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]
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.