Consider a random experiment in which a die made of a heterogeneous material is tossed. Its center of gravity is not at the geometric center. In this case, we cannot consider the outcomes (losing a one, two, etc.) to be equally probable. It is known from physics that the bone will more often fall on the face that is closer to the center of gravity. How to determine the probability of getting, for example, three points? The only thing you can do is roll this die n times (where n is a fairly large number, say n=1000 or n=5000), count the number of three points rolled n 3 and consider the probability of the outcome of rolling three points to be n 3 /n – relative frequency of three points. In a similar way, you can determine the probabilities of other elementary outcomes - one, two, four, etc. In theory, this course of action can be justified by introducing a statistical definition of probability.
Probability P(wi) is defined as the limit on the relative frequency of occurrence of the outcome w i in the process of an unlimited increase in the number of random experiments n, that is
where m n (wi) is the number of random experiments (out of the total number n of random experiments performed) in which the occurrence of an elementary outcome w i was recorded.
Since no evidence is given here, we can only hope that the limit in the last formula exists, basing our hope on life experience and intuition.
In practice, very often problems arise in which it is impossible or extremely difficult to find any other way to determine the probability of an event, other than a statistical determination.
Continuous probability space.
As mentioned earlier, the set of elementary outcomes can be more than countable (that is, uncountable). Thus, an experiment consisting of randomly throwing a point onto a segment has an innumerable number of outcomes. One can imagine that an experiment consisting of measuring temperature at a given moment at a given point also has an uncountable number of outcomes (indeed, temperature can take any value from a certain interval, although in reality we can only measure it with a certain accuracy, and the practical implementation of such experiment will give a finite number of outcomes). In the case of an experiment with an uncountable set W of elementary outcomes, any subset of the set W cannot be considered an event. It should be noted that subsets of W that are not events are mathematical abstractions and do not occur in practical problems. Therefore, in our course this paragraph is optional.
To introduce the definition of a random event, consider a system (finite or countable) of subsets of the space of elementary outcomes W.
If two conditions are met:
1) from A’s membership in this system, it follows that A belongs to this system;
2) from belonging to this system it follows that A i A j belongs to this system
such a system of subsets is called algebra.
Let W be some space of elementary outcomes. Make sure that the two subset systems are:
1) W, Æ; 2) W, A, , Æ (here A is a subset of W) are algebras.
Let A 1 and A 2 belong to some algebra. Prove that A 1 \ A 2 belong to this algebra.
Let us call an s-algebra a system I of subsets of a set W that satisfies condition 1) and condition 2)¢:
2)¢ if the subsets A 1, A 2,¼, A n, ¼ belong to I, then their countable union (by analogy with summation, this countable union is briefly written by the formula) also belongs to I.
A subset A of the set of elementary outcomes W is an event if it belongs to some s-algebra.
It can be proven that if we choose any countable system of events belonging to some s-algebra and carry out with these events any operations accepted in set theory (union, intersection, difference and addition), then the result will be a set or event belonging to the same s-algebra algebra.
Let us formulate an axiom called the axiom of A.N. Kolmogorov.
Each event corresponds to a non-negative number P(A) that does not exceed one, called the probability of event A, and the function P(A) has the following properties:
2) if the events A 1 , A 2 ,..., A n , ¼ are inconsistent, then
If a space of elementary outcomes W, an algebra of events, and a function P defined on it that satisfies the conditions of the above axiom are given, then they say that a probability space is given.
This definition of a probability space can be extended to the case of a finite space of elementary outcomes W. Then the system of all subsets of the set W can be taken as an algebra.
Geometric probability
In one special case, we will give a rule for calculating the probability of an event for a random experiment with an uncountable set of outcomes.
If a one-to-one correspondence can be established between the set W of elementary outcomes of a random experiment and the set of points of some flat figure S (large sigma), and a one-to-one correspondence can also be established between the set of elementary outcomes favorable to event A and the set of points of the flat figure s ( sigma small), which is part of the figure S, then
where s is the area of the figure s, S is the area of the figure S. Here, naturally, it is assumed that the figures S and s have areas. In particular, for example, the figure s can be a straight line segment with an area equal to zero.
Note that in this definition, instead of a flat figure S, we can consider the interval S, and instead of its part s, we can consider the interval s, which entirely belongs to the interval s, and the probability can be represented as the ratio of the lengths of the corresponding intervals.
Example. Two people have lunch in the dining room, which is open from 12 to 13 hours. Each of them comes at a random time and has lunch within 10 minutes. What is the probability of their meeting?
Let x be the time the first one arrives at the dining room, and y the time the second one arrives.
One can establish a one-to-one correspondence between all pairs of numbers (x;y) (or set of outcomes) and the set of points of a square with side equal to 1 on the coordinate plane, where the origin corresponds to the number 12 on the X axis and on the Y axis, as shown in Figure 6. Here, for example, point A corresponds to the outcome that the first one arrived at 12.30, and the second at 13.00. In this case, obviously, the meeting did not take place.
If the first one arrived no later than the second one (y ³ x), then the meeting will take place under the condition 0 £ y - x £ 1/6 (10 minutes is 1/6 hour).
If the second one arrived no later than the first one (x³y), then the meeting will occur under the condition 0 £ x – y £ 1/6..
A one-to-one correspondence can be established between the set of outcomes favorable to the meeting and the set of points in the region s shown in shaded form in Figure 7.
The required probability p is equal to the ratio of the area of region s to the area of the entire square. The area of the square is equal to unity, and the area of the region s can be defined as the difference between unity and the total area of the two triangles shown in Figure 7. This implies:
Problems with solutions.
A coin with a radius of 1.5 cm is thrown onto a chessboard with a square 5 cm wide. Find the probability that the coin will not land on any cell boundary.
Task II.
A bridge spans the 100 m wide river. At some point, when there are two people on the bridge, the bridge collapses and both of them fall into the river. The first one knows how to swim and will be saved. The second one does not know how to swim, and will be saved only if it falls no further than 10 meters from the shore or no further than 10 meters from the first. What is the probability that the second person will be saved?
Task III.
Anti-tank mines are placed on a straight line 15 m apart. A tank 2 m wide drives perpendicular to this straight line. What is the probability that he will not be blown up by a mine?
Task VI.
