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Total Probability Formula and Bayes Formulas

In this lesson we will look at an important corollary addition and multiplication theorems of probabilities and learn how to solve typical problems on the topic. Readers who have read the article about dependent events, it will be simpler, since in it we have actually already begun to use the total probability formula. If you came from a search engine and/or don’t understand probability theory (link to 1st lesson of the course), then I recommend visiting these pages first.

Actually, let's continue. Let's consider dependent event, which can only occur as a result of the implementation of one of the incompatible hypotheses , which form full group. Let their probabilities and the corresponding conditional probabilities be known. Then the probability of the event occurring is:

This formula is called total probability formulas. In textbooks it is formulated as a theorem, the proof of which is elementary: according to algebra of events, (an event occurred And or an event occurred And after it came an event or an event occurred And after it came an event or …. or an event occurred And after it came an event). Since hypotheses are incompatible, and the event is dependent, then according the theorem of addition of probabilities of incompatible events (first step) And theorem of multiplication of probabilities of dependent events (second step):

Many people probably anticipate the content of the first example =)

Wherever you spit, there is an urn:

Problem 1

There are three identical urns. The first urn contains 4 white and 7 black balls, the second - only white and the third - only black balls. One urn is selected at random and a ball is drawn from it at random. What is the probability that this ball is black?

Solution: consider the event - a black ball will be drawn from a randomly chosen urn. This event can occur as a result of one of the following hypotheses:
- the 1st urn will be selected;
- the 2nd urn will be selected;
- the 3rd urn will be selected.

Since the urn is chosen at random, the choice of any of the three urns equally possible, hence:

Please note that the above hypotheses form full group of events, that is, according to the condition, a black ball can only appear from these urns, and, for example, cannot come from a billiard table. Let's do a simple intermediate check:
, OK, let's move on:

The first urn contains 4 white + 7 black = 11 balls, each classical definition:
- probability of drawing a black ball given that, that the 1st urn will be selected.

The second urn contains only white balls, so if chosen the appearance of the black ball becomes impossible: .

And finally, the third urn contains only black balls, which means the corresponding conditional probability extracting the black ball will be (the event is reliable).

- the probability that a black ball will be drawn from a randomly chosen urn.

Answer:

The analyzed example again suggests how important it is to delve into the CONDITION. Let's take the same problems with urns and balls - despite their external similarity, the methods of solution can be completely different: somewhere you only need to use classical definition of probability, somewhere events independent, somewhere dependent, and somewhere we are talking about hypotheses. At the same time, there is no clear formal criterion for choosing a solution - you almost always need to think about it. How to improve your skills? We decide, we decide and we decide again!

Problem 2

The shooting range has 5 rifles of varying accuracy. The probabilities of hitting the target for a given shooter are respectively equal and 0.4. What is the probability of hitting the target if the shooter fires one shot from a randomly selected rifle?

A short solution and answer at the end of the lesson.

In most thematic problems, the hypotheses are, of course, not equally probable:

Problem 3

There are 5 rifles in the pyramid, three of which are equipped with an optical sight. The probability that a shooter will hit the target when firing a rifle with a telescopic sight is 0.95; for a rifle without an optical sight, this probability is 0.7. Find the probability that the target will be hit if the shooter fires one shot from a rifle taken at random.

Solution: in this problem the number of rifles is exactly the same as in the previous one, but there are only two hypotheses:
- the shooter will select a rifle with an optical sight;
- the shooter will select a rifle without an optical sight.
By classical definition of probability: .
Control:

Consider the event: - a shooter hits a target with a rifle taken at random.
By condition: .

According to the total probability formula:

Answer: 0,85

In practice, a shortened way of formatting a task, which you are also familiar with, is quite acceptable:

Solution: according to the classical definition: - the probability of choosing a rifle with an optical sight and without an optical sight, respectively.

By condition, - the probability of hitting the target from the corresponding types of rifles.

According to the total probability formula:
- the probability that a shooter will hit a target with a randomly selected rifle.

Answer: 0,85

The following task is for you to solve on your own:

Problem 4

The engine operates in three modes: normal, forced and idle. In idle mode, the probability of its failure is 0.05, in normal operation mode - 0.1, and in forced mode - 0.7. 70% of the time the engine operates in normal mode, and 20% in forced mode. What is the probability of engine failure during operation?

Just in case, let me remind you that to get the probability values, the percentages must be divided by 100. Be very careful! According to my observations, people often try to confuse the conditions of problems involving the total probability formula; and I specifically chose this example. I'll tell you a secret - I almost got confused myself =)

Solution at the end of the lesson (formatted in a short way)

Problems using Bayes' formulas

The material is closely related to the content of the previous paragraph. Let the event occur as a result of the implementation of one of the hypotheses . How to determine the probability that a particular hypothesis occurred?

Given that that event has already happened, hypothesis probabilities overrated according to the formulas that received the name of the English priest Thomas Bayes:

- the probability that the hypothesis took place;
- the probability that the hypothesis took place;
…
- the probability that the hypothesis took place.

At first glance it seems completely absurd - why recalculate the probabilities of hypotheses if they are already known? But in fact there is a difference:

This a priori(estimated before tests) probability.

This a posteriori(estimated after tests) probabilities of the same hypotheses, recalculated in connection with “newly discovered circumstances” - taking into account the fact that the event definitely happened.

Let's look at this difference with a specific example:

Problem 5

2 batches of products arrived at the warehouse: the first - 4000 pieces, the second - 6000 pieces. The average percentage of non-standard products in the first batch is 20%, and in the second - 10%. The product taken from the warehouse at random turned out to be standard. Find the probability that it is: a) from the first batch, b) from the second batch.

First part solutions consists of using the total probability formula. In other words, calculations are carried out under the assumption that the test not yet produced and event “the product turned out to be standard” not yet.

Let's consider two hypotheses:
- a product taken at random will be from the 1st batch;
- a product taken at random will be from the 2nd batch.

Total: 4000 + 6000 = 10000 items in stock. According to the classical definition:
.

Control:

Let's consider the dependent event: - a product taken at random from the warehouse will be standard.

In the first batch 100% - 20% = 80% standard products, therefore: given that that it belongs to the 1st party.

Similarly, in the second batch 100% - 10% = 90% of standard products and - the probability that a product taken at random from a warehouse will be standard given that that it belongs to the 2nd party.

According to the total probability formula:
- the probability that a product taken at random from a warehouse will be standard.

Part two. Let a product taken at random from a warehouse turn out to be standard. This phrase is directly stated in the condition, and it states the fact that the event happened.

According to Bayes formulas:

a) - the probability that the selected standard product belongs to the 1st batch;

b) - the probability that the selected standard product belongs to the 2nd batch.

