Download - Lecture 04[Math Preliminaries]
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Mathematical Preliminaries
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Mathematical Preliminaries
Set Theory
Combinatorics
Mathematical Functions
Summations
Probability Theory
Topics
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Set Theory
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Set Theory
A Setis a collection of objects called members. If a, b, c, d are members of a set Swe write :S ={a, b, c, d}
where a, b, c, d S ( read a, b, c, d belong to S )
The set builder notation can be used to define a set. For example, the set
S={1, 3, 5, 7, 11, 13}
can be defined as
S = {x: where x is prime number and x
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Set Theory
A setA is called asubsetof setB if every element ofA is also a memberB. Symbolically:
A
B, ifx thenxA B
The setA isproper subsetofB, if A contains only some of the elements ofB. Symbolicallythe relationship is represented as:
A B
A
B
A=B
A B
A
Subsets
The relationship among sets is usually shown by a picture, which is called Venn diagram.
Figures (a), (b) show Venn diagrams for a subset and proper subset
B
(a) A is proper subset of B (b) A is subset of B
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Set Operations
B, ifx or xA BA
Union The union of setA and setB is set of all elementsx such thatx is in A or x is in B.
Symbolically:
Example (1): A={ 1, 3, 4, 5},B={ 3, 4, 7, 8, 9}. The Union of sets A andB is the
set { 1, 3, 4, 5, 7,8, 9}.
U
Example(2): Union operation can be used to add elements to a set. Consider sets {a, b}, {c, d}
(i) Setting S to empty set S=
(ii) To insert elements of first set into S, we perform union of Swith {a, b}
S = S {a, b} = {a, b}
(iii) To insert elements of second set {b, c} into S, we perform union of Swith {c, d}S = S {c, d} = {a, b, c, d}U
U
The Union operation is pictorially represented by Venn diagram, as shown below:
A B A BU
Union set
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Set OperationsIntersection
B, ifx and xA BA
The intersection of setA and setB is set of all elementsx,such that x is in A and
x is in B. Symbolically:
I
Example: Let A={1,2, 3, 4, 5} and B={2, 4, 7, 8} then
A B = {2, 4}
The intersection operation is pictorially represented by Venn Diagram, as shown below.
A B A BI
Set intersection
I
Two sets are called disjoint, if their intersection is empty set. Thus, ifA andB are disjoint,
A B=I
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Set OperationsDifference Set
The difference of setA and setB is set of all elementsx, such thatx is not in B if x is in A.Symbolically:
B A , if x A then x B The difference operation is illustrated by Venn Diagram, as shown below.
Example(1): IfA={1,2,3,4,5} andB={4, 6, 8, 10} thenA B = {1, 2, 3, 5}
Example(2): We can use difference operation to delete elements from a set. Consider
S= {a, b, c, d, e, f }
(i) To delete element a we perform difference operation on S and set {a}.
S = S {a} = { b, c, d, e, f }
(ii) To delete element dwe perform difference operation on set S and set {d}.
S = S-{d} = { b, c, e, f }
A B B - ADifference set
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Set TheoryCartesian Product
The Cartesian productof two sets A and B, denoted byAXB, is a set of all
ordered pairs such that the first element of the pair is in setA and the second element
is in setB. It is also called cross product. Mathematically,
Example: Let A = {a, b, c} andB = {g, h}, then
AXB = {(a, g), (a, h), (b, g), (b, h), (c, g), (c, h) }
AXB = { (a , b) : a A and b B }
The cardinality of Cartesian product is the product of cardinalities of sets
|AXB | = | A | .| B |
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Set TheoryBinary Relation
A binary relation R on setsA andB is a subset of the Cartesian productAXB of the sets .In set notation, the relation is represented as:
R AXB
Example: Consider the sets A = {a, b, c}, B = {x, y}
Their cross product is set AXB = {(a, x), (a, y), (b, x), (b, y), (c, x), (c, y) }
(1) A binary relation R1 is
R1 = { (a, y), (b, y), (c, y) }
(2) Another binary relation R2 is
R2 = {(a, x), (a, y) }
Since a cross product set can have one or more subsets there can be many binary relations
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Binary RelationExample
Agraph G is a pair (V, E ), where V is a finite set, called vertex set, andEis a binaryrelation on V, called edge set
Example: Consider the graph G = (V, E) , and V = { a, b, c }, vertex set for the graph
A binary relation is
E = {(a, a), (a, b), (a, c), (b, c), (c, b) }
The setErepresents edges of graph G. The graph is shown pictorially in the diagram below.