In the interval (0; 2), two numbers are randomly selected. Find the probability that the square of the larger number is smaller than the smaller number
Two points are thrown at random onto a segment. They break the segment into three parts. What is the probability that a triangle can be formed from the resulting segments?
Task VI.
Three points are thrown at random onto a segment, one after the other. What is the probability that the third point will fall between the first two?
Problem I. The position of a coin on a chessboard is completely determined by the position of its geometric center. The entire set of outcomes can be depicted as a square S with side 5. The entire set of favorable outcomes is then depicted as a square s lying inside the square S, as shown in Figure 1.
The desired probability is then equal to the ratio of the area of the small square to the area of the large square, that is, 4/25
Task II. Let us denote by x the distance from the left bank of the river to the point of fall of the first person, and by y the distance from the left bank to the point of fall of the second person. Obviously, both x and y belong to the interval (0;100). Thus, we can conclude that the entire set of outcomes can be mapped onto a square, the lower left corner of which lies at the origin of coordinates, and the upper right corner lies at the point with coordinates (100;100). Two lanes: 0
Task III.
According to the conditions of the problem, the position of the tank in the gap between two adjacent mines is completely determined by the position of a straight line equidistant from the sides of the tank. This line is perpendicular to the line along which the mines are laid, and the tank is blown up by a mine if this line is located closer than 1 meter from the edge of the gap. Thus, the entire set of outcomes is mapped to an interval of length 15, and the set of favorable outcomes is mapped to an interval of length 13, as shown in Figure 3. The desired probability is 13/15.
Task IV.
Let's denote one of the numbers as x and the other as y. The entire set of possible outcomes is mapped into a square OBCD, two sides of which coincide with the coordinate axes and have a length equal to 2, as shown in Figure 4. Let us assume that y is a smaller number. Then the set of outcomes is mapped into triangle OCD with area equal to 2. The chosen numbers must satisfy two inequalities:
at<х, у>x 2
The set of numbers that satisfy these inequalities is displayed in the shaded area in Figure 4. The area of this area is determined as the difference between the area of the triangle OEG, equal to 1/2, and the area of the curvilinear triangle OFEG. The area s of this curvilinear triangle is given by the formula
and is equal to 1/3. From this we find that the area of the shaded figure OEF is 1/6. Thus, the desired probability is 1/12.
Let the length of the segment be l. If we take x and y to be the distances from the left end of the segment to the points mentioned in the problem statement, then the set of all outcomes can be mapped onto a square with side l, one of the sides of which lies on the x coordinate axis, and the other on the y coordinate axis . If we accept the condition y>x, then the set of outcomes will be mapped onto the triangle OBC shown in Figure 5. The area of this triangle is l 2 /2. The resulting segments will have lengths: x, y–x and l-y. Now let's remember geometry. A triangle can be formed from three segments if and only if the length of each segment is less than the sum of the lengths of the other two segments. This condition in our case leads to a system of three inequalities
The first inequality is transformed to the form x
 |
Task VI. Let's take the length of the segment to be l. Let the distance from the left end of the segment to the first point be x, to the second point – y, and to the third point – z. Then the entire set of outcomes is mapped into a cube, three edges of which lie on the x, y and z axes of the rectangular coordinate system, and with an edge of length l. Let's assume that y>x. Then the set of outcomes will be mapped into the direct prism ABCA 1 B 1 C 1 shown in Figure 6. The condition z>x means that all outcomes will be mapped to the region lying above the plane AD 1 C 1 B shown in Figure 7. This plane is now all valid outcomes will be mapped into a pyramid with a square AA 1 B 1 B at the base and a height B 1 C 1 . All outcomes satisfying the condition z Problems for independent solution. 1. Two ships must approach the same pier. The arrival times of both ships are independent and equally possible during a given day. Determine the probability that one of the steamships will have to wait for the berth to be cleared if the first steamer's stay time is one hour, and the second one is two hours. Answer: 139/1152. 2. An automatic traffic light is installed at the intersection, in which the light is green for one minute and red for half a minute, then green again for one minute and red for half a minute, etc. At a random moment in time, a car approaches the intersection. What is the probability that he will cross the intersection without stopping? Answer: 2/3 3. A coin of radius 1.5 cm is thrown onto an infinite chessboard with a square 5 cm wide. Find the probability that a coin will be located in no more than two squares of the chessboard. Answer: 16/25. 4. A triangle is randomly fit into the circle. What is the probability that it is acute? Answer: 1/4. 5. A triangle is randomly fit into a circle. What is the probability that it is rectangular? Answer: 0. 6. A rod of length a is randomly broken into three parts. Find the probability that the length of each part is greater than a/4. Answer: 1/16.
The concept of probability of an event refers to the fundamental concepts of probability theory. Probability is a quantitative measure of the possibility of the occurrence of a random event A. It is denoted by P(A) and has the following properties.
Probability is a positive number ranging from zero to one:
The probability of an impossible event is zero
The probability of a reliable event is equal to one
Classic definition of probability. Let = (1, 2,…, n) be the space of elementary events that describe all possible elementary outcomes and form a complete group of incompatible and equally possible events. Let event A correspond to a subset of m elementary outcomes
these outcomes are called favorable to event A. In the classical definition of probability, it is believed that the probability of any elementary outcome
and the probability of event A favored by m outcomes is equal to
Hence the definition:
The probability of event A is the ratio of the number of outcomes favorable to this event to the total number of all equally possible incompatible elementary outcomes that form the complete group. The probability is given by the formula
where m is the number of elementary outcomes favorable to event A, and is the number of all possible elementary outcomes of the test.
The classical definition of probability makes it possible in some problems to analytically calculate the probability of an event.
Let an experiment be carried out as a result of which certain events may occur. If these events form a complete group of pairwise incompatible and equally possible events, then experience is said to have symmetry of possible outcomes and is reduced to a “scheme of cases.” For experiments that are reduced to a case scheme, the classical probability formula is applicable.
