Problems in probability theory with solutions
1. Combinatorics
Problem 1 . There are 30 students in the group. It is necessary to choose a headman, a deputy headman and a trade union organizer. How many ways are there to do this?
Solution. Any of the 30 students can be chosen as a headman, any of the remaining 29 students can be chosen as a deputy, and any of the remaining 28 students can be chosen as a trade union organizer, i.e. n1=30, n2=29, n3=28. According to the multiplication rule, the total number N of ways to select a headman, his deputy and a trade union leader is N=n1´n2´n3=30´29´28=24360.
Problem 2 . Two postmen must deliver 10 letters to 10 addresses. In how many ways can they distribute the work?
Solution. The first letter has n1=2 alternatives - either it is taken to the addressee by the first postman, or by the second. For the second letter there are also n2=2 alternatives, etc., i.e. n1=n2=…=n10=2. Therefore, by virtue of the multiplication rule, the total number of ways to distribute letters between two postmen is equal to
Problem 3. There are 100 parts in the box, of which 30 are 1st grade parts, 50 are 2nd grade, the rest are 3rd grade. How many ways are there to remove one grade 1 or grade 2 part from a box?
Solution. A part of the 1st grade can be extracted in n1=30 ways, a part of the 2nd grade can be extracted in n2=50 ways. According to the sum rule, there are N=n1+n2=30+50=80 ways to extract one part of the 1st or 2nd grade.
Problem 5 . The order of performance of the 7 participants in the competition is determined by lot. How many various options Is it possible to draw lots in this case?
Solution. Each variant of the draw differs only in the order of the participants in the competition, i.e. it is a permutation of 7 elements. Their number is equal
Problem 6 . 10 films are participating in the competition in 5 categories. How many options for prize distribution are there if the following rules are established for all categories? various awards?
Solution. Each of the prize distribution options is a combination of 5 films out of 10, differing from other combinations both in composition and in their order. Since each film can receive awards in one or several categories, the same films can be repeated. Therefore, the number of such combinations is equal to the number of placements with repetitions of 10 elements of 5:
Problem 7 . 16 people participate in a chess tournament. How many games must be played in a tournament if one game must be played between any two participants?
Solution. Each game is played by two participants out of 16 and differs from the others only in the composition of the pairs of participants, i.e., it is a combination of 16 elements of 2 each. Their number is equal to 
Problem 8 . In the conditions of task 6, determine how many options for the distribution of prizes exist if for all nominations identical prizes?
Solution. If the same prizes are established for each nomination, then the order of films in a combination of 5 prizes does not matter, and the number of options is the number of combinations with repetitions of 10 elements of 5, determined by the formula
Task 9. The gardener must plant 6 trees within three days. In how many ways can he distribute his work over the days if he plants at least one tree a day?
Solution. Suppose a gardener plants trees in a row and can take various solutions regarding which tree to stop on the first day and which tree to stop on the second. Thus, one can imagine that the trees are separated by two partitions, each of which can stand in one of 5 places (between the trees). The partitions must be there one at a time, because otherwise not a single tree will be planted on some day. Thus, you need to select 2 elements out of 5 (no repetitions). Therefore, the number of ways .
Problem 10. How many four-digit numbers (possibly starting with zero) are there whose digits add up to 5?
Solution. Let's imagine the number 5 as a sum of consecutive ones, divided into groups by partitions (each group in total forms the next digit of the number). It is clear that 3 such partitions will be needed. There are 6 places for partitions (before all units, between them and after). Each space can be occupied by one or more partitions (in the latter case there are no ones between them, and the corresponding sum is zero). Let's consider these places as elements of a set. Thus, you need to select 3 elements out of 6 (with repetitions). Therefore, the required number of numbers 
Problem 11 . In how many ways can a group of 25 students be divided into three subgroups A, B and C of 6, 9 and 10 people respectively?
Solution. Here n=25, k=3, n1=6, n2=9, n3=10..gif" width="160" height="41">
Problem 1 . There are 5 oranges and 4 apples in a box. 3 fruits are selected at random. What is the probability that all three fruits are oranges?
Solution. The elementary outcomes here are sets that include 3 fruits. Since the order of the fruits is indifferent, we will consider their choice to be unordered (and non-repetitive)..gif" width="21" height="25 src=">. The number of favorable outcomes is equal to the number of ways to choose 3 oranges from the available 5, i.e.. gif" width="161 height=83" height="83">.
