Bayes' theorem is the theory of the probability of an event. A simple explanation of Bayes' theorem

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 us to reassess the probabilities of the 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. 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?

Decision. 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.

Decision. 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 particular 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.

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

- Bernoulli's 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.

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!).

- the probability of the event occurring B, depending on a number of hypotheses A i if the degrees of reliability of these hypotheses are known (for example, measured experimentally);

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 (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 further trained). 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, used only when absolutely necessary), or old filtering methods - "blacklists" and regular expressions (since such letters often have a stereotypical form).

Notes

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

Bayes' theorem is described in detail in a separate article. This is a wonderful work, but it has 15,000 words. The very essence of the theorem is briefly explained in the same translation of the article by Kalid Azad.

The results of research and testing are not events. There is a method for diagnosing cancer, but there is the event itself - the presence of the disease. The algorithm checks if the message contains spam, but the event (spam really came to the mail) must be considered separately from the result of its work.
There are errors in test results. Often our research methods reveal what is not (false positive) and do not reveal what is (false negative).
With the help of trials, we get the probabilities of a certain outcome. We too often look at test results by themselves and do not take into account method errors.
False positive results distort the picture. Suppose you are trying to detect some very rare phenomenon (1 in 1,000,000). Even if your method is accurate, it is likely that its positive result will actually be a false positive.
It is more convenient to work with natural numbers. Better to say: 100 out of 10,000, not 1%. With this approach, there will be fewer errors, especially when multiplying. Let's say we need to work on that 1% further. Reasoning in percentages is clumsy: "in 80% of cases out of 1% got a positive outcome." Much easier information is perceived as follows: "in 80 cases out of 100, a positive outcome was observed."
Even in science, any fact is just the result of applying some method. From a philosophical point of view, a scientific experiment is just a test with a probable error. There is a method that reveals a chemical substance or some phenomenon, and there is an event itself - the presence of this phenomenon. Our test methods can give a false result, and any equipment has an inherent error.

Bayes' theorem turns test results into probabilities of events.

If we know the probability of an event and the probability of false positives and false negatives, we can correct for measurement errors.
The theorem correlates the probability of an event with the probability of a certain outcome. We can relate Pr(A|X): the probability of an event A given an outcome X, and Pr(X|A): the probability of an outcome X given an event A.

Understanding the Method

The article referenced at the beginning of this essay discusses the diagnostic method (mammogram) that detects breast cancer. Let's consider this method in detail.

1% of all women have breast cancer (and, accordingly, 99% do not get sick)
80% of mammograms detect the disease when it really is (and, accordingly, 20% do not detect)
9.6% of studies detect cancer when there is none (and thus 90.4% correctly report a negative result)

Now let's create a table like this:

How to work with this data?

1% of women get breast cancer
if the patient has a disease, look in the first column: there is an 80% chance that the method gave the correct result, and a 20% chance that the result of the study is incorrect (false negative)
if the patient has not been diagnosed with the disease, look at the second column. With a probability of 9.6% we can say that the positive result of the study is incorrect, and with a 90.4% probability we can say that the patient is really healthy.

How accurate is the method?

Now let's look at the positive test result. What is the probability that a person is really sick: 80%, 90%, 1%?

Let's think:

There is a positive result. We will analyze all possible outcomes: the result obtained can be both true positive and false positive.
The probability of a true positive result is equal to: the probability of getting sick multiplied by the probability that the test actually detected the disease. 1% * 80% = .008
The probability of a false positive result is equal to: the probability that the disease is not present, multiplied by the probability that the method detected the disease incorrectly. 99% * 9.6% = .09504

Now the table looks like this:

What is the probability that a person is really sick if a positive mammogram result is obtained? The probability of an event is the ratio of the number of possible outcomes of an event to the total number of all possible outcomes.

Event Probability = Event Outcomes / All Possible Outcomes

The probability of a true positive result is .008. The probability of a positive outcome is the probability of a true positive outcome + the probability of a false positive.

(.008 + 0.09504 = .10304)

So, the probability of a disease with a positive result of the study is calculated as follows: .008 / .10304 = 0.0776. This value is about 7.8%.

That is, a positive mammogram result only means that the probability of having a disease is 7.8%, and not 80% (the latter value is only the estimated accuracy of the method). Such a result seems incomprehensible and strange at first, but you need to consider: the method gives a false positive result in 9.6% of cases (and this is quite a lot), so there will be many false positive results in the sample. For a rare disease, most positive results will be false positives.

Let's run our eyes over the table and try to intuitively grasp the meaning of the theorem. If we have 100 people, only one of them has the disease (1%). In this person, with 80% probability, the method will give a positive result. Of the remaining 99%, 10% will have positive results, which gives us, roughly speaking, 10 out of 100 false positives. If we consider all positive results, then only 1 out of 11 will be true. Thus, if a positive result is obtained, the probability of the disease is 1/11.

Above, we calculated that this probability is equal to 7.8%, i.e. the number is actually closer to 1/13, but here, with simple reasoning, we were able to find a rough estimate without a calculator.

Bayes' theorem

Now let's describe the course of our thoughts with a formula, which is called Bayes' theorem. This theorem allows you to correct the results of the study in accordance with the distortion that false positive results introduce:

Pr(A|X) = probability of disease (A) with a positive result (X). This is exactly what we want to know: what is the probability of an event in the event of a positive outcome. In our example, it is equal to 7.8%.
Pr(X|A) = probability of a positive result (X) in the case when the patient is really sick (A). In our case, this is the value of true positive - 80%
Pr(A) = probability of getting sick (1%)
Pr(not A) = probability of not getting sick (99%)
Pr(X|not A) = probability of a positive outcome of the study if there is no disease. This is the value of false positives - 9.6%.