After revaluation hypotheses, of course, still form full group:
(examination;-))

Answer:

Ivan Vasilyevich, who again changed his profession and became the director of the plant, will help us understand the meaning of the revaluation of hypotheses. He knows that today the 1st workshop shipped 4,000 products to the warehouse, and the 2nd workshop - 6,000 products, and comes to make sure of this. Let's assume that all products are of the same type and are in the same container. Naturally, Ivan Vasilyevich preliminarily calculated that the product that he would now remove for inspection would most likely be produced by the 1st workshop and most likely by the second. But after the chosen product turns out to be standard, he exclaims: “What a cool bolt! “It was rather released by the 2nd workshop.” Thus, the probability of the second hypothesis is overestimated for the better, and the probability of the first hypothesis is underestimated: . And this revaluation is not unfounded - after all, the 2nd workshop not only produced more products, but also works 2 times better!

Pure subjectivism, you say? In part - yes, moreover, Bayes himself interpreted a posteriori probabilities as trust level. However, not everything is so simple - there is also an objective grain in the Bayesian approach. After all, the likelihood that the product will be standard (0.8 and 0.9 for the 1st and 2nd workshops, respectively) This preliminary(a priori) and average assessments. But, speaking philosophically, everything flows, everything changes, including probabilities. It is quite possible that at the time of the study the more successful 2nd workshop increased the percentage of standard products produced (and/or the 1st workshop reduced), and if you check a larger number or all 10 thousand products in the warehouse, then the overestimated values ​​will turn out to be much closer to the truth.

By the way, if Ivan Vasilyevich extracts a non-standard part, then on the contrary - he will be more “suspicious” of the 1st workshop and less of the second. I suggest you check this out for yourself:

Problem 6

2 batches of products arrived at the warehouse: the first - 4000 pieces, the second - 6000 pieces. The average percentage of non-standard products in the first batch is 20%, in the second - 10%. The product taken from the warehouse at random turned out to be Not standard. Find the probability that it is: a) from the first batch, b) from the second batch.

The condition is distinguished by two letters, which I have highlighted in bold. The problem can be solved from scratch, or using the results of previous calculations. In the sample, I carried out a complete solution, but in order to avoid any formal overlap with Problem No. 5, the event “a product taken at random from a warehouse will be non-standard” indicated by .

The Bayesian scheme for reestimating probabilities is found everywhere, and it is also actively exploited by various types of scammers. Let’s consider a three-letter joint stock company that has become a household name, which attracts deposits from the public, supposedly invests them somewhere, regularly pays dividends, etc. What's happening? Day after day, month after month passes, and more and more new facts, conveyed through advertising and word of mouth, only increase the level of trust in the financial pyramid (posteriori Bayesian reestimation due to past events!). That is, in the eyes of investors there is a constant increase in the likelihood that “this is a serious company”; while the probability of the opposite hypothesis (“these are just more scammers”), of course, decreases and decreases. What follows, I think, is clear. It is noteworthy that the earned reputation gives the organizers time to successfully hide from Ivan Vasilyevich, who was left not only without a batch of bolts, but also without pants.

We will return to equally interesting examples a little later, but for now the next step is perhaps the most common case with three hypotheses:

Problem 7

Electric lamps are manufactured at three factories. The 1st plant produces 30% of the total number of lamps, the 2nd - 55%, and the 3rd - the rest. The products of the 1st plant contain 1% of defective lamps, the 2nd - 1.5%, the 3rd - 2%. The store receives products from all three factories. The purchased lamp turned out to be defective. What is the probability that it was produced by plant 2?

Note that in problems on Bayes formulas in the condition Necessarily there is a certain what happened event, in this case the purchase of a lamp.

Events have increased, and solution It’s more convenient to arrange it in a “quick” style.

The algorithm is exactly the same: in the first step we find the probability that the purchased lamp will turn out to be defective.

Using the initial data, we convert percentages into probabilities:
- the probability that the lamp was produced by the 1st, 2nd and 3rd factories, respectively.
Control:

Similarly: - the probability of producing a defective lamp for the corresponding factories.

According to the total probability formula:

- the likelihood that the purchased lamp will be defective.

Step two. Let the purchased lamp turn out to be defective (the event occurred)

According to Bayes' formula:
- the probability that the purchased defective lamp was manufactured by a second plant

Answer:

Why did the initial probability of the 2nd hypothesis increase after revaluation? After all, the second plant produces lamps of average quality (the first is better, the third is worse). So why did it increase a posteriori Is it possible that the defective lamp is from the 2nd plant? This is no longer explained by “reputation”, but by size. Since plant No. 2 produced the largest number of lamps (more than half), the subjective nature of the overestimation is at least logical (“most likely, this defective lamp is from there”).

It is interesting to note that the probabilities of the 1st and 3rd hypotheses were overestimated in the expected directions and became equal:

Control: , which was what needed to be checked.

By the way, about underestimated and overestimated estimates:

Problem 8

In the student group, 3 people have a high level of training, 19 people have an average level and 3 people have a low level. The probabilities of successfully passing the exam for these students are respectively equal to: 0.95; 0.7 and 0.4. It is known that some student passed the exam. What is the probability that:

a) he was prepared very well;
b) was moderately prepared;
c) was poorly prepared.

Perform calculations and analyze the results of re-evaluating the hypotheses.

The task is close to reality and is especially plausible for a group of part-time students, where the teacher has virtually no knowledge of the abilities of a particular student. In this case, the result can cause quite unexpected consequences. (especially for exams in the 1st semester). If a poorly prepared student is lucky enough to get a ticket, then the teacher is likely to consider him a good student or even a strong student, which will bring good dividends in the future (of course, you need to “raise the bar” and maintain your image). If a student studied, crammed, and repeated for 7 days and 7 nights, but was simply unlucky, then further events can develop in the worst possible way - with numerous retakes and balancing on the brink of elimination.

Needless to say, reputation is the most important capital; it is no coincidence that many corporations bear the names of their founding fathers, who led the business 100-200 years ago and became famous for their impeccable reputation.

Yes, the Bayesian approach is to a certain extent subjective, but... that’s how life works!

Let’s consolidate the material with a final industrial example, in which I will talk about hitherto unknown technical intricacies of the solution:

Problem 9

Three workshops of the plant produce the same type of parts, which are sent to a common container for assembly. It is known that the first workshop produces 2 times more parts than the second workshop, and 4 times more than the third workshop. In the first workshop the defect rate is 12%, in the second - 8%, in the third - 4%. For control, one part is taken from the container. What is the probability that it will be defective? What is the probability that the extracted defective part was produced by the 3rd workshop?

Ivan Vasilyevich is on horseback again =) The film must have a happy ending =)

Solution: unlike Problems No. 5-8, here a question is explicitly asked, which is resolved using the total probability formula. But on the other hand, the condition is a little “encrypted”, and the school skill of composing simple equations will help us solve this puzzle. It is convenient to take the smallest value as “x”:

Let be the share of parts produced by the third workshop.

