# PROBABILITY_ FINITE FIELD OF PROBABILITY. Frequency Consider an by linzhengnd

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```									                       PROBABILITY, FINITE FIELD OF PROBABILITY.

Frequency
Consider an experience and an event A appropriate this experience. Repeat that experience n times
nder identical conditions, note with  number of achievements of the event A and with (n   )
number of failures of A .

Number f n         is called frequency .
n
For example take a coin one hundred times and acknowledge that the containing arms appeared 51
times .
51
Number f100        is the frequency of emblem in the 100 experiments. The frequency
100
varies from experience to experience. It is an exploratory empirical data. Number  can vary from
0 to n including. We have  =0, when the n consecutive repetitions of experience,A event has not
been done ever. Conversely, if the n consecutive experiments, A event took place in all experiences,
results   n .

Experience shows that for many phenomena of mass frequency f n for n growing steadily,
approaching ever more than a certain value. This property is called the law of large numbers.

Events equally possible

A and B two events on the same experience.
If for the sake of perfect symmetry, we can say that both events have the same chance to be
realized, say the events are equally possible.
For example :
1. Experience consists of throwing a coin.
A and B events that leave one side or the other. If the coin is perfect, we have no reason to
assume that one side has a greater chance of occurrence than the other. This is confirmed
experimentally, in that currency throwing a large number of times the two sides appear
about as often. A and B events are equally possible.
2. The experience this time consists of throwing a perfectly cubical dice and build a
homogeneous material, such as center of symmetry coincides with the center of gravity . We
also assumed that the disposal is made on a perfectly flat. In these circumstances we have
no reason to assume that a large number of throws will come mainly from some of the dice.
Elementary events {1},{2},{3},{4},{5},{6} have the same chance to check. We say the events
are equally possible.
Elementary events, where all are equal-when possible, we will call here before equally possible
cases.
Probability

Definition. The probability of an event is equal to the ratio of the number of cases equally possible
carrying out the event and the number of cases equally possible.
In short, we say that the probability of the event A is equal to the ratio of the number of cases
favorable completion event m and n the number of cases equally possible. We write:
P(A) = m / n

For example :

1. An urn contains 20 identical balls numbered
1, 2, ... , 20.
What is the probability that a bile extraction to obtain a numbered with a perfect square?
Note with A the event that we want to calculate the probability. The number of cases equally
possible is 20. The number of cases favorable completion event is 3. These three cases are:
the extraction of the ball 4, ball or 9 ball 16. We therefore:

P(A)=3/20.

2. What probabitatea as throwing two dice to get a "double" ie to get to each of the two dice the
same number of points?
Noting with one of the dice A and B on the other, we have 36 possible cases, as illustrated by
the following matrix  ij  with :
.i  1, 6, j  1, 6 .
3. Suppose we have two urns, the first containing 8 balls numbered 1,2,...,8, and the second 7
balls, numbered 1,2,...,7.
a) What is the probability of making one extraction of each ballot box to get the first ballot
box number?
b) But the probability of obtaining two different parity numbers?
c) But the probability that the number entered on the first ball drawn from urn to be lower
than the second ball drawn from urn?
a) positive cases in the first of these problems can appear in 1,2), (1,4), (1,6), (3,2),
(3,4), (3,6), (5,2), (5,4), (5,6), (7,2), (7,4), (7,6). What number equals 56 possible
cases, probability is 12/56 = 3/14.
b) The second issue we have more favorable cases. In addition to the above we have and
(2,1), (2,3), (2,5), (2,7), (4,1), (4,3), (4,5), (4,7), (6,1), (6,3), (6,5), (6,7), (8,1), (8,3),
(8,5), (8,7). The probability is 28/56 = 1/2.

Properties of probability
The probability of an event A on a note by P(A) enjoys the following properties:

1. 0 ≤ P(A) ≤ 1

2. P(E) = 1

3. P(Ǿ) = 0

4. P( A     B ) = P(A) + P(B), dacă A∩B = Ǿ.

5. P( A ) = 1 - P(A).

Relationship 1. follows immediately from the obvious fact 0  m  n and so

m
0     1
n

To demonstrate the relationship 2., We see that the event m<n means that there are basic events for
which event E not be verified, Ce is impossible, since we know that the event E is always done.

Relationship 3. results from the fact that the event Ǿ we have m = 0. Indeed if we had m > 0 means
that there is at least one case that made possible the Ǿ , Which is absurd.

To demonstrate the relation 4 note the number n equal number of cases possible, by m the number
of cases favorable completion event A , by s the number of cases favorable completion of the event B.

Because A∩B=Ǿ means that events A and B are mutually incompatible. None of the cases favorable
completion event can not achieve on B and vice versa. It follows that m + s it is equally possible
number of cases favorable event A B.

As defined probability, we have:

m          s
P( A)      , P( B)  ,
n          n

ms
P( A B) 
n

So

P( A  B)  P( A)  P( B),

If A      B = Ǿ.

To demonstrate the relationship 5., We see that
A   A  E, A      A Ǿ

On the basis of equality 4., It

P ( A)  P ( A )  P( E ).

Considering 2., Obtain the equality 5.

That relationship often have interest in solving problems. If we want to calculate the probability of
event A note that the reasoning and calculations are more difficult than calculating the probability of
the event contrary A first calculate the probability P ( A ) of it and then calculate:

P ( A)   P ( A ),

Independent Events

Let A and B two events, if

P(A∩B)=P(A)xP(B)

events A and B are, by definition, independent.

For example :

1. Consider the experience consisting of throwing two dice, one red and one green. Be the event
that the red dice experience to show the event as B 1 and green dice to show the 5. A and B
are independent events?
To answer the question must calculate the probability value P(A ∩ B), P(A), P(B).
Elementary events are (j,k) , cu (j = 1, 2, 3, 4, 5, 6; k = 1, 2, 3, 4, 5, 6) where f means the
number of dots on the dice k for the red and the green dice.
All these events are equally possible. We therefore equal to 36 possible cases.We favor a
single case A∩B, namely (1.5). Therefore
1
P( A  B) 
36

For A we have six cases favorable:

(1,1), (1,2), (1,3), (1,4), (1,5), (1,6).

So

6 1
P( A)        .
36 6

Positive cases for B we have:

(1,5), (2,5), (3,5), (4,5), (5,5), (6,5).

Follows
6 1
P( B)         .
36 6

Relationship

P( A  B)  P( A) xP( B) .

is satisfied and so A and B are independent events.

Multiplication rule of probability

In the general case of two events A and B corresponding to two experiments S1 and S2 which have
no connection between them, possible equal number of cases and is favorable considering the
events A and B independent. Therefore if the p probability of completion means the event A, and q
the probability of the event B, the probability to achieve the event A ∩ B is pq.

For example :

1. U urn contains three red and seven white balls and urn V contains three white balls. Extract
one ball from each urn. What is the probability that both balls drawn are white?
Whether the event to get a white ball from urn U. We have 10 possible cases of which 7 are
equally favorable. So:
7
p      .
10

Either event removing a white ball from urn U. We all possible equal to 10 cases of which
only 3 are favorable. Therefore:
3
q        .
10
Notice that even I drew a white ball of U, whether I took a red ball, the number of cases
equally possible in all 10 urn V remains,and the number of cases equally possible favorable
all 3. Events A and B are independent, so we can apply the rule of multiplication of
probabilities. The probability of removing two white balls is:
7 3   21
 
10 10 100
The calculation of this probability can be made without applying the multiplication theorem
of probability, but is more complicated.

OSMAN HAYATI
EM 21

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