w3 probability

8/2/2019 W3 Probability

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Centre for Computer Technology

ICT114Mathematics for

Computing

Week3

Probability


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March 20, 2012

Objectives

Review of Week2

Experiments

Multiplication Rule Permutations

Combinations

Probability

Conditional Probability


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March 20, 2012

Measures of Central Tendency

x1+x2+x3+.....+xn

Mean, = ------------------------

n

Median is the middlemost number

Mode of a data set is the value that occursmost often


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March 20, 2012

Measures of Dispersion

Variance,

(x1- )2 + (x2- )2 + .....+ (xn- )22 = -------------------------------------------

n

Standard Deviation = (positive square root)


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March 20, 2012

Weighted Mean

For a given set of data, X= { x1, x2, ..., xn}

and corresponding non-negative weights,

W= { w1, w2, ..., wn}the weighted mean/average, is given by

w1x1+w2x2+w3x3+.....+wnxn

X = ---------------------------------------w1+w2+w3++wn


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March 20, 2012

Mean and Variance from

Frequency TableInterval mid point (x) frequency (f) f.X f.X2

a0- a1 x1 f1 f1.x1 f1.x1.x1

a1- a2 x2 f2 f2.x2 f2.x2.x2

an-1an xn fn fn.xn fn.xn.xn

All Total f Total f.x Total f.x.x

Mean = total (f.x) / total f

Variance = total (f.x.x)/total f (mean)2


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Centre for Computer Technology

Probability


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March 20, 2012

Experiments

Experiment is a term used to describe anyprocess that generates a data set.

Experiments when performed under verynearly identical conditions, give resultsthat are identical.

In other words, the variables that affect theoutcome of the result are controlled.


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March 20, 2012

Random Experiments

In some experiments we cannot controlcertain variables that affect the outcome,

even though most of the conditions are thesame.

Such experiments are called random

experiments. In most cases the outcome is affected by

chance.


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March 20, 2012

Random Experiments

In a random experiment, we obtain anentire set of possible outcomes.

Example: Tossing a die.

The result of the experiment is that it will

come up with one of the numbers thatbelongs to the set {1, 2, 3, 4, 5, 6}


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March 20, 2012

Sample Space

The set of all possible outcomes of astatistical experiment is called Sample

Space. It is generally denoted by the symbol, S,

which corresponds to the universal set.

Each element in a sample space is calledan element or a member or a samplepoint.


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March 20, 2012

Sample Space

Frequently there are more than onesample space possible.

Generally, there is one set that providesthe most information.

Example: Tossing a die

Possible outcomes are

{1, 2, 3, 4, 5, 6}

{even, odd}


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March 20, 2012

Sample Space

If the sample space has a finite number ofelements, it is called a finite sample space

or a discrete sample space.

If the sample space has a non countable

number of elements, it is called an infinitesample space or a nondiscrete samplespace.


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March 20, 2012

Sample Space

Example:

Sample space resulting from the tossing ofa coin, yields a discrete sample space.

Picking any number, not just integers, from1 to 10 yields a nondiscrete sample space.


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March 20, 2012

Sample Space

One form to representa sample space is bythe use of a tree

diagram.Example:Consider theselection of two itemsat random from aproduction line. Eachitem is classifieddefective (D) or nondefective (N).


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March 20, 2012

Event

An event is a subset of the samplespace S, generally denoted by A.

In other words it is a set of possibleoutcomes for the experiment.

If the outcome is an element of A,

then the event A has occurred.


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March 20, 2012

Event

An event which consists of a single pointof S is called a simple or elementary

event. S itself is called the sure or certain event,

since an element of S must occur.

The empty set (null) is called theimpossible element because an element of cannot occur.


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March 20, 2012

Event

Let A and B represent two events in S.

1. A U Bis the event either A or B or both.

2. A Bis the event both A and B3. A Bis the event A but not B.

4. A/is the event not A.

5. If A and B are disjoint, i.e., A B = ,then the events are mutually exclusive


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March 20, 2012

Counting Sample Points

In statistics, there is a need to evaluate thechance linked with the occurrence of

certain elements during an experiment. Sometimes all the sample points are

counted without actually listing them.

The basic principle of counting is calledthe multiplication rule.


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March 20, 2012

Multiplication Rule

If an operation can be performed in n1ways, and if for each of these a secondoperation can be performed in n2 ways,

then the two operations can be performedtogether in n1.n2 ways.

Example: Number of sample points whena pair of dice is thrown is (6).(6) = 36ways.


