers claim? A probability distribution

governs the behavior of the sample mean to tell whether

or not (given a significance level) the evidence is

significant to support the claim.

1.2 Classical Probability

Probability theory originated in games of chance. If we toss an

unbiased coin independently ntimes, where n is very large, then the

ratio of heads to the total number of tosses will be very close to 0.5.

Similarly, the relative frequency of getting a hearts in a standard deck

of 52 cards is 0.25.

The classical definition of probability relies at the assumption that allpossible outcomes of the activity are equally likely. The ratios are

obtained a priorieven without doing the actual experiment, unlike

the relative frequency approach or obtaining the probability a

posteriori.

1.2.1 Random Experiment

Consider the magic 8-ball toy, a novelty item at which its primary use

is to give random advices, making it a popular toy for fortune telling.

Inside the ball is a buoyant icosahedron. All of the twenty faces of

this polyhedron hold an answer:

Affirmative answers: Neutral answers:

It is certain (a1) Reply hazy try again (o1)

It is decidedly so (a2) Ask again later (o2)

Without a doubt (a3) Better not tell you now (o3)

Yes definitely (a4) Cannot predict now (o4)

You may rely on it (a5) Concentrate and ask again (o5)

As I see it, yes (a6)

Most likely (a7) Negative answers:

Outlook good (a8)

Yes (a9) My reply is no (n1)

Signs point to yes (a10) My sources say no (n2)


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Don't count on it (n3)

Outlook not so good (n4)

Very doubtful (n5)

Shaking the magic 8-ball and waiting for an answer to show up is an

example of a random experiment. A random experiment is an activity

which can be repeated many times under the same conditions, but

whose outcome cannot be predicted with certainty. Thus in a random

experiment, the outcome can no way be predicted by any previous

outcomes.

Here are some other examples of a random experiment:

The tossing of a coin

Rolling a die

Selecting a numbered ball in an urn

Spinning a bottle.

1.2.2 Sample Space, Outcomes and Events

A random experiment has a set of realizations or sample points. The

realizations of a random experiment is called an outcome. All possible

outcomes belong to a set called the sample space. Any subset of the

sample space is an event.

Definitions: Outcomes, denoted by are realizations of a random

experiment. They are the elements of the sample space.

Sample Space, denoted by is the set of all possible

outcomes for a random experiment listed in a mutually

exclusive and exhaustive way.

No magic. The proprietary toy, Magic 8-Ball is

nothing but an instrument used in a random

experiment.


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Events are any subsets of the sample space and are

denoted by capital latin letters.

Example: In the magic 8-ball experiment, there is a total of 20

outcomes. Consider the following:

= {a1, a2, a3, a4, a5, a6, a7, a8, a9, a10, o1, o2, o3,

o4, o5, n1, n2, n3, n4, n5}

A = the event of observing the response my sources say

no = {n2}

B = the event of observing a neutral response

= {o1, o2, o3, o4, o5}

In the definition of a sample space, the phrase mutually exclusive

means that there should be no overlap of outcomes. For example, the

set

= {H, T, heads, tails}

is not appropriate since the outcome of seeing heads in the experiment

of tossing a coin is represented by two outcomes in the set, hence the

outcomes are not mutually exclusive. On the other hand, the word

exhaustive means that all outcomes must be included in the sample

space. The set

* = {1, 2, 4, 5, 6}

is not a valid sample space for the rolling of a die experiment because

the outcome of three dots in the upper face is not represented in the

set.

Example: Consider the random experiment of tossing a coin thrice

= {(HHH), (HHT), (HTH), (HTT), (THH), (THT),

(TTH), (TTT)}1


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= {=(1, 2, 3)| i{H, T}, i = 1, 2, 3}

A = event of observing 2 heads

= {(HHT), (HTH), (THH)}

B = observing at least 2 tails

= {(HTT),(THT),(TTH),(TTT)C = observing the same face on the 1st& 3rdtosses

= {(HHH), (HTH), (THT), (TTT)}

= {=(1, 2, 3)| i{H, T}, i = 1, 2, 3

and 1 = 3}

Writing the sample space is not unique in terms of the outcomes. For

example, consider again the experiment tossing a coin thrice, the

sample space

1= {0, 1, 2, 3}

can be used as a sample space especially when the attribute of interest

is the number of heads.

