The probabilities P(H i) of the hypotheses H i are called a priori probabilities - the probabilities before the experiments.
The probabilities P(A/H i) are called a posteriori probabilities - the probabilities of the hypotheses H i refined as a result of the experiment.

Example #1. The device can be assembled from high quality parts and from parts of ordinary quality. About 40% of devices are assembled from high quality parts. If the device is assembled from high quality parts, its reliability (probability uptime) over time t is 0.95; if from parts of ordinary quality - its reliability is 0.7. The device was tested for time t and worked flawlessly. Find the probability that it is assembled from high quality parts.
Solution. Two hypotheses are possible: H 1 - the device is assembled from high-quality parts; H 2 - the device is assembled from parts of ordinary quality. The probabilities of these hypotheses before the experiment: P(H 1) = 0.4, P(H 2) = 0.6. As a result of the experiment, event A was observed - the device worked flawlessly for time t. Conditional probabilities of this event under hypotheses H 1 and H 2 are: P(A|H 1) = 0.95; P(A|H 2) = 0.7. Using formula (12), we find the probability of the hypothesis H 1 after the experiment:

Example #2. Two shooters independently shoot at the same target, each firing one shot. The probability of hitting the target for the first shooter is 0.8, for the second 0.4. After shooting, one hole was found in the target. Assuming that two shooters cannot hit the same point, find the probability that the first shooter hit the target.
Solution. Let event A be one hole found in the target after shooting. Before the start of the shooting, hypotheses are possible:
H 1 - neither the first nor the second shooter will hit, the probability of this hypothesis: P(H 1) = 0.2 0.6 = 0.12.
H 2 - both shooters will hit, P(H 2) = 0.8 0.4 = 0.32.
H 3 - the first shooter will hit, and the second will not hit, P(H 3) = 0.8 0.6 = 0.48.
H 4 - the first shooter will not hit, but the second will hit, P (H 4) = 0.2 0.4 = 0.08.
The conditional probabilities of the event A under these hypotheses are:

After experience, the hypotheses H 1 and H 2 become impossible, and the probabilities of the hypotheses H 3 and H 4
will be equal:

Example #3. In the assembly shop, an electric motor is connected to the device. Electric motors are supplied by three manufacturers. There are 19.6 and 11 electric motors of the named plants in the warehouse, respectively, which can work without failure until the end of the warranty period, respectively, with probabilities of 0.85, 0.76 and 0.71. The worker randomly takes one engine and mounts it to the device. Find the probability that the electric motor, mounted and working without fail until the end of the warranty period, was supplied by the first, second or third manufacturer, respectively.
Solution. The first test is the choice of the electric motor, the second is the operation of the electric motor during the warranty period. Consider the following events:
A - the electric motor works flawlessly until the end of the warranty period;
H 1 - the fitter will take the engine from the products of the first plant;
H 2 - the fitter will take the engine from the products of the second plant;
H 3 - the fitter will take the engine from the products of the third plant.
The probability of event A is calculated by the formula full probability:

Conditional probabilities are specified in the problem statement:

Let's find the probabilities

Example #4. The probabilities that during the operation of the system, which consists of three elements, the elements with numbers 1, 2 and 3 will fail, are related as 3: 2: 5. The probabilities of detecting failures of these elements are 0.95, respectively; 0.9 and 0.6.

b) Under the conditions of this task, a failure was detected during the operation of the system. Which element is most likely to fail?

Solution.
Let A be a failure event. Let us introduce a system of hypotheses H1 - failure of the first element, H2 - failure of the second element, H3 - failure of the third element.
We find the probabilities of hypotheses:
P(H1) = 3/(3+2+5) = 0.3
P(H2) = 2/(3+2+5) = 0.2
P(H3) = 5/(3+2+5) = 0.5

According to the condition of the problem, the conditional probabilities of event A are:
P(A|H1) = 0.95, P(A|H2) = 0.9, P(A|H3) = 0.6

a) Find the probability of detecting a failure in the system.
P(A) = P(H1)*P(A|H1) + P(H2)*P(A|H2) + P(H3)*P(A|H3) = 0.3*0.95 + 0.2*0.9 + 0.5 *0.6 = 0.765

b) Under the conditions of this task, a failure was detected during the operation of the system. Which element is most likely to fail?
P1 = P(H1)*P(A|H1)/ P(A) = 0.3*0.95 / 0.765 = 0.373
P2 = P(H2)*P(A|H2)/ P(A) = 0.2*0.9 / 0.765 = 0.235
P3 = P(H3)*P(A|H3)/ P(A) = 0.5*0.6 / 0.765 = 0.392

The maximum probability of the third element.

