w3 probability
TRANSCRIPT
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Centre for Computer Technology
ICT114Mathematics for
Computing
Week3
Probability
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Objectives
Review of Week2
Experiments
Multiplication Rule Permutations
Combinations
Probability
Conditional Probability
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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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Measures of Dispersion
Variance,
(x1- )2 + (x2- )2 + .....+ (xn- )22 = -------------------------------------------
n
Standard Deviation = (positive square root)
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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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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Combinations
The number of combinations of n distinctobjects taken r at a time is
n!ncr = --------------
r! (n-r)!
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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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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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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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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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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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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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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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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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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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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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Lets do this together
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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Lets do this together
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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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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Lets do this together
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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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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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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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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(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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Lets do this together
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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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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References
Spiegel, Schiller, Srinivasan : Probabilityand Statistics
M R Spiegel : Theory and Problems ofStatistics, Schaum's Outline Series,McGraw Hill
http://mathworld.wolfram.com