We say that an event Aoccurs when one of its elements is the outcome

of the experiment. That is, if we say A= event of observing 2 heads

occurs, then either HHT, HTHor THHis the outcome. Events having

only one element are called elementary events while events having

more than one elements are called compound events. Elementary and

compound events will be formally defined later.

So far, the examples above are all examples of discrete sample space,

in particular, finite sample spaces. Suppose someone wants to measure

the height of a person. The sample space for this experiment is

= {| (0, )}.

Definitions: Discrete Sample Space is a sample space which isfinite or countably infinite.

Continuous Sample Spaceis a sample space which

is neither finite nor countably infinite.


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Example: Consider a random experiment of observing the number

of people going inside a mall in a particular day and the

average time in seconds of the waiting time between

each person going inside the mall. The sample space can

be represented as follows:

={=(x,y)| x=0,1,2, and y 0}.

Here, an outcome (37,000, 0.394) is a point in the sample

space wherein there were 37,000 people who went inside

the mall in a particular day with an average waiting

time of 0.394 seconds between each person going inside

the mall.

Exercise: Specify the correct sample space for the following

random experiments and find their respective

cardinalities:

i. 6 balls are drawn without replacement from an

urn filled with balls labeled 1 to 42, ordering is

not important

ii. 6 balls are drawn with replacement from an urn

filled with balls labeled 1 to 42 and observing

which element was selected at each draw

iii. Measuring the weight of a newborn baby

iv. Tossing a coin until a head comes up

v. Rolling a die until the face with six dots comes

up

1.2.3 Basic Concepts of Set Theory

Set theory concepts are important in learning probability. The sample

space can be viewed as a set having all elements, or a universal set.Events are sets while outcomes are elements. Here is a review of the

basic concept definitions in set theory.

The occurrence of an event Acan be translated in the language of sets

and can be schematically represented by its corresponding Venn


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diagram. Consider the following below. Note that the shaded regions

are the regions of interest.

The sample space : universal set

Event Aoccurs : set A

Event Adoes not occur : complement

of A, Ac

At least one of Aand B : union

Occur A B

Both Aand Boccur : intersection

A B

Aand B cannot occur : A and B

simultaneously disjoint

A B=

Aoccurs but notB : difference

A-B = ABc

Either in Aor in B : symmetric

but not both difference

(A B) (A B) = A B

BA

A

Ac

A

A B

A B

A B

A B


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Definition: An event Eis said to be a simpleor elementary event

if and only if

i. E and

ii.

For every event A ,E A=or E A=E.

A compound event is an event that can be expressed

as the union of distinct elementary events.

These set operations can now be used to form new events from the

old or existing ones. The process is called event composition.

Exercise: Consider the random experiment of tossing a coin thrice.

Let Bi= event that the ith toss is a head, i = 1,2,3

Express the following events in terms of the B is:

C = the 1sttoss is a head

B = the 1sttwo tosses are heads

D = there is exactly one toss which results as a head

E = there is at least one toss which results as a headF = there is at most one toss which results as a head

G = the second toss is a tail and the third toss is a head

H = the second toss is a tail and there is at most one

head

I = the second toss is a tail and there is exactly one

head

J = all tosses are tails

Exercise: Let A, B and Cbe arbitrary events in . Define the

following events in terms of A, Band C:

D1= event that (et) at least two of the events

A, B, C occur

D2= et at exactly two of the events A, B, C occur

D3= et at least one of the events A, B, C occur


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D4= et exactly one of the events A, B, C occur

D5= et at most two of the events A, B, C occur

Now suppose that all A, B and C are disjoint, find

suitable expressions for D1to D5.

1.2.4 Generalized Union and Intersection of Events

The union of events B1, B2, , Bnis the event consisting of outcomes

which belong to at least one of the events B1, B2, , Bn.