Brief theory

If an event occurs only if one of the events forming a complete group of incompatible events occurs, then it is equal to the sum of the products of the probabilities of each of the events and the corresponding conditional probability wallet .

In this case, the events are called hypotheses, and the probabilities are called a priori. This formula is called the total probability formula.

The Bayes formula is used in solving practical problems, when an event that appears together with any of the events forming a complete group of events has occurred and it is required to carry out a quantitative reassessment of the probabilities of hypotheses. A priori (before experience) probabilities are known. It is required to calculate a posteriori (after experience) probabilities, i.e. Essentially, you need to find the conditional probabilities. Bayes formula looks like this:

The next page deals with the problem on .

Problem solution example

Condition of task 1

In the factory, machines 1, 2 and 3 produce 20%, 35% and 45% of all parts, respectively. In their products, the defect is respectively 6%, 4%, 2%. What is the probability that a randomly selected item is defective? What is the probability that it was produced: a) by machine 1; b) machine 2; c) machine 3?

Problem 1 solution

Denote by the event that the standard product turned out to be defective.

An event can only occur if one of three events occurs:

The product is produced on machine 1;

The product is produced on machine 2;

The product is produced on machine 3;

Let's write the conditional probabilities:

Total Probability Formula

If an event can only occur when one of the events that form a complete group of incompatible events occurs, then the probability of the event is calculated by the formula

Using the total probability formula, we find the probability of an event:

Bayes formula

Bayes' formula allows you to "rearrange cause and effect": according to known fact event to calculate the probability that it was caused by a given cause.

Probability that a defective item was produced on machine 1:

Probability that a defective item was produced on machine 2:

Probability that a defective item was produced on machine 3:

Condition of task 2

The group consists of 1 excellent student, 5 good students and 14 mediocre students. An excellent student answers 5 and 4 with equal probability, a good student answers 5, 4 and 3 with equal probability, and a mediocre student answers 4,3 and 2 with equal probability. A randomly selected student answered 4. What is the probability that a mediocre student was called?

Problem 2 solution

Hypotheses and conditional probabilities

The following hypotheses are possible:

The excellent student answered;

Answered good;

– answered mediocre student;

Let event -student get 4.

Answer:

The price is strongly influenced by the urgency of the decision (from days to several hours). Online help in the exam / test is carried out by appointment.

The application can be left directly in the chat, having previously thrown off the condition of the tasks and informing you of the deadlines for solving it. The response time is several minutes.

Objective: to form skills for solving problems in probability theory using the total probability formula and the Bayes formula.

Total Probability Formula

Event Probability BUT, which can occur only if one of the incompatible events occurs B x, B 2 ,..., B n, forming a complete group is equal to the sum of the products of the probabilities of each of these events and the corresponding conditional probability of event A:

This formula is called total probability formula.

Probability of hypotheses. Bayes formula

Let the event BUT can occur if one of the incompatible events occurs B b B 2 ,...,B p, forming a complete group. Since it is not known in advance which of these events will occur, they are called hypotheses. Probability of an event occurring BUT is determined by the total probability formula:

Assume that a test has been performed, as a result of which an event has occurred BUT. It is required to determine how they have changed (due to the fact that the event BUT already arrived) probabilities of hypotheses. Conditional probabilities of hypotheses are found by the formula

In this formula, the index / = 1.2

This formula is called the Bayes formula (after the English mathematician who derived it; published in 1764). The Bayes formula allows you to re-estimate the probabilities of hypotheses after the result of the test becomes known, as a result of which the event appeared BUT.

Task 1. The plant produces a certain type of part, each part has a defect with a probability of 0.05. The part is inspected by one inspector; it detects a defect with a probability of 0.97, and if no defect is found, it passes the part into the finished product. In addition, the inspector may mistakenly reject a part that does not have a defect; the probability of this is 0.01. Find the probabilities of the following events: A - the part will be rejected; B - the part will be rejected, but erroneously; C - the part will be skipped to the finished product with a defect.