Problem 2 . The teacher asks each of the three students to think of any number from 1 to 10. Assuming that each student’s choice of any given number is equally possible, find the probability that one of them will have the same number.
Solution. First, let's calculate the total number of outcomes. The first student chooses one of 10 numbers and has n1=10 possibilities, the second also has n2=10 possibilities, and finally, the third also has n3=10 possibilities. By virtue of the multiplication rule, the total number of ways is equal to: n= n1´n2´n3=103 = 1000, i.e. the entire space contains 1000 elementary outcomes. To calculate the probability of event A, it is convenient to move on to the opposite event, i.e., count the number of cases when all three students think of different numbers. The first one still has m1=10 ways to choose a number. The second student now has only m2=9 possibilities, since he has to take care that his number does not coincide with the intended number of the first student. The third student is even more limited in his choice - he has only m3=8 possibilities. Therefore, the total number of combinations of conceived numbers in which there are no matches is m=10×9×8=720. There are 280 cases in which there are matches. Therefore, the desired probability is equal to P = 280/1000 = 0.28.
Problem 3 . Find the probability that in an 8-digit number exactly 4 digits are the same and the rest are different.
Solution. Event A=(an eight-digit number contains 4 identical digits). From the conditions of the problem it follows that there are five different digits in the number, one of them is repeated. The number of ways to select it is equal to the number of ways to select one digit from 10 digits..gif" width="21" height="25 src="> . Then the number of favorable outcomes. The total number of ways to compose 8-digit numbers is |W|=108 . The required probability is
Problem 4 . Six clients randomly contact 5 firms. Find the probability that no one will contact at least one company.
Solution. Consider the opposite event https://pandia.ru/text/78/307/images/image020_10.gif" width="195" height="41">. The total number of ways to distribute 6 clients across 5 companies. Hence 
. Hence, .
Problem 5 . Let there be N balls in an urn, of which M are white and N–M are black. n balls are drawn from the urn. Find the probability that there will be exactly m white balls among them.
Solution. Since the order of the elements is unimportant here, the number of all possible sets of volume n of N elements is equal to the number of combinations of m white balls, n–m black balls, and, therefore, the required probability is equal to P(A) = https://pandia. ru/text/78/307/images/image031_2.gif" width="167" height="44">.
Problem 7 (meeting problem) . Two persons A and B agreed to meet at certain place between 12 and 1 p.m. The first person to arrive waits for the other person for 20 minutes and then leaves. What is the probability of a meeting between persons A and B, if the arrival of each of them can happen at random within the specified hour and the moments of arrival are independent?
Solution. Let us denote the moment of arrival of person A by x and person B by y. In order for the meeting to take place, it is necessary and sufficient that x-yô £20. Let's depict x and y as coordinates on a plane, choosing the minute as the scale unit. All possible outcomes are represented by the points of a square with a side of 60, and those favorable to the meeting are located in the shaded area. The desired probability is equal to the ratio of the area of the shaded figure (Fig. 2.1) to the area of the entire square: P(A) = (602–402)/602 = 5/9.
3. Basic formulas of probability theory
Problem 1 . There are 10 red and 5 blue buttons in a box. Two buttons are pulled out at random. What is the probability that the buttons will be the same color ?
Solution. The event A=(buttons of the same color are taken out) can be represented as a sum , where the events and mean the choice of red and blue buttons, respectively. The probability of pulling out two red buttons is equal, and the probability of pulling out two blue buttons https://pandia.ru/text/78/307/images/image034_2.gif" width="19 height=23" height="23">.gif" width="249" height="83">
Problem 2 . Among the company’s employees, 28% speak English, 30% speak German, 42% know French; English and German – 8%, English and French – 10%, German and French – 5%, all three languages – 3%. Find the probability that a randomly selected employee of the company: a) knows English or German; b) knows English, German or French; c) does not know any of the listed languages.
Solution. Let us denote by A, B and C the events that a randomly selected employee of the company speaks English, German or French, respectively. Obviously, the proportion of company employees who speak certain languages determines the probabilities of these events. We get:
a) P(AÈB)=P(A)+P(B) -P(AB)=0.28+0.3-0.08=0.5;
b) P(AÈBÈC)=P(A)+P(B)+P(C)-(P(AB)+P(AC)+P(BC))+P(ABC)=0.28+0, 3+0.42-
-(0,08+0,1+0,05)+0,03=0,8;
c) 1-P(AÈBÈC)=0.2.