We can conclude that to get the probability of an event, you need to divide the probability of a true positive outcome by the probability of all positive outcomes. Now we can simplify the equation:
Pr(X) is the normalization constant. She served us well: without her, a positive test outcome would give us an 80% chance of an event.
Pr(X) is the probability of any positive result, whether it is a true positive in a patient study (1%) or a false positive in a healthy study (99%).

In our example, Pr(X) is a rather large number because there is a high probability of false positive results.

Pr(X) produces a result of 7.8%, which at first glance seems counterintuitive.

The meaning of the theorem

We are testing to find out the true state of things. If our tests are perfect and accurate, then the probabilities of trials and probabilities of events will coincide. All positive results will be truly positive, and negative results will be negative. But we live in the real world. And in our world, tests give wrong results. Bayes' theorem accounts for skewed results, corrects for errors, reconstructs the population, and finds the probability of a true positive result.

Spam filter

Bayes' theorem is successfully applied in spam filters.

We have:

event A - in a spam email
the result of the test is the content of certain words in the letter:

The filter takes test results (content of certain words in the email) into account and predicts if the email contains spam. Everyone understands that, for example, the word "Viagra" is more common in spam than in regular emails.

The blacklist-based spam filter has the disadvantage of often producing false positives.

The Bayesian spam filter takes a measured and reasonable approach: it works with probabilities. When we analyze the words in an email, we can calculate the probability that the email is spam rather than making yes/no decisions. If there is a 99% chance that the email contains spam, then the email is indeed spam.

Over time, the filter trains on an ever larger sample and updates the probabilities. For example, advanced filters based on Bayes' theorem check many words in a row and use them as data.

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INFORMATION TECHNOLOGY, COMPUTER SCIENCE, AND MANAGEMENT

On the applicability of the Bayes formula

DOI 10.12737/16076

A. I. Dolgov**

1Joint-Stock Company "Design Bureau for Radio Monitoring of Control, Navigation and Communication Systems", Rostov-on-Don, Russian Federation

On applicability of Bayes" formula*** A. I. Dolgov1**

1"Design bureau on monitoring of control, navigation and communication systems" JSC, Rostov-on-Don, Russian Federation

The subject of this study is the Bayes formula. The purpose of this work is to analyze and expand the scope of the formula. The primary task is to study publications devoted to this problem, which made it possible to identify the shortcomings of the application of the Bayes formula, leading to incorrect results. The next task is to construct modifications of the Bayes formula that take into account various single evidences and obtain correct results. And, finally, on the example of specific initial data, the incorrect results obtained using the Bayes formula and the correct results calculated using the proposed modifications are compared. Two methods were used in the study. First, the analysis of the principles of constructing known expressions used to write the Bayes formula and its modifications was carried out. Secondly, a comparative evaluation of the results (including a quantitative one) was carried out. The proposed modifications provide a wider application of the Bayes formula in theory and practice, including in solving applied problems.

Key words: conditional probabilities, incompatible hypotheses, compatible and incompatible evidence, normalization.

Bayes" formula is the research subject. The work objective is to analyze the formula application and widen the scope of its applicability. The first-priority problem includes the identification of the Bayes" formula disadvantages based on the study of the relevant publications leading to incorrect results. The next task is to construct the Bayes" formula modifications to provide an accounting of various single indications to obtain correct results. And finally, the incorrect results obtained with the application of Bayes" formula are compared to the correct results calculated with the use of the proposed formula modifications by the example of the specific initial data. Two methods are used in studies. First, the analysis of the principles of constructing the known expressions used to record the Bayesian formula and its modifications is conducted. Secondly, a comparative evaluation of the results (including the quantitative one) is performed. The proposed modifications provide a wider application of Bayes" formula both in theory and practice including the solution of the applied problems.

Keywords: conditional probabilities, inconsistent hypotheses, compatible and incompatible indications, normalizing.

Introduction. The Bayes formula is increasingly used in theory and practice, including in solving applied problems with the help of computer technology. The use of mutually independent computational procedures makes it especially effective to apply this formula when solving problems on multiprocessor computing systems, since in this case the parallel implementation is performed at the level of the general scheme, and when adding the next algorithm or class of problems, there is no need to re-perform parallelization work.

The subject of this study is the applicability of the Bayes formula for the comparative estimation of posterior conditional probabilities of inconsistent hypotheses under various single evidences. As the analysis shows, in such cases, the normalized probabilities of incompatible combined events belonging to

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IS eö AND IS X X<и H

"The work was carried out as part of an initiative research project.

** Email: [email protected]

""The research is done within the frame of the independent R&D.

for different complete groups of events . At the same time, the compared results turn out to be inadequate to real statistical data. This is due to the following factors:

Incorrect normalization is used;

The presence or absence of intersections of the considered evidence is not taken into account.

In order to eliminate the identified shortcomings, cases of applicability of the Bayes formula are identified. If the specified formula is not applicable, the problem of constructing its modification is solved, which ensures that various single evidences are taken into account with obtaining correct results. On the example of specific initial data, a comparative assessment of the results was carried out:

Incorrect - obtained using the Bayes formula;

Correct - calculated using the proposed modification.

Starting positions. The following statements are based on the principle of preserving the probability ratios: “Correct processing of the probabilities of events is feasible only when normalizing using one common normalizing divisor that ensures the equality of the ratios of the normalized probabilities to the ratios of the corresponding normalized probabilities” . This principle represents the subjective basis of probability theory, but is not properly reflected in modern educational and scientific and technical literature.

