Bayes distribution. Formula of full probability, Bayes formula

Brief theory

If an event occurs only under the condition that one of the events of forming a complete group of incomplete events appear, it is equal to the amount of the probability of each of the events to the appropriate conditional probability of the wallet.

At the same time, events are called hypothesis, and probabilities - a priori. This formula is called a full probability formula.

The Bayes formula is used in solving practical tasks when an event appearing in conjunction with any of the events generating a full event group has occurred and a quantitative revaluation of probabilities of hypotheses is required. A priori (before experience) is known to be known. It is required to calculate the posteriori (after experience) of the probability, i.e. Essentially you need to find conditional probabilities. Bayes formula looks like this:

The next page addresses the task of.

An example of solving the problem

Task condition 1.

At the factory, machines 1,2 and 3 produce 20%, 35% and 45% of all parts. In their products, marriage is respectively 6%, 4%, 2%. What is the likelihood that the randomly chosen product turned out to be defective? What is the probability that it was produced: a) machine 1; b) machine 2; c) Machine 3?

Solution of problem 1.

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

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

The product is made on the machine 1;

The product is made on the machine 2;

The product is made on the machine 3;

We write conditional likelihoods:

Formula full probability

If an event can occur only when performing one of the events that form a commission of incomprehensible events, the probability of an event is calculated by the formula

According to the full probability formula, we find the likelihood of an event:

Formula Bayes.

The Bayes formula allows you to "rearrange the cause and consequence": according to the known fact, the event calculates the likelihood that it was caused by this reason.

The likelihood that the defective product is made on the machine 1:

The likelihood that the defective product is made on the machine 2:

The probability that the defective product is made on the machine 3:

Task condition 2.

The group consists of 1 excellent student, 5 students and 14 students who have time to have time. The excellent student responds to 5 and 4 with an equal probability, the good one responds to 5, 4 and 3 with an equal probability, and a mediocre successive student responds to 4.3 and 2 with equal probability. Randomly chosen student answered 4. What is the likelihood that a mediocre student has been called?

Task Solution 2.

Hypothesis and conditional probabilities

The following hypotheses are possible:

Answered an excellent student;

Answered a good one;

- made a mediocre student;

Let the event receive 4.

Answer:

The price strongly affects the urgency of the solution (from day to several hours). Online assistance on the exam / standings is carried out by appointment.

The application can be left directly in the chat, having previously throwing the condition of tasks and informing the decision you need. Answer time - a few minutes.

Let them probably be known and the corresponding conditional probabilities. Then the likelihood of events is equal to:

This formula was named formulas full probability. In textbooks, it is formulated by theorem, the proof of which is elementary: according to algebra of events, (event occurred and or Event happened and After it has ever orevent happened and After it has ever or …. or Event happened and after it has come an event). Since hypothesis inconsistent, and an event - depending on the theorem of the addition of probability of incomplete events (first step) and the intention theorem of probability dependent events (second step):

Probably many anticipate the content of the first example \u003d)

Wherever I spit - everywhere urn:

Task 1.

There are three identical urns. In the first urn there are 4 whites and 7 black balls, in the second - only white and in the third - only black balls. Maudoku is chosen one urn and a ball is extracted from her at random. What is the probability that this ball is black?

Decision: Consider an event - a black ball will be extracted from the atmosphere of the chosen urn. This event may occur or do not occur as a result of the implementation of one of the following hypotheses:
- 1st urn will be chosen;
- 2nd urn will be chosen;
- The 3rd urn will be chosen.

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

Please note that listed hypotheses form full group of events, that is, by condition, a black ball may appear only from these urns, and for example, do not fly from the billiard table. We will draw a simple intermediate check:
, OK, go further:

In the first urn 4 white + 7 black \u003d 11 balls, by classical definition:
- the probability of the extraction of a black ball given thatthat the 1st urn will be chosen.

In the second urn, only white balls, so in case of its choice The appearance of a black ball becomes impossible: .

And finally, in the third urn one black balls, and therefore corresponding conditional probability Black ball extract will be (Event is reliably).

- The likelihood that a black ball will be extracted from the random of the chosen urn.

Answer:

The disassembled example again suggests how important it is to delve into the condition. Take the same tasks with urns and balls - when they are external similarity, solutions can be completely different: somewhere you need to apply only classical probability definition, somewhere events independentsomewhere dependent, and somewhere we are talking about hypotheses. At the same time, there is no clear formal criterion to select the solution of the solution - it almost always needs to think over it. How to improve your qualifications? We decide, we decide and solve again!

Task 2.

In the dash there are 5 different rifle battle accuracy. The probabilities of entering the target for this arrow are respectively 0.5; 0.55; 0.7; 0.75 and 0.4. What is the likelihood of hitting the target, if the shooter makes one shot of a randomly selected rifle?

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

In most the thematic tasks, the hypothesis, of course, are not evenly equal:

Task 3.

In the pyramid 5 rifles, three of which are equipped with an optical sight. The likelihood that the shooter will hit the target when shot of a rifle with an optical sight, equal to 0.95; For a rifle without an optical sight, this probability is 0.7. Find the chance that the target will be amazed if the shooter produces one shot of a row taken rifle.

Decision: In this task, the number of rifles is exactly the same as in the previous one, but there are only two hypothesis:
- the shooter will choose a rifle with an optical sight;
- The shooter will choose a rifle without an optical sight.
By classical definition of probability: .
Control:

Consider the event: - the shooter will hit the target from at random rifles taken.
By condition: .

According to the full probability formula:

Answer: 0,85

In practice, quite a shortened way of registration of the task, which you also know too:

Decision: By classical definition: - probabilities of choosing a rifle with optical and without optical sight, respectively.

By condition, - The probabilities of entering the target from the corresponding types of rifles.

According to the full probability formula:
- The likelihood that the shooter will hit the target from the ramp of the selected rifle.

Answer: 0,85

Next task for self solutions:

Task 4.

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

Just in case I remind you - to get the probability interests percentages need to be divided into 100. Be very careful! According to my observations, the conditions for the tasks on the full probability formula are often trying to take up; And I specifically picked up such an example. I will tell you a secret - I have not confused almost \u003d)

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

Tasks for Bayes Formulas

Material is closely related to the content of the previous paragraph. Let the event occurred as a result of the implementation of one of the hypotheses . How to determine the likelihood that there was a place of such a hypothesis?

Given thatthat event already happenedprobability hypothesis overestimated According to the formulas that got the name of the English priest Thomas Bayes:

- the likelihood that there was a hypothesis;
- the likelihood that there was a hypothesis;
…
- The likelihood that there was a hypothesis.

At first glance, it seems to be complete nonsense - why recalculate the probability of hypotheses, if they are so famous? But in fact, there is a difference:

- this is a priori (Rated before Tests) probability.

- this is aPAPERY (Rated after Tests) The likelihood of the same hypotheses, recalculated due to "with newly discovered circumstances" - taking into account the fact that the event reliable happened.

Consider this distinction on a specific example:

Task 5.

The warehouse received 2 batch products: 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%. Raduchi taken from the warehouse the product turned out to be standard. Find the likelihood that it: a) from the first batch, b) from the second batch.

First part solutions It consists in using full probability formula. In other words, the calculations are held under the assumption that the test not yet produced Event "The product turned out to be standard" until it came.

Consider two hypotheses:
- the right-handed product will be from the 1st Party;
- The border taken the product will be from the 2nd Party.

Total: 4000 + 6000 \u003d 10,000 products in stock. By classical definition:
.

Control:

Consider the dependent event: - Magnifying the product taken from the warehouse will be Standard.