According to the condition, the first workshop produces 4 times more than the third workshop, so the share of the 1st workshop is .

In addition, the first workshop produces 2 times more products than the second workshop, which means the share of the latter: .

Let's create and solve the equation:

Thus: - the probability that the part removed from the container was produced by the 1st, 2nd and 3rd workshops, respectively.

Control: . In addition, it would not hurt to look at the phrase again “It is known that the first workshop produces products 2 times more than the second workshop and 4 times more than the third workshop.” and make sure that the obtained probability values ​​actually correspond to this condition.

Initially, one could take the share of the 1st or the share of the 2nd workshop as “X” - the probabilities would be the same. But, one way or another, the most difficult part has been passed, and the solution is on track:

From the condition we find:
- the probability of manufacturing a defective part for the relevant workshops.

According to the total probability formula:
- the likelihood that a part randomly removed from a container will turn out to be non-standard.

Question two: what is the probability that the extracted defective part was produced by the 3rd workshop? This question assumes that the part has already been removed and it turned out to be defective. We re-evaluate the hypothesis using Bayes' formula:
- the desired probability. Completely expected - after all, the third workshop not only produces the smallest proportion of parts, but also leads in quality!

If the event A can only happen when one of the events that form a complete group of incompatible events , then the probability of the event A calculated by the formula

This formula is called total probability formula .

Let us again consider the complete group of incompatible events, the probabilities of which . Event A can only happen together with any of the events that we will call hypotheses . Then, according to the total probability formula

If the event A happened, this may change the probabilities of the hypotheses .

By the probability multiplication theorem

.

Similarly, for the remaining hypotheses

The resulting formula is called Bayes formula (Bayes formula ). The probabilities of the hypotheses are called posterior probabilities , whereas - prior probabilities .

Example. The store received new products from three factories. The percentage composition of these products is as follows: 20% - products of the first enterprise, 30% - products of the second enterprise, 50% - products of the third enterprise; further, 10% of the products of the first enterprise are of the highest grade, at the second enterprise - 5% and at the third - 20% of the products of the highest grade. Find the probability that a randomly purchased new product will be of the highest quality.

Solution. Let us denote by IN the event that the highest grade products will be purchased, let us denote the events that consist in the purchase of products belonging to the first, second and third enterprises, respectively.

You can apply the total probability formula, and in our notation:

Substituting these values ​​into the total probability formula, we obtain the desired probability:

Example. One of the three shooters is called to the firing line and fires two shots. The probability of hitting the target with one shot for the first shooter is 0.3, for the second - 0.5; for the third - 0.8. The target is not hit. Find the probability that the shots were fired by the first shooter.

Solution. Three hypotheses are possible:

The first shooter is called to the line of fire,

The second shooter is called to the line of fire,

A third shooter is called to the firing line.

Since calling any shooter into the line of fire is equally possible, then

As a result of the experiment, event B was observed - after the shots were fired, the target was not hit. The conditional probabilities of this event under the hypotheses made are equal to:

Using the Bayes formula, we find the probability of the hypothesis after the experiment:

Example. Three automatic machines process parts of the same type, which after processing are transferred to a common conveyor. The first machine produces 2% of defects, the second - 7%, the third - 10%. The productivity of the first machine is 3 times greater than the productivity of the second, and the third is 2 times less than the second.

a) What is the defect rate on the assembly line?

b) What is the proportion of parts from each machine among the defective parts on the conveyor?

Solution. Let's take one part at random from the assembly line and consider event A - the part is defective. It is associated with hypotheses regarding where this part was processed: - a part taken at random was processed on the th machine.

Conditional probabilities (in the problem statement they are given in the form of percentages):

The dependencies between machine productivity mean the following:

And since the hypotheses form a complete group, then .

Having solved the resulting system of equations, we find: .

a) The total probability that a part taken at random from the assembly line is defective:

In other words, among the mass of parts coming off the assembly line, defects amount to 4%.

b) Let it be known that the part taken at random is defective. Using Bayes' formula, we find the conditional probabilities of the hypotheses:

Thus, in the total mass of defective parts on the conveyor, the share of the first machine is 33%, the second – 39%, the third – 28%.

Practical tasks

Exercise 1

Solving problems in the main branches of probability theory

The goal is to gain practical skills in solving problems in

branches of probability theory

Preparation for practical assignment

Familiarize yourself with theoretical material on this topic, study the content of the theoretical material, as well as the relevant sections in literary sources

Procedure for completing the task

Solve 5 problems according to the number of the task option given in Table 1.

Source data options

Table 1

	task number

Composition of the report on task 1

5 solved problems according to option number.

Problems to solve independently

1.. Are the following groups of events cases: a) experience - tossing a coin; events: A1- appearance of the coat of arms; A2- appearance of a number; b) experiment - tossing two coins; events: IN 1- the appearance of two coats of arms; AT 2 - the appearance of two numbers; AT 3- the appearance of one coat of arms and one number; c) experience - throwing a dice; events: C1 - the appearance of no more than two points; C2 - the appearance of three or four points; C3 - appearance of at least five points; d) experience - shooting at a target; events: D1- hit; D2- miss; e) experience - two shots at a target; events: E0- not a single hit; E1- one hit; E2- two hits; f) experience - removing two cards from the deck; events: F1 - the appearance of two red cards; F2- the appearance of two black cards?

2. In the urn A are white and B black balls. One ball is drawn at random from the urn. Find the probability that this ball is white.

3. In urn A white and B black balls. One ball is taken from the urn and set aside. This ball turned out to be white. After this, another ball is taken from the urn. Find the probability that this ball will also be white.

4. In urn A white and B black balls. One ball was taken out of the urn and, without looking, it was put aside. After that, another ball was taken from the urn. He turned out to be white. Find the probability that the first ball put aside is also white.

5. From an urn containing A white and B black balls, take out one by one all the balls except one. Find the probability that the last ball remaining in the urn will be white.

6. From the urn in which A white balls and B black, take out all the balls in it in a row. Find the probability that the white ball will be drawn second in order.

7. There are A white and B black balls in an urn (A > 2). Two balls are taken from the urn at once. Find the probability that both balls are white.

8. In the urn A are white and B black balls (A > 2, B > 3). Five balls are taken from the urn at once. Find probability R that two of them will be white and three black.

9. In a game consisting of X products available I defective. Selected from the batch for control I products. Find probability R which of them is exactly J products will be defective.

10. The die is rolled once. Find the probability of the following events: A - appearance of an even number of points; IN- appearance of at least 5 points; WITH- appearance no more than 5 points.

11. The dice are rolled twice. Find probability R that the same number of points will appear both times.

12. Two dice are thrown at the same time. Find the probabilities of the following events: A- the sum of the points drawn is 8; IN- the product of the rolled points is 8; WITH- the sum of the rolled points is greater than their product.

13. Two coins are tossed. Which of the following events is more likely: A - the coins will lie on the same sides; IN - will the coins end up on different sides?