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Combinatorics
P i
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Permutations
Apermutation is an arrangement of n objects in some order, such that each object appearsexactly once.
Elements of a set of n objects can be arranged in the following ways
n.(n-1).(n-2)..3.2.1
because, the first place can be filled in n ways, and the second place in n-1 ways. Thus, first
and second places together can be filled in n(n-1) ways. The first, second and third places can
be filled in n.(n-1).(n-2) ways, and so on
The product n(n-1)(n-2).3.2.1 is called n-factorial, which is denoted as n!
Example: The elements of Set S={ a, b, c } can be permuted in six ways because
3! = 3.2.1 = 6
The six arrangements are: abc, acb, bac, bca, cab, cba
It is assumed that 0! = 1 (factorial of zero is one ). The factorial of a negative integeris
not defined
Definition
i l i
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Factorial FunctionStirlings Approximation
It can be shown that 1000! is an integer consisting of 2,500 digits.
The analysis of algorithms sometimes involves large inputs. For large values of n, the
n! is approximated by the following formula,:
n! ( 2 n ) ( n / e) n , where e = 2.718
The above formula is referred to as Stirlings approximation
2.6 x 10 3532
20,922,789,888,00016
40,3208
244
22
n!n
The n! factorial increases rapidly with the increase of n. The table below shows the
growth.
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CombinationsDefinition
The arrangement of n objects taken k ( k n ) at a time, without regard to the order,is called k-combination.
Example: The objects a, b, c, dhave following 2-combinations:
ab, ac, ad, bc, bd, cd
Observe that the combination ab ba i.e, order does not matter
The number of k-combinations of n objects is denoted by the symbol
Another notation is C(n ,k).
It can be shown that :
The k-combinations are called binomial coefficientsbecause they occur in
binomial theorem for the expansion of the expression:
C(n,k) =
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Mathematical Functions
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Mathematical Functions
Several mathematical functions are used to analyze algorithms and study their behavior.We briefly examine the properties of some these functions From analysis perspective, the
following functions are particularly useful:
Use in Analysis
i. y = x (Floor function)
ii. y = x (Ceil function )
iii. y = log ax (Logarithm function)
iv. y = x (Linear function )
v. y = x2 (Quadratic function)
vi. y = x3 (Cubic function)
vii. y = 2x (Exponential function with base 2)
viii. y =nn (Exponential)
When a mathematical function is used to represent the running time of an algorithm, the
function is referred to asgrowth function. The behavior (growth rate) of a function can berepresented pictorially by plotting agraph of (x,y)points given by the functional relation
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Floor Function
Thefloor function y(x) is defined as:
y = x wherex is a real numberandy is the largestinteger which issmaller than or equal to
numberx.
Example: 4.9= 4,
4.0 = 4
It follows from the definition : x -1
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Ceil Function
The ceil function y(x) is defined as:
y = x
where x is real number andy is thesmallestinteger larger than or equal tox.
Example: 4.9= 5
4.0 = 4
It follows from the definition : x x < x +1
Definition
The ceil function is used for mapping a real valuedfunctions into integerfunction
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The logarithm function y= logax is the inverse of exponential function ay, where a is some
positive constant.
Thus, if y = log a x ,
then
x = a y
Example: Consider, y= log 5 25 . Since 25= 52, it follows y=2 .