In the first batch 100% - 20% \u003d 80% of standard products, so: given thatthat it belongs to the 1st Party.

Similarly, in the second batch 100% - 10% \u003d 90% of standard products and - the probability that the product taken in the warehouse will be standard given thatthat it belongs to the 2nd Party.

According to the full probability formula:
- The likelihood that the product taken in the warehouse will be standard.

Part two. Let the muddy taken from the warehouse the product turned out to be standard. This phrase is directly spelled out in the condition, and it states the fact that the event occurred.

According to Bayes formulas:

a) - the likelihood that the chosen standard product belongs to the 1st batch;

b) - the likelihood that the chosen standard product belongs to the 2nd batch.

After revaluation hypotheses, of course, are still formed full group:
(Check ;-))

Answer:

Ivan Vasilyevich will help us to understand the meaning of revaluation of the hypotheses, who again changed his profession and became the director of the plant. He knows that today the 1st workshop shipped 4000 to the warehouse, and the 2nd shop is 6000 products, and comes to make sure that. Suppose all the products of the same type are in one container. Naturally, Ivan Vasilyevich preliminarily calculated that the product that he now removes to check is probably released by the 1st workshop and with a probability - the second. But after the selected product turns out to be standard, it exclaims: "What a cool bolt! - He was most likely released the 2nd shop. " Thus, the probability of the second hypothesis is revalued for the better, and the probability of the first hypothesis is underestimated :. And this revaluation is not displaced - because the 2nd shop produced not only more products, but also works 2 times better!

Do you say pure subjectivism? Partly yes, moreover, Bayes himself interpreted aPAPERY probability as trust level. However, not everything is so simple - there is 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 is preliminary (a priori) and middleestimates. But, expressing philosophically - everything flows, everything changes, and probabilities including. It is possible that at the time of research A more successful 2nd workshop raised the percentage of standard products (and / or the 1st workshop reduced)And if you check more or all 10 thousand products in stock, then revalued values \u200b\u200bwill be much closer to the truth.

By the way, if Ivan Vasilyevich removes a non-standard detail, then on the contrary - it will be more "suspecting" the 1st workshop and less - the second. I propose to make sure that:

Task 6.

The warehouse received 2 batch products: 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%. Raduch taken from the warehouse product turned out notstandard. Find the likelihood that it: a) from the first batch, b) from the second batch.

The condition is distinguished by two letters that I highlighted bold font. The task can be solved with a "clean sheet", or take advantage of the results of previous calculations. In the sample, I had a complete solution, but in order not to have a formal overlay with task number 5, an event "The product taken from the warehouse will be non-standard" marked through.

The Bayesian chart of transaction of probabilities occurs everywhere, and it is actively exploited by various kinds of scammers. Let us consider the negative JSC into three letters that attracts the deposits of the population, allegedly invests somewhere, it is properly paid dividends, etc. What's happening? It takes day after day, month after month and more and more new facts, reported by advertising and "sorrolhed radio", only increase the level of confidence in the financial pyramid (A posteriori Bayesov Revaluation in connection with the events that occurred!). That is, in the eyes of depositors there is a constant increase in the likelihood that "This is a serious office"; In this case, the probability of the opposite hypothesis ("These are the next chisels"), of course, decreases and decreases. Further, I think, understandable. It is noteworthy that earned reputation gives the organizers the time to successfully hide from Ivan Vasilyevich, who remained not only without a bolt party, but also without pants.

To no less interesting examples, we will return a little later, as well as the queue, perhaps the most common case with three hypotheses:

Task 7.

Electrollamps are made on three factories. The 1st plant produces 30% of the total number of lamps, 2nd - 55%, and the 3rd is the rest. Products of the 1st plant contains 1% defective lamps, 2nd - 1.5%, 3rd - 2%. The store comes the products of all three plants. The purchased lamp was marriage. What is the probability that it is produced by the 2nd plant?

Note that in the tasks for Bayes formulas in the condition before Some appears what happenedevent in this case - buying a lamp.

Events added, and decision It is more convenient to arrange in the "fast" style.

The algorithm is exactly the same: in the first step we find the likelihood that the purchased lamp at all will it turn out defective.

Using the source data, we translate interest in the probability:
- The likelihood that the lamp is produced by the 1st, 2nd and 3rd plants, respectively.
Control:

Similarly: - probabilities of manufacturing a defective lamp for relevant factories.

According to the full probability formula:

- The likelihood that the purchased lamp will be with a marriage.

Step second. Let the purchased lamp be defective (the event happened)

By Bayes formula:
- the likelihood that the purchased defective lamp is made by the second plant

Answer:

Why is the initial probability of the 2nd hypothesis after revaluation increased? After all, the second plant produces medium in the quality of the lamp (first - better, the third is worse). So why has increased apottery The probability that defective lamp is from the 2nd plant? This is not explained by the "reputation", but the size. Since Plant No. 2 has released the largest amount of lamps, then on it (at least subjectively) and foam: "Most likely, this defective lamp is from there".

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

Control: What was required to check.

By the way, about understated and overestimated estimates:

Task 8.

In the student group, 3 people have a high level of preparation, 19 people - medium and 3 - low. The probabilities of successful passing exam for students' data are respectively equal: 0.95; 0.7 and 0.4. It is known that some student passed the exam. What is the probability that:

a) it was prepared very well;
b) the average was prepared;
B) was well prepared.

Perform calculations and analyze the results of the revaluation of the hypotheses.

The task is close to reality and is particularly believable for a group of architect students, where the teacher practically does not know the abilities of a student. In this case, the result may cause a pretty of unexpected consequences. (especially for exams in the 1st semester). If a poorly prepared student was lucky to the ticket, the teacher is likely to consider it well or even a strong student who will bring good dividends in the future (Naturally, you need to "raise the bar" and maintain your image). If a student of 7 days and 7 nights taught, shut, repeated, but he was simply not lucky, then further events can develop in the very bad veneer - with numerous renovation and balancing on the verge of departure.

What to say, reputation is the most important capital, it's not by chance that many corporations wear the names-names of their founding fathers, who led a job 100-200 years ago and became famous for their impeccable reputation.

Yes, the Bayesian approach to a certain extent is subjective, but ... life is so arranged!

Fulfill the material by the final industrial example in which I will tell you about the still not met technical understanding of the decision:

Task 9.

Three factory workshops produce similar items that enter the assembly into a common container. It is known that the first workshop produces 2 times more details than the second workshop, and 4 times more than the third workshop. In the first workshop, the marriage is 12%, in the second - 8%, in the third - 4%. To control the container is taken one detail. What is the probability that it will be defective? What is the probability that the extracted defective item released the 3rd shop?

Taki Ivan Vasilyevich again on horseback \u003d) there must be a happy finish from the movie)

Decision: Unlike tasks No. 5-8, there is explicitly given a question that is permitted using the full probability formula. But on the other hand, the condition is a bit "encrypted", and to solve this rebus will help us will help the school skill to form the simplest equations. For "X" convenient to accept the smallest meaning:

Let - the proportion of details produced by the third workshop.

By condition, the first shop produces 4 times the third workshop, therefore the share of the 1st workshop is.

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

Let us and solve the equation:

Thus,: - the likelihood that the part extracted from the container is released 1st, 2nd and 3rd shops, respectively.

Control: . In addition, it will not be superfluous to look at the phrase "It is known that the first shop 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 \u200b\u200bdo correspond to this condition.

For "X" initially it was possible to take a share of the 1st or the share of the 2nd workshop - probabilities will come out by the same. But, one way or another, the most difficult plot was passed, and the decision is included in the rolled rut:

From the condition we find:
- probabilities of manufacturing a defective part for the respective workshops.