14. In urn A white and B black balls (A > 2; B > 2). Two balls are drawn from the urn at the same time. Which event is more likely: A- balls of the same color; IN - balls of different colors?

15. Three players are playing cards. Each of them was dealt 10 cards and two cards were left in the draw. One of the players sees that he has 6 cards of diamonds and 4 non-diamonds in his hands. He discards two of these four cards and takes a draw for himself. Find the probability that he will buy two diamonds.

16. From an urn containing P numbered balls, all the balls in it are taken out at random, one after another. Find the probability that the numbers of the drawn balls will be in order: 1, 2,..., P.

17. The same urn as in the previous problem, but after each ball is taken out, it is put back in and mixed with others, and its number is written down. Find the probability that a natural sequence of numbers will be written: 1, 2,..., n.

18. A full deck of cards (52 sheets) is divided at random into two equal packs of 26 sheets each. Find the probabilities of the following events: A - each pack will contain two aces; IN- one of the packs will not contain a single ace, and the other will not have all four; S-v one of the packs will have one ace, and the other will have three.

19. 18 teams participate in the basketball championship, from which two groups of 9 teams each are randomly formed. There are 5 teams among the competition participants

extra-class. Find the probabilities of the following events: A - all top-class teams will be in the same group; IN- two top-class teams will fall into one of the groups, and three - into the other.

20. Numbers are written on nine cards: 0, 1, 2, 3, 4, 5, 6, 7, 8. Two of them are taken out at random and placed on the table in the order of appearance, then the resulting number is read, for example 07 (seven), 14 ( fourteen), etc. Find the probability that the number will be even.

21. The numbers are written on five cards: 1, 2, 3, 4, 5. Two of them, one after the other, are taken out. Find the probability that the number on the second card will be greater than the number on the first.

22. The same question as in problem 21, but after the first card is taken out, it is put back and mixed with the rest, and the number on it is written down.

23. In urn A white, B black and C red balls. All the balls in it are taken out of the urn one by one and their colors are recorded. Find the probability that white appears in this list before black.

24. There are two urns: in the first A white and B black balls; in the second C white and D black. A ball is drawn from each urn. Find the probability that both balls are white.

25. In the conditions of problem 24, find the probability that the drawn balls will be of different colors.

26. There are seven slots in the revolver drum, five of them contain cartridges, and two are left empty. The drum is rotated, as a result of which one of the nests randomly appears against the trunk. After this, the trigger is pressed; if the cell was empty, the shot does not occur. Find probability R the fact that, having repeated this experiment two times in a row, we will not shoot both times.

27. Under the same conditions (see problem 26), find the probability that the shot will occur both times.

28. The urn contains A; balls marked with numbers 1, 2, ..., To From the urn I one ball at a time is taken out (I<к), The ball number is recorded and the ball is placed back in the urn. Find probability R that all recorded numbers will be different.

29. The word “book” is made up of five letters of the split alphabet. A child who cannot read scattered these letters and then collected them in random order. Find probability R that he came up with the word “book” again.

30. The word “pineapple” is made from the letters of the split alphabet. A child who cannot read scattered these letters and then collected them in random order. Find probability R that he has the word “pineapple” again

31. Several cards are drawn from a full deck of cards (52 sheets, 4 suits). How many cards must be taken out in order to say with a probability greater than 0.50 that among them there will be cards of the same suit?

32. N people are randomly seated at a round table (N> 2). Find probability R that two fixed persons A And IN will be nearby.

33. The same problem (see 32), but the table is rectangular, and N people are seated randomly along one of its sides.

34. Lotto barrels have numbers from 1 to N. Of these N Two barrels are randomly selected. Find the probability that both barrels contain numbers less than k (2

35. Lotto barrels have numbers from 1 to N. Of these N Two barrels are randomly selected. Find the probability that one of the barrels contains a number greater than k , and on the other - less than k . (2

36. Battery from M guns fires at a group consisting of N goals (M< N). The guns choose their targets sequentially, randomly, provided that no two guns can fire at the same target. Find probability R that targets numbered 1, 2,... will be fired upon M.

37.. Battery consisting of To guns, fires at a group consisting of I aircraft (To< 2). Each weapon chooses its target randomly and independently of the others. Find the probability that everything To guns will fire at the same target.

38. In the conditions of the previous problem, find the probability that all guns will fire at different targets.

39. Four balls are scattered randomly across four holes; each ball falls into one or another hole with the same probability and independently of the others (there are no obstacles to several balls falling into the same hole). Find the probability that there will be three balls in one of the holes, one in the other, and no balls in the other two holes.

40. Masha quarreled with Petya and does not want to ride on the same bus with him. There are 5 buses from the hostel to the institute from 7 to 8. Anyone who does not catch these buses is late for the lecture. In how many ways can Masha and Petya get to the institute on different buses and not be late for the lecture?

41. The information technology department of the bank employs 3 analysts, 10 programmers and 20 engineers. For overtime on a holiday, the head of the department must allocate one employee. In how many ways can this be done?

42. The head of the bank's security service must place 10 guards at 10 posts every day. In how many ways can this be done?

43. The new president of the bank must appoint 2 new vice presidents from among 10 directors. In how many ways can this be done?

44. One of the warring parties captured 12 and the other 15 prisoners. In how many ways can 7 prisoners of war be exchanged?

45. Petya and Masha collect video discs. Petya has 30 comedies, 80 action films and 7 melodramas, Masha has 20 comedies, 5 action films and 90 melodramas. In how many ways can Petya and Masha exchange 3 comedies, 2 action films and 1 melodrama?

46. ​​Under the conditions of problem 45, in how many ways can Petya and Masha exchange 3 melodramas and 5 comedies?

47. Under the conditions of problem 45, in how many ways can Petya and Masha exchange 2 action films and 7 comedies?

48. One of the warring parties captured 15 and the other 16 prisoners. In how many ways can 5 prisoners of war be exchanged?

49. How many cars can be registered in 1 city if the number has 3 numbers and 3 letters (only those whose spelling matches the Latin ones - A, B, E, K, M, N, O, R, S, T, U, X )?

50. One of the warring parties captured 14, and the other - 17 prisoners. In how many ways can 6 prisoners of war be exchanged?

51. How many different words can you form by rearranging the letters in the word “mother”?

52. There are 3 red and 7 green apples in a basket. One apple is taken out of it. Find the probability that it will be red.

53. There are 3 red and 7 green apples in a basket. One green apple was taken out and put aside. Then 1 more apple is taken out of the basket. What is the probability that this apple will be green?

54. In a batch of 1000 products, 4 are defective. For control, a batch of 100 products is selected. What is the probability of LLP that the control lot will not contain any defective ones?

56. In the 80s, the game “Sports Loto 5 out of 36” was popular in the USSR. The player marked 5 numbers on a card from 1 to 36 and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player did not guess a single number.