Solution

Let's denote the hypotheses:

H= (a standard part will be sent for inspection);

H= (a non-standard part will be sent for inspection).

Event A =(part will be rejected).

From the condition of the problem we find the probabilities

P H (A) = 0,01; Pfi(A) = 0,97.

According to the total probability formula, we obtain

The probability that a part will be rejected by mistake is

Let's find the probability that the part will be skipped into the finished product with a defect:

Answer:

Task 2. The product is checked for standardity by one of three commodity experts. The probability that the product will get to the first merchandiser is 0.25, to the second - 0.26 and to the third - 0.49. The probability that the product will be recognized as a standard by the first merchandiser is 0.95, by the second - 0.98, by the third - 0.97. Find the probability that the standard product is checked by the second inspector.

Solution

Let's denote the events:

L. =(the product for verification will go to the /-th commodity manager); / = 1, 2, 3;

B =(the product will be recognized as standard).

According to the condition of the problem, the probabilities are known:

We also know the conditional probabilities

Using the Bayes formula, we find the probability that the standard product is checked by the second controller:

Answer:“0.263.

A task 3. Two machines produce parts that go to a common conveyor. The probability of obtaining a non-standard part on the first machine is 0.06, and on the second - 0.09. The performance of the second machine is twice that of the first. A non-standard part was taken from the conveyor. Find the probability that this part is produced by the second machine.

Solution

Let's denote the events:

A. =(the part taken from the assembly line is produced by the i-th machine); / = 1.2;

AT= (the part taken will be non-standard).

We also know the conditional probabilities

Using the total probability formula, we find

Using the Bayes formula, we find the probability that the non-standard part taken is produced by the second automaton:

Answer: 0,75.

Task 4. A device is tested, consisting of two nodes, the reliability of which is 0.8 and 0.9, respectively. Nodes fail independently of each other. The device failed. Find, taking into account this, the probabilities of hypotheses:

• a) only the first node is faulty;
• b) only the second node is faulty;
• c) both nodes are faulty.

Solution

Let's denote the events:

D = (7th node will not fail); i = 1,2;

D - corresponding opposite events;

BUT= (during the test, the device will fail).

From the condition of the problem we obtain: P(D) = 0.8; P(L 2) = 0,9.

By the property of the probabilities of opposite events

Event BUT is equal to the sum of products independent events

Using the addition theorem for the probabilities of incompatible events and the theorem for multiplying the probabilities of independent events, we obtain

Now we find the probabilities of the hypotheses:

Answer:

Task 5. At the factory, bolts are made on three machines, which produce respectively 25%, 30% and 45% of the total number of bolts. In the production of machine tools, the defect is respectively 4%, 3% and 2%. What is the probability that a bolt, randomly taken from an incoming product, will be defective?

Solution

Let's denote the events:

4 = (a randomly taken bolt was made on the i-th machine); i = 1, 2, 3;

AT= (a bolt taken at random will be defective).

From the condition of the problem, using the classical probability formula, we find the probabilities of the hypotheses:

Also, using the classical probability formula, we find the conditional probabilities:

Using the total probability formula, we find

Answer: 0,028.

Task 6. The electronic circuit belongs to one of the three batches with a probability of 0.25; 0.5 and 0.25. The probability that the circuit will work beyond the warranty period for each of the parties, respectively, is 0.1; 0.2 and 0.4. Find the probability that a randomly chosen circuit will work beyond the warranty period.

Solution

Let's denote the events:

4 \u003d (randomly taken scheme from r-th party); i = 1, 2, 3;

AT= (a randomly taken circuit will work beyond the warranty period).

According to the condition of the problem, the probabilities of the hypotheses are known:

We also know the conditional probabilities:

Using the total probability formula, we find

Answer: 0,225.

Task 7. The device contains two blocks, the serviceability of each of which is necessary for the operation of the device. The probabilities of failure-free operation for these blocks are 0.99 and 0.97, respectively. The device is out of order. Determine the probability that both units failed.

Solution

Let's denote the events:

D = ( z block will fail); i = 1,2;

BUT= (device will fail).

From the condition of the problem, according to the property of the probabilities of opposite events, we obtain: DD) = 1-0.99 = 0.01; DD) = 1-0.97 = 0.03.