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March 20, 2012

In the previous example,

The first dice can land in n1=6 ways

The second dice can also land in n2

=6 ways

Using the Multiplication rule,

The pair of dice can land inn1.n2= (6).(6) = 36 ways


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March 20, 2012

Example: A developer of a new subdivision

offers prospective home buyers a choice ofTudor, rustic, colonial, and traditional

exterior styling in ranch, two-storey, and

split-level floor plans. In how many differentways can a buyer order one of these

homes?

The buyer has to choose fromn1=4 and n2=3 homes.

n1n2=(4).(3)=12 possible homes


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March 20, 2012

Generalized Multiplication Rule

If an operation can be performed in n1ways, and if for each of these a second

operation can be performed in n2 ways,and for each of the first two a thirdoperation can be performed in n3 ways,and so forth, then the sequence of k

operations can be performed inn1.n2.n3nk ways.


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March 20, 2012

Permutations

When dealing with a group of objects,there are a number of possible

arrangements. For example, a number of different

arrangements possible when a group ofsix people are sitting around a table.

A permutation is an arrangement of all orpart of a set of objects.


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March 20, 2012

Permutations

Example: What are the differentarrangements possible with the alphabets

a, b, c taking all of them at a time.

The possible permutations are

abc, acb, bca, bac, cab, cba


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March 20, 2012

There are n1=3 choices for the first

position, n2=2 choices for the secondposition and n3=1 choice for the lastposition.

The total permutations/arrangements aren1. n2. n3 = 3.2.1 = 6

n distinct objects can be arranged in

n.(n-1).(n-2)1 ways


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March 20, 2012

Permutations

n distinct objects can be arranged in

n.(n-1).(n-2)1 ways

(the number of permutations for n distinct objects)The product is represented by n!read as n factorial.

By definition 0! = 1! = 1


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March 20, 2012

Permutations

The number of permutations for n distinctobjects taken r at a time

n.(n-1).(n-2)(n-r+1)

The above product is represented by

npr = n! / (n-r)!


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March 20, 2012

Example: Three awards (research, teaching,

service) will be given one year for a class of 25graduate students in a statistics department. Ifeach student can receive at most one award,how many possible selections are there?

Since the awards are distinguishable, the totalpossible selections are

25p3 = 25! / (25-3)! = 25!/22! = 25.24.23

= 13800


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March 20, 2012

Lets do this together

A president and a treasurer are to bechosen from a student club consisting of50 people. How many different choices of

officers are possible ifa. there are no restrictions

b. A will serve only if he is president

c. B and C will serve together or not at alld. D and E will not serve together


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March 20, 2012

Some more concepts

The number of permutations of n distinct objectsarranged in a circle is (n-1)!

The number of distinct permutations of n thingsof which n1 are of one kind, n2 of a secondkind,.nk of a k

th kind is

n!

-------------------

(n1!)(n2!)(nk!)


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March 20, 2012

Some more concepts

The number of arrangements of a set of nobjects into r cells with n1 elements in thefirst cell, n2 elements in the second, and

so forth, isn!

-----------------------

(n1!)(n2!)..(nr!)n1+n2++nr = n


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March 20, 2012

Combinations

A combination is a way of selecting robjects from n without regard to

order. It is a partition with two cells.

One cell contains ther objectsselected and the other cell containing

the (n-r) objects that are left.


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March 20, 2012

Combinations

The number of combinations of n distinctobjects taken r at a time is

n!ncr = --------------

r! (n-r)!


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March 20, 2012

Example

Example:

A young boy asks his mother to get fivegame-boy cartridges from his collection of10 arcade and 5 sports games. How manyways are there that his mother will get 3arcade and 2 sports games, respectively?


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March 20, 2012

Example (contd)

The number of ways of selecting 3 cartridgesfrom 10 is 10c3= 120.

The number of ways of selecting 2 cartridgesfrom 5 is 5c2 = 10

Using the multiplication rule with n1=120 and n2= 10, there are

(120).(10) = 1200 ways


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March 20, 2012

Concept of Probability

In a random experiment, there isuncertainty if a particular event will or willnot occur.

As a measure of chance or probability, avalue between 0 and 1 is assigned.

If an event certainly occurs, the probabilityof the event happening is 100% or 1.Similarly for other degrees of uncertainty.


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March 20, 2012

Two important procedures

Classical ApproachIf an event can occur in h different ways

out of a total of n possible ways, all ofwhich are equally likely, then theprobability of that event occurring is h/n.