57. In the 80s, the game “Sports Loto 5 out of 36” was popular in the USSR. The player marked 5 numbers on a card from 1 to 36 and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guessed one number.

58. In the 80s, the game “Sports Loto 5 out of 36” was popular in the USSR. The player marked 5 numbers on a card from 1 to 36 and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guessed 3 numbers.

59. In the 80s, the game “Sports Loto 5 out of 36” was popular in the USSR. The player marked 5 numbers on a card from 1 to 36 and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player did not correctly match all 5 numbers.

60. In the 80s, the game “Sports Loto 6 out of 49” was popular in the USSR. The player marked 6 numbers from 1 to 49 on a card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guessed 2 numbers.

61. In the 80s, the game “Sports Loto 6 out of 49” was popular in the USSR. The player marked 6 numbers from 1 to 49 on a card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player did not guess a single number.

62.In the 80s, the game “Sports Loto 6 out of 49” was popular in the USSR. The player marked 6 numbers from 1 to 49 on a card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guessed all 6 numbers.

63. In a batch of 1000 products, 4 are defective. For control, a batch of 100 products is selected. What is the probability of LLP that the control lot will contain only 1 defective one?

64. How many different words can you form by rearranging the letters in the word “book”?

65. How many different words can you form by rearranging the letters in the word “pineapple”?

66. 6 people entered the elevator, and the hostel has 7 floors. What is the probability that all 6 people will exit on the same floor?

67. 6 people entered the elevator; the building has 7 floors. What is the probability that all 6 people will exit on different floors?

68. During a thunderstorm, a wire broke in the section between 40 and 79 km of the power line. Assuming that a break is equally possible at any point, find the probability that the break occurred between the 40th and 45th kilometers.

69. On a 200-kilometer section of the gas pipeline, a gas leak occurs between compressor stations A and B, which is equally possible at any point in the pipeline. what is the probability that the leak occurs no further than 20 km from A

70. On a 200-kilometer section of the gas pipeline, a gas leak occurs between compressor stations A and B, which is equally possible at any point in the pipeline. What is the probability that the leak occurs closer to A than to B?

71. The traffic police inspector’s radar has an accuracy of 10 km/hour and rounds to the nearest direction. What happens more often - rounding in favor of the driver or the inspector?

72. Masha spends 40 to 50 minutes on the way to the institute, and any time in this interval is equally probable. What is the probability that she will spend 45 to 50 minutes on the road?

73. Petya and Masha agreed to meet at the Pushkin monument from 12 to 13 o’clock, but no one could indicate the exact time of arrival. They agreed to wait for each other for 15 minutes. What is the probability of their meeting?

74. Fishermen caught 120 fish in the pond, 10 of them were ringed. What is the probability of catching a ringed fish?

75. From a basket containing 3 red and 7 green apples, all the apples are taken out one by one. What is the probability that the 2nd apple will be red?

76. From a basket containing 3 red and 7 green apples, all the apples are taken out one by one. What is the probability that the last apple will be green?

77. Students believe that out of 50 tickets, 10 are “good”. Petya and Masha take turns drawing one ticket each. What is the probability that Masha got a “good” ticket?

78. Students believe that out of 50 tickets, 10 are “good”. Petya and Masha take turns drawing one ticket each. What is the probability that they both got a “good” ticket?

79. Masha came to the exam knowing the answers to 20 questions out of 25 in the program. The professor asks 3 questions. What is the probability that Masha will answer 3 questions?

80. Masha came to the exam knowing the answers to 20 questions out of 25 in the program. The professor asks 3 questions. What is the probability that Masha will not answer any questions?

81. Masha came to the exam knowing the answers to 20 questions out of 25 in the program. The professor asks 3 questions. What is the probability that Masha will answer 1 question?

82. The statistics of loan requests from the bank are as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of non-repayment of loans is 0.01, 0.05 and 0.2, respectively. What percentage of loans are not repaid?

83. the probability that the weekly turnover of an ice cream merchant will exceed 2000 rubles. is 80% in clear weather, 50% in partly cloudy weather and 10% in rainy weather. What is the probability that the turnover will exceed 2000 rubles. if the probability of clear weather is 20%, and partly cloudy and rainy - 40% each.

84. In urn A there are white (b) and B black (h) balls. Two balls are drawn (simultaneously or sequentially) from the urn. Find the probability that both balls are white.

85. In urn A white and B

86. In ballot box A white and B

87. In ballot box A white and B black balls. One ball is taken from the urn, its color is noted and the ball is returned to the urn. After this, another ball is taken from the urn. Find the probability that these balls will be different colors.

88. There is a box of nine new tennis balls. To play, take three balls; After the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games there will be no unplayed balls left in the box?

89. Leaving the apartment, N each guest will wear his own galoshes;

90. Leaving the apartment, N guests who have the same shoe sizes wear galoshes in the dark. Each of them can distinguish the right galosh from the left, but cannot distinguish his own from someone else's. Find the probability that Each guest will wear galoshes belonging to the same pair (maybe not their own).

91. In the conditions of problem 90, find the probability that everyone will leave in their galoshes if the guests cannot distinguish the right galoshes from the left and simply take the first two galoshes they come across.

92. Shooting is underway at an aircraft, the vulnerable parts of which are two engines and the cockpit. In order to hit (disable) an aircraft, it is enough to hit both engines together or the cockpit. Under these firing conditions, the probability of hitting the first engine is equal to p1 second engine p2, cockpit p3. Aircraft parts are affected independently of each other. Find the probability that the plane will be hit.

93. Two shooters, independently of one another, fire two shots (each at its own target). Probability of hitting the target with one shot for the first shooter p1 for the second p2. The winner of the competition is the shooter whose target has the most holes. Find probability Rx that the first shooter wins.

94. behind a space object, the object is detected with probability R. Object detection in each cycle occurs independently of the others. Find the probability that when P cycles the object will be detected.

95. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the word "end" will appear.

96. Two balls are scattered randomly and independently of each other into four cells located one after the other in a straight line. Each ball has an equal probability of 1/4 landing in each cell. Find the probability that the balls will fall into neighboring cells.

97. Fire is fired at the aircraft with incendiary shells. The aircraft's fuel is concentrated in four tanks located in the fuselage, one after the other. The areas of the tanks are the same. In order to set the plane on fire, it is enough to hit two shells either in the same tank or in adjacent tanks. It is known that two shells hit the tank area. Find the probability that the plane will catch fire.

98. From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards will be of different suits.

99. From a full deck of cards (52 sheets), four cards are taken out at once, but each card is returned to the deck after removal. Find the probability that all four of these cards will be of different suits.

100. When the ignition is turned on, the engine starts to run with probability R.

101. The device can operate in two modes: 1) normal and 2) abnormal. Normal mode is observed in 80% of all cases of device operation; abnormal - in 20%. Probability of device failure over time t in normal mode it is 0.1; in abnormal - 0.7. Find the total probability R failure of the device.