If this principle is violated, information about the degree of possibility of the events under consideration is distorted. The results obtained on the basis of distorted information and the decisions made turn out to be inadequate to the real statistical data.

The following concepts will be used in this article:

An elementary event is an event that is not divisible into elements;

Combined event - an event representing one or another combination of elementary events;

Compatible events - events that in some cases of a comparative assessment of their probabilities may be incompatible, and in other cases joint;

Incompatible events are events that are incompatible in all cases.

According to the probability multiplication theorem, the probability P (U ^ E) of the product of elementary events U ^ and

E is calculated as a product of probabilities P(Uk E) = P(E)P(U^E) . In this regard, the Bayes formula is often

is written in the form Р(Ик\Е) = - - - , describing the definition of a posteriori conditional probabilities

P(U-E) of the hypotheses Uk (k = 1,...n) based on the normalization of a priori probabilities P(U-E) of the considered combined incompatible events I to E. Each of these events represents a product whose factors are one of the considered hypotheses and one accountable evidence. At the same time, everything is considered

uIKE events (k = 1,...n) form a complete group of uIKE incompatible combined events, due to

with which their probabilities P(Ik E) should be normalized taking into account the total probability formula, according to which

swarm P(E) = 2 P(Uk)P(E\Uk). Therefore, the Bayes formula is most often written in the most commonly used form:

P(Uik) P(EIK)

P(Uk \ E) \u003d -. (one)

^ cation of the Bayes formula.

th Analysis of the features of the construction of the Bayes formula, aimed at solving applied problems, as well as examples

“and its practical application allow us to draw an important conclusion regarding the choice of a complete group of combined events compared in terms of the degree of possibility (each of which is the product of two elementary events - one of the hypotheses and the evidence taken into account). Such a choice is made subjectively by the decision maker, on the basis of objective initial data inherent in the typical conditions of the situation: the types and number of hypotheses evaluated and the evidence specifically taken into account.

Incomparable probabilities of hypotheses with single inconsistent evidence. The Bayes formula is traditionally used in the case of determining posterior conditional probabilities that are not comparable in terms of the degree of possibility.

the probability of hypotheses H^ with single incompatible evidence, each of which can "appear

only in combination with any of these hypotheses. In this case, full groups and HkE are selected, combined

bath events in the form of products, the factors of which are one of the evidence of c. (1=1,...,m) and one

of the n hypotheses under consideration.

The Bayes formula is used for a comparative assessment of the probabilities of combined events of each such complete group, which differs from other complete groups not only in the evidence e taken into account, but also in the general case in the types of hypotheses H ^ and (or) their number n (see, for example, )

RNky = P(Hk) P(eH)

% P(Hk) P(Er\Hk) k = 1

In a special case for n = 2

RNk\E,~ P(Hk) P(EN)

% P(Hk) P(E,\H k) k = 1

and the results obtained are correct, due to the observance of the principle of conservation of probability ratios:

P(H1E,) _ P(H 1)P(E,\H1) / P(H2) P(E,\H2) = P(H 1) P(E,\H1)

P(H 2 = % PW1!)

The subjectivity of the choice of a complete group of combined events compared in terms of the degree of possibility (with

certain variable elementary events) allows you to select a complete group of events and Hk E ■ s

by negating the elementary event E ■ () and write the Bayes formula (1 = 1,.. ., m) as follows:

P(Hk \ E) -= - RNSh ±.

% P(Hk)P(E, Hk)

Such a formula is also applicable and makes it possible to obtain the correct results if the calculated to

the normalized probabilities are compared under the various hypotheses considered, but not under various

authorities. ¡^

Comparable probabilities of hypotheses under single inconsistent evidence. Judging by well-known publica-^

is used for a comparative assessment of a posteriori conditional probabilities of hypotheses for various single evidences.

authorities. At the same time, attention is not paid to the following fact. In these cases, the normalized ^ probabilities of incompatible (incompatible) combined events belonging to different complete groups n of events are compared. However, in this case, the Bayes formula is not applicable, since combined events that are not included in one complete group are compared, the normalization of the probabilities of which is carried out using different n normalizing divisors. The normalized probabilities of incompatible (incompatible) combined events can be compared only if they belong to the same complete group of events and are normalized by ¡3 using a common divisor equal to the sum of the probabilities of all normalized events included in the complete §

In general, the following can be considered incompatible evidence:

Two evidence (for example, evidence and its denial); ^

Three evidence (for example, in a game situation, win, lose and draw); ^

Four testimonials (particularly in sports, winning, losing, drawing and replaying), etc. ^

Consider a fairly simple example (corresponding to the example given in ) of applying the Bayes formula ^ to determine the posterior conditional probabilities of the hypothesis H ^ for two incompatible events in

in the form of evidence L]- and its denial L]

P(H, k) - ^ . ^ P(A^k" , (2)

] E P(Hk> P(A]\vk> k - 1

■ _ P(HkA ]) P(Hk> P(A ]\nk>

P(H,\A,) ----k-]-. (3)

V k\A]> P(A > n

] E P(Hk) P(A]\Hk) k -1

In cases (2) and (3), subjectively selected full groups compared in terms of the degree of possibility of com-

binned events are, respectively, the sets and H to A and and H to A. This is the case when the formula

k-1 k ] k-1 k ]

Bayes is inapplicable, since the principle of preserving the ratios of probabilities is violated - the equality of the ratios of normalized probabilities to the ratios of their corresponding normalized probabilities is not observed:

P(H to A]] P(Hk) P(A]\Hk) / P(Hk) P(A]\Hk) P(Hk) P(A] Hk)