Example 1.13. The lottery draws 1000 tickets, including 5 winning ones. Determine the probability that when purchasing one lottery ticket you will receive a win
The elementary event of this experience is the purchase of a ticket. Each lottery ticket is unique, as it has its own number, and the purchased ticket is not returned. Event A is that the winning ticket is purchased. When buying one of 1000 tickets, all possible outcomes of this experience will be = 1000, the outcomes form a complete group of incompatible events. The number of outcomes favorable to event A will be equal to =5. Then the probability of winning by buying one ticket is equal to
P(A) = = 0.005
To directly calculate probabilities, it is convenient to use combinatorics formulas. Let us demonstrate this using the example of a sampling control problem.
Example 1.14 Let there be a batch of products, some of which are defective. A portion of the products is selected for control. What is the probability that among the selected products there will be exactly defective ones?
The elementary event in this experiment is the selection of an elemental subset from the original elemental set. The selection of any part of products from a batch of products can be considered equally possible events, so this experience is reduced to a scheme of cases. To calculate the probability of the event A = (among defective products, if they were selected from a batch of defective products), you can apply the classical probability formula. The number of all possible outcomes of the experiment is the number of ways in which products can be selected from batch in, it is equal to the number of combinations of elements by: . An event favorable to event A consists of the product of two elementary events: (from defective products _ are selected (from _ standard products _ are selected). The number of such events, in accordance with the multiplication rule of combinatorics, will be
Then the desired probability
For example, let =100, =10, =10, =1. Then the probability that among the selected 10 products there will be exactly one defective product is equal to
Statistical definition of probability. In order to apply the classical definition of probability in the conditions of a given experiment, it is necessary that the experiment correspond to the pattern of cases, and for most real problems these requirements are practically impossible to meet. However, the probability of an event is an objective reality that exists regardless of whether the classical definition is applicable or not. There is a need for another definition of probability, applicable when experience does not correspond to the pattern of cases.
Let the experiment consist of conducting a series of tests repeating the same experiment, and let event A occur once in a series of experiments. The relative frequency of the event W(A) is the ratio of the number of experiments in which event A occurred to the number of all experiments performed
It has been experimentally proven that frequency has the property of stability: if the number of experiments in a series is large enough, then the relative frequencies of event A in different series of the same experiment differ little from each other.
The statistical probability of an event is the number to which the relative frequencies tend if the number of experiments increases without limit.
Unlike a priori (calculated before experiment) classical probability, statistical probability is a posteriori (obtained after experiment).
Example 1.15 Meteorological observations over 10 years in a certain area showed that the number of rainy days in July in different years was: 2; 4; 3; 2; 4; 3; 2; 3; 5; 3. Determine the probability that any particular day in July will be rainy
Event A is that it will rain on a certain day in July, for example July 10th. The statistics provided do not contain information about which specific days in July it rained, so we can assume that all days are equally possible for this event. Let one year be one series of tests of 31 one days. There are 10 series in total. The relative frequencies of the series are:
The frequencies are different, but they are observed to group around the number 0.1. This number can be taken as the probability of event A. If we take all the days of July for ten years as one series of tests, then the statistical probability of event A will be equal to
Geometric definition of probability. This definition of probability generalizes the classical definition to the case when the space of elementary outcomes includes an uncountable set of elementary events, and the occurrence of each of the events is equally possible. The geometric probability of event A is the ratio of the measure (A) of the region favorable for the occurrence of the event to the measure () of the entire region
If the areas represent a) lengths of segments, b) areas of figures, c) volumes of spatial figures, then the geometric probabilities are respectively equal
Example 1.16. Advertisements are posted at intervals of 10 meters along the shopping row. Some customers have a viewing width of 3 meters. What is the probability that he will not notice the advertisement if he moves perpendicular to the shopping row and can cross the row at any point?
The section of the shopping row located between two advertisements can be represented as a straight segment AB (Fig. 1.6). Then, in order for the buyer to notice the advertisements, he must pass through straight segments AC or DV equal to 3 m. If he crosses the shopping row at one of the points of the segment SD, the length of which is 4 m, then he will not notice the advertisement. The probability of this event will be
Classical and statistical definition of probability. Geometric probability.
The main concept of probability theory is the concept of a random event. A random event is an event that, if certain conditions are met, may or may not occur. For example, hitting a certain object or missing when shooting at this object from a given weapon is a random event.
An event is called reliable if it definitely occurs as a result of the test. An event that cannot happen as a result of a test is called impossible.
Random events are said to be inconsistent in a given trial if no two of them can occur together.
Random events form a complete group if during each trial any of them can appear and no other event inconsistent with them can appear.
Let us consider the complete group of equally possible incompatible random events. We will call such events outcomes. An outcome is said to favor the occurrence of event A if the occurrence of this event entails the occurrence of event A.
The probability of event A is the ratio of the number m of outcomes favorable to this event to the total number n of all equally possible incompatible elementary outcomes that form a complete group
Geometric probability is one way to specify probability; let Ω be a bounded set of Euclidean space having volume λ(Ω) (respectively length or area in a one- or two-dimensional situation), let ω be a point taken randomly from Ω, let the probability that a point will be taken from a subset be proportional to its volume λ (x), then the geometric probability of a subset is defined as the ratio of volumes: The geometric definition of probability is often used in Monte Carlo methods, for example, to approximate the values of multiple definite integrals.
Probability addition and multiplication theorems
Probability addition and multiplication theorems
The sum of two events A and B is an event C, consisting of the occurrence of at least one of the events A or B.
Probability addition theorem
The probability of the sum of two incompatible events is equal to the sum of the probabilities of these events:
P (A + B) = P (A) + P (B).
In the case when events A and B are joint, the verity of their sum is expressed by the formula
P (A + B) = P (A) + P (B) – P (AB),
where AB is the product of events A and B.
Two events are called dependent if the probability of one of them depends on the occurrence or non-occurrence of the other. in the case of dependent events, the concept of conditional probability of an event is introduced.
The conditional probability P(A/B) of event A is the probability of event A, calculated under the condition that event B occurred. Similarly, P(B/A) denotes the conditional probability of event B, provided that event A has occurred.
The product of two events A and B is an event C, consisting of the joint occurrence of event A and event B.
Probability multiplication theorem
The probability of two events occurring is equal to the probability of one of them multiplied by the conditional probability of the other in the presence of the first:
P (AB) = P (A) · P (B/A), or P (AB) = P (B) · P (A/B).
Consequence. The probability of the joint occurrence of two independent events A and B is equal to the product of the probabilities of these events:
P (AB) = P (A) · P (B).