Problem 3 . The family has two children. What is the probability that the eldest child is a boy if it is known that the family has children of both sexes?
Solution. Let A=(the eldest child is a boy), B=(the family has children of both sexes). Let us assume that the birth of a boy and the birth of a girl are equally probable events. If the birth of a boy is denoted by the letter M, and the birth of a girl by D, then the space of all elementary outcomes consists of four pairs: . In this space, only two outcomes (MD and DM) correspond to event B. Event AB means that the family has children of both sexes. The eldest child is a boy, therefore the second (youngest) child is a girl. This event AB corresponds to one outcome – MD. Thus, |AB|=1, |B|=2 and
Problem 4 . The master, having 10 parts, of which 3 are non-standard, checks the parts one by one until he comes across a standard one. What is the probability that he will check exactly two details?
Solution. Event A=(the master checked exactly two parts) means that during such a check the first part turned out to be non-standard, and the second was standard. This means, where =(the first part turned out to be non-standard) and =(the second part was standard). Obviously, the probability of event A1 is also equal to 
, since before taking the second part the master had 9 parts left, of which only 2 were non-standard and 7 were standard. By the multiplication theorem
Problem 5 . One box contains 3 white and 5 black balls, another box contains 6 white and 4 black balls. Find the probability that a white ball will be drawn from at least one box if one ball is drawn from each box.
Solution. The event A=(a white ball is taken out of at least one box) can be represented as a sum , where the events mean the appearance of a white ball from the first and second boxes, respectively..gif" width="91" height="23">..gif " width="20" height="23 src=">.gif" width="480" height="23">.
Problem 6 . Three examiners take an exam in a certain subject from a group of 30 people, with the first examining 6 students, the second - 3 students, and the third - 21 students (students are selected randomly from a list). The attitude of the three examiners towards those who are poorly prepared is different: the chances of such students passing the exam with the first teacher are 40%, with the second - only 10%, and with the third - 70%. Find the probability that a poorly prepared student will pass the exam .
Solution. Let us denote by the hypotheses that the poorly prepared student answered the first, second and third examiner, respectively. According to the conditions of the problem
, 
, 
.
Let event A=(poorly prepared student passed the exam). Then again, due to the conditions of the problem
, 
, 
.
Using the total probability formula we get:
Problem 7 . The company has three sources of supply of components - companies A, B, C. The share of company A accounts for 50% of the total supply volume, B - 30% and C - 20%. It is known from practice that among the parts supplied by company A, 10% are defective, by company B - 5%, and by company C - 6%. What is the probability that a part taken at random will be suitable?
Solution. Let event G be the appearance of a suitable part. The probabilities of the hypotheses that the part was supplied by companies A, B, C are equal to P(A)=0.5, P(B)=0.3, P(C)=0.2, respectively. The conditional probabilities of the appearance of a good part are equal to P(G|A)=0.9, P(G|B)=0.95, P(G|C)=0.94 (as the probabilities of opposite events to the appearance of a defective part). Using the total probability formula we get:
P(G)=0.5×0.9+0.3×0.95+0.2×0.94=0.923.
Problem 8 (see task 6). Let it be known that the student did not pass the exam, i.e. received an “unsatisfactory” grade. Which of the three teachers was he most likely to answer? ?
Solution. The probability of getting a “failure” is equal to . You need to calculate conditional probabilities. Using Bayes' formulas we get:
https://pandia.ru/text/78/307/images/image059_0.gif" width="183" height="44 src=">, 
.
It follows that, most likely, the poorly prepared student took the exam to a third examiner.
4. Repeated independent tests. Bernoulli's theorem
Problem 1 . The die is thrown 6 times. Find the probability that a six will be rolled exactly 3 times.
Solution. Rolling a die six times can be thought of as a sequence of independent trials with a probability of success ("sixes") of 1/6 and a probability of failure of 5/6. We calculate the required probability using the formula 
.
Problem 2 . The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 2 times.
Solution. The required probability is equal to the sum of the probabilities of three events, consisting in the fact that the coat of arms will not appear even once, or once, or twice:
P(A) = P6(0) + P6(1) + P6(2) = https://pandia.ru/text/78/307/images/image063.gif" width="445 height=24" height= "24">.
Problem 4 . The coin is tossed 3 times. Find the most probable number of successes (coat of arms).