P(Hk E P(Hk) P(A]\Hk)/ E P(Hk) P(A]\Hk) P(Hk) P(A] Hk)

k - 1 /k - 1 According to the principle of conservation of probability ratios, correct processing of event probabilities is feasible only when normalizing using one common normalizing divisor equal to the sum of all compared normalized expressions. So

E P(Hk)P(A]\Hk) + E P(Hk)P(A]\Hk) - E P(Hk)[P(A]\Hk) + P(Hk) P(A]\Hk )] - EP (Hk) - 1. to -1 to -1 to -1 to -1

Thus, the fact is revealed that there are varieties of the Bayes formula that differ from

known for the lack of a normalizing divisor:

A,) - P(H) P(A]\Hk), P(Hk A,) - P(H) P(A, Hk). (4)

J to I ■> to

In this case, the equality of the ratios of the normalized probabilities to the ratios of the corresponding normalized probabilities is observed:

m^A^ P(Hk) P(A]\Hk)

A,) P(H k) P(A, Hk)

Based on the subjective choice of non-traditionally recorded complete groups of incompatible combined events, it is possible to increase the number of modifications of the Bayes formula that include evidence, as well as one or another number of their denials. For example, the most complete group of combined events

u and Hk /"./ ^ u and Hk E\ corresponds (taking into account the absence of a normalizing divisor) the modification formula; =1 A"=1; \u003d 1 Bayesian

P(Hk\~) - P(Hk) ПЁ^^^

where an elementary event in the form of evidence E \ e II II / "/ is one of the elements of the indicated set

o In the absence of denials of evidence, that is, when E\ \u003d // e and /"./,

^ P(H\E) P(Hk) P(E,\Hk)

E P(Hk) P(E \ Hk) k - 1

Thus, the modification of the Bayes formula, intended to determine the conditional probabilities of hypotheses that are comparable in terms of the degree of possibility for single incompatible evidence, is as follows. The numerator contains the normalized probability of one of the combined incompatible events that form a complete group, expressed as a product of a priori probabilities, and the denominator contains the sum of all

normalized probabilities. At the same time, the principle of preserving the ratios of probabilities is observed - and the result obtained is correct.

Probabilities of hypotheses under single compatible evidence. Bayesian formulas are traditionally used to determine the posterior conditional probabilities of the hypotheses Hk (k = 1,...,n) compared in terms of the degree of possibility for one of several considered compatible evidence EL (1 = 1,...,m). In particular (see

for example, and ), when determining the a posteriori conditional probabilities Р(Н 1Е^) and Р(Н 1 Е2) for each of the two compatible evidences Е1 and Е2, formulas of the form are used:

P(H 1) PE\H1) P(Hj) P(E2Hj) P(H J E1) = --1-and P(H J E 2) =--1-. (5)

I P(Hk) PE\Hk) I P(Hk) P(E2 Hk)

k = 1 k = 1 Note that this is another case where the Bayes formula is not applicable. Moreover, in this case, two shortcomings must be eliminated:

The illustrated normalization of the probabilities of combined events is incorrect, due to belonging to different complete groups of the events under consideration;

The symbolic records of the combined events HkEx and HkE2 do not reflect the fact that the considered evidence E x and E 2 are compatible.

To eliminate the last drawback, a more detailed record of combined events can be used, taking into account the fact that compatible evidence E1 and E2 in some cases may be incompatible, and in others joint:

HkE1 = HkE1 E2 and HkE2 = HkE 1E2+HkE1 E2, where E1 and E 2 are evidence opposite to E1 and E 2.

It is obvious that in such cases the product of events Hk E1E2 is taken into account twice. In addition, it can be taken into account again separately, but this does not happen. The fact is that in the situation under consideration, the assessed situation is influenced by three probable incompatible combined events: HkE1E2, HkE 1E2 and

Hk E1E2. At the same time, for the decision maker, it is of interest to assess the degree of possibility only

two incompatible combined events: HkE1 E2 and HkE 1E2, which corresponds to considering only g

single evidence. ¡C

Thus, when constructing a modification of the Bayes formula for determining a posteriori conditional values,

The probability of hypotheses with single compatible evidence must be based on the following. Person accepting ^

decision, we are interested in exactly what elementary event, represented by one or another piece of evidence from

number considered actually happened in specific conditions. If another elementary event occurs in K

in the form of a single certificate, a reconsideration of the decision is required, due to the results of a comparative assessment of n

a posteriori conditional probabilities of hypotheses with the indispensable consideration of other conditions affecting the real general

setting. 3

Let us introduce the following notation: HkE- for one (and only one) incompatible combined co- ^

being, which consists in the fact that out of m > 1 considered elementary events Ei (i = 1,...,m) together with the hypothesis “

Hk, one elementary event Ex occurred and no other elementary events occurred. se"

In the simplest case, two single incompatible evidence is considered. If confirmed

one of them is expected, the conditional probability of evidence in a general form is expressed by the formula l

P(Hk E-) = P(Ei\Hk) -P(EjE^Hk) = P(Ei\Hk) -P(M^Hk)P(M^Hk) , i = 1, -2 (6) g

The validity of the formula can be clearly seen (Fig. 1).