According to the full probability formula:
- The likelihood that the whole detail extracted from the container will be non-standard.

The question is the second: What is the likelihood that the extracted defective part released the 3rd shop? This question suggests that the item is already extracted, and it turned out to be defective. Overestimate the hypothesis by the Bayes formula:
- The desired probability. Absolutely expected - because the third workshop produces not only the smallest share of details, but also leads in quality!

In this case I had to simplify a four-story shotthat in the tasks to the Bayes formulas have to do quite often. But for this lesson, I somehow have so accidentally picked up examples in which many calculations can be carried out without ordinary fractions.

Since soon, there are no points "A" and "BE", the answer is better to provide text comments:

Answer: - the likelihood that the part extracted from the container will be defective; - The likelihood that the retrieved defective item released the 3rd shop.

As you can see, the tasks on the formula for the full probability and the Bayes formula are quite simple, and, probably, for this reason, they are so often trying to make it difficult to make it difficult to think about what I mentioned at the beginning of the article.

Additional examples are in the file with ready-made solutions on F.P.V. and Bayes formulaIn addition, you will probably have wishes to more deeply familiarize yourself with this topic in other sources. And the topic is really very interesting - what is only one paradox Bayes.which justifies that everyday council that if a person is diagnosed with a rare disease, it makes sense to repeat and even two repeated independent surveys. It would seem that it makes exclusively from despair ... - But no! But we will not be about sad.

- The likelihood that an arbitrary chosen student will pass the exam.
Let the student pass the exam. According to Bayes formulas:
but) - The likelihood that the student who passed the exam was prepared very well. An objective initial probability is overpriced, since almost always some "middling" is lucky with questions and they respond very strongly, which causes the erroneous impression of impeccable training.
b) - The likelihood that the student who passed the exam was prepared by the average. The initial probability is slightly overpriced, because Students with an average level of preparation are usually most, in addition, the teacher will take the unsuccessful "excellent students", and occasionally and poorly spending a student who is very lucky with a ticket.
in) - The likelihood that a student who passed the exam was well prepared. The initial probability is revalued for the worse. Not surprising.
Check:
Answer :

An understanding (study) of probabilities begins where the classic course of probability theory ends. For some reason, the school and university teach the frequency (combinatorial) probability, or the likelihood of what is determined. The human brain works differently. We have theories (opinions) about everything in the world. We subjectively assess the likelihood of certain events. We can also change your opinion if something unexpected happened. This is what we do every day. For example, if you meet with a girlfriend at the monument to Pushkin, you understand whether it will be in time, late for 15 minutes or half an hour. But going to the area of \u200b\u200bthe subway, and seeing 20 cm of fresh snow, you update your probabilities to take into account new data.

Such an approach was first described by Bayes and Laplace. Although Laplace, I think he was not familiar with the work of Bayes. According to the incomprehensible reason, the Bayesian approach is quite poorly represented in Russian-speaking literature. For comparison, I will note that at the request of Bayes Ozon gives 4 references, and Amazon is about 1000.

The present note is a translation of a small English book, and will give you an intuitive understanding of how to use the Bayes theorem. It begins with definition, and then uses examples in Excel, which will allow tracking the entire course of reasoning.

Scott Hartshorn. Bayes' Theorem Examples: A Visual Guide for Beginners. - 2016, 82 p.

Download note in format or, examples in format

Definition of the Bayes Theorem and an intuitive explanation

Theorem Bayes.

where a and b is events, P (a) and p (b) - probabilities a and b excluding each other, p (a | b) is the conditional probability of event A, provided that B is true, P (b | a) - Conditional probability B, if and truly.

In fact, the equation is somewhat more complicated, but for most applications it is enough. The result of calculations is simply normalized weighted value based on the initial assumption. So, take the initial assumption, weigh it in relation to other initial possibilities, normalize on the basis of observation:

In the course of solving problems, we will perform the following steps (hereinafter they will become clearer):

1. Determine what kind of probabilities we want to calculate, and what we observe.
2. Evaluate the initial probabilities for all possible options.
3. Assuming the truth of a certain initial option, calculate the likelihood of our observation; And so for all initial options.
4. Find a suspended value as the work of the initial probability (step 2) and the conditional probability (step 3), and so for each of the initial options.
5. Normalize Results: Divide each weighted probability (step 4) for the sum of all suspended probabilities; The sum of normalized probabilities \u003d 1.
6. Repeat steps 2-5 for each new observation.

Example 1. A simple example with bones

Suppose your friend has 3 bones: from 4, 6 and 8 faces. He randomly chooses one of them, does not show you, throws and reports the result - 2. Calculate the likelihood that a 4-year-old was selected, a 6-year-old, 8-year-old.

Step 1. We want to calculate the likelihood of a 4-graded choice, a 6-year-old or 8-year-old. We observe the dropped number - 2.

Step 2. Since the bones were 3, the initial probability of choosing each of them is 1/3.

Step 3. Observation - the bone fell as a face 2. If a 4-year-old was taken, the chances of this are equal to 1/4. For a 6-graded chance of 2-ki dropping chances - 1/6. For 8-graded - 1/8.

Step 4. Loss of 2-ki for a 4-year \u003d 1/3 * 1/4 \u003d 1/12, for a 6-grade \u003d 1/3 * 1/6 \u003d 1/18, for 8-year \u003d 1/3 * 1/8 \u003d 1/24.

Step 5. Overall probability of losing 2-ki \u003d 1/12 + 1/18 + 1/24 \u003d 13/72. This is less than 1, because the chances of throwing 2-ku less than 1. But we know that you have already threw out 2-ku. Thus, we need to divide the chances of each version from step 4 to 13/72, so that the sum of all chances for all bones to lie down the 2nd 1. This process is called normalization.

Normalizing each weighted probability, we find the likelihood that this bone was chosen:

• 4-year \u003d (1/12) / (13/72) \u003d 6/13
• 6-year \u003d (1/18) / (13/72) \u003d 4/13
• 8-year \u003d (1/24) / (13/72) \u003d 3/13

And this is the answer.

When we started solving the task, we suggested that the probability of choosing a certain bone is 33.3%. After falling out 2-ki, we calculated the chances that the 4-graded 4-graded chance was chosen to 46.1%, the chances of choosing a 6-year-old decreased to 30.8%, and the chances that the 8-year-old was chosen and fell at all up to 23.1%.

If you make another throw, we could use new calculated interest as our initial assumptions and clarify the probabilities based on the second observation.

If you have the only observation, all the steps are convenient to present in the form of a table:

Table. 1. Step-by-step solution in the form of a table (forms, see Excel file on a sheet Example 1.)

Note:

• If instead of 2-ki fell out, for example, 7-ka, then the chances in step 3 would be zero, and after normalization, the chances of the 8-year-old would be 100%.
• As an example includes only three bones and one throw, we used simple fractions. For most problems with a large number of options and events, it is easier to work with decimal fractions.

Example 2. More bones. More throws

This time we have 6 bones with 4, 6, 8, 10, 12 and 20 faces. We choose one of them randomly and throw 15 times. What is the likelihood that a certain bone was chosen?

I use the model in Excel (Fig. 1; see Leaf Example 2.). Random numbers are generated in column B using function \u003d rationing (1; $ B $ 9). In this case, an 8-year-bound cell is selected in the B9 cell, so random numbers can receive values \u200b\u200bfrom 1 to 8. Since Excel updates random numbers after each change on the sheet, I copied the column to the buffer and inserted only the values \u200b\u200bin the C column. Now the values \u200b\u200bare not Change and will be used for subsequent drawings. (I added you the opportunity to "play" with the choice of the number of faces and random throws on the sheet Example 2 game. Especially curious results are obtained, if in the cell B9 set the number 13 🙂 - Approx. Baguzin.)