Logarithm FunctionDefinition
The figure shows a plot of logarithm function to the base 2
The logarithm function will frequently arise in the analysis of algorithms
i h i
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Logarithm FunctionBases
The choice of base depends on the nature of application and data type.
The base 10 is normally used in computations involving decimal numbers. Here are some
examples
log10
1 =0 because 1= 10 0 by definition
log 10 10 =1 10= 10 1log 10 100 = 2 100=10
2
log 10 1000000 = 6 1000000=106
The base 2 is often used in the analysis of algorithms. Some examples of base 2 are
log 2 1=0 because 1= 20
by definitionlog 2 2= 1 2=21
log 2 16= 4 16= 24
log 2 64= 6 64= 26
The logarithm to base 2 are referred to as binary logarithm. A special notation lg is used
to represent a binary algorithm, as followslg (n) log 2 (n)
The base e 2.718 is used in calculus for purposes of differentiation and integration It iscalled natural logarithm. A special notation ln is used to represent natural logarithms.
ln (n) log e (n)
L ith F ti
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Logarithm FunctionProperties
log bx . y = logbx + logby (Sum rule)
logbx / y = logbx logby (Difference rule)
logbxn = n logbx (Exponentiation rule)
x logby = y logbx ( Symmetry rule )
The following properties of logarithms are often used in several applications
The last property is particularly useful forinterchanging logarithmic exponentialand
base of exponentiation. Here are some examples.
2log 2n = n log 2
2 = n
8log 2n = n log 2
8 = n3
3log 2 n = n log 2 3 = n1.585 ( Interchanging 3 and n and using lg 3 = 1.585)
(Interchanging 2 and n)
(Interchanging 8 and n)
L i h F i
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Logarithm FunctionChange of Base
Example : lg(binary logarithm) can be converted to ln (natural logarithm):
lg x = log2x
= log ex / loge2
= ln x / ln 2
loga x = logbx / logb a
Sometimes it is necessary to change the base of a logarithm. The formula for thetransformation from base a to b is as follows:
( ln 2 = 0.6931)
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Mathematical Functions
3.856 x 10 215
1.269 x 10 89
2.631 x 10 35
20,922,789,888,000
40,320
24
2
n!
3.4 x 10 38
1.8 x 10 19
4,294,967,296
65,536
256
16
4
2n
5.283 x 10 269
3.940 x 10 115
1.461 x 10 48
1.845 x 1019
16,777,216
256
2
nn
7
6
5
4
3
2
1
lg n
128
64
32
16
8
4
2
n
2,097,15216,38489611
262,1444,0963848
32,7681,0241605.7
4,096256644
51264242.8
641682
8421.4
n3n2n lg nn
The table shows the growth rate of common mathematical functions, which are useful in theanalysis of algorithms
Growth Rates
The logarithm lg n has the lowest growth rate , and the exponential function n n
the highest
The relationship of functions in terms of their growth rates is symbolically represented as
lg n < n < n < n lgn < n2
< n3
< 2n
< n! < nn
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Growth FunctionsGraph
n
f(n)
The plot of mathematical functions lg n, n, n lg n n2, n3, 2n is shown below.
Because of extremely rapid growth of n! and nn, these functions are not shown in theranges shown in the graph scale
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Summations
Summation
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Summation
The arithmetic summation consist of sum natural numbers (1, 2, 3.n)Arithmetic
= 1 + 2 + 3+..+ n = n ( n+1 ) / 2
The sum can be simply evaluated, as under.
Let Sbe sum of first n terms of the series.Therefore,
S= 1+ 2 + 3 + + (n-2)+(n-1)+ n [ n terms] (1)
Writing the summation in reverse order
S= n + (n-1)+(n-2)++3 + 2 + 1 [n terms] . .(2)
Adding (1) and (2)
2S= (n+1)+(n+1)+(n+1)+.(n+1) + (n+1) + (n+1) [ n terms] ...(3)
= n(n+1)
Thus, S = n ( n + 1) / 2
Another summation sometimes used in analysis is sum of squares of natural numbers. It has
the sum:
S= 12 + 22 + 32+ +n2 = n(n+1)(2n+1)/6
=
n
k
k1
S ti
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Summation
The geometric summation consists of sum ofpowers of some fixed number.