Rice. 1. Geometric interpretation of the calculation of P(Hk E-) for / = 1,...,2 With conditionally independent evidence

P(K1K2\Hk) = p(E\Hk)P(E2\Hk),

therefore, taking into account (6)

P(Hk E-) = PE Hk) - P(E1 Hk) P(E21Hk) , = 1,.,2. (7)

Similarly, the probability P(HkE-) of one of the three (/ = 1,...,3) incompatible events HkE^ is expressed by the formula

For example, for i = 1:

p(HkEl) = P(Ei\Hk)-[ S P(Ei\Hk)P(Ej\Hk) ] + P(EiE2E3Hk)

p(HkE-) = P(E7|Hk)- P(E]E^Hk)- P(E7EjHk) + P(E]E2E3\Hk)

The validity of this formula is clearly confirmed by the geometric interpretation presented in Fig.

Rice. 2. Geometric interpretation of the calculation of P(Hk E-) for / = 1,...,3

Using the method of mathematical induction, one can prove the general formula for the probability P(Hk E-) for any number of evidence e, 0=1,...,m):

P(HkE-) = P(E, Hk) - m PE\Hk) P(E]\Hk) + 1 P(E\Hk) P(E]\Hk) P(E^Hk) + ■■■ + (-1)

] = 1(] * 0 ],1 * 1

Using the probability multiplication theorem, we write the conditional probability Р(НкЕ~-) in two forms:

^ from which it follows that

P(Hk E -) = P(Hk) P(E-|Hk) = P(E-) P(Hk

E-)= P(HkE-) "" P(E-)

Using the total probability formula P(Ei) = S P(H£) P(Ei Hk) it turns out that

E-) \u003d P (HkET)

2 P(HkE-) k \u003d 1

Substituting the expressions for Р(НкЕ-) in the form of the right part of (8) into the obtained formula, we obtain the final form of the formula for determining the a posteriori conditional probabilities of the hypotheses H^ (k = 1, ..., n) for one of several considered incompatible single evidence : (E^\Hk)

P(Hk)[P(E,\Hk) - 2 P(E,\Hk) P(Ep k) +...+ (-1)m-1 P(P P(Erk)] P(H, E~) =-] = 1(] * ■----(9)

k 1 p t t t

2 P(Hk) 2 [P(E,\Hk) - 2 P(EgHk) P(E^Hk) + ...+ (-1)m-1 P(P P (Ep k)]

k=1 , = 1 ) = 1() *,) ■! =1

Comparative estimates. Quite simple, but illustrative examples are considered, limited to the analysis of calculated a posteriori conditional probabilities of one of two hypotheses with two single evidences. 1. Probabilities of hypotheses under incompatible single evidence. Let's compare the results obtained using the Bayes formulas (2) and (3), using the example of two evidences L. = L and L. = L with the initial data:

P(H1 = 0.7; P(H2) = 0.3; P(L| H^ = 0.1; P(L\n 1) = 0.9; P(L\H2) = 0.6 P(A\H2) = 0.4 In the considered examples with the hypothesis H1, the traditional formulas (2) and (3) lead to the following results:

P(N.) P(A\No 0 07

P (N, L) \u003d - 11 \u003d - \u003d 0.28,

2 P(Hk) P(A\Hk)k = 1

R(N L R(A\N 1) 0 63

P (N, L) \u003d - 11 \u003d - \u003d 0.84,

2 P(Hk) P(A\Hk) k = 1

forming divides P (H 1 L) \u003d P (H ^ P (L \ Hp \u003d 0.07; P (H ^ A) \u003d P (H 1) P (n | H ^ \u003d 0.63. 1 of the proposed formulas with respect to:

R<Н)Р(АНА-Р(А|Н1) _ 0,07

and with the proposed formulas (4) that do not have normalizing divisors: “and

Thus, in the case of applying the proposed formulas, the ratio of normalized probabilities is equal to the ratio of normalized probabilities: K

rm f P(H 1) P(A\H 1) A11 |

When using known formulas with the same ratio -;-=-= 0.11 normalized verons

P(H 1) P(A\H 1) «§

ratios indicated in the numerators, the ratio of the resulting normalized probabilities: 2

P(H 1) P(A\H 1) P(A\H 1) 0.63

P (H1 L) \u003d 0.28 P (H 1 L) \u003d 0.84

That is, the principle of conservation of probability ratios is not observed, and incorrect results are obtained. In this case, £

in the case of applying known formulas, the value of the relative deviation of the ratio (11) of the a posteriori conditional and conditional probabilities of the hypotheses from the correct results (10) turns out to be very significant, since it is

°, 33 - °, P x 100 \u003d 242%.. I

2. Probabilities of hypotheses under compatible single evidence. Let us compare the results obtained using the Bayes formulas (5) and the constructed correct modification (9), using the following initial data:

P(H1 = 0.7; P(H2) = 0.3; P(E1H1) = 0.4; P(E2H1) = 0.8; P(E1\H2) = 0.7; P(E^ H2) = 0.2.113

In the examples under consideration with the hypothesis H 2 in the case of using traditional formulas (5):

P(H 2) P(E1 H 2) Q, 21

P(H 2 E1) =-2-!-2- = - = Q,429,

p(Hk) p(El Hk) k = 1

P(H 2) P(E 2 H 2) Q,Q6

P(H 2 E 2) \u003d -2-- \u003d - \u003d 0.097.

I P(Hk) P(E 2 Hk) k = 1

In the case of applying the proposed formula (9), taking into account (7), P(H

P(H2) 0.168

E.) ----- 0.291,

Z P(Hk) Z "

P(H2) 0.018

E0) ----- 0.031.

Z P(Hk) Z k - 1 i - 1

When using the proposed correct formulas, due to the same denominators, the ratio P(H2) -

Normalized probabilities, indicated in numerators, is equal to the ratio

P(H2)

normalized probabilities:

That is, the principle of conservation of probability ratios is observed.

However, in the case of applying known formulas with the ratio of the normalized probabilities indicated in the numerators

P (H 2) P (E1 \ H 2) _ 0.21 _3 5 P (H 2) P (E 2 H 2) 0.06,

ratio of normalized probabilities:

P (H 2 \u003d 0.429 \u003d 4.423. (13)

P(H 2 \e2) 0.097

That is, the principle of conservation of probability ratios, as before, is not respected. In this case, in the case of applying the known formulas, the value of the relative deviation of the ratio (13) of the a posteriori conditional probabilities of the hypotheses from the correct results (12) also turns out to be very significant:

9.387 4.423 x 100 = 52.9%.

Conclusion. An analysis of the construction of specific formula relations that implement the Bayes formula and its modifications, proposed for solving practical problems, allows us to state the following. The full group of comparable 2 possible combined events can be selected subjectively by the decision maker. This choice is based on the considered objective initial data characteristic of a typical situation (specific types and number of elementary events - estimated hypotheses and evidence). Of practical interest is the subjective choice of other options of the full group compared in terms of the degree of possibility.

combined events - thus, a significant variety of formula ratios is provided when constructing non-traditional variants of modifications of the Bayes formula. This, in turn, can be the basis for improving the mathematical support of software implementation, as well as expanding the scope of new formula relations for solving applied problems.

Bibliographic list

1. Gnedenko, B. V. An elementary introduction to the theory of probability / B. V. Gnedenko, A. Ya. Khinchin. - 114 New York: Dover Publications, 1962. - 144 rubles.

2. Venttsel, E. S. Probability theory / E. S. Venttsel. - 10th ed., erased. - Moscow: Higher School, 2006. - 575 p.

3. Andronov. A. M., Probability theory and mathematical statistics / A. M. Andronov, E. A. Kopytov, L. Ya. Gringlaz. - St. Petersburg: Peter, 2004. - 481 p.

4. Zmitrovich, A. I. Intelligent information systems / A. I. Zmitrovich. - Minsk: TetraSistems, 1997. - 496 p.

5. Chernorutsky, I. G. Decision-making methods / I. G. Chernorutsky. - St. Petersburg: BHV-Petersburg, 2005. - 416 p.

6 Naylor, C.-M. Build Your Own Expert System / C.-M. Naylor. - Chichester: John Wiley & Sons, 1987. - 289 p.

7. Romanov, V.P. Intelligent information systems in the economy / V.P. Romanov. - 2nd ed., erased.

Moscow: Exam, 2007. - 496 p.

8. Economic efficiency and competitiveness / D. Yu. Muromtsev [and others]. - Tambov: Tambov Publishing House. state tech. un-ta, 2007.- 96 p.

9. Dolgov, A. I. Correct modifications of the Bayes formula for parallel programming / A. I. Dolgov // Supercomputer technologies: materials of the 3rd All-Russian. scientific-technical conf. - Rostov-on-Don. - 2014.- Vol. 1 - S. 122-126.

10. A. I. Dolgov, On the correctness of modifications of the Bayes formula / A. I. Dolgov, Vestnik Don. state tech. university

2014. - V. 14, No. 3 (78). - S. 13-20.

1. Gnedenko, B.V., Khinchin, A.Ya. An elementary introduction to the theory of probability. New York: Dover Publications, 1962, 144 p.

2 Ventsel, E.S. Teoriya veroyatnostey. 10th ed., reimpr. Moscow: Vysshaya shkola, 2006, 575 p. (in Russian).

3. Andronov, A.M., Kopytov, E.A., Gringlaz, L.Y. Teoriya veroyatnostey i matematicheskaya statistika. St. Petersburg: Piter, 2004, 481 p. (in Russian).

4. Zmitrovich, A.1. Intellektual "nye informatsionnye sistemy. Minsk: TetraSistems, 1997, 496 p. (in Russian).

5. Chernorutskiy, I.G. Methodology prinyatiya resheniy. St. Petersburg: BKhV-Peterburg, 2005, 416 p. (in Russian).

6 Naylor, C.-M. Build Your Own Expert System. Chichester: John Wiley & Sons, 1987, 289 p.

7. Romanov, V.P. Intellektual "nye informatsionnye sistemy v ekonomike. 2nd ed., reimpr. Moscow: Ekzamen, 2007, 496 p. (in Russian).

8. Muromtsev, D.Y., et al. Ekonomicheskaya effektivnost" i konkurentosposobnost". Tambov: Izd-vo Tamb. gos. technology un-ta, 2007, 96 p. (in Russian). IB

9. Dolgov, A1. Korrektnye modifikatsii formuly Bayesa dlya parallel "nogo programmirovaniya. Superkomp" yuternye tekhnologii: mat-ly 3-y vseros. scientific-techn. conf. Rostov-on-Don, 2014, vol. 1, pp. 122-126 (in Russian). ^

10. Dolgov, A1. O korrektnosti modifikatsiy formuly Bayesa. ^ Vestnik of DSTU, 2014, vol. 14, no. 3 (78), pp. 13-20 (in Russian). *

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

This formula is called total probability formula .

Consider again the complete group of incompatible events , whose probabilities of occurrence are . Event BUT can only occur together with any of the events that we will call hypotheses . Then according to the total probability formula

If the event BUT happened, it can change the probabilities of the hypotheses .

According to the probability multiplication theorem

Similarly, for other 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 enterprises. 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 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.

Decision. Denote by AT the event consisting in the fact that the premium product will be purchased, let us denote the events consisting in the purchase of products belonging to the first, second and third enterprises, respectively.

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

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

Example. One of the three shooters is called to the line of fire 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.

Decision. 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 was called to the line of fire.

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

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

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

Example. On three automatic machines, parts of the same type are processed, which arrive after processing on a common conveyor. The first machine gives 2% rejects, 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 are the proportions of the parts of each machine among the defective parts on the conveyor?

Decision. Let's take one part at random from the assembly line and consider event A - the part is defective. It is associated with hypotheses as to where this part was machined: - a randomly selected part was machined on the th machine,.

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

The dependencies between machine performances 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, in the mass of parts coming off the assembly line, the defect is 4%.

b) Let it be known that a part taken at random is defective. Using the 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 sections of probability theory

The goal is to gain practical skills in solving problems on

sections of probability theory

Preparation for the practical task

To get acquainted with the theoretical material on this topic, to study the content of the theoretical, as well as the relevant sections in the literature

Task execution order

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

Initial data options

Table 1

	task number

The composition of the report for task 1

5 solved problems according to the variant number.

Tasks for independent solution

1.. Are the following groups of events cases: a) experience - tossing a coin; events: A1- the appearance of the coat of arms; A2- the appearance of a number; b) experience - tossing two coins; events: IN 1- the appearance of two coats of arms; IN 2 - the appearance of two digits; IN 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 - the appearance of at least five points; d) experience - a shot at a target; events: D1- hit; D2- miss; e) experience - two shots at the target; events: E0- not a single hit; E1- one hit; E2- two hits; f) experience - drawing two cards from the deck; events: F1- the appearance of two red cards; F2- the appearance of two black cards?

2. Urn A contains 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 whites and B black balls. One ball is taken out of the urn and set aside. This ball is white. After that, another ball is taken from the urn. Find the probability that this ball is also white.

4. In urn A whites and B black balls. One ball was taken out of the urn and put aside without looking. 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 whites and B black balls, take out one by one all the balls except one. Find the probability that the last ball left in the urn is white.

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

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

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

9. In a party consisting of X products, there is I defective. From the batch is selected for control I products. Find probability R which of them exactly J products will be defective.

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

11. A die is thrown 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: BUT- the sum of the dropped points is equal to 8; AT- the product of the dropped points is equal to 8; WITH- the sum of the dropped points is greater than their product.

13. Two coins are tossed. Which of the following events is more likely: BUT - coins will lie on the same sides; AT - Do the coins lie on different sides?

14. In urn A whites and B black balls (A > 2; B > 2). Two balls are taken out of the urn at the same time. Which event is more likely: BUT- balls of the same color; AT - balls of different colors?

15. Three players are playing cards. Each of them is dealt 10 cards and two cards are left in the draw. One of the players sees that he has 6 cards of a diamond suit and 4 cards of a non-diamond suit. He discards two of those four cards and takes the draw. Find the probability that he buys two diamonds.

16. From an urn containing P numbered balls, randomly take out one by one all the balls in it. 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 taking out each ball is put back in and mixed with others, and its number is written down. Find the probability that the natural sequence of numbers will be written down: 1, 2,..., n.

18. A full deck of cards (52 sheets) is divided at random into two equal packs of 26 sheets. Find the probabilities of the following events: BUT - in each of the packs there will be two aces; AT- in one of the packs there will be no aces, and in the other - all four; S-in one of the packs will have one ace, and the other pack 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 participants of the competition

extra class. Find the probabilities of the following events: BUT - all extra-class teams will fall into the same group; AT- two extra-class teams will get 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 is even.

21. 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 is greater than the number on the first.

22. The same question as in problem 21, but the first card after being drawn 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. One by one, all the balls in it are taken out of the urn and their colors are written down. Find the probability that white appears before black in this list.

24. There are two urns: in the first one A whites 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. Under the conditions of Problem 24, find the probability that the drawn balls will be of different colors.

26. There are seven nests in the drum of a revolver, five of them are loaded with cartridges, and two are left empty. The drum is set in rotation, as a result of which one of the sockets is randomly placed against the barrel. After that, the trigger is pressed; if the cell was empty, the shot does not occur. Find probability R the fact that, having repeated such an experiment twice in a row, we will not shoot both times.

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

28. There is an A in the urn; balls labeled 1, 2, ..., to From the urn I once one ball is drawn (I<к), the number of the ball is written down and the ball is put back into the urn. Find probability R that all recorded numbers will be different.

29. The word "book" is composed of five letters of the split alphabet. A child who could not read scattered these letters and then put them together in random order. Find probability R the fact that he again got the word "book".

30. The word "pineapple" is made up of the letters of the split alphabet. A child who could not read scattered these letters and then put them together in random order. Find probability R the fact that he again has the word "pineapple

31. From a full deck of cards (52 sheets, 4 suits), several cards are taken out at once. 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 faces BUT and AT will be nearby.

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

34. Numbers from 1 to N. Of these N two barrels are randomly selected. Find the probability that numbers less than k are written on both barrels (2

35. Numbers from 1 to N. Of these N two barrels are randomly selected. Find the probability that one of the barrels has a number greater than k , and on the other - less than k . (2

36. Battery out M guns firing 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 the fact that targets with numbers 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 selects its target randomly and independently of the others. Find the probability that all to guns will fire at the same target.

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

39. Four balls are randomly scattered over four holes; each ball hits one or another hole with the same probability and independently of the others (there are no obstacles to getting several balls 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 with him in the same bus. There are 5 buses from the hostel to the institute from 7 to 8. Those who do not have time for these buses are 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. There are 3 analysts, 10 programmers and 20 engineers in the information technology department of the bank. 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 security service of the bank must daily place 10 guards in 10 posts. In how many ways can this be done?

43. The new president of the bank must appoint 2 new vice presidents from among the 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 digits and 3 letters )?

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 be formed 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 of it and set aside. Then 1 more apple is taken out of the basket. What is the probability that this apple is green?

54. In a batch of 1,000 items, 4 are defective. For control, a batch of 100 products is selected. What is the probability of LLP that the control lot will not be defective?

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

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

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

59. In the 80s, the sportloto 5 out of 36 game was popular in the USSR. The player noted on the card 5 numbers from 1 to 36 and received prizes of various denominations if he guessed a different number of numbers announced by the draw commission. Find the probability that the player did not guess all 5 numbers.

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

61. In the 80s, the game "sportloto 6 out of 49" was popular in the USSR. The player noted on the card 6 numbers from 1 to 49 and received prizes of various denominations if he guessed a different number of numbers announced by the draw commission. Find the probability that the player did not guess any number.

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

63. In a batch of 1,000 items, 4 are defective. For control, a batch of 100 products is selected. What is the probability of LLP that only 1 defective will be in the control lot?

64. How many different words can be formed by rearranging the letters in the word "book"?

65. How many different words can be formed 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 exit on the same floor?

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

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

69. On the 200-kilometer section of the gas pipeline, there is a gas leak between compressor stations A and B, which is equally possible at any point of the pipeline. What is the probability that the leak occurs within 20 km of A

70. On the 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 is closer to A than to B?

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

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

73. Petya and Masha agreed to meet at the monument to Pushkin from 12 to 13 hours, 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, take out all the apples in turn. What is the probability that the 2nd apple is red?

76. From a basket containing 3 red and 7 green apples, take out all the apples in turn. What is the probability that the last apple is green?

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

78. Students consider that out of 50 tickets 10 are “good”. Petya and Masha take turns pulling 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 of the program out of 25. 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 of the program out of 25. The professor asks 3 questions. What is the probability that Masha will not answer any of the questions?

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

82. The statistics of bank loan requests is as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of loan default is 0.01, 0.05 and 0.2, respectively. What proportion of loans are non-refundable?

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 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. White (b) and C are in urn A black (h) balls. Two balls are taken out of the urn (simultaneously or sequentially). Find the probability that both balls are white.

85. In urn A whites and B

86. In urn A whites and B

87. In urn A whites and B black balls. One ball is taken out of the urn, its color is marked and the ball is returned to the urn. After that, another ball is taken from the urn. Find the probability that these balls will be of different colors.

88. There is a box with nine new tennis balls. Three balls are taken for the game; after the game they are put back. When choosing balls, they do not distinguish between played and unplayed balls. What is the probability that after three games there will be no unplayed balls in the box?

89. Leaving the apartment, N each guest will put on their own galoshes;

90. Leaving the apartment, N guests with the same shoe size put on 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 put on galoshes belonging to one pair (maybe not their own).

91. Under 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 that come across.

92. Shooting is underway at the aircraft, the vulnerable parts of which are two engines and the cockpit. In order to hit (disable) the aircraft, it is enough to hit both engines together or the cockpit. Under given firing conditions, the probability of hitting the first engine is p1 second engine p2, cockpit p3. Parts of the aircraft 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 their 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, in the target of which there will be more holes. Find probability Rx what the first shooter wins.

94. behind a space object, the object is detected with a 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 alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the word "end" will be obtained.

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

97. Incendiary projectiles are being fired at the aircraft. The fuel on the aircraft is concentrated in four tanks located in the fuselage one after the other. Tank sizes are the same. In order to ignite the aircraft, 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 are of the same suit.

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 being taken out. Find the probability that all four cards are of the same suit.

100. When the ignition is turned on, the engine starts with a 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 in time t in normal mode is 0.1; in the abnormal - 0.7. Find Total Probability R failure of the device.

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

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

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

105. The store receives goods from 3 suppliers: 55% from the 1st, 20 from the 2nd and 25% from the 3rd. The share of marriage 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 alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability of getting the word "book".

107. The store receives goods from 3 suppliers: 55% from the 1st, 20 from the 2nd and 25% from the 3rd. The share of marriage 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 over four cells located one after the other in a straight line. Each ball with the same probability 1/4 hits each cell. Find the probability that 2 balls fall into the same cell

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

110. Incendiary projectiles are fired at the aircraft. The fuel on the aircraft is concentrated in four tanks located in the fuselage one after the other. Tank sizes are the same. In order to ignite the aircraft, 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. Incendiary projectiles are fired at the aircraft. The fuel on the aircraft is concentrated in four tanks located in the fuselage one after the other. Tank sizes are the same. In order to ignite the aircraft, it is enough to hit two shells in neighboring tanks. It is known that two shells hit the tank area. Find the probability that the plane will catch fire

112. In urn A whites and B black balls. One ball is taken out of the urn, its color is marked and the ball is returned to the urn. After that, another ball is taken from the urn. Find the probability that both balls drawn are white.

113. In urn A whites and B black balls. Two balls are taken out of the urn at once. Find the probability that these balls will be of different colors.

114. Two balls are scattered randomly and independently of each other over four cells located one after the other in a straight line. Each ball with the same probability 1/4 hits 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 of the program out of 25. The professor asks 3 questions. What is the probability that Masha will answer 2 questions?

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

117. The statistics of bank loan requests is as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of loan default is 0.01, 0.05 and 0.2, respectively. What proportion of loans are non-refundable?

118. 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the word "end" will be obtained.

119 The statistics of bank loan requests is as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of loan default is 0.01, 0.05 and 0.2, respectively. What proportion of loans are non-refundable?

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 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.