Probability and Stochastic Processes for Engineers, 525.414, 1/4/05

Module 1 The Axioms of Probability

SET THEORY We begin our discussion of set theory with a number of definitions and specifications of nomenclature. A set is a collection of objects called elements. The subset, B, of a set A is such that all of the elements of B are contained in A. The set notation is as follows: { }1 2, , nA = " , where the Greek letters zeta, , are the elements of the set A. Alternatively one can write ; " is an element of "i iA A or ; " is not an element of "i iA A We denote the empty set, the one that contains no elements as { } . This is also known as the null set. If a set, A, contains n elements, then the total number of subsets is 2n. This is without regard to order. For example, if the set consists of { },a b , then the possible subsets are { } { } { } { }, , , ,a b a b . Note that the subset { },a b is contained within A and is indistinguishable from { },b a . For example, denote the faces of a die by fi. These faces are elements of the set

{ }1 2 6, , ,S f f f= " . Here , thus S has subsets, 6n = 62 64= { } { } { } { } { }1 6 1 2 1 2 3, , , , , , , , .f f f f f f f S " " " These subsets are represented by their elements. Another example is to toss a coin twice. The outcomes are the set { }, , ,S hh ht th tt= . There are subsets, for example 42 16=

{ } { }{ } { }{ }{ } { }

heads at first toss , ;

only one head , ;

head at first toss, tail on second ;

no head or tail , .

A hh

B th ht

C h

D e

= =

= =

= =

= = { }

ht

t

tc

1

These subsets are represented by their properties. The above examples dealt with discrete events; an example of a continuum of events might be the set of all points on the square of dimension

, where (0,T ) { }set of all points in squareS = (see sketch).

T

T

x

y

00

x Ty T

T

T

x

y

00

x Ty T

A possible subset of S might be the points such that x y , the shaded portion in this sketch:

T

T

yx y x

y < x

X

Y

20

200

x = y

y > x

y < x

From the preceding, its obvious that ( ) 200 .400

P C =

16

Finally, we ask, What is the probability that the trains meet (the commuter knows from experience that this probability is zero)? To attack this problem consider the sketches below. These time lines are in fact, variations on the theme of Venn diagrams.

Train X in station Train Y in station

x x + 4 y + 5y

Train X in station Train Y in station

x x + 4 y + 5y To have any overlap between these two periods, we must satisfy the two following constraints:

x x + 4

y + 5y

x x + 4x x + 4

y + 5y y + 5y

4x y+

x x + 4

y + 5y

x x + 4x x + 4

y + 5y y + 5y

5y x+ The combination of these two requirements gives the event { }4D x y= < 5 . This region is shown shaded in the sketch below,

17

X

Y

20

200

y = x + 4

y = x - 5

5

4

X

Y

20

200

y = x + 4

y = x - 5

5

4

and the resulting probability of this event is

{ } 159.54 5400

P x y < = .

End Module 1

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