Finite Union : U

=

=

Similarly, define the union of a countably infinite sequence of events

B1, B2, as the event whose outcomes belong to at least one of the

events B1, B2, .

Countable Union : U

=

=

The intersection of events B1, B2, , Bn is the event consisting of

outcomes which belong to all of the events B1, B2, , Bn.

Finite Intersection : I

=

=

Similarly, define the intersection of a countably infinite sequence of

events B1, B2, as the event whose outcomes belong to all of the

events B1, B2, .

Countable Intersection : I

==

Exercise: Prove that De Morgans Laws are valid for a finite

collection of sets. The same argument can be extended

to a countable series of sets by considering an infinite

sequence of events rather than a finite one.


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Example: A couple plans to have 7 children. Define the event

Gi= event that the ith child is a girl.

The following events can be expressed in terms of Gias

follows:

A = et all children are girls

I

=

=

B = et all children are boys

==

==

UI

C = et only the first child is a girl

U

=

=

D = et there is only 1 girl among the seven children

U U

=

=

=

Exercise: Define the sample space as the set of nonnegative real

numbers. Let

K

=

=

=

Evaluate the following sets:

1.

U

=

2. I

=


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1.2.5 Methods of Assigning Probabilities

Let us now introduce the notation P()for probability. For example,

the probability of an event Ais P(A).

Definition: The classical probability or a priori approach in

assigning probabilities considers a random experiment

with n() equally likely outcomes. Consider an event

with n(A)outcomes, then

( ) ( )

( )=

.

Classical probability assumes that all the outcome are equally likely

and the sample space to be finite. The sample space should therefore

be defined in such a way that the outcomes are equally likely to

happen. As will be discussed later, classical probability definition is

too restrictive and circular.

Example: A bingo shaker contains three B chips labeled B1, B2

and B3, two N chips labeled N31 and N32, and O61

chip. A second bingo shaker has a B4 chip, two N chips

labeled N33 and N34, and three O chips labeled O62,

O63 and O64. One chip is selected randomly from each

shaker. The sample space in this random experiment is

as follows:

={=(x,y)| x{B1,B2,B3,N31,N32,O61} and

y{B3,N33,N34,O62,O63,O64}}

All outcomes are equally likely because selection is done

at random. Also, n() = (6)(6) = 36.

Define the following events:

A = event of selecting 2 B chips

B = event of selecting 2 N chips


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Now,

P(A)= n(A)/n() = (3)(1)/36 = 3/36

P(B)= n(B)/n() = (2)(2)/36 = 4/36

The probabilities in the example above is possible to acquire since the

sample space is appropriately defined with equally likely outcomes. If

the sample space is defined in this manner: ={=(x,y)| x{B,N,O}

and y{B,N,O}} the outcomes are not anymore equally likely to

occur. We have just seen a case where P({(BB)})


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2000000 333275 0.1666375

Notice that as the number of trials increase, the relative

frequency approaches a certain value, 1/6. Thus, the

relative frequency is a good probability estimate when nis large enough.

Doing the experiment and solely relying on the relative frequency does

not tell exactly that the probability is 1/6. The value could have been

0.16666793. But classical probability gives an evidence that the limit

of the relative frequency as the number of trials approach infinity is

indeed 1/6.

Definition: In Subjective Probability,P(A)is derived from an

individuals personal judgment. If the individual feels

that the event is more likely to occur then P(A)is close

to 1. If it is less likely to occur then P(A)is close to 0.

1.3 The Axiomatic Definition of Probability

Classical probability definition has two main flaws: it is too restrictive

and circular. For probabilities to be defined, classical probability

restricts the outcomes of a random experiment to be equally likely

(e.g. assume the coin is fair, the dice is balanced, etc.). The equally

likely assumption also makes the definition circular. Equally likely is

equally probable the very concept being defined is used as an

assumption. Thus there is a need to provide a definition of probability

which is free from these flaws while still not contradictory to the

classical one.