Consequence. When n identical independent trials are performed, in each of which event A appears with probability p, the probability of event A appearing at least once is 1 - (1 - p)n
The probability of at least one event occurring. Example. Bayes formula.
The probability of making at least one mistake on a notebook page is p=0.1. The notebook has 7 written pages. What is the probability P that there is at least one mistake in the notebook?
The probability of the occurrence of event A, consisting of events A1, A2,..., Аn, independent in the aggregate, is equal to the difference between unity and the product of the probabilities of opposite events Ǡ1, Ǡ2, ... Ǡn.
P(A) = 1 - q1q2…qn
The probability of the opposite event is q = 1 - p.
In particular, if all events have the same probability equal to p, then the probability of the occurrence of at least one of these events is equal to:
Р(А) = 1 – qn = 1 – (1 – p)n = 1 – (1 – 0.1)7 = 0.522
Answer: 0.522
Bayes formula.
Let us assume that some experiment is being carried out, and n uniquely possible and incompatible hypotheses with probabilities can be expressed about the conditions for its conduct. Let as a result of the experiment event A may or may not occur, and it is known that if the experiment occurs when the hypothesis is fulfilled, then the question is, how will the probability of hypotheses if it becomes known that event A has occurred? In other words, we are interested in the probability values. Based on relations (4) and (5), we have 
But according to the formula of total probability Therefore 
Formula (12) is called Bayes' formula*.
6.Bernoulli formula. Examples.
Bernoulli's formula is a formula in probability theory that allows you to find the probability of the occurrence of event A during independent trials. Bernoulli's formula allows you to get rid of a large number of calculations - addition and multiplication of probabilities - with a sufficiently large number of tests. Named after the outstanding Swiss mathematician Jacob Bernoulli, who derived the formula.
Formulation
Theorem: If the Probability p of the occurrence of event A in each trial is constant, then the probability that event A will occur k times in n independent trials is equal to: where. .
Proof
Since, as a result of independent tests conducted under identical conditions, an event occurs with probability, therefore the opposite event with probability 
Let us denote the occurrence of an event in a trial with a number. Since the conditions of the experiments are the same, these probabilities are equal. Let an event occur once as a result of experiments, then the other times this event does not occur. An event can appear once in trials in various combinations, the number of which is equal to the number of combinations of elements by This number of combinations is found by the formula: 
In this case, the probability of each combination is equal to the product of the probabilities: 
Applying the theorem of addition of probabilities of incompatible events, we obtain the final Bernoulli Formula:
Local and integral theorems of Laplace. Examples.
Local and integral theorems of Laplace
Local Laplace theorem. The probability that in n independent trials, in each of which the probability of an event occurring is equal to p(0< р < 1), событие наступит ровно k раз (безразлично, в какой последовательности), приближенно равна (тем точнее, чем больше n) 
To determine the values of φ(x), you can use a special table.
Laplace's integral theorem. The probability that in n independent trials, in each of which the probability of an event occurring is equal to p(0< р < 1), событие наступит не менее k1 раз и не более k2 раз, приближенно равна
P(k1;k2)=Φ(x"") - Φ(x")
Here 
-Laplace function The values of the Laplace function are found using a special table.
Example. Find the probability that event A will occur exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.
Solution. According to the condition, n=243; k = 70; p =0.25; q= 0.75. Since n=243 is a fairly large number, we use Laplace’s local theorem: 
where x = (k-np)/ √npq.
Let's find the value of x From the table n we find f(1.37) = 0.1561. Required probability
P(243)(70) = 1/6.75*0.1561 =0.0231.
Numerical characteristics of discrete quantities. Examples
Numerical characteristics of discrete random variables
The distribution law fully characterizes the random variable. However, when it is impossible to find the distribution law, or this is not required, you can limit yourself to finding values called numerical characteristics of a random variable. These values determine some average value around which the values of the random variable are grouped, and the degree to which they are scattered around this average value.
Definition. The mathematical expectation of a discrete random variable is the sum of the products of all possible values of the random variable and their probabilities.
The mathematical expectation exists if the series on the right side of the equality converges absolutely.
From the point of view of probability, we can say that the mathematical expectation is approximately equal to the arithmetic mean of the observed values of the random variable.
Theoretical points. Examples.
The idea of this method is to equate theoretical and empirical points. So we'll start by discussing these concepts.
Let 
-- independent sampling from a distribution depending on an unknown parameter 
The theoretical moment of the -th order is the function 
where is a random variable with a distribution function. We especially note that the theoretical moment is a function of unknown parameters, since the distribution depends on these parameters. We will assume that mathematical expectations exist, at least for the Empirical moment of the th order is called 
Note that, by definition, empirical moments are functions of the sample. notice, that 
-- this is the well-known sample mean.
In order to find estimates of unknown parameters using the method of moments, you should:
explicitly calculate the theoretical moments, and compose the following system of equations for unknown variables
In this system, parameters are considered as fixed.
solve system (35) with respect to variables Since the right side of the system depends on the sample, the result will be functions of 
These are the required parameter estimates using the method of moments.
12. Chebyshev’s inequality. Law of large numbers.
The Chebyshev inequality, also known as the Bienaime–Chebyshev inequality, is a common inequality in measure theory and probability theory. It was first obtained by Bienaime (French) in 1853, and later also by Chebyshev. The inequality used in measure theory is more general; probability theory uses its corollary.
Chebyshev's inequality in measure theory
Chebyshev's inequality in measure theory describes the relationship between the Lebesgue integral and the measure. An analogue of this inequality in probability theory is Markov's inequality. Chebyshev's inequality is also used to prove the embedding of a space in a weak space
Formulations
Let be a space with measure. Let also
Summable by function
Then the following inequality is true:
More generally:
If is a non-negative real measurable function non-decreasing on the domain of definition then In terms of space Let Then
Chebyshev's inequality in probability theory
Chebyshev's inequality in probability theory states that a random variable generally takes values close to its mean. More precisely, it gives an estimate of the probability that a random variable will take a value far from its mean. Chebyshev's inequality is a consequence of Markov's inequality.
Formulations
Let a random variable be defined on a probability space, and its mathematical expectation and variance are finite. Then 
where If , where is the standard deviation and , then we get 
In particular, a random variable with finite variance deviates from the mean by more than standard deviations with probability less than It deviates from the mean by standard deviations with probability less than .