Solution. Possible values for the number of successes in the three trials under consideration are m = 0, 1, 2 or 3. Let Am be the event that the coat of arms appears m times in three coin tosses. Using Bernoulli's formula it is easy to find the probabilities of events Am (see table):
From this table it can be seen that the most probable values are the numbers 1 and 2 (their probabilities are 3/8). The same result can be obtained from Theorem 2. Indeed, n=3, p=1/2, q=1/2. Then
, i.e. .
Task 5. As a result of each visit of the insurance agent, the contract is concluded with probability 0.1. Find the most likely number of concluded contracts after 25 visits.
Solution. We have n=10, p=0.1, q=0.9. The inequality for the most probable number of successes takes the form: 25×0.1–0.9£m*£25×0.1+0.1 or 1.6£m*£2.6. This inequality has only one integer solution, namely, m*=2.
Problem 6 . It is known that the defect rate for a certain part is 0.5%. The inspector checks 1000 parts. What is the probability of finding exactly three defective parts? What is the probability of finding at least three defective parts?
Solution. We have 1000 Bernoulli tests with a probability of “success” p=0.005. Applying the Poisson approximation with λ=np=5, we obtain
2) P1000(m³3)=1-P1000(m<3)=1-»1-,
and P1000(3)"0.14; Р1000(m³3)»0.875.
Problem 7 . The probability of a purchase when a customer visits a store is p=0.75. Find the probability that with 100 visits the client will make a purchase exactly 80 times.
Solution. In this case, n=100, m=80, p=0.75, q=0.25. We find 
, and determine j(x)=0.2036, then the required probability is equal to Р100(80)= 
.
Task 8. The insurance company concluded 40,000 contracts. The probability of an insured event for each of them during the year is 2%. Find the probability that there will be no more than 870 such cases.
Solution. According to the task conditions, n=40000, p=0.02. We find np=800,. To calculate P(m £ 870) we use the Moivre-Laplace integral theorem:
P(0
We find from the table of values of the Laplace function:
P(0 Problem 9
.
The probability of an event occurring in each of 400 independent trials is 0.8. Find a positive number e such that, with probability 0.99, the absolute value of the deviation of the relative frequency of occurrence of an event from its probability does not exceed e. Solution. According to the conditions of the problem, p=0.8, n=400. We use a corollary from the Moivre-Laplace integral theorem: 5.
Discrete random variables Problem 1
.
In a set of 3 keys, only one key fits the door. The keys are searched until a suitable key is found. Construct a distribution law for the random variable x – the number of tested keys .
Solution. The number of keys tried could be 1, 2 or 3. If only one key was tried, this means that this first key immediately matched the door, and the probability of such an event is 1/3. So, Next, if there were 2 tested keys, i.e. x=2, this means that the first key did not work, but the second did. The probability of this event is 2/3×1/2=1/3..gif" width="100" height="21"> The result is the following distribution series: Problem 2
.
Construct the distribution function Fx(x) for the random variable x from problem 1. Solution. The random variable x has three values 1, 2, 3, which divide the entire numerical axis into four intervals: . If x<1, то неравенство x£x невозможно (левее x нет значений случайной величины x) и значит, для такого x функция Fx(x)=0. If 1£x<2, то неравенство x£x возможно только если x=1, а вероятность такого события равна 1/3, поэтому для таких x функция распределения Fx(x)=1/3. If 2£x<3, неравенство x£x означает, что или x=1, или x=2, поэтому в этом случае вероятность P(x And finally, in the case of x³3 the inequality x£x holds for all values of the random variable x, so P(x So we got the following function: Problem 3.
The joint distribution law of random variables x and h is given using the table Calculate the particular laws of distribution of the component quantities x and h. Determine whether they are dependent..gif" width="423" height="23 src=">; https://pandia.ru/text/78/307/images/image086.gif" width="376" height="23 src=">. The partial distribution for h is obtained similarly: https://pandia.ru/text/78/307/images/image088.gif" width="229" height="23 src=">. The obtained probabilities can be written in the same table opposite the corresponding values of random variables: Now let’s answer the question about the independence of random variables x and h..gif" width="108" height="25 src="> in this cell. For example, in the cell for the values x=-1 and h=1 there is a probability of 1/ 16, and the product of the corresponding partial probabilities 1/4 × 1/4 is equal to 1/16, i.e., it coincides with the joint probability. This condition is also checked in the remaining five cells, and it turns out to be true in all. Therefore, random variables x. and h are independent. Note that if our condition were violated in at least one cell, then the quantities should be recognized as dependent. To calculate the probability Let's mark the cells for which the condition https://pandia.ru/text/78/307/images/image092.gif" width="574" height="23 src="> Problem 4
.
