1 Events and operations over events.
1)Let A,B-events. Define the union A,B -event or sets
(denote by AuB) as the set consisting of choose points belonging to the either A or B or both
(оба
закрашены)
2) Define
the intersection of A and B (
B)
As the set of points that belong to both A and B.
A,B
“
=”
+”
=”*”
Def 2
2 events in a sample space are said to be disjoined if A and b have no points in common that if it is impossible that both A and B occur doing the same performance at the experiment.
A
2 The classical definition of probability.
Def1
Suppose we have an experiment whose by a capital letter such as X called a random variable. Sample space of the experiment is the set of all possible outcomes . IS the sample space is either finite or countably infinite the random variables is said to be discrete .
We generally denote a sample space by the capital greek letter Omega. The sample space Omega corresponds to the set of possible outcomes of the experiment.
Def 2
Let
us choose Omega = {
},
A=
}
When we have P(A) is Probability of A is equal numbers of favorable
divided to numbers of all outcomes corresponds to the classical
probability .
Def 3
Let X be a random variables which denotes the value of the outcomes of certain experiment and a sum that this experiment has only finitely
Many possible outcomes .
A
distribution function for X is a real valued function m whose domain
is Omega and which satisfied : 1)m(
)
for each i.
2)
For
any subset E 
Omega we define the probability of E to be the number P(E) given by
the next P(E) =
Example :
Omega ={1,2,3,4,5,6 }
We find the probability of event - choose the event numbers
C={2,4,6}
Solution
P(c) = 1/6 + 1/6 + 1/6=1/2
If we have the simple event
3)Geometric definition of probability. The problem of meeting. Geometric Probability
Let us choose the area G and g. A- event at random gets the on point
P(A)=S(g)/S(G)
There is g-favorable outcomes , G- Omega
Meeting problem is a typical application of geometric probability.
Two persons A and B agreed to meet at a given place between 11 and 12 o’clock
If
1 person come first he should wait 
.
Find the probability of event C={Person A and B had meet}
Solution:
x-1st person
Y-2nd person
|x-y|
S(G)=3600
S(g)=3600- (2 *40*40)/2=2000
P(C) =2000/3600=5/9
y
x
4)Elements of the combinatory
Ordered Sample of size K, with replacement.
The
Number of ordered sequence {
}
where 
belong to 
}
Is n*n*n….*n (k times) or 
Ordered Sample of size K, without replacement.
The
numberof ordered sequence {
} where the
belong to 
But repetition is not allowed
The
element 
can be appear more then once in the sequence
n(n-1
)….. (n-k+1)=
k from 1to n
Unordered Sample of size k without replacement
The number of unordered sets { } where belong to } k from 1to n are distaned elements selecting k distined object out of n
If order doesn’t count
=
Unordered Sample of size k with replacement
We wish ti find the number of unordered sets { } where belong to } and repetition is allowed
5) A permutation. The number of permutations of n objects.
The notion of permutation is used with several slightly different meanings, all related to the act of permuting objects or values. Informally, a permutation of a set of objects is an arrangement of those objects into a particular order. For example, there are six permutations of the set {1,2,3}, namely (1,2,3), (1,3,2), (2,1,3), (2,3,1), (3,1,2), and (3,2,1).
The number of permutations of n distinct objects taken k at a time, denoted by Pnk, where repetitions are not allowed, is given by
Pnk=n(n−1)(n−2)...(n−r+1)=n!/(n−k)!