Law of Large Numbers
The basic concepts of probability theory are the concepts of a random event and a random variable. At the same time, it is impossible to predict in advance the result of a test in which this or that event or any specific value of a random variable may or may not appear, since the outcome of the test depends on many random reasons that cannot be taken into account.
However, when tests are repeated several times, patterns characteristic of massive random phenomena are observed. These patterns have the property of stability. The essence of this property is that the specific features of each individual random phenomenon have almost no effect on the average result of a large mass of similar phenomena, and the characteristics of random events and random variables observed in tests, with an unlimited increase in the number of tests, become practically non-random.
Let a large series of experiments of the same type be carried out. The outcome of each individual experience is random and uncertain. However, despite this, the average result of the entire series of experiments loses its random character and becomes natural.
For practice, it is very important to know the conditions under which the combined action of many random causes leads to a result that is almost independent of chance, since it allows one to foresee the course of phenomena. These conditions are indicated in theorems, which are generally called the law of large numbers.
The law of large numbers should not be understood as any one general law associated with large numbers. The law of large numbers is a generalized name for several theorems, from which it follows that with an unlimited increase in the number of trials, average values tend to certain constants.
These include the theorems of Chebyshev and Bernoulli. Chebyshev's theorem is the most general law of large numbers, Bernoulli's theorem is the simplest.
The proof of the theorems, united by the term “law of large numbers,” is based on Chebyshev’s inequality, which establishes the probability of deviation from its mathematical expectation: 
Mathematical formulation
It is necessary to determine the maximum of the linear objective function (linear form) under the conditions 
Sometimes a certain set of restrictions in the form of equalities is also imposed on it, but you can get rid of them by sequentially expressing one variable in terms of others and substituting it in all other equalities and inequalities (as well as in the function). Such a problem is called a “basic” or “standard” problem in linear programming.
A geometric method for solving linear programming problems for two variables. Example.
Solution domain for a linear inequality in two variables 
is a half-plane. In order to determine which of the two half-planes corresponds to this inequality, it is necessary to reduce it to the form or Then the desired half-plane in the first case is located above the straight line a0 + a1x1 + a2x2 = 0, and in the second - below it. If a2=0, then inequality (8) has the form 
; in this case we obtain either a right half-plane or a left half-plane.
The solution domain of a system of inequalities is the intersection of a finite number of half-planes described by each individual inequality. This intersection represents a polygonal region G. It can be either bounded or unbounded, and even empty (if the system of inequalities is inconsistent). 
Rice. 2
The solution domain G has the important property of convexity. A region is called convex if any two of its points can be connected by a segment entirely belonging to the given region. In Fig. 2 shows the convex region G1 and the non-convex region G2. In the region G1, two of its arbitrary points A1 and B1 can be connected by a segment, all points of which belong to the region G1. In the region G2, one can choose two of its points A2 and B2 such that not all points of the segment A2B2 belong to the region G2.
A reference line is a line that has at least one common point with the region, and the entire region is located on one side of this line. In Fig. Figure 2 shows two support lines l1 and l2, i.e. in this case the lines pass through the vertex of the polygon and through one of its sides, respectively.
Similarly, we can give a geometric interpretation of a system of inequalities with three variables. In this case, each inequality describes a half-space, and the entire system is the intersection of half-spaces, i.e., a polyhedron that also has the property of convexity. Here the reference plane passes through a vertex, edge, or face of a polyhedral region.
Based on the introduced concepts, we will consider a geometric method for solving a linear programming problem. Let a linear objective function f = c0 + c1x1 + c2x2 of two independent variables be given, as well as some joint system of linear inequalities describing the solution domain G. It is required to find among the feasible solutions one at which the linear objective function f takes the smallest value.
Let us set the function f equal to some constant value C: f = c0 + c1x1 + c2x2 = C. This value is achieved at points of the line that satisfy the equation. When this line is parallelly transferred in the positive direction of the normal vector n(c1,c2), the linear function f will increase, and when it is transferred in the opposite direction, it decreases.
Let us assume that the straight line written in the form (9), with parallel translation in the positive direction of the vector n, first encounters the region of feasible solutions G at some of its vertices, and the value of the objective function is equal to C1, and the straight line becomes the reference one. Then the value of C1 will be minimal, since further movement of the line in the same direction will lead to an increase in the value of f.
Thus, optimization of a linear objective function on a polygon of feasible solutions occurs at the points of intersection of this polygon with the reference lines corresponding to this objective function. In this case, the intersection can be at one point (at the vertex of the polygon) or at an infinite number of points (at the edge of the polygon).
Simplex method algorithm for solving a general linear programming problem. Table.
Solution algorithms
The most famous and widely used in practice for solving a general linear programming (LP) problem is the simplex method. Despite the fact that the simplex method is a fairly effective algorithm that has shown good results in solving applied LP problems, it is an algorithm with exponential complexity. The reason for this is the combinatorial nature of the simplex method, which sequentially enumerates the vertices of the polyhedron of feasible solutions when searching for the optimal solution.
The first polynomial algorithm, the ellipsoid method, was proposed in 1979 by the Soviet mathematician L. Khachiyan, thus solving a problem that had remained unsolved for a long time. The ellipsoid method has a completely different, non-combinatorial nature than the simplex method. However, from a computational point of view, this method turned out to be unpromising. Nevertheless, the very fact of polynomial complexity of problems led to the creation of a whole class of effective LP algorithms - interior point methods, the first of which was N. Karmarkar's algorithm, proposed in 1984. Algorithms of this type use a continuous interpretation of the LP problem, when instead of enumerating the vertices of the polyhedron for solutions to the LP problem, a search is carried out along trajectories in the space of problem variables that do not pass through the vertices of the polyhedron. The interior point method, which, unlike the simplex method, traverses points from the interior of the feasible region, uses log-barrier nonlinear programming methods developed in the 1960s by Fiacco and McCormick.
24.Special cases in the simplex method: degenerate solution, infinite set of solutions, lack of solution. Examples.
Using the artificial basis method to solve a general linear programming problem. Example.
Artificial basis method.