102. A store receives goods from 3 suppliers: 55% from the 1st, 20 from the 2nd and 25% from the 3rd. The percentage of defects is 5, 6 and 8 percent, respectively. What is the probability that the purchased defective product came from a second supplier.

103.The flow of cars past gas stations consists of 60% trucks and 40% cars. What is the probability of a truck being at a gas station if the probability of refueling it is 0.1, and the probability of a passenger car is 0.3

104. The flow of cars past gas stations consists of 60% trucks and 40% cars. What is the probability of a truck being at a gas station if the probability of refueling it is 0.1, and the probability of a passenger car is 0.3

105. A store receives goods from 3 suppliers: 55% from the 1st, 20 from the 2nd and 25% from the 3rd. The percentage of defects is 5, 6 and 8 percent, respectively. What is the probability that the purchased defective product came from the 1st supplier.

106. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the word “book” will appear.

107. A store receives goods from 3 suppliers: 55% from the 1st, 20 from the 2nd and 25% from the 3rd. The percentage of defects is 5, 6 and 8 percent, respectively. What is the probability that the purchased defective product came from the 1st supplier.

108. Two balls are scattered randomly and independently of each other into four cells located one after the other in a straight line. Each ball has an equal probability of 1/4 landing in each cell. Find the probability that 2 balls will fall into one cell

109. When the ignition is turned on, the engine starts to run with probability R. Find the probability that the engine will start running the second time the ignition is turned on;

110. Fire is fired at the aircraft with incendiary shells. The aircraft's fuel is concentrated in four tanks located in the fuselage, one after the other. The areas of the tanks are the same. In order to set the plane on fire, it is enough to hit two shells in the same tank. It is known that two shells hit the tank area. Find the probability that the plane will catch fire

111. Fire is fired at the aircraft with incendiary shells. The aircraft's fuel is concentrated in four tanks located in the fuselage, one after the other. The areas of the tanks are the same. In order to set the plane on fire, it is enough to hit the adjacent tanks with two shells. It is known that two shells hit the tank area. Find the probability that the plane will catch fire

112.In urn A white and B black balls. One ball is taken from the urn, its color is noted and the ball is returned to the urn. After this, another ball is taken from the urn. Find the probability that both drawn balls will be white.

113. In ballot box A white and B black balls. Two balls are drawn from the urn at once. Find the probability that these balls will be different colors.

114. Two balls are scattered randomly and independently of each other into four cells located one after the other in a straight line. Each ball has an equal probability of 1/4 landing in each cell. Find the probability that the balls will fall into neighboring cells.

115. Masha came to the exam knowing the answers to 20 questions out of 25 in the program. The professor asks 3 questions. What is the probability that Masha will answer 2 questions?

116. Students believe that out of 50 tickets, 10 are “good”. Petya and Masha take turns drawing one ticket each. What is the probability that they both got a “good” ticket?

117. The statistics of loan requests from the bank are as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of non-repayment of loans is 0.01, 0.05 and 0.2, respectively. What percentage of loans are not repaid?

118. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the word "end" will appear.

119 Statistics on loan requests from the bank are as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of non-repayment of loans is 0.01, 0.05 and 0.2, respectively. What percentage of loans are not repaid?

120. the probability that the weekly turnover of an ice cream merchant will exceed 2000 rubles. is 80% in clear weather, 50% in partly cloudy weather and 10% in rainy weather. What is the probability that the turnover will exceed 2000 rubles. if the probability of clear weather is 20%, and partly cloudy and rainy - 40% each.

Goal of the work: develop skills in solving problems in probability theory using the total probability formula and Bayes formula.

Total Probability Formula

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

This formula is called the total probability formula.

Probability of hypotheses. Bayes formula

Let the event A may occur subject to the occurrence of one of the incompatible events V b 2 ,..., V n, forming a complete group. Since it is not known in advance which of these events will occur, they are called hypotheses. Probability of event occurrence A determined by the total probability formula:

Let us assume that a test was carried out, as a result of which an event occurred A. It is necessary to determine how the changes (due to the fact that the event A has already arrived) the probability of the hypotheses. Conditional probabilities of hypotheses are found using the formula

In this formula, index / = 1.2

This formula is called Bayes' formula (named after the English mathematician who derived it; published in 1764). Bayes' formula allows us to reestimate the probabilities of hypotheses after the result of the test that resulted in the event becomes known. A.

Task 1. The factory 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 detected, 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 incorrectly; C - the part will be passed into the finished product with a defect.

Solution

Let us denote the hypotheses:

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

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

Event A =(the part will be rejected).

From the problem conditions we find the probabilities

R N (A) = 0,01; Pfi(A) = 0,97.

Using the total probability formula we get

The probability that a part will be rejected incorrectly is

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

Answer:

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

Solution

Let's denote the events:

L. =(the product will go to the/th merchandiser for inspection); / = 1, 2, 3;

B =(the product will be considered standard).

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

Conditional probabilities are also known

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

Answer:“0.263.

Task 3. Two machines produce parts that go onto a common conveyor. The probability of receiving a non-standard part on the first machine is 0.06, and on the second - 0.09. The productivity of the second machine is twice that of the first. A non-standard part was taken from the assembly line. Find the probability that this part was produced by the second machine.

Solution

Let's denote the events:

A. =(a part taken from the conveyor was produced by the /th machine); / = 1.2;

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

Conditional probabilities are also known

Using the total probability formula we find

Using the Bayes formula, we find the probability that the selected non-standard part was produced by the second machine:

Answer: 0,75.

Task 4. A device consisting of two units is being tested, the reliability of which is 0.8 and 0.9, respectively. Nodes fail independently of each other. The device failed. Taking this into account, find the probability of the 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;

A= (during testing there will be a device failure).

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

By the property of the probabilities of opposite events

Event A equal to the sum of the products of independent events

Using the theorem for adding 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 produced on three machines, which produce 25%, 30% and 45% of the total number of bolts, respectively. In machine tool products, defects are 4%, 3% and 2%, respectively. What is the probability that a bolt randomly taken from an incoming product will be defective?

Solution

Let's denote the events:

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

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

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

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

Using the total probability formula we find

Answer: 0,028.

Task 6. The electronic circuit belongs to one of three parties with probabilities of 0.25; 0.5 and 0.25. The probability that the circuit will operate beyond the warranty service life for each batch is 0.1; 0.2 and 0.4. Find the probability that a randomly chosen circuit will operate beyond its warranty period.

Solution

Let's denote the events:

4 = (randomly taken circuit from the ith batch); i = 1, 2, 3;

IN= (a randomly chosen circuit will work beyond the warranty period).

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

Conditional probabilities are also known:

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 has failed. Determine the probability that both units failed.

Solution

Let's denote the events:

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

A= (device will fail).

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

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

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

Answer:

Problems to solve independently Task 1. In the warehouse of the television studio there are 70% of picture tubes manufactured by plant No. 1; the remaining picture tubes were manufactured by plant No. 2. The probability that the picture tube will not fail during the warranty service life is 0.8 for picture tubes from factory No. 1 and 0.7 for picture tubes from factory No. 2. The picture tube survived the warranty service life. Find the probability that it was manufactured by plant No. 2.

Task 2. Parts are received for assembly from three machines. It is known that the 1st machine gives 0.3% of defects, the 2nd - 0.2%, the 3rd - 0.4%. Find the probability of receiving a defective part for assembly if 1000 parts were received from the 1st machine, 2000 from the 2nd, 2500 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 productivity of the second machine is three times greater than the productivity of the first. Find the probability that a part taken at random from a conveyor that receives parts from both machines will be standard.

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 cargo for the first, second and third firms are equal to 0.05, respectively; 0.1 and 0.07. Having compared these data with data on the safety of cargo transportation, the manager came to the conclusion that the choice was equivalent and decided to make it by lot. Find the probability that the shipped 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 has failed. Determine the probability that the second unit failed.

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

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

Task 8. Defects in factory products due to a defect A is 5%, and among those rejected based on A products are defective in 10% of cases R. And in products free from defects 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 randomly selected car will work properly.

Problem 10. Two sensors send signals into a common communication channel, with the first one sending 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?

Problem 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 selected at random, and a part is taken from this batch. Determine the probability that the part taken will be standard.

Problem 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 a part from the first plant is of excellent quality is 0.7; for parts from the second and third factories, 0.8 and 0.9, respectively. The part taken at random turned out to be of excellent quality. Find the probability that the part was manufactured by the first plant.

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

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

Problem 15. The weightlifting coach calculated that in order to receive team points in a given weight category, an athlete must push a barbell of 200 kg. Ivanov, Petrov and Sidorov are vying for a place on the team. During training, Ivanov tried to lift such a weight in 7 cases, and lifted it in 3 of them. Petrov lifted in 6 out of 13 cases, 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 scoring points to the team.
• b) The team did not receive any scoring points. Find the probability that Sidorov performed.

Problem 16. There are 12 red and 6 blue balls in a white box. In black there are 15 red and 10 blue balls. Throwing a dice. If a number of points is a multiple of 3, then a ball is taken at random from the white box. If any other number of points is rolled, a ball is taken at random from the black box. What is the probability of a red ball appearing?

Problem 17. Two boxes contain radio tubes. The first box contains 12 lamps, 1 of which is non-standard; in the second there are 10 lamps, of which 1 is non-standard. A lamp is taken at random from the first box and placed in the second. Find the probability that a lamp taken at random from the second box will be non-standard.

Problem 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 extracted ball will be white if all possible assumptions about the initial composition of the balls (based on color) are equally possible.

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

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

1. Total probability formula.

Let event A occur subject to the occurrence of one of the incompatible events B 1, B 2, B 3, ..., B n, which form a complete group. Let the probabilities of these events and conditional probabilities be knownP(A/B 1), P(A/B 2), ..., P(A/B n) event A. You need to find the probability of event A.

Theorem:The probability of event A, which can occur only if one of the incompatible events occurs B 1, B 2, B 3, ..., B n , forming a complete group, is equal to the sum of the products of the probabilities of each of these events by the corresponding conditional probability of event A:

– Total probability formula.

Proof:

According to the condition, event A can occur if one of the incompatible events occursB 1, B 2, B 3, ..., B n. In other words, the occurrence of event A means the occurrence of one (no matter which) of the incompatible events:B 1 *A, B 2*A, B 3*A, ..., B n*A. Using the addition theorem, we get:

According to the theorem of multiplication of probabilities of dependent events, we have:

etc.

Example: There are 2 sets of parts. The probability that a part from the first set is standard is 0.8, and for the second set it is 0.9. Find the probability that a part taken at random (from a set taken at random) is standard.

Solution: Event A - “The extracted part is standard.” Event - “They removed a part manufactured by 1 plant.” Event - “A part manufactured by the second plant was removed.” R( B 1 )=P(B 2)= 1/2.P(A / B 1 ) = 0.8 - probability that the part manufactured at the first plant is standard. P(A / B 2 )=0.9 - probability that the part manufactured at the second plant is standard.

Then, according to the total probability formula, we have:

Example: The assembler received 3 boxes of parts manufactured by plant No. 1 and 2 boxes of parts manufactured by plant No. 2. The probability that a part manufactured by plant No. 1 is standard is 0.8. For plant No. 2 this probability is 0.9. The assembler randomly removed a part from a randomly selected box. Find the probability that a standard part is removed.

Solution: Event A - “Standard part removed.” Event B 1 - “The part was removed from the box of factory No. 1.” Event B 2 - “The part was removed from the box of factory No. 2.” R( B 1)= 3/5. P(B 2 )= 2/5.

P(A / B 1) = 0.8 - probability that the part manufactured at the first plant is standard. P(A /B 2) = 0.9 - probability that the part manufactured at the second plant is standard.

Example:The first box contains 20 radio tubes, of which 18 are standard. The second box contains 10 radio tubes, of which 9 are standard. One radio tube was randomly transferred from the second box to the first. Find the probability that a lamp drawn at random from the first box will be a standard one.

Solution:Event A - “A standard lamp was removed from 1 box.” EventB 1 - “A standard lamp was transferred from the second to the first box.” EventB 2 - “A non-standard lamp was transferred from the second to the first box.” R( B 1 )= 9/10. P(B 2)= 1/10.P(A / B 1)= 19/21 - probability of getting a standard part out of the first box, provided that the same standard part was put into it.

P(A / B 2 )= 18/21 - probability of taking a standard part out of the first box, provided that a non-standard part was placed in it.

2. Formulas of hypotheses of Thomas Bayes.

Let event A occur subject to the occurrence of one of the incompatible events B 1, B 2, B 3, ..., B n, forming a complete group. Since it is not known in advance which of these events will occur, they are called hypotheses. The probability of occurrence of event A is determined by the total probability formula discussed earlier.

Let us assume that a test was carried out, as a result of which event A occurred. Let us set our task to determine how the probabilities of the hypotheses have changed (due to the fact that event A has already occurred). In other words, we will look for conditional probabilitiesP(B 1 /A), P(B 2 /A), ..., P(B n /A)

Let's find the conditional probability P(B 1/A) . By the multiplication theorem we have:

This implies:

Similarly, formulas are derived that determine the conditional probabilities of the remaining hypotheses, i.e. conditional probability any hypothesis B k (i =1, 2, …, n ) can be calculated using the formula:

Thomas Bayes hypothesis formulas.

Thomas Bayes (English mathematician) published the formula in 1764.

These formulas make it possible to reestimate the probabilities of hypotheses after the result of the test that resulted in event A becomes known.