Event BUT occurs only when at least one of the events D or A 2 . Therefore, this event is equal to the sum of the events BUT= D + BUT 2 .

By the addition theorem for the probabilities of joint events, we obtain

Using the Bayes formula, we find the probability that the device failed due to the failure of both blocks.

Answer:

Tasks for independent solution Task 1. In the warehouse of the television studio there are 70% of kinescopes manufactured by plant No. 1; the remaining kinescopes were manufactured by plant No. 2. The probability that the kinescope will not fail during the warranty period is 0.8 for kinescopes of plant No. 1 and 0.7 for kinescopes of plant No. 2. The kinescope has withstood the warranty period. Find the probability that it was produced by plant number 2.

Task 2. Parts from three automatic machines come to the assembly. It is known that the 1st machine gives 0.3% of defects, the 2nd - 0.2%, the 3rd - 0.4%. Find the probability of receipt of a defective part for assembly, if 1000 parts were received from the 1st machine, 2000 from the 2nd, and 2500 parts from the 3rd.

Task 3. Two machines produce identical parts. The probability that a part produced on the first machine will be standard is 0.8, and on the second - 0.9. The performance of the second machine is three times that of the first. Find the probability that the standard part will be taken at random from the conveyor, which receives parts from both machines.

Task 4. The head of the company decided to use the services of two of the three transport companies. The probabilities of untimely delivery of goods for the first, second and third firms are 0.05, respectively; 0.1 and 0.07. Comparing these data with data on the safety of cargo transportation, the manager came to the conclusion that the choice was equitable and decided to make it by lot. Find the probability that the sent cargo will be delivered on time.

Task 5. The device contains two blocks, the serviceability of each of which is necessary for the operation of the device. The probabilities of failure-free operation for these blocks are 0.99 and 0.97, respectively. The device is out of order. Determine the probability that the second unit failed.

A task 6. The assembly shop receives parts from three machines. The first machine gives 3% of the marriage, the second - 1% and the third - 2%. Determine the probability of a non-defective part getting into the assembly if 500, 200, 300 parts were received from each machine, respectively.

Task 7. The warehouse receives the products of three firms. Moreover, the production of the first firm is 20%, the second - 46% and the third - 34%. It is also known that the average percentage of non-standard products for the first firm is 5%, for the second - 2% and for the third - 1%. Find the probability that a product chosen at random was produced by the second company if it turned out to be standard.

Task 8. Marriage in the production of the plant due to a defect a is 5%, and among those rejected on the basis of a products in 10% of cases there is a defect R. And in defect-free products a, defect R occurs in 1% of cases. Find the probability of encountering a defect R in all products.

Task 9. The company has 10 new cars and 5 old ones that were previously under repair. The probability of proper operation for a new car is 0.94, for an old one - 0.91. Find the probability that a car chosen at random will work properly.

Task 10. Two sensors send signals to a common communication channel, and the first of them sends twice as many signals as the second. The probability of receiving a distorted signal from the first sensor is 0.01, from the second - 0.03. What is the probability of receiving a distorted signal in a common communication channel?

Task 11. There are five batches of products: three batches of 8 pieces, of which 6 are standard and 2 non-standard, and two batches of 10 pieces, of which 7 are standard and 3 are non-standard. One of the batches is chosen at random, and a detail is taken from this batch. Determine the probability that the selected part will be standard.

Task 12. The assembler receives, on average, 50% of the parts from the first plant, 30% from the second plant, and 20% from the third plant. The probability that the part of the first factory is of excellent quality is 0.7; for parts of the second and third plants, respectively, 0.8 and 0.9. Randomly taken part turned out to be of excellent quality. Find the probability that the part is made by the first factory.

Task 13. Customs inspection of cars is carried out by two inspectors. On average, out of 100 cars, 45 pass through the first inspector. The probability that during the inspection a car that complies with customs rules will not be detained is 0.95 for the first inspector and 0.85 for the second. Find the probability that a car that complies with the customs rules will not be detained.

Task 14. The parts needed to assemble the device come from two automatic machines, the performance of which is the same. Calculate the probability of a standard part entering the assembly if one of the automata gives an average of 3% violation of the standard, and the second - 2%.