Fig. 1. Generator random numbers

Step 2. Since only six cubes, the probability of choosing one random is 1/6 or 0.167.

Steps 3 and 4. We write the equation for the likelihood of the initial choice of a certain bone after the corresponding throw. As we have seen at the end of Example 1, some throws may not correspond to one or another bones. For example, 9-ki loss makes the probability of 4-, 6- and 8-face bones equal to zero. If the "legitimate" number fell, then its probability for this bone is equal to a unit divided by the number of faces. For convenience, we combined steps 3 and 4, so we immediately recover the formula for the likelihood of a throw multiplied by the normalized probability after the previous throw (Fig. 2):

If (throw\u003e the numbers of the faces; 0; 1 / number of faces * Previous normalized probability)

If you carefully use, you can drag this formula to all lines.

Fig. 2. The probability equation; To enlarge image Click on it right-click and select Open picture in a new tab

Step 5. The last step is the normalization of the results after each throw (the region L11: R28 in Fig. 3).

Fig. 3. Normalization of results

So, after 15 shots with a probability of 96.4%, we can assume that the 8-graded bone was originally chosen. Although there are chances that the bone was chosen with b about by the number of faces: 3.4% for 10-graded bone, 0.2% - for 12-graded, 0.0001% - for 20-graded. But the probability of 4- and 6-graded bones is zero, since among the dropped numbers were 7 and 8. This, naturally, it corresponds to the fact that we have entered the number 8 into the cell B9, limiting the values \u200b\u200bfor the generator of random numbers.

If we construct a chart of the likelihood of each option of the initial choice of bones, throwing throwing, we will see (Fig. 4):

• After the first throw, the probability of choosing a 4-face dice drops to zero, since 6-ka immediately dropped. Therefore, the leadership seized the version of the 6-face bone.
• For several first shots, the 6-face bone has the greatest probability, as it contains the least faces among bones that can respond to the fallen values.
• On the fifth throw fell 8-ka, the probability of the 6-year-old drops to zero, and the 8-year-old becomes the leader.
• The probabilities of 10-, 12- and 20-graded bones at the first throws smoothly decreased, and then experienced a splash when the 6-faced bone fell out of the race. This is due to the fact that the results were normalized in a much smaller sample.

Fig. 4. Change of probability throw by throw

Note:

• The Bayes Theorem for Multiple Events is simply repeated multiplication on sequentially updated data. The final answer does not depend on how the events occurred.
• It is not necessary to normalize the probabilities after each event. You can do it once at the very end. The problem is that if not to be normalized constantly, the probabilities become so small that Excel can work incorrectly due to rounding errors. Thus, it is more practical to normalize at every step than to check if you did not come to the border of the Excel accuracy.

Bayes Theorem. Terminology

• The initial probability is the likelihood of every possibility before the observation occurred is called a priori.
• Normalized response after calculating the probability for each data point (for each observation) is called a posteriorio.
• The total probability used to normalize the response is constant Normalization.
• Conditional probability, i.e. The probability of each event is called playing.

Here's how these terms look for the first example (comparing Fig. 1).

Fig. 5. The terms of the Bayes theorem

The Bayes Theorem itself in new definitions looks like this (comparing with formula 2):

Example 3. Unfair coin

You have a coin that, as you suspect, is not honest. You throw it 100 times. Calculate the likelihood that a dishonest coin will fall up with an eagle up with a probability of 0%, 10%, 20%, 30%, 40%, 50%, 60%, 70%, 80%, 90%, 100%.

Refer to the Excel file, sheet Example 3.. In cells B13: B112, I generated a random number from 0 to 1, and with the help of a special insert, I moved the value to the C column in the cell B8, I indicated the expected percentage of eagle falling for this dishonest coin. In column D using a function if I turned the likelihood in units (eagles, for probability r from 0.35 to 1) or in zeros (ripes for r from 0 to 0.35).

Fig. 6. Initial data for dishonest coins

I got 63 eagles and 37 making, which corresponds well to the generator of random numbers, if we have installed the probability of eagles 65%.

Step 1. We want to calculate the likelihoods that the eagles refer to the baskets of 0%, 10%, ... 100%, watching 63 eagles and 37 rivers with 100 shots.

Step 2. There are 11 initial capabilities: probabilities 0%, 10%, ... 100%. We will naive that all initial capabilities have an equal likelihood, that is, 1 chance of 11 (Fig. 7). (We might more realistically give the initial probabilities located in the area of \u200b\u200b50% large weights than probabilities at the edges - 0% and 100%. But the most remarkable is that, since we have a whole 100 throws, the initial probabilities are not so much Important!)

Step 3 and 4. Calculation of truth. To calculate the probability after each tossing in Excel, the function is used if. In case the eagle fell, plausibility is equal to the work of the possibility of the previous normalized probability. If the river fell out, plausibility is equal to (1 minus the ability) * the previous normalized probability (Fig. 8).

Fig. 8. Relapidity

Step 5. Normalization is performed as in the previous example.

The results are most visualized in the form of a series of histograms. The initial schedule is a priori probability. Then every new schedule is the situation after the next 25 shots (Fig. 9). Since we set the probability of the Eagle 65% at the entrance, the graphs presented do not cause surprise.

Fig. 9. Probability of options after a series of throws

What actually means a 70% chance for the possibility of 0.6? This is not a 70% chance that the coin accurately falls by 60%. Since we had a pitch of 10% between options, we estimate that there is a 70% chance that this coin will fall into the range between 55 and 65%. The solution to use 11 initial options, with a 10% increment was completely arbitrary. We could use 101 initial possibility with 1% increments. In this case, we would receive a result with a maximum at 63% (since we had 63 eagle) and a smoother drop in the graph.

Please note that in this example we observed a slower convergence compared to an example 2. This is due to the fact that the difference between the coin, turning 60% against 70%, less than between cubes with 8 and 10 faces.

Example 4. More bones. But with errors in the data stream

Let's go back for example 2. A friend in the bone bag from 4, 6, 8, 10, 12, 20 grands. He takes out one bone randomly and throws it 80 times. He records the dropped numbers, but in 5% of cases is mistaken. In this case, a random number appears from 1 and 20 instead of the actual throw result. After 80 throws, what do you think, what kind of bone was chosen?

As input in Excel (sheet Example 4.) I entered the number of parties (8), as well as the likelihood that the data contains an error (0.05). Formula for throw value (Fig. 10):

If (adhesis ()\u003e error probabilities; permanent (1; number of faces); rationing (1; 20))

If the random number is greater than the error probability (0.05), then there was no error throw, so that the generator of random numbers selects the value between 1 and the "mounted" sides of the cube, otherwise you should generate a random integer between 1 and 20.

Fig. 10. Calculation of throw value

At first glance, we could solve this problem in the same way as in Example 2. But, if you do not consider the likelihood of errors, we will get a probability schedule as in Fig. 11. (The easiest way to get it in Excel is to first generate throws in the column in the value of the error 0.05; then transfer the values \u200b\u200bof the throws into the column C, and finally change the value in the cell B11 to 0; since the formula for calculating attribution in the D14 range : J94 refer to the B11 cell, the effect of not accounting errors will be achieved.)

Fig. 11. Processing the values \u200b\u200bof the throws without taking into account the probability of error presence

Since the likelihood of error is small, and the random number generator is configured to the 8-year-old, the probability of the latter with each throw becomes dominant. Moreover, since an error can with a probability of 40% (eight of twenty) to give a value within 8, then the value of the error that affects the result, appeared only on the 63rd throw. However, if errors are not taken into account, the probability of the 8-year one will turn to zero, and 100% will receive a 20-year-old. Note that by the 63rd throw the likelihood of a 20-grade was only 2 * 10 -25.

The chances of getting an error - 5%, and the likelihood that the error will give a value greater than 8, is 60%. Those., 3% of the throws will give an error with a value of more than 8, which happened on the throw 63, when recording was made 17. If the formula likelihood does not take into account possible errors, we get the taking off of the probability of a 20-year-grade from 2 * 10 -25 up to 1, as in Fig. eleven.

If a person scrupulously monitors the data, it can detect this error and not to take erroneous values. To automate the process, add the equation of truth-like by error check. Never install the zero probabilities of errors, if you admit that they cannot be completely excluded. If you take into account the probabilities of errors, then hundreds of "correct" data will not allow separate erroneous values \u200b\u200bto spoil the picture.

We supplement the equation of the credibility of the error check (Fig. 12):

If ($ C15\u003e F $ 13; $ B $ 11 * 1/20 * N14; ($ B $ 11 * 1/20 + (1- $ B $ 11) / F $ 13) * N14)

Fig. 12. The function of loudness with errors

If the recorded value of the throw is greater than the number of faces ($ C15\u003e F $ 13), the conditional probability is not reset, but reduce, taking into account the probability of the error ($ B $ 11 * 1/20 * N14). If a recorded number is less than the number of faces, the conditional probability is not in full, as well as taking into account the possible error ($ B $ 11 * 1/20 + (1- $ B $ 11) / F $ 13) * N14). In the latter case, we believe that the recorded number could be as a consequence of the error ($ B $ 11 * 1/20) and the result of the correct entry (1- $ B $ 11) / F $ 13).

The change in the normalized probability becomes more resistant to possible errors (Fig. 13).

Fig. 13. Changing the normalized probability from the throw to the throw

In this example, the 6-graded bone is initially a favorite, because the first 3 throw - 5, 6, 1. Then the 7-ka and the probability of the 8-year one goes upwards. However, the appearance of 7-ki does not reset the likelihood of a 6-year-old, because the 7-ka may be an error. And the next nine shots seem to confirm this when values \u200b\u200bare not more than 6: the probability of a 6-year-old begins to grow again. Nevertheless, on the 14th and 15th throws, 7-ki fall out again, and the probability of a 6-face bone approaches zero. Later, the values \u200b\u200bof 17 and 19 appear, which "system" defines how clearly erroneous.

Example 4A. What if you really have a high frequency of errors?

This example is similar to the previous one, but the frequency of errors is increased from 5% to 75%. Since the data has become less relevant, we increased the number of shots up to 250. By applying the same equations as in example 4, we obtain the following chart:

Fig. 14. Normalized probability at 75% of erroneous entries

With such a high frequency of errors, much more throws took. In addition, the result is less defined, and the 6-year-old periodically becomes more likely. If you have an even higher error rate, for example, 99%, you can still get the correct answer. Obviously, the higher the frequency of errors, the more throws need to be done. For 75% of errors, we obtain one correct value of four. If the probability of an error is 99%, we would get only one correct value from one hundred. We are probably needed 25 times more data to identify the dominant option.

And what if you do not know the likelihood of error? I recommend "play" with examples 4 and 4A, setting various values \u200b\u200bfrom very small in the cell B11 (for example, 2 * 10 -25 for example 4) to very large (for example, 90% for example 4A). Here are the main conclusions:

• If the error frequency estimate is higher than the actual error frequency, the results will converge slowly, but still converge to the correct answer.
• If you evaluate the error frequency too low, there is a risk that the results will not be correct.
• The smaller the actual frequency of errors, the greater the place for the maneuver you have in guessing the frequency of errors.
• The higher the actual frequency of errors, the more data you need.

Example 5. The problem of the German tank

In this task you are trying to evaluate how many tanks were produced, based on the serial numbers of captured tanks. The Bayes Theorem was used by allies during the Second World War, and ultimately gave results lower than those reported intelligence. After the war, the record has shown that statistical estimates using the Bayes Theorem were more accurate. (It is curious that I wrote a note on this topic, not yet knowing what probability by Bayesu; see. - Approx. Baguzin.)

So, you analyze serial numbers removed from broken or captured tanks. The goal is to evaluate how many tanks were produced. That's what you know about the serial numbers of tanks:

• They start with 1.
• These are integers without skipping.
• You found the following serial numbers: 30, 70, 140, 125.

We are interested in the answer to the question: what is the maximum number of tanks? I will start with 1000 tanks. But someone else could start with 500 tanks or 2000 tanks, and we can get different results. I am going to analyze every 20 tanks, which means that I have 50 initial opportunities for the number of tanks. You can complicate the model, and analyze for each individual in Excel, but the answer will not change much, and the analysis will complicate significantly.

I assume that all the possibility of the number of tanks is equal (i.e., the likelihood of 50 tanks is the same as 500). Please note that the Excel file has more columns than shown in the figure. The conventional probability for the likelihood function is very similar to the conditional probability of Example 2:

• If the observed serial number is greater than the maximum serial number for this group, the probability of the presence of such a number of tanks is 0.
• If the observed serial number is less than the maximum serial number for this group, the probability is a unit divided by the number of tanks multiplied by the normalized probability in the previous step (Fig. 15).

Fig. 15. Conditional probabilities of distribution of tanks by groups

Normalized probabilities look as follows (Fig. 16).

Fig. 16. Normalized probability of tanks

There is a large splash of probability for the maximum observed serial number. After that, an asymptotic decrease to zero occurs. For 4 detected serial numbers, the maximum responds to 140 tanks. But, despite the fact that this number is the most likely answer, this is not the best assessment, since it almost certainly underestimates the number of tanks.

If you take the weighted average number of tanks, i.e. To summarize multiplied groups and their probabilities for four tanks, applying the formula:

Rounded (BD9: DA9; BD14: DA14); 0)

we get the best estimate of 193.

If we originally proceeded out of 2000 tanks, there would be a weighted average of 195 tanks, which essentially does not change anything.

Example 6. Drug Testing

You know that 0.5% of the population uses drugs. You have a test that gives 99% of the true positive results for the drug, and 98% of true negative results for non-consuming. You randomly choose a person, spend the test and get a positive result. What is the likelihood that a person actually uses drugs?

For our random individual initial probability The fact that it is a drug consumer is 0.5%, and the likelihood that it is not a drug consumer is 99.5%.

The next step is the calculation of the conditional probability:

• If the subject consumes drugs, the test will be positive in 99% of cases and negative in 1% of cases.
• If the subject does not use drugs, then the test will be positive in 2% of cases and negative in 98% of cases.

The functions of believing for consumed and non-drugs are presented in Fig. 17.

Fig. 17. Functions of believing: (a) for drug use; (b) for non-drugs

After normalization, we see that, despite the positive feedback, the likelihood that this random person uses drugs is only 0.1992 or 19.9%. This result surprises many people, because in the end, the accuracy of the test is quite high - as much as 99%. Since the initial probability was only 0.5%, even a large increase in this probability was not enough to make the response really large.

The intuition of most people does not take into account the initial probability. Even if the conditional probability is really high, a very low initial probability can lead to a low finite probability. The intuition of most people is configured around the initial probability of 50/50. If this is the case, the test result is positive, then the normalized probability will be the expected 98%, confirming that a person uses drugs (Fig. 18).

Fig. 18. Test result with initial probability 50/50

An alternative approach to the explanation of such situations, see.

Look the bibliography on the Bayes theorem at the end of the notes.

If the event BUT can occur only when performing one of the events that form complete group of incomplete events then the probability of an event BUT Calculated by formula

This formula is called formula full probability .

We again consider a complete group of incomplete events, the probability of the appearance of which . Event BUT can happen only with any of the events that will be called hypothesis . Then according to the formula of the full probability

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

By probability multiplication theorem

.

Similarly, for the remaining hypotheses

The resulting formula is called bayes formula (bayes formula ). The probability of hypotheses is called a posteriori probabilities , whereas - a priori probabilities .

Example. The store received new products from three enterprises. The percentage of this product is as follows: 20% - the products of the first enterprise, 30% - the products of the second enterprise, 50% are the products of the Third Enterprise; Further, 10% of the products of the first enterprise of the highest grade, at the second enterprise - 5% and on the third - 20% of the highest grade products. Find the likelihood that randomly purchased new products will be the highest grade.

Decision. Denote by IN The event, which consists in the fact that the products of the highest grade will be purchased, through the events to purchase products belonging to the first, second and third enterprises according to the first, second and third enterprises.

You can apply a full probability formula, and in our notation:

Substituting these values \u200b\u200bin the formula of the full probability, we obtain the desired probability:

Example. One of the three shooters is called to the line of fire and produces two shots. The probability of hitting the target with one shot for the first arrow is 0.3, for the second - 0.5; For the third - 0.8. The target is not amazed. Find the chance that the shots are made by the first shooter.

Decision. Three hypotheses are possible:

The first shooter is called on the line of fire,

The second shooter is called on the line of fire,

The third shooter is caused to the line of fire.

Since the challenge on the line of fire of any arrow is equilibrium, then

As a result of the experience, an event was observed in - after the shooters made, the target was not amazed. The conditional probabilities of this event with the hypotheses made are equal:

according to the Bayes formula, we find the likelihood of hypothesis after experience:

Example. On three machines machines are processed by the same type of part arriving after processing on a common conveyor. The first machine gives 2% of marriage, the second - 7%, the third - 10%. The performance of the first machine is 3 times more performance of the second, and the third is 2 times less than the second.

a) What is the percentage of marriage on the conveyor?

b) What are the stake of the details of each machine among defective parts on the conveyor?

Decision. Take from the conveyor to bring one detail and consider the event A - the detail is defective. It is associated with hypotheses as to where this detail was processed: - the detail taken by the Machine was processed by the machine.

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

The relationship between machine manufacturers means the following:

And since hypothesis form a complete group, then.

Deciding the resulting system of equations, we find :.

a) the full chance that the detail taken from the conveyor is defective:

In other words, in the mass of parts converging from the conveyor, the marriage is 4%.

b) Let it know that the detail taken is defective. Using the Bayes formula, we will find the conditional probability of hypotheses:

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

Practical tasks

Exercise 1

Solving tasks for the main sections of the theory of probability

The goal is to obtain practical skills in solving problems

sections of probability theory

Preparation for the implementation of a practical task

Read the theoretical material on this topic, to explore the content of theoretical, as well as the relevant sections in literary sources

Procedure for performing the task

Solve 5 tasks according to the number of the task version given in Table 1.

Options for source data

Table 1

	task number

The composition of the report on the task 1

5 solved tasks according to the number of the option.

Tasks for self solutions

1 .. Are the following events groups: a) experience - throwing coins; events: A1.- the appearance of the coat of arms; A2.- the appearance of the figures; b) experience - throwing two coins; events: IN 1- the appearance of two coats of arms; AT 2 -the appearance of two digits; IN 3- the appearance of one coat of arms and one digit; c) experience - throwing a playing bone; 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 - target shot; events: D1.- hit; D2 -slip; e) experience - two target shots; events: E0- Not a single hit; E1- one hit; E2.- two hit; e) experience - remove two cards from the deck; events: F1 -the emergence of two red cards; F2.- The appearance of two black cards?

2. In the black and b urn black balls. One ball is removed from the urns. Find the chance that this ball is white.

3. In the urn A white I. B. black balls. From the urns take one ball and lay aside. This ball was white. After that, another ball takes from the urn. Find the chance that this ball will also be white.

4. In the urn A white and B. black balls. One ball was taken out of the urns and, without looking, put aside aside. After that, another ball took from the urn. He was white. Find the chance that the first ball deflected aside is also white.

5. From the urn containing a white and B. black balls, take out one after another all the balls except one. Find the likelihood that the last remaining ball in the ball will be white.

6. From the urn in which A white balls and b black, remove all the balls in it in it. Find the chance that the second balloon will be taken out in order.

7. In the urn A white and b black balls (A. > 2). Two balls are removed from the urns. Find the likelihood that both balls will be white.

8. In the black and b urn black balls (a\u003e 2, b\u003e 3). Five balls are removed from the urns. Find probability rthat two of them will be white, and three black.

9. In a party consisting of x products, available I.defective. From the party is selected for control I products. Find probability rthat of them exactly j products will be defective.

10. Playing bone rushes once. Find the probability of the following events: BUT -the emergence of an even number of points; IN- the appearance of at least 5 points; FROM-the appearance of no more than 5 points.

11. Playing bone rushes twice. Find probability rthe fact that both times the same number of points will appear.

12. Two playing bones are thrown at the same time. Find the probabilities of the following events: BUT- the sum of the dropped points is 8; IN- the product of the dropped points is 8; FROM-the amount of points dropped more than their work.

13. Two coins rush. Which of the events is more likely: BUT -coins will be lying with the same parties; IN -coins will lie in different sides?

14. In the urn A white and B. black balls (A. > 2; B. > 2). From the urns at the same time two balls. What event is more likely: BUT- balls of the same color; IN -balls of different colors?

15. Three players play cards. Each of them handed over 10 cards and two cards are left in a bona. One of the players sees that he has 6 maps of a bubnova suit and 4 - not a bubnova. He resets two cards from these four and takes himself a bona. Find the likelihood that he will buy two tambourous cards.

16. From the urn containing ppurpose balls, at random takes out one after another all the balls in it. Find the likelihood that the numbers of the cut balls will go in order: 1, 2, ..., p.

17. The same urn as in the previous task, but each ball after removing is embedded back and mixed with others, and its number is recorded. Find the chance that the natural sequence of numbers will be recorded: 1, 2, ..., p.

18. Full deck of cards (52 sheets) is divided into two equal packs of 26 sheets. Find the probabilities of the following events: BUT -in each of the packs will be two aces; IN- In one of the packs there will not be a single ace, and in the other - all four; C-B.one buckle will be one ace, and in the other - three.

19. In the drawing of the basketball championship, 18 teams are involved, of which two groups of 9 teams are randomly formed in each. Among the participants of the competition there are 5 teams

extra class. Find the probabilities of the following events: BUT -all Extra-Class commands will fall into the same group; IN- Two emergency teams will fall into one of the groups, and three to another.

20. Nine cards are written numbers: 0, 1, 2, 3, 4, 5, 6, 7, 8. Two of them are removed at random and stacked on the table in order of appearance, then the resulting number is read, for example 07 (seven), 14 ( Fourteen), etc. Find the likelihood that the number is even.

21. On five cards, numbers are written: 1, 2, 3, 4, 5. Two of them, one after another, removed. Find the likelihood that the number on the second card will be greater than on the first.

22. The same question is that in the task 21, but the first card after removing is placed back and mixed with the rest, and the number that stands on it is written.

23. In the urn A white, B. black and with red balls. From the urns take out one after another all the balls in it and write their colors. Finding the chance that in this list the white color will appear before black.

24. There are two urns: in the first A white and B. black balls; In the second C. white and D. black. From each urn removed over the ball. Find the likelihood that both balls will be white.

25. Under the conditions of Tasks 24, find the likelihood that the balls will be different colors.

26. In the Revolver drum, seven nests, of which five were laid, and two were left empty. The drum is driven by rotation, as a result of which one of the nests turns out against the trunk. After that, the trigger is pressed; If the cell was empty, the shot does not occur. Find probability rthe fact that, repeating such an experience two times in a row, we are both not shifted.

27. In the same conditions (see Task 26) find the likelihood that both times the shot will occur.

28. In the urn there is a; balls marked with numbers 1, 2, ..., toFrom urns I.once removed one ball (I.<к), the ball number is written and the ball holds back into the urn. Find probability rthat all recorded numbers will be different.

29. Of the five letters of a split alphabet, the word "book" is drawn up. A child who does not know how to read, scattered these letters and then collected in random order. Find probability rthe fact that he again turned out the word "book."

30. The word "pineapple" was drawn up from the letters of a split alphabet. A child who does not know how to read, scattered these letters and then collected in random order. Find probability rthat he again has the word "pineapple

31. Multiple maps are removed from the full deck of cards (52 sheets, 4 suites). How many cards need to be removed in order to be more likely than 0.50, to argue that there will be the same cards among them?

32. N.a man randomly disappear over the round table (N\u003e2). Find probability rthat two fixed faces BUTand INwill be near.

33. The same task (cm 32), but the table is rectangular, and n a person is rapidly accidentally along one of his sides.

34. The lotto barrels are written numbers from 1 to N.Of these N.barrels are accidentally chosen two. Find the likelihood that the numbers smaller than K are written on both kegs (2

35. On the barrels lotto written numbers from 1 to N.Of these N.barrels are accidentally chosen two. Find the chance that one of the kegs written the number greater than K , and on the other - less than k . (2

36. Battery out M.the guns leads fire on a group consisting of N.goals (M.< N). The guns choose their goals consistently, randomly, provided that no two guns can shoot one target. Find probability rwhat will be fired by targets with numbers 1, 2, ..., M.

37 .. Battery, consisting of toguns, leads fire on a group consisting of I.aircraft (to< 2). Each instrument chooses a goal accidentally and regardless of others. Find the likelihood that all tothe guns will shoot at the same purpose.

38. In the conditions of the previous task, find the likelihood that all guns will shoot different purposes.

39. Four balls randomly scatter on four holes; Each ball enters that or another well with the same probability and regardless of others (obstacles to getting into the same one and the same well of several balls). Find the chance that three balls will be in one of the wells, to the other one, and there will be no balls in the other left holes.

40. Masha quarreled with Peter and does not want to go with him in one bus. 5 buses leave from the hostel to the institute from 7 to 8. Who did not have time for these buses is late for a lecture. How many ways to Masha and Peter can get to the institute for different buses and not be late for a lecture?

41. There are 3 analytics, 10 programmers and 20 engineers in the information management of the bank. For overtime on a holiday, the head of the Department should allocate one employee. How many ways can this be done?

42. The head of the Bank's Security Service should separate 10 guards for 10 posts. 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. How many ways can this be done?

44. One of the warring parties captured 12, and the other - 15 captives. How many ways can I exchange 7 prisoners of war?

45. Petya and Masha collect a video video. Petit has 30 comedies, 80 militants and 7 meloders, Masha - 20 comedies, 5 militants and 90 meloders. How many ways Peter and Masha can exchange 3 comedies, 2 militants and 1 melodrama?

46. \u200b\u200bIn terms of problems with 45, in some ways, Peter and Masha can exchange 3 melodramas and 5 comedies?

47. Under the conditions of tasks 45, in some ways, Peter and Masha can exchange 2 militants and 7 comedies.

48. One of the warring parties captured 15, and the other - 16 prisoners. How many ways can I exchange 5 prisoners of war?

49. How many cars can be registered in 1 city if the number has 3 digits and 3 letters (only those whose writing coincides with Latin - A, B, E, K, M, N, O, P, C, T, Y, X )?

50. One of the warring parties captured 14, and the other - 17 captives. How many ways can I exchange 6 prisoners of war?

51. How many different words can be made by rearring letters in the word "mother"?

52. In a basket of 3 red and 7 green apples. It takes one apple from it. Find the likelihood that it will be red.

53. In a basket of 3 red and 7 green apples. It was taken out of it and postponed one green apple. After that, 1 apple is taken out of the basket. What is the likelihood that this apple will be green?

54. In a batch consisting of 1000 products, 4 have defects. To control, choose a batch of 100 products. What is the likelihood of TOO that in the checklist will not be defective?

56. In the 80s, the Sportoto 5 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 advantages if he guess a different number of numbers announced by the circulation commission. Find the chance that the player has not guessed a single number.

57.In the 80s in the USSR, the game "Sportlot 5 of 36" was popular. The player noted on the card 5 numbers from 1 to 36 and received prizes of various advantages if he guess a different number of numbers announced by the circulation commission. Find the chance that the player guess one number.

58.In the 80s in the USSR, the game "Sportoto 5 of 36" was popular. The player noted on the card 5 numbers from 1 to 36 and received prizes of various advantages if he guess a different number of numbers announced by the circulation commission. Find the chance that the player guess 3 numbers.

59.In the 80s in the USSR, the game "Sportoto 5 of 36" was popular. The player noted on the card 5 numbers from 1 to 36 and received prizes of various advantages if he guess a different number of numbers announced by the circulation commission. Find the chance that the player has not guessed all the 5 numbers.

60. In the 1980s, the Gameloto 6 of 49 was popular in the USSR. The player noted on the card 6 numbers from 1 to 49 and received prizes of various advantages if he guess a different number of numbers declared by the circulation commission. Find the chance that the player guess 2 numbers.

61. In the 1980s, the Greyloto 6 of 49 was popular in the USSR. The player noted on the card 6 numbers from 1 to 49 and received prizes of various advantages if he guess a different number of numbers declared by the circulation commission. Find the chance that the player has not guessed a single number.

62. In the 80s, the Greyloto 6 of 49 was popular in the USSR. The player noted on the card 6 numbers from 1 to 49 and received prizes of various advantages if he guess a different number of numbers declared by the circulation commission. Find the likelihood that the player gave all the 6 numbers.

63. In a batch consisting of 1000 products, 4 have defects. To control, choose a batch of 100 products. What is the likelihood of TOO that only 1 defective will be in the checklist?

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

65. How many different words can be made by rearring letters in the word "pineapple"?

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

67. 6 people entered the elevator, the building has 7 floors. What is the likelihood that all 6 people will come out on different floors?

68. During a thunderstorm on a plot between 40 and 79 km, the power lines occurred. Considering that the break is equally possible anywhere, find the likelihood that the broken occurred between the 40th and 45th kilometers.

69. Per 200 kilometer section of the gas pipeline occurs a gas leak between compressor stations A and B, which is equally possible at any point of the pipeline. What is the likelihood that leakage takes place no further than 20 km from A

70. Per 200 kilometer section of the gas pipeline occurs gas leak between compressor stations A and B, which is equally possible anywhere in the pipeline. What is the likelihood that the leak occurs closer to and than k in

71. Radar DPS inspector has an accuracy of 10 km / hour and rounds in the next direction. What happens more often - rounding in favor of the driver or inspector?

72. Masha spends on the road to the institute from 40 to 50 minutes, and any time in this gap is equivalent. What is the likelihood 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 specify the time of arrival. They agreed to wait 15 minutes. What is the probability of their meeting?

74. Fishermen caught 120 fish in the pond, of which 10 were heated. What is the probability of catching the ocelled fish?

75. From the basket containing 3 red and 7 green apples remove all the apples in turn. What is the likelihood that the 2nd apple will be red?

76. From the basket containing 3 red and 7 green apples remove all the apples in turn. What is the likelihood that the last apple will be green?

77. Students consider that out of 50 tickets 10 are "good." Petya and Masha take turns pulling one ticket. 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. What is the likelihood that about both of them got a "good" ticket?

79. Masha came to the exam knowing the answers to 20 questions of the program from 25. Professor Specifies 3 questions. What is the likelihood that Masha will answer 3 questions?

80. Masha came to the exam knowing the answers to 20 questions of the program from 25. Professor Specifies 3 questions. What is the likelihood that Masha will not answer any question?

81. Masha came to the exam knowing the answers to 20 questions of the program from 25. Professor Specifies 3 questions. What is the likelihood that Masha will answer 1 question?

82. Statistics of credit requests in the bank is as follows: 10% - state. Authorities, 20% - other banks, the rest are individuals. The probability of no return of loans, respectively, 0.01, 0.05 and 0.2. What share of loans does not return?

83. The likelihood that the weekly turnover of the ice cream trader will exceed 2000 rubles. It is 80% with clear weather, 50% with cloud variable and 10% with rainy weather. What is the probability that turnover exceeds 2000 rubles. If the probability of clear weather is 20%, and variable cloudiness and rain is 40%.

84. In the urn and white (b) and in black (h) balls. From the urns are taken out (at the same time or sequentially) two balls. Find the likelihood that both balls will be white.

85. In the urn A white and B.

86. In urn A. white and B.

87. In urn A. white and B. black balls. One ball is removed from the urn, its color and the ball returns to the urn. After that, one more ball is taken from the urn. Find the likelihood that these balls will be different colors.

88. There is a box with nine new tennis balls. For the game take three goals; After the game they are put back. When choosing balls, the players are not distinguished from non-chairs. What is the likelihood that after three games in the box will not be left for the balls?

89. Leaving the apartment, N. each guest will put on its helos;

90. Leaving the apartment, N.guests having the same shoe sizes are put on a mandrel in the dark. Each of them can distinguish the right calico from the left, but cannot distinguish its strangers. Find the likelihood that each guest, puts on the Kalosh, belonging to one pair (maybe not their own).

91. In the conditions of the task of the 90Night, the likelihood of what everyone will leave in their calories if guests can not distinguish the right calosa from the left and just take the first two caloshs.

92. Archery is conducted by the aircraft whose vulnerable parts are two engines and a pilot cabin. In order to hit (output) an airplane, it is enough to hit both engines together or a pilot cabin. Under these conditions of shooting, the likelihood of damage to the first engine is equal to p1second Engine p2,pilot cabins p3.Parts of the aircraft are affected independently of each other. Find the chance that the plane will be amazed.

93. Two arrows, regardless of each other, make two shots (each by their target). The likelihood of hitting the target with one shot for the first arrow p1for the second p2.The winning competition is considered to be the shooter, in the target of which will be more printed. Find probability RHwhat will win the first arrows.

94. Behind the space object, the object is detected with a probability r.The object detection in each cycle occurs independently of others. Find the likelihood that when pcycles object will be detected.

95. 32 The letters of the Russian alphabet are written on the circle of split alphabet. Five cards are removed at random one after the other and stacked on the table in the order of appearance. Find the chance that the word "end" will be.

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

97. Archery is made by aircraft incendiary shells. Fuel on the plane is concentrated in four tanks located in the fuselage one by one. Plaza tanks are the same. In order to ignite the plane, it is enough to get two shells either in the same tank or in the neighboring tanks. It is known that two projectiles fell into the tanks area. Find the chance that the plane will light up.

98. Four cards are removed from the full deck of maps (52 sheets). Find the likelihood that all these four cards will be different textures.

99. Four cards are removed from the full deck of cards (52 sheets), but each card after removing returns to the deck. Find the likelihood that all these four cards will be different textures ..

100. When the ignition is turned on, the engine begins to work 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 the device; Abnormal - 20%. The probability of the failure of the device in order t.in normal mode is 0.1; In abnormal - 0.7. Full probability rthe failure of the device is out.

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

103.The vehicles by gas station consists of 60% of cargo and 40% of passenger cars. What is the probability of finding a gas station of a truck, if the probability of its refueling 0.1, and passenger - 0.3

104. The flow of cars by gas station consists of 60% of cargo and 40% of passenger cars. What is the probability of finding a gas station of a truck, if the probability of its refueling 0.1, and passenger - 0.3

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

106. 32 The letters of the Russian alphabet are written on the circle of split alphabet. Five cards are removed at random one after the other and stacked on the table in the order of appearance. Find the likelihood that the word "book" will be.

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

108. Two balls are scattered by chance and independently of each other in four cells located one after another in a straight line. Each ball with the same probability of 1/4 enters each cell. Find the likelihood that 2 balls will fall into one cell

109. When switching on ignition, the engine begins to work with a probability r. Find the likelihood that the engine will start working with the second ignition switch;

110. The aircraft shooting is made by the incendial shells. Fuel on the plane is concentrated in four tanks located in the fuselage one by one. Plaza tanks are the same. In order to light the plane, it is enough to get two projectiles into the same tank. It is known that two projectiles fell into the tanks area. Find the likelihood that the plane will light up

111. The aircraft shooting is made by the incendial shells. Fuel on the plane is concentrated in four tanks located in the fuselage one by one. Plaza tanks are the same. In order to ignite the aircraft, it is enough to get two projectiles in the adjacent tanks. It is known that two projectiles fell into the tanks area. Find the likelihood that the plane will light up

112.In urn A. white and B. black balls. One ball is removed from the urn, its color and the ball returns to the urn. After that, one more ball is taken from the urn. Find the likelihood that both cut balls will be white.

113. In urn A. white and B. black balls. Two balls are removed from the urn. Find the likelihood that these balls will be different colors.

114. Two balls are scattered by chance and independently from each other in four cells located one another in a straight line. Each ball with the same probability of 1/4 enters each cell. Find the chance that the balls will fall into the neighboring cells.

115. Masha came to the exam knowing the answers to 20 questions of the program from 25. Professor Specifies 3 questions. What is the likelihood 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. What is the likelihood that about both of them got a "good" ticket?

117. The statistics of credit requests in the bank is as follows: 10% - state. Authorities, 20% - other banks, the rest are individuals. The probability of no return of loans, respectively, 0.01, 0.05 and 0.2. What share of loans does not return?

118. 32 The letters of the Russian alphabet are written on the circle of split alphabet. Five cards are removed at random one after the other and stacked on the table in the order of appearance. Find the chance that the word "end" will be.

119 Statistics of requests for loans in the bank is as follows: 10% - state. Authorities, 20% - other banks, the rest are individuals. The probability of no return of loans, respectively, 0.01, 0.05 and 0.2. What share of loans does not return?

120. The likelihood that the weekly turnover of the ice cream will exceed 2000 rubles. It is 80% with clear weather, 50% with cloud variable and 10% with rainy weather. What is the probability that turnover exceeds 2000 rubles. If the probability of clear weather is 20%, and variable cloudiness and rain is 40%.