Let the random variable ξ have the following distribution law: Calculate the mathematical expectation Mx, variance Dx and standard deviation s. Solution. By definition, the mathematical expectation of x is equal to Standard deviation https://pandia.ru/text/78/307/images/image097.gif" width="51" height="21">. Solution. Let's use the formula Problem 6
.
For a pair of random variables from Problem 3, calculate the covariance cov(x, h). Solution. In the previous problem the mathematical expectation was already calculated .
It remains to calculate And .
Using the partial distribution laws obtained in solving Problem 3, we obtain ; ;
and that means which was to be expected due to the independence of the random variables. Task 7.
The random vector (x, h) takes values (0,0), (1,0), (–1,0), (0,1) and (0,–1) equally likely. Calculate the covariance of random variables x and h. Show that they are dependent. Solution. Since P(x=0)=3/5, P(x=1)=1/5, P(x=–1)=1/5; Р(h=0)=3/5, P(h=1)=1/5, P(h=–1)=1/5, then Мx=3/5´0+1/5´1+1 /5´(–1)=0 and Мh=0; М(xh)=0´0´1/5+1´0´1/5–1´0´1/5+0´1´1/5–0´1´1/5=0. We obtain cov(x, h)=М(xh)–МxМh=0, and the random variables are uncorrelated. However, they are dependent. Let x=1, then the conditional probability of the event (h=0) is equal to P(h=0|x=1)=1 and is not equal to the unconditional probability P(h=0)=3/5, or the probability (ξ=0,η =0) is not equal to the product of probabilities: Р(x=0,h=0)=1/5¹Р(x=0)Р(h=0)=9/25. Therefore, x and h are dependent. Problem 8
. Random increments in the stock prices of two companies for days x and h have a joint distribution given by the table: Find the correlation coefficient. Solution. First of all, we calculate Mxh=0.3-0.2-0.1+0.4=0.4. Next, we find the particular laws of distribution of x and h: We define Mx=0.5-0.5=0; Mh=0.6-0.4=0.2; Dx=1; Dh=1–0.22=0.96; cov(x, h)=0.4. We get Task 9.
Random increments in stock prices of two companies per day have variances Dx=1 and Dh=2, and their correlation coefficient r=0.7. Find the variance of the price increment of a portfolio of 5 shares of the first company and 3 shares of the second company. Solution. Using the properties of dispersion, covariance and the definition of the correlation coefficient, we obtain: Problem 10
.
The distribution of a two-dimensional random variable is given by the table: Find the conditional distribution and conditional expectation h at x=1. Solution. The conditional mathematical expectation is From the conditions of the problem we find the distribution of the components h and x (last column and last row of the table). A.V.: Yulia, you did your diploma in Sergei Choban’s studio “Movement Coordination”, where your design object was the D-1 block in Skolkovo. As far as I can tell, your work was probably the most specific: you were designing for a place whose context had not yet been formed. How does it feel? Yu.A.: Working without an existing context was indeed a bit strange. In the Skolkovo area, the master plan for which was developed by Sergei Choban's Speech bureau together with David Chipperfield's company, we were allocated a plot of land, and we had to figure out what we could do with it. In the first semester, we were divided into 3 groups of 4 people and a competition was announced between us planning solution one block. We had to place on the piece of land that we got twelve, on average five-story, houses, according to the number of students in the group. It so happened that our team won the competition: Anya Shevchenko, Dima Stolbovoy, Artem Slizunov and me. We came up with a fairly strict plan, which was limited not only by some cadastral parameters, but also terms of reference, and design code. What is your master plan? We changed the structure that was in the original version of the general plan: in order to reduce the scale of the environment, we divided our block into 4 sub-blocks with a public space inside each. In addition, each subdivision had its own function: housing, startups, a subdivision with a sports function and a major building, and an area with a dormitory, a hotel, a museum, and the main square is also located here. What restrictions did you write in the design code? The quarter is very small, and the intentions of each of the participants could greatly influence the others. Therefore, we did not prescribe specific materials, but regulated possible changes in shape by setting the “footprint” and FAR. For example, if you “gnaw out”, your number of storeys increases, which in turn is also limited to a certain level. What was the next step? Next, each of us had to develop one of the buildings on the site, but which one, with what function, was determined by lot, we drew papers with “lots”. This was Sergei Tchoban's plan. And this situation is fundamentally different from the one when you yourself choose the topic of your diploma and design a building with a certain function, which, perhaps, you dreamed of designing for all six years of study. Here we had to come to terms with what we were given by lot, and, on the one hand, it was quite painful, but on the other hand, this was a situation close to life. What did you get? I was lucky, in my opinion. I designed a startup building. With certain dimensions, which were not possible to change. The most important principle from which I proceeded was both ideological and functional: today it is a startup, but tomorrow, quite