1.3.1 The Event Space

Not all subsets of the sample space can be considered as an event.These events are not of interest or cannot be measured. This is not

apparent when considering random experiments like tossing a coin or

rolling a die. But when measuring length where the sample space is

the set of positive real numbers, we can consider events like observing

a length between 48 to 90 inches, but not events such as observing


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irrational numbers between 48 to 90 inches. The events under

consideration will form a class which is called the event space.

Definition: Event space F is the class of all events associated with

a given experiment. For mathematical consistency, weconsider the event space to be a -algebra. That is,

i. F. The sample space is an event. Since it

contains all possible outcomes, is called the sure

event.

ii. Closure under complementation. If F

then F.

iii. Closure under countable union. If A1, A2,

A3 is a sequence of events belonging in F, then

U

= F.

Theorem: F. The empty set is also called as the impossible

event. This holds from i and ii.

Theorem: F is closed under finite union. Consider a sequence of

events A1, A2, A3, Ak, Ak+1, Ak+2, where Ak+i =

for i = 0, 1, 2,.

Theorem: F is closed under finite intersection. (pf. exercise)

Theorem: F is closed under countable intersection. (pf. exercise)

1.3.2 Mutually Exclusive Events and Partition of a Set

Definition: Define a sequence of events A1, A2, A3, in F. The

events in the sequence are mutually exclusive if all pairs

Aiand Ajare mutually exclusive. That is,

= .


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It is impossible for any two or any number of events in

the sequence to happen at the same time. Mutually

exclusive events are also called pairwise disjoint events.

Consequently, the definition of mutually exclusive

events also hold for a finite number of events.

Example: In a standard deck of 52 cards, the event of observing

an Ace and the event of observing a king are two

mutually exclusive events since there is no card which

is both an Ace and a king. On the other hand, the event

of observing a king and the event of observing a spade

suit are not mutually exclusive events because of the

fact that there is a king of spades.

Definition: The sets A1, A2, A3, Anin Fis said to be a partition

of a set Aif and only if the following conditions hold:

i. Nonempty: K = ,

ii. U

=

= and

iii. A1, A2, A3, Anare pairwise disjoint.

Exercise: Let A and Bbe sets in F. Show that the sets (A-B), AB

and (B-A)form a partition of . Assume that iin

the definition is true2.

1.3.3 The Axiomatic Definition of Probability

Published in his book Foundations of the Theory of Probability,

Andrey N. Kolmogorov has laid the groundwork for the modern

probability.

Definition: A probability function P() is a set function which

assigns a number P(A) for every A in the -field ,

satisfying the following axioms:

i. P(A) 0 for every F,


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( ) ( ) ( )

=

+=

( )

=

=

Since the probability measure is nonnegative (axiomi), the

last line holds if and only if = 0.

Theorem: Finite Additivity.Let A1, A2,,Anbe a collection of

mutually exclusive events in F, then

( )==

=

U .

Proof: Let A1, A2,,Anbe a collection of mutually exclusive

events in F. Define a countably infinite sequence {Ai}

= A1, A2, , An, An+1, in Fsuch that Ai= for

i>n. By the bound law and the assumption on A1,

A2,,An, the events in the sequence {Ai}are mutually

exclusive. Also by the bound law,

UUUUU K

=

+====

===

.

Now, (supply justifications)

( )

( ) ( ) ( ) ( )

( )

+==

+==

+==

=

==

+=

+=+=

=

=

UU

( )=

=

The example below shows that the classical probability assumption of

equally likely outcomes need not hold when obtaining probabilities

using the axiomatic definition. Another example shows that the


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axiomatic probability definition is consistent with the classical

probability definition.

Example: Consider tossing a biased coin where the event of

observing a head is twice as likely as the event ofobserving a tail. Find the probability of observing a tail.

This is a random experiment where the outcomes are

not equally probable. Still, = {H,T}. Define eventA

= {T}. If we let P(A) = p, then P(Ac) = P({H}) = 2p.

Now to find the value of p,

( ) ( ) ==

( ) ( ) ( )

=

=

+=+=

Example: Consider a sample space with nequally likely outcomes.