Example: Parts manufactured by the factory workshop are sent to one of two inspectors to check their standardness. The probability that the part will reach the first inspector is 0.6, and the second one is 0.4. The probability that a suitable part will be recognized as standard by the first inspector is 0.94, for the second inspector this probability is 0.98. During inspection, the acceptable part was recognized as standard. Find the probability that the first inspector checked this part.

Solution: Event A - “The good part is recognized as standard.” Event B 1 - “The part was checked by the first inspector.” EventB 2 - “The part was checked by the second inspector.” R( B 1 )=0.6. P(B 2 )=0.4.

P(A / B 1) = 0.94 - probability that the part checked by the first inspector is recognized as standard.

P(A / B 2) = 0.98 - probability that the part checked by the second inspector is recognized as standard.

Then:

Example:To participate in student qualifying sports competitions, 4 people were allocated from the first group of the course, 6 people from the second group, and 5 people from the third group. The probability that a student in the first group will be included in the national team is 0.9; for students in the second and third groups, these probabilities are 0.7 and 0.8, respectively. As a result of the competition, a randomly selected student ended up in the national team. Which group most likely does he belong to?

Solution: Event A - “A student chosen at random got into the institute’s team.” Event B 1 - “A student from the first group was selected at random.” Event B 2 - “A student from the second group was selected at random.” Event B 3 - “A student from the third group was selected at random.” R( B 1)= 4/15 . P(B 2) = 6/15. P(B 3)= 5/15.

P(A / B 1)=0.9 is the probability that a student from the first group will make it to the national team.

P(A / B 2) = 0.7 is the probability that a student from the second group will make it to the national team.

P(A / B 3 )=0.8 is the probability that a student from the third group will make it to the national team.

Then:

The probability that a student from the first group made it to the team.

The probability that a student from the second group made it to the team.

The probability that a student from the third group made it to the team.

Most likely, a student from the second group will make it to the team.

Example:If the machine deviates from the normal operating mode, the C 1 alarm will go off with a probability of 0.8, and the C 2 alarm will go off with a probability of 1. The probability that the machine is equipped with a C 1 or C 2 alarm is 0.6 and 0.4, respectively. A signal has been received to cut the machine gun. What is more likely: the machine is equipped with a signaling device C 1 or C 2?

Solution:Event A - “A signal to cut the machine gun has been received.” Event B 1 - “The machine is equipped with a C1 signaling device. EventB 2 - “The machine is equipped with a C2 signaling device. R( B 1 )= 0.6. P(B 2) = 0.8.

P(A / B 1) = 0.8 is the probability that a signal will be received, provided that the machine is equipped with a signaling device C1.

P(A / B 2 )=1 - probability that a signal will be received, provided that the machine is equipped with a C2 signaling device.

Then:

There is a possibility that upon receiving a signal to cut the machine, the C1 alarm went off.

There is a possibility that upon receiving a signal to cut the machine, the C2 alarm went off.

Those. It is more likely that when cutting the machine, a signal will be received from signaling device C1.

A consequence of the two main theorems of probability theory - the theorems of addition and multiplication - are the formulas of total probability and the Bayes formulas.

In the language of event algebra, the set , , ¼, is called full group of events, If:

1. Events are pairwise incompatible, i.e. , , ;.

2. The sum totals the entire probability space .

Theorem 5 (Total probability formula). If the event A can occur only if one of the events (hypotheses) , ,¼, appears, forming a complete group, then the probability of the event A equal to

Proof. Since hypotheses , ,¼, are the only possible ones, and the event A according to the conditions of the theorem can only occur together with one of the hypotheses, then . From the incompatibility of hypotheses follows incompatibility .

We apply the probability addition theorem in the form (6):

By the multiplication theorem. Substituting this representation into formula (13), we finally have: , which is what needed to be proved.

Example 8. An export-import company is about to enter into a contract for the supply of agricultural equipment to one of the developing countries. If the company's main competitor does not simultaneously bid for a contract, then the probability of receiving a contract is estimated at 0.45; otherwise – at 0.25. According to company experts, the probability that a competitor will put forward proposals for concluding a contract is 0.40. What is the probability of concluding a contract?

Solution. A -“the company will enter into a contract”, - “the competitor will put forward its proposals”, - “the competitor will not put forward its proposals”. According to the conditions of the problem , . Conditional probabilities of concluding a contract for a firm , . According to the total probability formula

A corollary of the multiplication theorem and the total probability formula is Bayes' formula.

Bayes formula allows you to recalculate the probability of each of the hypotheses, provided that the event occurred. (It applies when the event A, which can appear with only one of the hypotheses forming a complete group of events, has occurred and it is necessary to quantitatively re-evaluate the prior probabilities of these hypotheses known before the test, i.e. it is necessary to find the posterior (obtained after the test) conditional probabilities of the hypotheses) , ,…, .

Theorem 6 (Bayes Formula). If the event A happened, then the conditional probabilities of the hypotheses are calculated using a formula called Bayes' formula:

Proof. To obtain the required formula, we write the theorem for multiplying the probabilities of events A and in two forms:

where Q.E.D.

The meaning of Bayes' formula is that when an event occurs A, those. As we receive new information, we can test and adjust the hypotheses put forward before testing. This approach, called Bayesian, makes it possible to correct management decisions in economics, estimates of unknown parameters of the distribution of the characteristics being studied in statistical analysis, etc.

Task 9. The group consists of 6 excellent students, 12 well-performing students and 22 mediocre performing students. An excellent student answers 5 and 4 with equal probability, an excellent 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 performing student was called?

Solution. Let's consider three hypotheses:

Event in question. From the problem statement it is known that

, , .

Let's find the probabilities of the hypotheses. Since there are only 40 students in the group, and 6 excellent students, then . Likewise, , . Applying the total probability formula, we find

Now we apply Bayes’ formula to the hypothesis:

Example 10. An economist-analyst conditionally divides the economic situation in a country into “good”, “mediocre” and “bad” and estimates their probabilities for a given point in time at 0.15; 0.70 and 0.15 respectively. Some index of economic condition increases with probability 0.60 when the situation is “good”; with a probability of 0.30 when the situation is mediocre, and with a probability of 0.10 when the situation is “bad”. Let the economic condition index increase at the moment. What is the probability that the country's economy is booming?

Solution. A= “the country’s economic condition index will increase”, H 1= “the economic situation in the country is “good””, H 2= “the economic situation in the country is “mediocre””, N 3= “the economic situation in the country is “bad”.” By condition: , , . Conditional probabilities: ,, . You need to find the probability. We find it using Bayes' formula:

Example 11. The trading company received televisions from three suppliers in a ratio of 1:4:5. Practice has shown that TVs coming from the 1st, 2nd and 3rd suppliers will not require repairs during the warranty period in 98%, 88% and 92% of cases, respectively.