The fixed number is referred to as geometric ratio. Hers is a common form of exponential
summation:
Geometric
The summation is evaluated as follows. Let S be the sum of first n terms of the series. Then,
S= r 0 + r 1 + r 2+r3 ..+ r n-1 + r n .(1)
Multiplying both sides of (1) with r
r .S = r 1 + r 2+r 3+ +rn-1 + rn +rn+1 ..(2)
Subtracting (1) from (2)(r-1).S = rn+1-r0
= rn+1 -1
Or S= (rn+1-1) / ( r -1) (3)
= r 0 + r 1+ r 2 + ..+ rn-1 + r n = ( rn+1 - 1) / (r - 1)
where r is thegeometric ratio.
Example(1) : Let r = 2, the sum of geometric series is as under
20 + 21+ 22+..+2n =2n+1 - 1
Example (2): Let r = 2/3, the sum is as under
(2/3)0 + (2/3)1 + +(2/3)n = ( (2/3)n+1 -1)/( 2/3 - 1 ) = 3 3(2/3)n+1
=
n
k
r0
k
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Summation
=
n
k
k0
2.
Arithogeometric
Another useful series, referred to as arithogeometric . It consists of sum of the product ofnatural numbers with exponentials of a constant. :
= 1.21 + 2.22 + 3.23+ .+ n.2n = (n-1) 2n+1 + 2k
= 1.m1 + 2.m2 + 3.m3+ .+ n.mnk
=
=
nk
k
km0
where m is some constant
The following summation of arithogeomatric series, with base 2, arises in the analysis of
some algorithms.
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SummationHarmonic
The harmonic summation consists of reciprocals of natural numbers (1,2,..n).
The Harmonic series does not have an exact formula for the sum . The approximate
sum is obtained by converting thesummation into an integral
By using mathematical analysis it can be shown that upperand lowerbounds of
harmonic series are given by the relation
= 1/1 + 1/2 + 1/3 +.+1/n ln n + + 1 / 2n - 1 / 12n
=0.57721 is called Eulers constant
=+
n
k
ln(n)1k11
/ln(n+1)
=
n
k
k1
/1
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Summation
The logarithmic summation includes logarithms of natural numbers (1,2,,n ). Assumingbase 2 for the logarithm, the series is expressed as follows:
= lg(1) + lg(2) + lg(3)+..+lg(n) . = lg(1.2.3..n) =lg( n! )
Logarithmic
An estimate for the above summation is obtained, by using the Stirling s approximation
for large factorial. Since n! ( 2 n ) ( n / e) n where e = 2.718, itfollows that
lg n! n lg( n) ( ignoring lower order terms and constants)
Therefore, the approximate summation for logarithm series for large n is given by :
lg(1) + lg(2) + lg(3)+..+lg(n) n lg( n )
=
n
k
k1
)lg(
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Sample Space
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Sample SpaceDefinition
The set of all possible outcomes of an experiment are calledsample space orpopulation..
Let e1, e2, e3, en are the probable n outcomes of an experiment. Then the sample space
is the set = { e1, e2, e3, en}
The Probability Theory is concerned with the study of experiments whose outcome is notknown in advance ( a priori). However, all likely outcomes are assumed to be finite and
known before the conduct of the experiment
For example, tossing of a coin is an experiment. The outcome of tossing cannot be
predicted in advance, but outcome is likely be Head or Tail
Sample Space
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Sample SpaceExamples
Example (1): Afair coin is tossed. The result can beHead( H ) or Tail( T ). The samplespace for the tossing experiment is:
= {H, T}.
Example (2) : Afair coin is tossed twice ( or two coins are tossed simultaneously) the sample
space for the toss is: = { HH, HT, TH, TT}
Example (3): Asix-faced die is thrown. The sample space for the event is
= {1, 2, 3, 4, 5, 6 }. where digits represent number of dots on each face of the die.