The artificial basis method is used to find an admissible basis solution to a linear programming problem when the condition contains equality-type constraints. Let's consider the problem:
max(F(x)=∑cixi|∑ajixi=bj, j=1,m; xi≥0).
The so-called “artificial variables” Rj are introduced into the constraints and into the goal function as follows:
∑ajix+Rj=bj, j=1,m;F(x)=∑cixi-M∑Rj
When introducing artificial variables in the artificial basis method into the objective function, they are assigned a sufficiently large coefficient M, which has the meaning of a penalty for introducing artificial variables. In the case of minimization, artificial variables are added to the goal function with a coefficient M. The introduction of artificial variables is permissible if, in the process of solving the problem, they successively vanish.
A simplex table, which is compiled during the solution process using the artificial basis method, is called extended. It differs from the usual one in that it contains two lines for the goal function: one for the component F = ∑cixi, and the other for the component M ∑Rj. Let's consider the procedure for solving the problem using a specific example.
Example 1. Find the maximum of the function F(x) = -x1 + 2x2 - x3 under the restrictions:
x1≥0, x2≥0, x3≥0.
Let's use the artificial basis method. Let's introduce artificial variables into the problem constraints
2x1 + 3x2 + x3 + R1 = 3;
x1 + 3x3 + R2 = 2 ;
Objective function F(x)-M ∑Rj= -x1 + 2x2 - x3 - M(R1+R2).
Let's express the sum R1 + R2 from the system of restrictions: R1 + R2 = 5 - 3x1 - 3x2 - 4x3, then F(x) = -x1 + 2x2 - x3 - M(5 - 3x1 - 3x2 - 4x3) .
When compiling the first simplex table (Table 1), we will assume that the original variables x1, x2, x3 are non-basic, and the introduced artificial variables are basic. In maximization problems, the sign of the coefficients for non-basic variables in the F- and M-rows is reversed. The sign of the constant value in the M-line does not change. Optimization is carried out first along the M-line. The selection of leading columns and rows, all simplex transformations when using the artificial basis method are carried out as in the usual simplex method. 
The maximum negative coefficient in absolute value (-4) determines the leading column and the variable x3, which will go into the basis. The minimum simplex ratio (2/3) corresponds to the second row of the table, therefore, the variable R2 must be excluded from the basis. The leading element is outlined. 
In the artificial basis method, artificial variables excluded from the basis are no longer returned to it, so the columns of elements of such variables are omitted. Table 2. decreased by 1 column. Carrying out a recalculation of this table, we move on to table. 3., in which the line M has been reset, it can be removed. After eliminating all artificial variables from the basis, we obtain an admissible basis solution to the original problem, which in the example under consideration is optimal:
x1=0; x2=7/9; Fmax=8/9.
If, when eliminating the M-string, the solution is not optimal, then the optimization procedure continues and is performed using the usual simplex method. Let's consider an example in which there are restrictions of all types: ≤,=,≥
Dual symmetric linear programming problems. Example.
Definition of a dual problem
Each linear programming problem can be associated in a certain way with some other problem (linear programming), called dual or conjugate with respect to the original or direct problem. Let us define the dual problem in relation to the general linear programming problem, which, as we already know, consists of finding the maximum value of a function under the conditions
is called dual to problem (32)–(34). Problems (32) – (34) and (35) – (37) form a pair of problems, called a dual pair in linear programming. Comparing the two formulated problems, we see that the dual problem is composed according to the following rules:
1. The target function of the original problem (32) – (34) is set to the maximum, and the target function of the dual problem (35) – (37) is set to the minimum.
2. Matrix 
composed of coefficients for unknowns in the system of constraints (33) of the original problem (32) – (34), and a similar matrix 
in the dual problem (35) – (37) are obtained from each other by transposition (i.e., replacing rows with columns and columns with rows).
3. The number of variables in the dual problem (35) – (37) is equal to the number of restrictions in the system (33) of the original problem (32) – (34), and the number of restrictions in the system (36) of the dual problem is the number of variables in the original problem.
4. The coefficients of the unknowns in the objective function (35) of the dual problem (35) – (37) are the free terms in the system (33) of the original problem (32) – (34), and the right-hand sides in the relations of the system (36) of the dual problem are coefficients for the unknowns in the objective function (32) of the original problem.
5. If the variable xj of the original problem (32) – (34) can only take positive values, then the jth condition in the system (36) of the dual problem (35) – (37) is an inequality of the form “? " If the variable xj can take both positive and negative values, then 1 – the relationship in system (54) is an equation. Similar connections take place between the restrictions (33) of the original problem (32) – (34) and the variables of the dual problem (35) – (37). If i – the relation in system (33) of the original problem is an inequality, then the i-th variable of the dual problem . Otherwise, the variable уj can take both positive and negative values.
Dual pairs of problems are usually divided into symmetric and asymmetric. In a symmetric pair of dual problems, constraints (33) of the direct problem and relations (36) of the dual problem are inequalities of the form “ ”. Thus, the variables of both problems can only take non-negative values.
Relationship between the variables of the direct and dual problems. Example.
30.Economic interpretation of dual problems. The significance of zero estimates in solving an economic problem. Examples.
The original task I had a specific economic meaning: the main variables xi denoted the amount of produced products of the i-th type, additional variables denoted the amount of surplus of the corresponding type of resource, each of the inequalities expressed the consumption of a certain type of raw material in comparison with the supply of this raw material. The objective function determined the profit from the sale of all products. Let us now assume that the enterprise has the opportunity to sell raw materials externally. What minimum price should be set for a unit of each type of raw material, provided that the income from the sale of all its reserves is not less than the income from the sale of products that can be produced from this raw material.
Variables y1, y2, y3 will denote the conditional expected price for resources of 1, 2, 3 types, respectively. Then the income from the sale of types of raw materials spent on the production of one unit of product I is equal to: 5y1 + 1·y3. Since the price of type I products is 3 units, then 5y1 + y3 3, due to the fact that the interests of the enterprise require that the income from the sale of raw materials be no less than from the sale of products. It is precisely because of this economic interpretation that the system of restrictions on the dual task takes the form: 
And the objective function G = 400y1 + 300y2 + 100y3 calculates the conditional total cost of all available raw materials. It is clear that by virtue of the first duality theorem, F(x*) = G(y*), equality means that the maximum profit from the sale of all finished products coincides with the minimum conditional price of resources. Conditional optimal prices уi show the lowest cost of resources at which it is profitable to convert these resources into products and produce.