possibly, it will no longer be. After all, what is Skolkovo anyway? No one can answer this question intelligibly. Studying the materials, I concluded that Skolkovo’s own development strategy is quite flexible. For me, this became the main condition that my project had to meet. Therefore, with a building width of 12 meters, it was important for me that there were no extra walls in my building. I left nothing except the stiffening cores, which are mandatory from a design point of view. Inside there is an open, free layout. As for the external appearance, I tried to design my building so that it was quite modest, but at the same time expressive. The main facade turned out to be a 12-meter end facing the boulevard. So I decided to sharpen its shape. The gable roof, which has become the visual accent of the entire building, plays an important role. It is an intermediate link between two “neighbors” of my object, different in height and expressiveness. It changed during the work. At first, the ideological context was a bit overwhelming. And then we began to perceive Skolkovo no longer as a phenomenon on a Russian scale, but to carefully consider the problems of the place itself. After all, today it can be an Innovation Center, and tomorrow it can be something else. So is your building supposed to be demolished? Good architecture can live longer than its original context. She also forms a new one. Was it difficult to work in a group? How were the relationships built within the studio when each of you started working on your own project? Yes, of course it's hard. After all, we succeeded in such a way that the wishes of each person could radically change the situation as a whole. The area is quite small, and someone’s idea to make, say, a console or something else, could affect, for example, insolation standards. And then we all sat down and started discussing whether it was right or wrong. The final version pleasantly surprised me. At first it seemed to me that everyone’s desire to do a wow-thesis-project would outweigh harmonious group work. But in the end the master plan turned out to be quite balanced. I think we managed to find " golden mean"between personal ambitions and the need to follow certain rules of the game. What features did training with Sergei Choban have? All the heads of our studio were very pleasant to work with. In addition to Sergei, these were Alexey Ilyin and Igor Chlenov from the Speech bureau; related specialists also came to help sort out certain components. Educational process was built in an amazingly precise manner, literally minute by minute. Although Sergei, to some extent, probably found it difficult with us. It seems to me that he was counting on the fact that we were almost professionals. And I can’t say that we are still children, but the difference between an office employee and a student is still incredibly great. He shared his knowledge with us not as a teacher, but as a practicing architect and managed to make us work more independently and with each other than with teachers. It really was “coordination of movements.” What did your two years of study at MARCH give you overall? I cannot say that the third eye has opened. But some doubts were resolved, some positions were strengthened. Now I am more responsible for what I do and what I say. Perhaps thank you very much for this MARSH, perhaps thank you very much for this time. I can say that the most valuable thing that MARCH has, the main resource of the school, is the people and some kind of special atmosphere. Mainly I went there for the people. I went to Sergei Sitar, to Kiril Assu, to Evgeniy Viktorovich, to Narine Tyutcheva. Besides, I had you, comrades, who inspired and supported me. I hope we will communicate in the future, I hope we will do something together. Where did you study before this? I defended my bachelor's degree at the Moscow Architectural Institute with the most wonderful teacher Irina Mikhailovna Yastrebova. And I can add that I have a very good attitude towards MARCHI and do not think that it is some kind of Soviet relic. He gives the academic basics, and everyone subsequently decides for themselves what they want to do. What do you want to do now? In all my years of existence in architecture, I have written about it, read about it, talked about it, but I have never created it in the full sense of the word. I was essentially engaged in paper architecture, with such, you know, pretensions to conceptual art. And if earlier I was completely confident that theory determines practice, now I cannot believe in it until I check it. Therefore, I now need to visit a construction site, I need to understand what it is like - when you have done something on paper, then fought for it, argued, agreed, and in the end you stand, look and understand: here it is, it happened! This is my fix idea. Therefore, I plan to practice for the next two years and will try to make my path to construction and implementation as short as possible.
. Hence, 
..gif" width="587" height="41">
. Namely, in each cell of the table we multiply the corresponding values and , multiply the result by the probability pij, and sum all this over all cells of the table. As a result we get:
.
Have you formed your own attitude towards the very idea of the Skolkovo Information Center during the work process?
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