Example(4): Two six-faced dice are tossed. The sample space consists of 36 pairs of
integers:
={(1,1),(1,2)(1,3),(1,4),(1,5), (1,6), (2,1),(2,2)(2,3,),(2,4),(2,5), (2,6),(3,1),(3,2)(3,3),(3,4),(3,5), (3,6), (4,1),(4,2)(4,3,),(4,4),(4,5), (4,6),
(5,1),(5,2)(5,3),(5,4),(5,5), (5,6), (6,1),(6,2)(6,3),(6,4),(6,5), (6,6) }
Example(5): The sample space for weather forecast can be = (fair, rainy, cloudy, partly cloudy, hailstorm, snow}
E t S
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Events SpaceDefinition
An event set consisting of asingle element is called elementary event.
The empty set is referred to as impossible event
A six-faced die is thrown. The sample space consists of all numbers that are likely tobe shown i. e,
={1, 2, 3, 4, 5, 6 }
Example(1): The event space that an even numberis shown:
E1={2, 4, 6}
Example(2): The event space that an of odd number is shown:
E2=(1, 3, 5}
Example(3) : The event space that a numbergreater than 4 is shown
E3={ 5, 6}
A subset E of the sample space is called event space . Symbolically,
E
P b biliti
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ProbabilitiesDefinition
LetEbe an event space consisting of events e1, e2, e3..en. Each of the events in the is
assigned a real value (number) denoted byP(e1), P(e2) , P(e3)..P(en), calledprobabilities, such that
P(e1) + P(e2) + P(e3)+.+ P(en) = 1
The functionP(e) is calledprobability function.
The probabilities are generally assigned on the basis of experience orjudgment. Some
times the probabilities are expressed as percentages, or in terms frequencies of happening of
different events.
Example(1): Chance that a flight will be on time is 90% and would be late is 10% . Thus,
P(on time)= 90/100=0.9 andP(late)=10/100=0.1
Example(2): Based on experience, we say that chance of finding a document on Yahoo is twice that of
finding on MSN, and four times on Google. Ifp is probability of finding the document on MSN then,
we have :P (msn ) = p ; P(yahoo ) = 2p ; P(google)) = 4p
By probability axiom, p + 2p + 4p =1 or p=1/7
Thus, we have P(msn)= 1 / 7, P(yahoo)= 2 / 7, andP(google)=4 / 7
Probability Theory
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Probability TheoryEquiprobable Events
In several cases theprobabilities are not known,but it is expected that all events are equally
likely to occur . Therefore, all the elementary events are assigned equal probabilities.
Let E be event space for equiprobable events
E ={e1, e2, .en}
Then, P( e1) = P( e2) = P(e3) =.= P( en) = p , say, wherep is some constant
Since by definition
P(e1)+P(e2)+P(e3)++P(en)=1
it follows that
p +p + p+..+ p ( n terms ) = 1. i.e n.p= 1 or p = 1/n
Thus, for equiprobable n possible events, P(ej) = 1 / n ( 1 j n)
The functionP(n) is called uniform probability function.