Let us once again pay attention to the fact that yi are only conditional, estimated, and not real prices for raw materials. Otherwise, the reader may find it strange that, for example, y1* = 0. This fact does not mean at all that the real price of the first resource is zero; nothing is free in this world. If the conditional price is equal to zero, it only means that this resource has not been completely used up, is available in excess, and is not in short supply. Indeed, let's look at the first inequality in the system of constraints of Problem I, in which the consumption of the first resource is calculated: 5x1* + 0.4x2* + 2x3* + 0.5x4* = 66< 400. его избыток составляет х5 = 334 ед. при данном оптимальном плане производства. Этот ресурс имеется в избытке, и поэтому для производителя он недефицитен, его условная цена равна 0, его не надо закупать. Наоборот, ресурс 2 и 3 используются полностью, причем у3 = 4 а у2 = 1, т. е. сырье третьего вида более дефицитно, чем второго, его условная цена больше. Если производитель продукции имел бы возможность приобретать дополнительно сырье к уже имеющемуся, с целью получения максимального дохода от производства, то увеличив сырье второго вида на единицу, он бы получил дополнительно доход в у2 денежных единиц, с увеличением на единицу сырья третьего вида, значение целевой функции увеличилось бы еще на у3 единицы.
If the manufacturer is faced with the question, “is it profitable to produce any product, provided that the cost per unit of production is 3, 1, 4 units of 1, 2, 3 types of raw materials, respectively, and the profit from sales is equal to 23 units,” then Due to the economic interpretation of the problem, it is not difficult to answer this question, since the costs and conditional prices of resources are known. Costs are equal to 3, 1, 4, and prices y1* = 0, y2* = 1, y3* = 4. This means that we can calculate the total conditional cost of the resources needed to produce this new product: 3 0 + 1 1 + 4 · 4 = 17< 23. значит продукцию производить выгодно, т. к. прибыль от реализации превышает затраты на ресурсы, в противном случае ответ бы на этот вопрос был отрицательным.
31.Use of the optimal plan and simplex table to determine the sensitivity intervals of the initial data.
32.Use of optimal plan and simplex table to analyze the sensitivity of the objective function. Example.
Transport problem and its properties. Example.
Classic definition of probability.
Various definitions of probability.
Algebra of events.
In order to quantitatively compare events with each other according to the degree of their possibility, obviously, it is necessary to associate a certain number with each event, the more possible the event, the greater the number. We will call this number the probability of an event. Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, probability of an event is a numerical measure of the degree of objective possibility of this event.
The first definition of probability should be considered classical, which arose from the analysis of gambling and was initially applied intuitively.
The classical method of determining probability is based on the concept of equally possible and incompatible events, which are the outcomes of a given experience and form a complete group of incompatible events.
The simplest example of equally possible and incompatible events forming a complete group is the appearance of one or another ball from an urn containing several balls of the same size, weight and other tangible characteristics, differing only in color, thoroughly mixed before being removed.
For this reason, a test whose outcomes form a complete group of incompatible and equally possible events is said to reduce to a pattern of urns, or case scheme, or fits into the classical scheme.
We will simply call the equally possible and incompatible events that make up the complete group cases or chances. Moreover, in each experiment, along with cases, more complex events can occur.
Example: When throwing a dice, along with the cases A i - the loss of i-points on the upper side, we can consider such events as B - the loss of an even number of points, C - the loss of a number of points that are a multiple of three...
In relation to each event that can occur during the experiment, cases are divided into favorable, in which this event occurs, and unfavorable, in which the event does not occur. In the previous example, event B is favored by cases A 2, A 4, A 6; event C – cases A 3, A 6.
Classical probability the occurrence of a certain event is usually called the ratio of the number of cases favorable to the occurrence of this event to the total number of equally possible, incompatible cases that make up the complete group in a given experiment:
Where P(A)– probability of occurrence of event A; m- the number of cases favorable to event A; n- total number of cases.
Examples:
1) (see example above) P(B)=, P(C)=.
2) The urn contains 9 red and 6 blue balls. Find the probability that one or two balls drawn at random will turn out to be red.
A- a red ball drawn at random:
m=9, n=9+6=15, P(A)=
B- two red balls drawn at random:
The following follows from the classical definition of probability: properties(show yourself):
1) The probability of an impossible event is 0;
2) The probability of a reliable event is 1;
3) The probability of any event lies between 0 and 1;
4) The probability of an event opposite to event A,
The classical definition of probability assumes that the number of outcomes of a trial is finite. In practice, very often there are tests, the number of possible cases of which is infinite. At the same time, the weakness of the classical definition is that very often it is impossible to present the test result as a set of elementary events. It is even more difficult to indicate the reasons for considering the elementary outcomes of a test to be equally possible. Usually, the equipossibility of elementary test outcomes is concluded from considerations of symmetry. However, such tasks are very rare in practice. For these reasons, along with the classical definition of probability, other definitions of probability are used.
Statistical probability event A is usually called the relative frequency of occurrence of this event in the tests performed:
where is the probability of occurrence of event A;
– relative frequency of occurrence of event A;
The number of trials in which event A appeared;
Total number of trials.
Unlike classical probability, statistical probability is an experimental characteristic.
Example: To control the quality of products from the batch, 100 products were selected at random, among which 3 products turned out to be defective. Determine the probability of marriage.
.
The statistical method of determining probability is applicable only to those events that have the following properties:
· The events under consideration should be the outcomes of only those tests that can be reproduced an unlimited number of times under the same set of conditions.
· Events must have statistical stability (or stability of relative frequencies). This means that in different series of tests the relative frequency of the event changes little.
· The number of trials resulting in event A must be large enough.
It is easy to verify that the properties of probability arising from the classical definition are also preserved in the statistical definition of probability.
Statistical definition of probability. - concept and types. Classification and features of the category "Statistical determination of probability." 2017, 2018.
Let N trials be performed, and event A occurs exactly M times. The ratio is called the relative frequency of event A and is denoted. The probability of event A is taken to be the number around which the observed relative frequency values are grouped: . ... .