Task 15. The weightlifting coach calculated that in order to receive team credits in this weight category, an athlete must push a 200 kg barbell. Ivanov, Petrov and Sidorov claim a place in the team. Ivanov during training tried to lift such weight in 7 cases, and lifted in 3 of them. Petrov lifted 6 times out of 13, and Sidorov has a 35% chance of successfully handling the barbell. The coach randomly selects one athlete for the team.

• a) Find the probability that the selected athlete will bring the team points.
• b) The team did not receive any points. Find the probability that Sidorov spoke.

Task 16. There are 12 red and 6 blue balls in a white box. In black - 15 red and 10 blue balls. Throw a dice. If the number of points is a multiple of 3, then a ball is randomly taken from the white box. If any other number of points falls out, then a ball is randomly taken from the black box. What is the probability of a red ball appearing?

Task 17. Two boxes contain radio tubes. The first box contains 12 lamps, of which 1 is non-standard; in the second one there are 10 lamps, 1 of which is non-standard. A lamp was taken at random from the first box and transferred to the second. Find the probability that a lamp drawn at random from the second box is non-standard.

Task 18. A white ball is dropped into an urn containing two balls, after which one ball is drawn at random. Find the probability that the drawn ball will be white if all possible assumptions about the initial composition of the balls (by color) are equally possible.

Task 19. A standard part is thrown into a box containing 3 identical parts, and then one part is drawn at random. Find the probability that a standard part is drawn if all possible guesses about the number of standard parts originally in the box are equally probable.

Task 20. To improve the quality of radio communication, two radio receivers are used. The probability of receiving a signal by each receiver is 0.8, and these events (signal reception by the receiver) are independent. Determine the probability of receiving a signal if the probability of failure-free operation during a radio communication session for each receiver is 0.9.

When deriving the total probability formula, it was assumed that the event BUT, the probability of which was to be determined, could happen to one of the events H 1 , N 2 , ... , H n, forming a complete group of pairwise incompatible events. The probabilities of these events (hypotheses) were known in advance. Let us assume that an experiment has been performed, as a result of which the event BUT has come. This Additional Information allows you to re-evaluate the probabilities of hypotheses H i , having calculated P(H i /A).

or, using the total probability formula, we get

This formula is called the Bayes formula or the hypothesis theorem. The Bayes formula allows you to "revise" the probabilities of hypotheses after the result of the experiment becomes known, as a result of which the event appeared BUT.

Probabilities Р(Н i) are the a priori probabilities of the hypotheses (they were calculated before the experiment). The probabilities P(H i /A) are the a posteriori probabilities of the hypotheses (they are calculated after the experiment). The Bayes formula allows you to calculate the posterior probabilities from their prior probabilities and from the conditional probabilities of the event BUT.

Example. It is known that 5% of all men and 0.25% of all women are color blind. A randomly selected person by the number of the medical card suffers from color blindness. What is the probability that it is a man?

Solution. Event BUT The person is colorblind. The space of elementary events for the experiment - a person is selected by the number of the medical card - Ω = ( H 1 , N 2 ) consists of 2 events:

H 1 - a man is selected,

H 2 - a woman is selected.

These events can be chosen as hypotheses.

According to the condition of the problem (random choice), the probabilities of these events are the same and equal to P(H 1 ) = 0.5; P(H 2 ) = 0.5.

In this case, the conditional probabilities that a person suffers from color blindness are equal, respectively:

P(A/N 1 ) = 0.05 = 1/20; P(A/N 2 ) = 0.0025 = 1/400.

Since it is known that the selected person is color blind, i.e., the event has occurred, we use the Bayes formula to reevaluate the first hypothesis:

Example. There are three identical boxes. The first box contains 20 white balls, the second box contains 10 white and 10 black balls, and the third box contains 20 black balls. A white ball is drawn from a box chosen at random. Calculate the probability that the ball is drawn from the first box.

Solution. Denote by BUT event - the appearance of a white ball. Three assumptions (hypotheses) can be made about the choice of the box: H 1 ,H 2 , H 3 - selection of the first, second and third boxes, respectively.

Since the choice of any of the boxes is equally possible, the probabilities of the hypotheses are the same:

P(H 1 )=P(H 2 )=P(H 3 )= 1/3.

According to the condition of the problem, the probability of drawing a white ball from the first box

Probability of drawing a white ball from the second box

Probability of drawing a white ball from the third box

We find the desired probability using the Bayes formula:

Repetition of tests. Bernoulli formula.