E i b bl E t
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Equiprobable EventsExamples
Example (1) In tossing a fair coin heads and tails have equal chance. The event set is {H, T}.,whereH refers to occurrence ofheadand T to occurrence oftail. Therefore,
P(H)=1/2;
P(T)=1/2
Example (2) When six-faced die is thrown, all numbers, 1 to 6, have equal chance of being
shown. Therefore, each event has probability 1/6. Thus, probabilities of showing different
numbers are
. P(1)=1/6, P(2)=1/6, P(3)=1/6, P(4)=1/6, P(5)=1/6, P(6)=1/6
Example(3): An array stores n random keys. Assuming that a search key has equal chance
of being in any of the array cells, the probability of finding the key in a given cell is 1/n
Example(4): A six-faced die is tossed. The event set that a odd number is shown consists ofthe elements 1, 3, 5. The probability that an odd number would be displayed is 3/6=1/2
P b bilit Th
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Probability TheoryLaw of Addition
According toLaw of Addition of Probabilities, If A andB are two mutually exclusive
events with probabilitiesP(A) andP(B) ,then probability thatA occurs ORB occurs is given
by thesum of probabilities P(A) + P(B). Simply stated ,
P(A ORB)= P(A)+ P(B)
In general, ifA1, A2, A3Akare mutually exclusive events then
P(A1
ORA2
ORA3..ORA
k)=P(A
1)+ P(A
2)+P(A
3)+.+P(A
k)
Two events A, B are said to be mutually exclusive provided that if A occurs thenB cannotoccur; conversely, if B occurs thenA cannot occur. For example, if a six-faced die is tossed
then showing of 1 and 6are mutually exclusive events because if 1 is shown then 6will not be
shown and vice versa.
Law of Addition
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Law of AdditionExamples
Example(2): A die is thrown. What is probability of showing aprime number or a number
greater than 5 ?
(1) Events set that aprime number would be shown is A= {2,3,5}. Each event has the
probability of 1/6. Since these are mutually exclusive events, probability that a prime
number would be shown is given byP( Prime Number) =P( 2 OR 3 OR 5) = 1/6+ 1/6+ 1/6 =1/2
(2)Events set that a number than greater 5 would be shown isB={6} . The Probability that
a number greater than 5 would be shown is
P( Number greater than 5) =1/6(3) Combining the above results, the probability that outcome would be aprime number or a
number greater than 5 is
P(Prime Number OR Number greater than 5)=P(Prime Number) + P(Number greater than 5)
=1/2 + 1/6
= 2/3
Example (1) : A six-faced die is tossed. What is the probability that an even number would
be show.?The likely even numbers are 2,4,6. Each has the probability of 1/6. Thus, according law
of addition
P( Even Number)= P(2 OR 4 OR 6)= 1/6 + 1/6+ 1/6 =1/2
Probability Theory
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Probability TheoryLaw of Multiplication
Two eventsA , B are said to be stochastically independentif occurrence ofA is notinfluenced by the occurrence ofB. Conversely, occurrence ofB is not influenced by
occurrence ofA.
For example, if a coin is tossed several times the result of previous toss does affect the
result of current toss, and result of current toss would not influence the result of next toss.
We can say that the events of successive tossing are stochastically independent events.
According toLaw of Multiplication, ifA andB are stochastically independent events then
the probability of occurrence A AND occurrence ofB is given by the product of theprobabilities P(A).P(B). Mathematically,
P( A AND B) = P(A).P(B)
In general, if A1, A2, A3, .Akare stochastically independent events then
P( A1 ANDA2 AND A3 AND Ak) = P(A1) .P(A2).P(A3). P(Ak)
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Probability Theory
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Probability TheoryExpected Cost
Let c1
, c2
, ..cn
be the costs associated with events e1
, e2
, en
. Let the probabilities for
the occurrence of events be P(e1), P(e2)..P(en) , then the expected cost (EC) is defined as
EC = c1.P( e1) + c2.P( e2)+cn.P (en)
Example: Suppose a coin is tossed 2 times. Assume that the costs associated with events
are :
if two heads appear cost is 2
if one head appears cost is 1
if no heads appear (equivalently only tails show) cost is 0 The event space Sfor the experiment is = { HH, HT, TH, TT}
Probability that event HH occurs = 1/2.1/2= 1/4
Probability that event HT occurs =1/2.1/2=1/4
Probability that event TH occurs = 1/2.1/2=1/4Probability that event TT occurs =1/2.1/2=1/4
The four events have associated costs of 2, 1,1, 0 with probabilities 1/4, 1/4. 1/4, 1/4.
Thus, the expected cost of the tossing experiment = 2.(1/4) + 1.(1/4) + 1.(1/4) + 0.(1/4)
= 1/ 2 + 1/ 4 + 1/ 4 = 1
By law of multiplication
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