Show that the probability of an event is

consistent with the classical definition of probability.

Define ={1, 2, , n}so that n() = n. Let E1,

E2, , Enbe nequally likely events where Ei = {i}, i= 1,2,,n. That is, the Eis form a partition of the

sample space. Define P(Ei) = pfor all i.

Now,

( ) ( )

( )

K

U

===

===

==

===

Using the result above, we get

( ) ( ) ( )

===

=

U


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( ) ( )

= .

Theorem: Probability of a Complement.For any event F,

P(Ac) = 1 P(A).

Proof: Let F. By the complement law, , so that

(supply justifications)

( ) ( )( ) ( )

=+

==

( ) ( )

=

Theorem: Probability of a difference. For any event F,

P(A-B) = P(A) P(AB).

Proof: Let, F. Note that ABand A-B form a partition

of A (verify) so that (supply missing steps or

justifications), P(A) = P(AB) + P(A-B). The result

immediately follows.

Theorem: Probability of a union of two events.For any event

F, P(AB) = P(A) + P(B) P(AB).

Proof: (Exercise. Hint: verify partitioning of some sets and use

the property on probability of a difference.)

Corollary: Probability of a Union of Three Events. For any

events F,

( ) ( ) ( ) ( ) ( ) ( ) ( )( )

+

++=

Proof: (Exercise. Use the property on the probability of a union

of two events.)


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Corollary: For any mutually exclusive events F, P(AB) =

P(A) + P(B). This corollary can also be a direct result

of finite additivity.

Theorem: Monotonicity Property. For any event F, If

then P(A) P(B).

Proof: (exercise)

Corollary: For any event , P(A) 1. To prove this, define

B=and use the previous theorem.

Example: If P(A) = 0, then P(AB) = 0.

Since ABis a subset of A, P(AB) P(A). By axiom (i),

this is possible only when P(AB) = 0.

1.3.4 Properties on Generalized Set Operations

The following special theorems show properties of probabilities of

events expressed as a generalized union or intersection.

Theorem: Inclusion Exclusion Formula. Let A1, A2,,Anbe

a collection of events in F. The probability of the union

of these nevents is given by the formula below:

( ) ( ) ( )

( )

+

+=

=

+


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( ) ( )

( ) ( )( )

++

=

+

=

++

+


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UU

=

=

=

The result follows.

Booles inequality also holds for a finite sequence of events A1,

A2,,An.

Example: Prove that if A and B are events then P(AB) 1

P(AC) P(Bc).

Proof: (supply justifications)

( ) ( )( )

=

=

But

( ) ( ) ( )( ) ( ) ( )( )

+

+

The result immediately follows.

Theorem: Bonferronis Inequality.Define a finite sequence of

eventsA1, A2, , An in F . The following inequality

holds:

( )==

I .

Another form of Bonferronis inequality is as follows:

( ) ( )

==

I .

Proof: (exercise)


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Theorem: Continuity from Below. If {An} is a monotone

nondecreasing sequence of events in Fand

F

then

( ) ( )

=

==

U

Proof: (exercise)

Theorem: Continuity from Above. If {An} is a monotone

nonincreasing sequence of events in F and

F

then

( ) ( )

=

==

I

Proof: (exercise)

1.3.4 Event Composition

Definition: Event Compositionis a way of defining an event interms of other events using set operations. This method

when paired with the properties of the probability

measure can be used to determine the probability of the

composed events.

The following are the steps in obtaining probabilities using event

composition and properties of the probability measure:

Step 1: Define the basic events.

Step 2: List (or compute) the probabilities of these basic events.

Step 3: Express events in question in terms of the basic events.

Step 4: Use the properties of the probability measure to obtain

the probabilities of these events.


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Example: Amelia Mintz and Tony Chu can detect if aspartame is

present in a particular meal. If aspartame is present,

Amelia can detect it with probability 0.95 while Tony

can detect it with probability 0.90. They can both detect

it 88% of the time. If a meal indeed has aspartamepresent, find the probability that

i. at least one of them will detect aspartame

presence.

ii. they will not detect aspartame in the meal.

iii. only Amelia will be able to detect aspartame in

the meal.

Solution: Define the following events:

A = et Amelia detects aspartame in the meal.

B = et Tony detects aspartame in the meal.

Given:

P(A) = 0.95

P(B) = 0.90

P(AB) = 0.88

i. =P(A) + P(B) P(AB)

= 0.95 + 0.90 0.88

= 0.97

ii. =1

= 1 0.97 = 0.03

iii.

= 0.95 0.88 = 0.07

1.3.4 Special Probability Examples

Example: This example shows that there can be nonempty setswhich have a probability of zero.

Suppose probabilities are assigned to the Borel sets of

=[0, 1]in such a way that for any real number aand


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bwhere 0a


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The event above is a match, thus the name matching

problem. The sample space is defined as:

( ) { } ===

KK

where n() = n(n-1)(n-2)(3)(2)(1) = n!.

The random experiment has equiprobable outcomes,

because the hatcheck person is absent-minded and

randomly gives back the hats. Classical probability can

be used in this example.

The following are the probabilities:

P(Ai) = (n-1)!/n! = 1/n

P(AiAj) = (n-2)!/n! = 1/n(n-1)

P(AiAjAk) = (n-3)!/n! = 1/n(n-1)(n-2)

P(A1A2An) = 1/n!

We therefore need to find

=U

, or the probability

of at least one match since the probability of no matchis just 1

=U

. Now, by the inclusion-exclusion

formula,

( ) ( ) ( )

( ) ( )

K

KU

+


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b. sampling is randomly done without replacement

4. Let Aand B be events. Establish the following:

a.

b.

c. Verify if is true or not.

5. Let A and Bbe disjoint events.

a. Are Acand Bcdisjoint?

b. Are and disjoint for any event nonempty set C?

c. Are and disjoint for any event nonempty set C?

6. A delegation of 4 students is selected each year from the School of

Statistics to attend the annual conference of the Philippine

Statistical Association.

a. In how many ways can the delegation be chosen if there are 12

eligible students?

b. There are 7 eligible males while there are 5 eligible females.

What is the probability that the delegation will be composed

of 2 male and 2 female students?

c. Two students are lovers and they are in agreement that they

should never part because love dominates everything they do.

What is the probability that this selection process will test

their supposedly unrelenting mutual pact, that is, find the

probability that the selection process includes either one of

them but not both?

7. After a typhoon, 50% of the residents of a particular municipality

in Camarines Sur were without electricity, 47% without water and

38% without telephone services. One of five residents still have all

three while 10% were without all three, 12% were without

electricity and water but still had a working telephone and 4%

were without electricity and a working telephone but still hadwater. A resident of the municipality is randomly selected for an

interview. Express the given events in set notation and calculate

the following probabilities:

a. the selected resident still has at least one of the utilities

b. the selected resident had water and telephone service but

without electricity


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12.Diseases A, B and C are prevalent among people in a certain

population. It is assumed that 10% of the population will contract

disease Asometime during their lifetime, 25% will contract disease

Band 20% will contract disease Ceventually. There are 5% who

will contract diseases Aand B, 5% who will contract diseases Aand Cand 5% who will contract diseases Band C. Lastly, 3% will

contract all three diseases sometime during their lifetime. A person

is chosen randomly from this population. Find the probability that

this selected person:

a. will contract at least one of the three diseases

b. will never contract any of the three diseases

c. will contract at most one disease

d. will contract only diseases Aand B

will contract exactly two diseases

13.Solve for the probability of total derangement in the old hats

problem where there are 10 guests who have checked their hats in

the counter.

14.There are 5 collectible stickers in pancit canton pouches. An

impulsive collector buys 15 pouches of the product. What is the

probability that the collector solves his problem and collects all 5

different stickers?

15.Prove Bonferronis inequality.

16.Prove continuity from above.

Lecture Notes in Statistics 121 Chapter 1 (Daquis)

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Transcript of Lecture Notes in Statistics 121 Chapter 1 (Daquis)