There are n trials, in each of which event A may or may not occur, and the probability of event A in each individual trial is constant, i.e. does not change from experience to experience. We already know how to find the probability of an event A in one experiment.

Of special interest is the probability of occurrence of a certain number of times (m times) of event A in n experiments. such problems are easily solved if the tests are independent.

Def. Several tests are called independent with respect to the event A if the probability of event A in each of them does not depend on the outcomes of other experiments.

The probability P n (m) of the occurrence of the event A exactly m times (non-occurrence n-m times, event ) in these n trials. Event A appears in a variety of sequences m times).

Bernoulli formula.

The following formulas are obvious:

P n (m less k times in n trials.

P n (m>k) = P n (k+1) + P n (k+2) +…+ P n (n) - probability of occurrence of event A more k times in n trials.1) n = 8, m = 4, p = q = ½,

Bayes formula

Bayes' theorem- one of the main theorems of elementary probability theory, which determines the probability of an event occurring under conditions when only some partial information about events is known based on observations. According to the Bayes formula, it is possible to more accurately recalculate the probability, taking into account both previously known information and data from new observations.

"Physical meaning" and terminology

Bayes' formula allows you to "rearrange the cause and effect": given the known fact of an event, calculate the probability that it was caused by a given cause.

Events reflecting the action of "causes" in this case are usually called hypotheses, because they are supposed the events leading up to it. The unconditional probability of the validity of a hypothesis is called a priori(How likely is the cause? generally), and conditional - taking into account the fact of the event - a posteriori(How likely is the cause? turned out to be taking into account the event data).

Consequence

An important consequence of the Bayes formula is the formula for the total probability of an event depending on several inconsistent hypotheses ( and only from them!).

Formula derivation

If an event depends only on causes A i, then if it happened, it means that some of the reasons necessarily happened, i.e.

By Bayes formula

transfer P(B) to the right, we obtain the desired expression.

Spam filtering method

A method based on Bayes' theorem has been successfully applied in spam filtering.

Description

When training the filter, for each word encountered in letters, its “weight” is calculated and stored - the probability that a letter with this word is spam (in the simplest case, according to the classical definition of probability: “appearances in spam / appearances of everything”).

When checking a newly arrived letter, the probability that it is spam is calculated according to the above formula for a set of hypotheses. In this case, "hypotheses" are words, and for each word "reliability of the hypothesis" -% of this word in the letter, and "dependence of the event on the hypothesis" P(B | A i) - previously calculated "weight" of the word. That is, the "weight" of the letter in this case is nothing but the average "weight" of all its words.

A letter is classified as "spam" or "non-spam" by whether its "weight" exceeds a certain bar set by the user (usually they take 60-80%). After a decision on a letter is made, the “weights” for the words included in it are updated in the database.

Characteristic

This method is simple (the algorithms are elementary), convenient (allows you to do without "black lists" and similar artificial tricks), effective (after training on a sufficiently large sample, it cuts off up to 95-97% of spam, and in case of any errors it can be retrained). In general, there are all indications for its widespread use, which is what happens in practice - almost all modern spam filters are built on its basis.

However, the method also has a fundamental drawback: it based on the assumption, what some words are more common in spam, while others are more common in regular emails, and is inefficient if this assumption is false. However, as practice shows, even a person is not able to determine such spam "by eye" - only after reading the letter and understanding its meaning.

Another, not fundamental, disadvantage associated with the implementation - the method works only with text. Knowing about this limitation, spammers began to put advertising information in the picture, while the text in the letter is either absent or does not make sense. Against this, one has to use either text recognition tools (an “expensive” procedure, it is used only when emergency), or old filtering methods - "black lists" and regular expressions (since such letters often have a stereotyped form).

see also

Notes

Links

Literature

• Byrd Kiwi. Rev. Bayes' theorem. // Computerra magazine, August 24, 2001
• Paul Graham. A plan for spam. // Personal website of Paul Graham.

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See what the "Bayes formula" is in other dictionaries:

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Books

• Probability Theory and Mathematical Statistics in Problems. More than 360 tasks and exercises, Borzykh D.A. The proposed manual contains tasks different levels difficulties. However, the main emphasis is placed on tasks of medium complexity. This is intentionally done to encourage students to…