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March 20, 2012

Two important procedures

Frequency ApproachIf after n repetitions of an experiment,

where n is very large, an event isobserved to occur in h of these, then theprobability of the event is h/n. This is also

called the empirical probability of theevent.


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March 20, 2012

Axiomatic approach

Let S be a sample space

Let C be a class of events and A be one of

them To each event A in C, a real number P(A)

is associated.

P is called the probability function andP(A) the probability of the element.


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March 20, 2012

Axioms of Probability

Axiom1: For every event A in class C

P(A) 0

Axiom2: For the sure or certain event S inthe class C, P(S) = 1

Axiom3: For any number of mutually exclusive events A1,A2, in the class C

P(A1UA2) = P(A1)+P(A2)+.


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March 20, 2012

Some Important Theorems

If A1A2, then

P(A1) P(A2)

P(A2-A1)=P(A1)-P(A2) For every event A,

0 P(A) 1

For , the empty set,

P() = 0


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March 20, 2012

Some Important Theorems

If A is the complement of A, then

P(A) = 1 P(A)

If A = A1 U A2 UU An, where A1,A2,An are mutually exclusive events,

then, P(A) = P(A1)+P(A2)++P(An)

If A and A are complementary events,then, P(A)+P(A) = 1


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March 20, 2012

Example: A coin is tossed twice. What is

the probability that at least one headoccurs?

The sample space for this experiment is

S = {HH, HT, TH, TT}

The number of times atleast one headoccurs is 3.

Therefore the probability that atleast onehead occurs is P(A) = .


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Draw one card from an ordinary deck ofcards.

a) What is the probability that it is theQueen of Hearts?

b) What is the probability that it is either

the King or the Queen of Hearts?c) What is the probability of getting boththe King and Queen of Hearts?

March 20, 2012


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March 20, 2012


A die is loaded in such a way that an evennumber is twice as likely to occur as anodd number. If E is the event that anumber less than 4 occurs on a singlethrow of the die, find P(E).

Hint : P(S) = P(A1)+P(A2)+.+P(An) = 1


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March 20, 2012

Addition Theorem of Probability

If A and B are any two events, then

P(AUB)=P(A)+P(B)-P(A B)


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March 20, 2012


John is a graduate who has beeninterviewed by two companies A and B.He assesses that his probability of gettingan offer from A is 0.8, probability of gettingan offer from B is 0.6. He also believesthat his chance of getting an offer from

both A and B is 0.5. What is the probabilityof he getting an offer from atleast one ofthese companies?


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March 20, 2012

Conditional Probability

The probability of an event B occurring, when anevent A has already occurred is calledconditional probability, represented by P(B/A).

The conditional probability of B, given A,denoted by P(B/A), is defined by

P(A B)

P(B/A) = ---------------- if P(A)>0P(A)


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March 20, 2012

Multiplication Theorem

If in an experiment the events A and B canboth occur, then

P(AB) = P(BA)

= P(B/A).P(A)

= P(A/B).P(B)


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March 20, 2012

Example: The probability that a regularly

scheduled flight departs on time isP(D)=0.83, the probability that that itarrives on time is P(A)=0.82, and the

probability that it departs and arrives ontime is P(DA)=0.78. Find the probability

that a plane (a) arrives on time given that

it departed on time, and (b) departed ontime given that it has arrived on time.


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March 20, 2012

(a) The probability that a plane arrives on

time given that it departed on time isP(A/D) = P(DA)/P(D) = 0.94

(b) The probability that a plane departedon time given that it has arrived on time is

P(D/A) = P(DA)/P(A) = 0.95


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March 20, 2012


Suppose we have a fuse box containing20 fuses, of which 5 are defective. If twofuses are selected at random andremoved from the box in successionwithout replacing the first, what is theprobability that both fuses are defective?


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March 20, 2012

Summary

Permutations, npr =n! / { (n-r)! }

Combinations, ncr = n! / { r! (n-r)! }

If an event occurs in h different ways out of a

total of n possible ways, then the probabilityis h/n.

Addition Theorem,

P(AUB) = P(A) + P(B) - P(A B) Conditional Probability,

P(B/A) = P(A B) / P(A)


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March 20 2012

References

Spiegel, Schiller, Srinivasan : Probabilityand Statistics

M R Spiegel : Theory and Problems ofStatistics, Schaum's Outline Series,McGraw Hill

http://mathworld.wolfram.com

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