Relative frequency. Let A be a random event that can occur in a given experiment. Let us recall that we are considering experiments that satisfy conditions a), b) of paragraph 2. Let us assume that after repeating the experiment N times, event A occurred M times. Definition... .
There is a large class of events whose probabilities cannot be calculated using the classical definition. First of all, these are events with unequally possible outcomes (for example, the die is “unfair”, the coin is flattened, etc.). In such cases, it can help... [read more].
Classic definition of probability. Subject of probability theory. Random events. Algebra of events. Relative frequency and probability of a random event. Complete group of events. Classic definition of probability. Basic properties of probability.... .
Probability manifests itself when the same random experiment is carried out many times, and in such a way that the results of already conducted experiments do not in any way affect subsequent ones. Under these conditions, the frequency of the occurrence of an event with an unlimited increase in the number of experiments tends to the probability of the event.
Consider a random experiment in which a die made of a heterogeneous material is tossed. Its center of gravity is not at the geometric center. In this case, we cannot consider the outcomes (losing a one, two, etc.) to be equally probable. It is known from physics that the bone will more often fall on the face that is closer to the center of gravity. How to determine the probability of getting, for example, three points? The only thing you can do is roll this die n times (where n-a fairly large number, say n=1000 or n=5000), count the number of three points rolled n 3 and consider the probability of the outcome of rolling three points to be equal to n 3/n- relative frequency of getting three points. In a similar way, you can determine the probabilities of other elementary outcomes - one, two, four, etc.
The classical definition of probability assumes that all elementary outcomes are equally possible. The equality of the outcomes of an experiment is concluded due to considerations of symmetry (as in the case of a coin or a dice). Problems in which symmetry considerations can be used are rare in practice. In many cases it is difficult to provide reasons for believing that all elementary outcomes are equally possible. In this regard, it became necessary to introduce another definition of probability, called statistical. To give this definition, the concept of relative frequency of an event is first introduced.
Definition 18.2.2. Relative frequency of an event, or frequency , called the relation
the number of experiments in which this event occurred to the number of all experiments performed. Let us denote the frequency of event A by W(A), then by definition W(A)= m/n ,
where m is the number of experiments in which event A appeared; n- the number of all experiments performed.
The event frequency has the following properties.
1. The frequency of a random event is the number between zero
and unit:
0< W(A)< 1
2. Frequency of reliable event Ω equal to one:
W(Ω)= 1
3. The frequency of the impossible event Ø is equal to:
W(Ø)=0.
4. The frequency of the sum of two incompatible events A and B is equal to the sum
frequencies of these events:
W(A+ B) = W(A)+ W(B)
Observations made it possible to establish that the relative frequency has the properties of statistical stability: in various series of polynomial tests (in each of which this event may or may not appear), it takes values quite close to some constant. this constant, which is an objective numerical characteristic of a phenomenon, is considered the probability of a given event.
Definition 18.2.3.( Statistical probability of an event is the number around which the frequency values of a given event are grouped in various series of a large number of tests.
More strictly statistical probability P( w i) defined as the limit on the relative frequency of occurrence of the outcome w i in the process of unlimited increase in the number of random experiments n, that is
Where m n(w i) – number of random experiments (out of the total number n random experiments performed) in which the occurrence of an elementary outcome was recorded w i.
In the case of a statistical definition, probability has the same properties as probability defined according to the classical scheme:
properties: 1) the probability of a reliable event is equal to one;
2) the probability of an impossible event is zero; 3) probability
random event lies between zero and one; 4) probability
the sum of two incompatible events is equal to the sum of the probabilities of these events.
Example. Out of 500 parts taken at random, 10 were defective. What is the frequency of defective parts?
W = 10/500 = 1/50 = 0.2
Geometric probability
The classical definition of probability assumes that the number of elementary outcomes is finite. In practice, there are experiments for which the set of such outcomes is infinite.
To overcome the disadvantage of the classical definition of probability, which is that it is not applicable to tests with an infinite number of outcomes, geometric probabilities are introduced - the probabilities of a point falling into a region.
Let the experiment consist of randomly selecting a point from a certain area. We assume that the choice of any point is equally possible. We denote a region defined in space by W. In an experiment involving a random selection of only one point from W, the set W is the space of elementary events. In this case, random events can be considered different subsets from W. We will say that a random event A has occurred if a randomly chosen point x belongs to a subset A, i.e.
Definition 18.2.4.
Let W be some segment, L its length. A – length segment l, belonging to W . Event A consists of a point thrown into a large segment in A hitting. Then
Similarly, if the set W of elementary outcomes of a random experiment is a figure on a plane with area S, and region A, its subset, where a point randomly thrown on W can fall, has area s, the corresponding probability of event A - falling into region A then
And finally, if we are talking about volumetric figures, respectively, W of volume V and the included region A of volume v
Remark 18.2.3.. Strictly speaking, the approach considered here requires the introduction of a more general characteristic (function) of a set - its measure ( mes(A)), special cases of which are length, area and volume, and then the probability of event A will be the ratio of the measure of set A to the measure of set W
Example 1. A circle is inscribed in a square. The dot is randomly thrown into the square. What is the probability that it will fall into the circle? According to the above formula, the corresponding probability will be the ratio of the area of the circle to the area of the square.
Example 2. Two people have lunch in a cafe during their lunch break, which starts at the same time and lasts 1 hour, from 12 to 13 hours. Each of them comes at a random time and has lunch within 10 minutes. What is the probability of their meeting?
Let x- time of arrival at the cafe first, and y- time of arrival of the second. They can only meet when they are both in a cafe.
If the second one arrives no later than the first one ( x ³ y), then the meeting will occur under the condition 0 £ x - y£1/6..
Thus, in the first case we will be satisfied by the condition y£ x+ 1/6, and in the second
y ≥ x- 1/6. The area that satisfies these two conditions is shaded in Fig. 2
In other words, in terms of geometric probability, the probability of meeting is the ratio of the area of the shaded “strip” between the straight lines y= x+ 1/6 and y = x- 1/6 inside the square to the area of the square itself.
Required probability p equal to the ratio of the area of the shaded area to the area of the entire square.. The area of the square is equal to one, and the area of the shaded area can be defined as the difference between one and the total area of the two triangles shown in Figure 7. It follows:
