Siberian State University of Telecommunications and Informatics

Department of Higher Mathematics

by discipline: "Probability theory and mathematical statistics"

"Formula of total probability and Bayes' (Bayesian) formula and their application"

Completed:

Head: Professor B.P. Zelentsov

Novosibirsk, 2010

Introduction 3

1. Formula of total probability 4-5

2. Bayes' formula (Bayes) 5-6

3. Problems with solutions 7-11

4. The main areas of application of Bayes' formula 11

Conclusion 12

Literature 13

Introduction

Probability theory is one of the classic branches of mathematics. It has a long history. The foundations of this branch of science were laid by great mathematicians. I will name, for example, Fermat, Bernoulli, Pascal.
Later development of the theory of probability was defined in the works of many scientists.
Scientists of our country made a great contribution to the theory of probability:
P.L. Chebyshev, A.M. Lyapunov, A.A. Markov, A.N. Kolmogorov. Probabilistic and statistical methods are now deeply embedded in applications. They are used in physics, engineering, economics, biology, and medicine. Their role has especially grown in connection with the development of computer technology.

For example, to study physical phenomena, observations or experiments are performed. Their results are usually recorded as values ​​of some observable quantities. When we repeat experiments, we find a scatter of their results. For example, repeating measurements of the same quantity with the same device while maintaining certain conditions (temperature, humidity, etc.), we get results that, at least slightly, but still differ from each other. Even multiple measurements do not make it possible to accurately predict the next measurement. In this sense, they say that the measurement result is a random quantity. An even more obvious example of a random variable is the number of a winning lottery ticket. There are many other examples of random variables. Nevertheless, in the world of accidents, certain patterns are found. The mathematical apparatus for studying such patterns is provided by the theory of probabilities.
Thus, probability theory deals with the mathematical analysis of random events and random variables associated with them.

1. Formula of total probability.

Let there be a group of events H 1 ,H 2 ,..., H n with the following properties:

1) all events are pairwise inconsistent: H i

2) their union forms the space of elementary outcomes W:

In this case, we will say that H 1 , H 2 ,...,H n form complete group of events... Such events are sometimes called hypotheses.

Let be A- some event: AÌW (Venn diagram is shown in Figure 8). Then there is total probability formula:

P(A) = P(A/H 1)P(H 1) + P(A/H 2)P(H 2) + ...+P(A/H n)P(H n) =

Proof. Obviously: A =

P(A) = P(

Considering that by the multiplication theorem P(

Example... The store sells electric lamps produced by three factories, with the share of the first plant being 30%, the second - 50%, and the third - 20%. Defect in their products is 5%, 3% and 2%, respectively. What is the likelihood that a lamp randomly chosen in a store was found to be defective.

Let the event H 1 is that the selected lamp comes from the first factory, H 2 on the second, H 3 - at the third plant. Obviously:

P(H 1) = 3/10, P(H 2) = 5/10, P(H 3) = 2/10.

Let the event A consists in the fact that the selected lamp turned out to be defective; A / H i means an event consisting in the fact that a defective lamp is selected from lamps manufactured in i th plant. It follows from the problem statement:

P (A/ H 1) = 5/10; P(A/ H 2) = 3/10; P(A/ H 3) = 2/10

Using the formula for total probability, we obtain

2. Bayes' formula (Bayes)

Let be H 1 ,H 2 ,...,H n- a complete group of events and AÌ W - some event. Then, according to the formula for the conditional probability

Here P(H k/A) - conditional probability of an event (hypothesis) H k or the likelihood that H k is implemented provided that the event A occurred.

By the probability multiplication theorem, the numerator of formula (1) can be represented as

P = P = P(A/H k)P(H k)

To represent the denominator of formula (1), you can use the total probability formula

P(A)

Now from (1) one can obtain a formula called Bayes' formula:

Bayes' formula calculates the probability of the hypothesis realization H k provided that the event A occurred. Bayes' formula is also called the formula for the probability of hypotheses. Probability P(H k) is called the prior probability of the hypothesis H k and the probability P(H k/A) - the posterior probability.

Theorem. The probability of a hypothesis after testing is equal to the product of the probability of a hypothesis before testing by the corresponding conditional probability of an event that occurred during testing, divided by the total probability of this event.

Example. Consider the above problem about electric lamps, just change the question of the problem. Suppose a customer bought a light bulb in this store, and it turned out to be defective. Find the probability that this lamp is manufactured at a second factory. The magnitude P(H 2) = 0.5 in this case, this is the a priori probability of an event that the purchased lamp is manufactured at a second plant. Having received information that the purchased lamp is defective, we can correct our estimate of the possibility of manufacturing this lamp at the second plant by calculating the posterior probability of this event.

Let's write out the Bayes formula for this case

From this formula we get: P(H 2 /A) = 15/34. As you can see, the information obtained has led to the fact that the probability of the event of interest to us is lower than the prior probability.

3. Problems with solutions.

Objective 1. The store has received new products from three enterprises. The percentage of these products is as follows: 20% - products of the first company, 30% - products of the second company, 50% - products of the third company; further, 10% of the products of the first enterprise are of the highest quality, at the second enterprise - 5% and at the third - 20% of the products of the highest quality. Find the likelihood that a new product accidentally purchased is of the highest grade.

Solution. Let us denote by V the event that a premium product is purchased through

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

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

Objective 2. One of the three shooters is called into 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.

Purpose of work: to form the skills of solving problems in the theory of probability using the formula of total probability and Bayes' formula.

Total Probability Formula

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

This formula is called total probability formula.

The probability of hypotheses. Bayes formula

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

Let's say that a test was performed, as a result of which the event appeared A... It is required to determine how they have changed (due to the fact that the event A has already occurred) the probabilities of hypotheses. Conditional probabilities of hypotheses are found by the formula

In this formula, the index / = 1.2

This formula is called Bayes' formula (after the name of the English mathematician who derived it; published in 1764). Bayes' formula allows you to overestimate the probabilities of hypotheses after the test result becomes known, as a result of which an event appeared A.

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

Solution

Let's designate the hypotheses:

N= (a standard part will be received for control);

N= (a non-standard part will be received for control).

Event A =(the part will be rejected).

From the condition of the problem, we find the probabilities

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

Using the formula for total probability, we obtain

The probability that a part will be rejected by mistake is

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

Answer:

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

Solution

Let's designate events:

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

B =(the product will be considered standard).

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

Also known are the conditional probabilities

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

Answer:"0.263.

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

Solution

Let's designate events:

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

V= (taken part will be non-standard).

Also known are the conditional probabilities

Using the formula for total probability, we find

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

Answer: 0,75.

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

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

Solution

Let's designate events:

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

D - corresponding opposite events;

A= (during the test there will be a failure of the device).

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

By the property of the probabilities of opposite events

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

Using the theorem of addition of the probabilities of inconsistent events and the theorem of multiplication of the probabilities of independent events, we obtain

Now we find the probabilities of hypotheses:

Answer:

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

Solution

Let's designate events:

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

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

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

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

Using the formula for total probability, we find

Answer: 0,028.

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

Solution

Let's designate events:

4 = (randomly taken scheme from the rth party); i = 1, 2, 3;

V= (a randomly chosen scheme will work beyond the warranty period).

By the condition of the problem, the probabilities of the hypotheses are known:

The conditional probabilities are also known:

Using the formula for total probability, we find

Answer: 0,225.

Task 7. The device contains two blocks, the serviceability of each of which is necessary for the operation of the device. The probabilities of no-failure operation for these units are 0.99 and 0.97, respectively. The device is out of order. Determine the probability that both units have failed.

Solution

Let's designate events:

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

A= (device will fail).

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

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

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

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

Answer:

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

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

Objective 3. The same parts are produced on two machines. The probability that the part produced on the first machine will be standard is 0.8, and on the second it is 0.9. The productivity of the second machine is three times that of the first. Find the probability that a standard part will be taken at random from a conveyor that receives parts from both machines.

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

Task 5. The device contains two blocks, the serviceability of each of which is necessary for the operation of the device. The probabilities of no-failure operation for these units are 0.99 and 0.97, respectively. The device is out of order. Determine the probability that the second unit failed.

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

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

Problem 8. Defect in the products of the plant due to defect a is 5%, and among those rejected on the basis of a products are defective in 10% of cases R. And in defect-free products a, defect R occurs in 1% of cases. Find the probability of meeting a defect R in all products.

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

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

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

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

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

Problem 14. The parts required for assembling the device come from two automatic machines, the performance of which is the same. Calculate the probability of receiving a standard part for assembly if one of the machines gives an average of 3% violation of the standard, and the other - 2%.

Problem 15. The weightlifting coach calculated that in order to receive team points in a given weight category, an athlete must push a barbell of 200 kg. Ivanov, Petrov and Sidorov are competing for a place in the team. During training, Ivanov tried to lift such a weight in 7 cases, and lifted in 3 of them. Petrov raised 6 times out of 13, and Sidorov has a 35% chance of successfully handling the bar. The coach will randomly draw one athlete to join the team.

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

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

Problem 17. There are radio tubes in two boxes. The first box contains 12 lamps, of which 1 is non-standard; in the second 10 lamps, of which 1 is non-standard. A lamp was taken at random from the first box and transferred to the second. Find the probability that the lamp taken out of the second box at random will be non-standard.

Problem 18. A white ball is dropped into an urn containing two balls, after which one ball is taken at random from it. Find the probability that the removed ball will turn out to be white if all possible assumptions about the initial composition of the balls (by color) are equally possible.

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

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

Let their probabilities and the corresponding conditional probabilities be known. Then the probability of the occurrence of the event is equal to:

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

Probably, many have a presentiment of the content of the first example =)

Wherever you spit - everywhere an urn:

Problem 1

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

Solution: consider the event - a black ball will be extracted from a randomly selected urn. This event may or may not occur as a result of the implementation of one of the following hypotheses:
- 1st urn will be selected;
- the 2nd urn will be selected;
- the 3rd urn will be selected.

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

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

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

There are only white balls in the second urn, so if chosen the appearance of the black ball becomes impossible: .

And, finally, in the third urn there are only black balls, which means that the corresponding conditional probability extracting the black ball will be (event valid).

- the probability that a black ball will be removed from a randomly selected urn.

Answer:

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

Task 2

There are 5 rifles of different accuracy in the shooting range. The probabilities of hitting the target for a given shooter are respectively equal to 0.5; 0.55; 0.7; 0.75 and 0.4. What is the probability of hitting a target if the shooter fires one shot with a randomly selected rifle?

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

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

Problem 3

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

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

Consider the event: - The shooter hits the target with a randomly taken rifle.
By condition: .

According to the formula of total probability:

Answer: 0,85

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

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

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

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

Answer: 0,85

The next task for an independent solution:

Problem 4

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

Just in case, let me remind you - to get the values ​​of the probabilities, the percentages must be divided by 100. Be very careful! According to my observations, the conditions of problems for the formula of total probability are often tried to be confused; and I specially selected such an example. I'll tell you a secret - I almost got confused myself =)

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

Bayesian Formula Problems

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

On condition that event already happened, probabilities of hypotheses overrated according to the formulas that received the surname of the English priest Thomas Bayes:

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

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

- this is a priori(estimated before tests) probabilities.

- this is a posteriori(estimated after tests) the probabilities of the same hypotheses, recalculated in connection with "newly discovered circumstances" - taking into account the fact that the event authentically happened.

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

Problem 5

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

First part solutions consists in using the formula for the total probability. In other words, the calculations are carried out under the assumption that the test not produced yet and event "The product turned out to be standard" until it came.

Consider two hypotheses:
- the product taken at random will be from the 1st batch;
- the product taken at random will be from the 2nd batch.

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

Control:

Consider a dependent event: - a random item taken from a warehouse will standard.

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

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

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

Part two. Let the product taken at random from the warehouse turn 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) is the probability that the selected standard product belongs to the 1st batch;

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

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

Answer:

Ivan Vasilievich will help us to understand the meaning of the reassessment of hypotheses, who again changed her profession and became the director of the plant. He knows that today the 1st workshop has shipped 4,000 items to the warehouse, and the 2nd workshop - 6,000 items, and he comes to make sure of this. Suppose all products are of the same type and are in one container. Naturally, Ivan Vasilyevich preliminary calculated that the product, which he would now extract for inspection, with probability will be produced by the 1st workshop and with the probability - by the second. But after the selected item is standard, he exclaims: “What a cool bolt! - it was rather released by the 2nd shop. " Thus, the probability of the second hypothesis is overestimated for the better, and the probability of the first hypothesis is underestimated:. And this reassessment is not unfounded - after all, the 2nd workshop produced not only more products, but also works 2 times better!

Pure subjectivity, you say? In part - yes, moreover, Bayes himself interpreted a posteriori probabilities 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 average estimates. But, philosophically speaking, everything flows, everything changes, including probabilities. It is possible that at the time of research more successful 2nd workshop increased the percentage of standard products (and / or the 1st workshop reduced), and if you check a larger quantity or all 10 thousand items in stock, then the overestimated values ​​will turn out to be much closer to the truth.

By the way, if Ivan Vasilyevich extracts a non-standard part, then on the contrary - he will more "suspect" the 1st shop and less - the second. I propose to make sure of this yourself:

Problem 6

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

The condition is distinguished by two letters, which I have highlighted in bold. The problem can be solved from scratch, or you can use the results of previous calculations. In the sample, I carried out a complete solution, but so that there was no formal overlap with Task No. 5, the event "The product taken at random from the warehouse will be non-standard" denoted by.

The Bayesian scheme for overestimating probabilities is ubiquitous, and various fraudsters also actively exploit it. Consider the three-letter joint-stock company that has become a household name, which attracts deposits from the population, allegedly invests them somewhere, regularly pays dividends, etc. What's happening? Day after day, month after month, and more and more new facts, conveyed through advertising and word of mouth, only increase the level of confidence in the financial pyramid. (Post hoc Bayesian reevaluation due to past events!)... That is, in the eyes of depositors, there is a constant increase in the likelihood that "This is a serious office"; while the probability of the opposite hypothesis is ("These are the next scammers"), of course, decreases and decreases. The rest, I think, is understandable. It is noteworthy that the reputation earned gives the organizers time to successfully hide from Ivan Vasilyevich, who was left not only without a batch of bolts, but also without pants.

We will return to no less curious examples a little later, but for now, the next in line is perhaps the most common case with three hypotheses:

Problem 7

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

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

The number of events increased, and solution it is more convenient to arrange in a “fast” style.

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

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

Likewise: - the likelihood of manufacturing a defective lamp for the respective factories.

According to the formula of total probability:

- the likelihood that the purchased lamp will be defective.

Step two. Suppose the purchased lamp turned out to be defective (the event happened)

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

Answer:

Why did the initial probability of the 2nd hypothesis increase after the reassessment? After all, the second plant produces lamps of average quality (the first is better, the third is worse). So why has it increased a posteriori the probability that the defective lamp is from the 2nd factory? This is no longer due to "reputation", but to size. Since plant number 2 produced the largest number of lamps, it is (at least subjectively) blamed: "Most likely, this defective lamp is from there".

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

Control: , which was required to be verified.

By the way, about underestimated and overestimated estimates:

Problem 8

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

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

Perform calculations and analyze the results of hypothesis reevaluation.

The task is close to reality and is especially plausible for a group of correspondence students, where the teacher practically does not know the abilities of this or that student. In this case, the result can cause quite unexpected consequences. (especially for exams in the 1st semester)... If a poorly prepared student is lucky enough to have a ticket, then the teacher is likely to consider him a good student or even a strong student, which will bring good dividends in the future. (naturally, you need to "raise the bar" and maintain your image)... If a student taught, crammed, repeated for 7 days and 7 nights, but he was simply unlucky, then further events can develop in the worst possible way - with numerous retakes and balancing on the verge of departure.

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

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

Let's consolidate the material with a final industrial example, in which I will talk about the technical subtleties of the solution that have not yet been encountered:

Problem 9

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

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

Solution: in contrast to Problems No. 5-8, here the question is explicitly asked, which is solved using the formula of total probability. But on the other hand, the condition is a little "encrypted", and the school skill of making up the simplest equations will help us to solve this puzzle. It is convenient to take the smallest value for "x":

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

By condition, the first shop produces 4 times more than the third shop, so the share of the 1st shop is.

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

Let's compose and solve the equation:

Thus: - the probabilities that the part taken out of the container was released by the 1st, 2nd and 3rd shops, respectively.

Control: . In addition, it will not be superfluous to look again at the phrase "It is known that the first shop produces products 2 times more than the second shop and 4 times more than the third shop." and make sure that the obtained values ​​of probabilities really correspond to this condition.

Initially, the share of the 1st or the share of the 2nd workshop could be taken for "X" - the probabilities will be the same. But, one way or another, the most difficult section has been passed, and the solution enters the knurled rut:

From the condition we find:
- the likelihood of manufacturing defective parts for the respective workshops.

According to the formula of total probability:
- the probability that a part taken at random from the container will be non-standard.

Question two: what is the probability that the recovered defective part was released by the 3rd workshop? This question assumes that the part has already been retrieved and is found to be defective. We reevaluate the Bayesian hypothesis:
Is the required probability. Quite expected - after all, the third workshop produces not only the smallest fraction of parts, but also leads in quality!

In this case, I had to simplify a four-story fraction, which in problems on Bayes' formulas has to be done quite often. But for this lesson, I somehow accidentally picked up examples in which many calculations can be done without ordinary fractions.

Since the condition does not contain items "a" and "b", then it is better to provide the answer with text comments:

Answer: - the probability that the part removed from the container will be defective; - the probability that the recovered defective part was released by the 3rd workshop.

As you can see, the problems for the total probability formula and Bayesian formulas are quite simple, and, probably, for this reason, they so often try to complicate the condition, which I already mentioned at the beginning of the article.

More examples are in the file with ready-made solutions for F.P.V. and Bayes' formulas, besides, there will probably be those who want to get more deeply acquainted with this topic in other sources. And the topic is really very interesting - which is only worth one Bayes paradox, which substantiates that everyday advice that if a person has been diagnosed with a rare disease, then it makes sense for him to conduct a second or even two repeated independent examinations. It would seem that this is done solely out of despair ... - but no! But let's not talk about sad things.

a) - the likelihood that the student who passed the exam was very well prepared. The objective initial probability turns out to be overestimated, since almost always some "middle peasants" are lucky with questions and they answer very strongly, which gives rise to the erroneous impression of impeccable preparation.

b) - the likelihood that the student who passed the exam was average prepared. The initial probability turns out to be slightly overestimated, since students with an average level of training are usually the majority, in addition, here the teacher will include unsuccessfully answered "excellent students", and occasionally a poorly performing student who was very lucky with a ticket.

v) - the likelihood that the student who passed the exam was poorly prepared. The original probability was overestimated for the worse. Not surprising.

Lesson number 4.

Topic: Formula of total probability. Bayes' formula. Bernoulli's scheme. Polynomial scheme. Hypergeometric diagram.

FORMULA OF TOTAL PROBABILITY

FORMULA BAYES

THEORY

Total probability formula:

Let there be a complete group of inconsistent events:

(, Then the probability of event A can be calculated by the formula

(4.1)

Events are called hypotheses. Hypotheses are put forward regarding the part of the experiment in which there is uncertainty.

, where are the prior probabilities of the hypotheses

Bayes' formula:

Let the experiment be completed and it is known that as a result of the experiment, event A. Then it is possible, taking into account this information overestimate the probabilities of hypotheses:

(4.2)

, where posterior probabilities of hypotheses

SOLUTION OF PROBLEMS

Objective 1.

Condition

In the 3 batches of parts received at the warehouse, suitable parts are 89 %, 92 % and 97 % accordingly. The number of parts in lots refers to both 1:2:3.

What is the probability that a part randomly selected from the warehouse will be defective? Let it be known that a randomly selected part is found to be defective. Find the probabilities that it belongs to the first, second and third parties.

Solution:

Let us denote by A the event that a randomly selected part turns out to be defective.

1st question - to the total probability formula

2nd question - to Bayes' formula

Hypotheses are put forward regarding the part of the experiment in which there is uncertainty. In this problem, the uncertainty lies in which batch the randomly selected part is from.

Let in the first batch a details. Then in the second batch - 2 a parts, and in the third - 3 a details. In total in three parties 6 a details.

(the percentage of marriage on the first line was converted to probability)

(the percentage of marriage on the second line was converted to probability)

(the percentage of marriage on the third line was converted to probability)

Using the formula for the total probability, we calculate the probability of an event A

-answer to 1 question

We calculate the probabilities that the defective part belongs to the first, second and third parties using the Bayes formula:

Objective 2.

Condition:

In the first urn 10 balls: 4 whites and 6 black. In the second urn 20 balls: 2 whites and 18 black. One ball is randomly selected from each urn and put into the third urn. Then one ball is randomly selected from the third urn. Find the probability that the ball taken from the third urn will be white.

Solution:

The answer to the problem question can be obtained using the total probability formula:

The uncertainty lies in which balls hit the third urn. We put forward hypotheses regarding the composition of the balls in the third urn.

H1 = (there are 2 white balls in the third urn)

H2 = (in the third urn there are 2 black balls)

H3 = (in the third urn there is 1 white ball and 1 black ball)

A = (the ball taken from the 3rd urn will be white)

Objective 3.

A white ball was dropped into an urn containing 2 balls of an unknown color. After that, we extract 1 ball from this urn. Find the probability that the ball removed from the urn will be white. The ball extracted from the above-described urn turned out to be white. Find the probabilities that there were 0 white balls, 1 white ball and 2 white balls in the urn before the transfer .

1 question c - on the formula of total probability

2 question–On the Bayes formula

Uncertainty lies in the original composition of the balls in the urn. We put forward the following hypotheses regarding the initial composition of the balls in the urn:

Hi = (the urn wasi-1 white ball),i = 1,2,3

, i = 1,2,3(in a situation of complete uncertainty, the prior probabilities of hypotheses are taken to be the same, since we cannot say that one option is more likely than another)

A = (the ball removed from the urn after transferring will be white)

Let's calculate the conditional probabilities:

Let's make a calculation using the formula of total probability:

Answer to 1 question

To answer the second question, we use the Bayes formula:

(decreased compared to the prior probability)

(unchanged from the prior probability)

(increased compared to the prior probability)

Conclusion from comparison of prior and posterior probabilities of hypotheses: the initial uncertainty has quantitatively changed

Task 4.

Condition:

When transfusing blood, it is necessary to take into account the blood groups of the donor and the patient. To a person who has fourth group blood any blood group can be transfused, man with the second and third group can be poured or blood of his group, or the first. To man with the first blood group blood transfusion possible only the first group. It is known that among the population 33,7 % have first group ny, 37,5 % have the second group, 20.9% have third group and 7.9% have a 4th group. Find the probability that a randomly taken patient can be transfused with the blood of a randomly taken donor.

Solution:

We put forward hypotheses about the blood group of a randomly taken patient:

Hi = (patienti-th blood group),i = 1,2,3,4

(Percentages converted to probabilities)

A = (can be transfused)

By the formula of total probability we get:

That is, transfusion can be carried out in about 60% of cases.

Bernoulli scheme (or binomial scheme)

Bernoulli's Trials - this is independent tests 2 the outcome, which we conditionally call success and failure.

p- success rate

q- the likelihood of failure

Probability of success does not change from experience to experience

The result of the previous test does not affect the following tests.

Performing the tests described above is called the Bernoulli scheme or binomial scheme.

Examples of Bernoulli tests:

Success - coat of arms

Failure- tails

Correct coin case

wrong coin case

p and q do not change from experience to experience, if during the experiment we do not change the coin

Success - drop "6"

Failure - all the rest

Correct dice case

Incorrect die case

p and q do not change from experience to experience, if during the experiment we do not change the dice

Success - hit

Failure - miss

p = 0.1 (shooter hits in one shot out of 10)

p and q do not change from experience to experience, if during the experiment we do not change the arrow

Bernoulli's formula.

Let be held n p. Consider the events

(vn Bernoulli trials with a success ratep will happenm successes),

- there is a standard notation for the probabilities of such events

<-Bernoulli's formula for calculating probabilities (4.3)

Explanation of the formula : the probability that m successes will occur (the probabilities are multiplied, since the tests are independent, and since they are all the same, a degree appears), - the probability that nm failures will occur (the explanation is the same as for successes), - the number of ways of implementation events, that is, in how many ways can m successes be placed on n places.

Consequences of the Bernoulli formula:

Corollary 1:

Let be held n Bernoulli trials with the probability of success p. Consider the events

A (m1,m2) = (number of successes inn Bernoulli tests will be in the range [m1;m2])

(4.4)

Explanation of the formula: Formula (4.4) follows from formula (4.3) and the theorem of addition of probabilities for inconsistent events, since - the sum (union) of inconsistent events, and the probability of each is determined by formula (4.3).

Corollary 2

Let be held n Bernoulli trials with the probability of success p. Consider an event

A = (inn Bernoulli trials will have at least 1 success}

(4.5)

Explanation of the formula: ={ there will be no success in n Bernoulli trials) =

(all n tests will fail)

Problem (on the Bernoulli formula and its consequences) example for problem 1.6-D. h.

Correct coin toss 10 times... Find the probabilities of the following events:

A = (the coat of arms will be drawn exactly 5 times)

B = (the coat of arms will be drawn no more than 5 times)

C = (the coat of arms will be dropped at least 1 time)

Solution:

Let us reformulate the problem in terms of Bernoulli tests:

n = 10 number of tests

success- coat of arms

p = 0.5 - probability of success

q = 1-p = 0.5 - probability of failure

To calculate the probability of event A, we use Bernoulli's formula:

To calculate the probability of event B, we use corollary 1 To Bernoulli's formula:

To calculate the probability of event C, we use corollary 2 To Bernoulli's formula:

Bernoulli's scheme. Calculation by approximate formulas.

APPROXIMATE FORMULA OF MUAVRE-LAPLACE

Local formula

p success and q failure then for everyone m the approximate formula is valid:

, (4.6)

m.

The value of the function can be found in the special table. It contains values ​​only for. But the function is even, i.e.

If, then it is believed

Integral formula

If the number of trials n in the Bernoulli scheme is large and the probabilities are also high p success and q failure, then the approximate formula is valid for all (4.7) :

The meaning of the function can be found in a special table. It contains values ​​only for. But the function is odd, i.e. .

If, then it is believed

APPROXIMATE POISSON FORMULAS

Local formula

Let the number of trials n according to the Bernoulli scheme, it is high, and the probability of success in one test is small, and the work is also small. Then it is determined by the approximate formula:

, (4.8)

The probability that the number of successes in n Bernoulli trials is m.

Function values can be viewed in a special table.

Integral formula

Let the number of trials n according to the Bernoulli scheme, it is high, and the probability of success in one test is small, and the work is also small.

Then determined by the approximate formula:

, (4.9)

The probability that the number of successes in n Bernoulli trials is in the range.

Function values can be viewed in a special table and then summed up by range.

You can see as examples for problems 1.7 and 1.8 D. z.

Calculation by Poisson's formula.

Problem (Poisson formula).

Condition:

The probability of distortion of one character when transmitting a message over a communication line is 0.001. The message is considered accepted if there is no distortion in it. Find the probability that a message consisting of 20 words by 100 characters each.

Solution:

Let us denote by A

-number of characters in the message

success: the character is not distorted

Probability of success

Let's calculate. See recommendations for using approximate formulas ( ) : for calculation you need to apply Poisson's formula

Probabilities for Poisson's formula with respect to andm can be found in a special table.

Condition:

The telephone exchange serves 1000 subscribers. The probability that within a minute some subscriber will need a connection is 0.0007. Calculate the probability that at least 3 calls will arrive at the telephone exchange in a minute.

Solution:

Let us reformulate the problem in terms of the Bernoulli scheme

success: a call arrives

Probability of success

- the range in which the number of successes should lie

A = (at least three calls will be received) -event, the probability of which is required. find in task

(less than three calls will be received) Go to add. event, since its probability is easier to calculate.

(calculation of terms see special table)

Thus,

Problem (local Muvre-Laplace formula)

Condition

Probability of hitting the target with one shot is equal to 0.8. Determine the probability that at 400 shots will occur exactly 300 hits.

Solution:

Let us reformulate the problem in terms of the Bernoulli scheme

n = 400 - number of tests

m = 300 - number of successes

success - hit

(Question of the problem in terms of the Bernoulli scheme)

Advance paynemt:

We carry out independent tests, in each of which we distinguish m options.

p1 - ​​probability of getting the first variant in one test

p2 is the probability of getting the second option in one test

…………..

pm is the probability of gettingm-th variant in one test

p1,p2, …………… ..,pm do not change from experience to experience

The sequence of tests described above is called polynomial scheme.

(for m = 2, the polynomial scheme turns into a binomial scheme), i.e., the binomial scheme described above is a special case of a more general scheme, called a polynomial scheme).

Consider the following events

А (n1, n2,…., Nm) = (in the n tests described above, option 1 appeared n1 times, option 2 appeared n2 times,… .., etc., nm times option m appeared)

Formula for calculating probabilities using a polynomial scheme

Condition

Dice thrown 10 times. It is required to find the probability that "6" will be rolled 2 times, and "5" will be dropped 3 times.

Solution:

Let us denote by A event the probability of which you want to find in the problem.

n = 10 - number of trials

m = 3

Option 1 - Drop 6

p1 = 1/6n1 = 2

Option 2 - Drop 5

p2 = 1/6n2 = 3

Option 3 - drop out of any face, except 5 and 6

p3 = 4/6n3 = 5

P (2,3,5) -? (the probability of the event referred to in the problem statement)

Polynomial Scheme Problem

Condition

Find the probability that among 10 randomly selected people will have four birthdays in the first quarter, three in the second, two in the third and one in the fourth.

Solution:

Let us denote by A event the probability of which you want to find in the problem.

Let us reformulate the problem in terms of a polynomial scheme:

n = 10 - number of trials = number of people

m = 4- the number of options that we distinguish in each trial

Option 1 - birth in the 1st quarter

p1 = 1/4n1 = 4

Option 2 - birth in the 2nd quarter

p2 = 1/4n2 = 3

Option 3 - birth in the 3rd quarter

p3 = 1/4n3 = 2

Option 4 - birth in the 4th quarter

p4 = 1/4n4 = 1

P (4,3,2,1) -? (the probability of the event referred to in the problem statement)

We assume that the probability of being born in any quarter is the same and is equal to 1/4. Let's carry out the calculation using the formula for the polynomial scheme:

Polynomial Scheme Problem

Condition

In the urn 30 balls: welcome back.3 white, 2 green, 4 blue and 1 yellow.

Solution:

Let us denote by A event the probability of which you want to find in the problem.

Let us reformulate the problem in terms of a polynomial scheme:

n = 10 - number of trials = number of balls selected

m = 4- the number of options that we distinguish in each trial

Option 1 - choosing a white ball

p1 = 1/3n1 = 3

Option 2 - the choice of the green ball

p2 = 1/6n2 = 2

Option 3 - choosing a blue ball

p3 = 4/15n3 = 4

Option 4 - choosing a yellow ball

p4 = 7/30n4 = 1

P (3,2,4,1) -? (the probability of the event referred to in the problem statement)

p1,p2, p3,p4 do not change from experience to experience as the choice is made with the return

Let's carry out the calculation using the formula for the polynomial scheme:

Hypergeometric scheme

Let there be n elements of k types:

n1 of the first type

n2 of the second type

nk type k

Of these n elements, randomly no return select m elements

Consider the event A (m1, ..., mk), which consists in the fact that among the selected m elements there will be

m1 of the first type

m2 of the second type

mk type k

The probability of this event is calculated by the formula

P (A (m1, ..., mk)) = (4.11)

Example 1.

The problem for the hypergeometric scheme (sample for the problem 1.9 D. h)

Condition

In the urn 30 balls: 10 white, 5 green, 8 blue and 7 yellow(balls differ only in color). 10 balls are randomly selected from the urn no return. Find the probability that among the selected balls there will be: 3 white, 2 green, 4 blue and 1 yellow.

We haven = 30,k = 4,

n1 = 10,n2 = 5,n3 = 8,n4 = 7,

m1 = 3,m2 = 2,m3 = 4,m4 = 1

P (A (3,2,4,1)) = = you can count to a number knowing the formula for combinations

Example 2.

An example of calculation according to this scheme: see calculations for the game Sportloto (topic 1)

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

This formula is called total probability formula .

Consider again the full group of incompatible events, the probabilities of which are ... Event A can happen only together with any of the events, which we will call hypotheses ... Then by the formula of total probability

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

By the probability multiplication theorem

.

Similarly, for other hypotheses

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

Example. The store has received new products from three enterprises. The percentage of these products is as follows: 20% - products of the first company, 30% - products of the second company, 50% - products of the third company; further, 10% of the products of the first enterprise are of the highest quality, at the second enterprise - 5% and at the third - 20% of the products of the highest quality. Find the likelihood that a new product accidentally purchased is of the highest grade.

Solution. Let us denote by V an event consisting in the fact that a premium product will be purchased, through we denote the events consisting in the purchase of products belonging to the first, second and third enterprises, respectively.

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

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

Example. One of the three shooters is called into 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.

Solution. Three hypotheses are possible:

The first shooter is called to the line of fire,

A second shooter has been called into the line of fire

A third shooter has been called into the line of fire.

Since the call to the line of fire of any shooter 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 come after processing to a common conveyor. The first machine gives 2% of scrap, the second - 7%, the third - 10%. The productivity of the first machine is 3 times more than the productivity of the second, and the third is 2 times less than the second.

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

b) What are the proportions of parts of each machine tool among the defective parts on the conveyor?

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

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

Dependencies between machine productivity mean the following:

And since hypotheses form a complete group, then.

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

a) The full probability that the part taken at random from the conveyor is defective:

In other words, in the mass of parts coming off the assembly line, the scrap is 4%.

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

The goal is to acquire practical skills in solving problems on

sections of probability theory

Preparation for the practical assignment

To get acquainted with the theoretical material on this topic, to study the theoretical content, as well as the corresponding sections in literary sources

The order of the assignment

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

Source data options

Table 1

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 incidents: a) experience - coin toss; developments: A1- the appearance of the coat of arms; A2- the appearance of a number; b) experience - throwing two coins; developments: IN 1- the appearance of two coats of arms; IN 2 - the appearance of two numbers; AT 3- the appearance of one coat of arms and one number; c) experience - throwing a dice; developments: 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; developments: D1- hit; D2 - miss; e) experience - two shots at the target; developments: E0- not a single hit; E1- one hit; E2- two hits; f) experience - removing two cards from the deck; developments: F1 - the appearance of two red cards; F2- the appearance of two black cards?

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

4. In urn A whites and B black balls. They took one ball out of the urn and put it aside without looking. After that, another ball was taken from the urn. It turned out to be white. Find the probability that the first ball set 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 remaining in the urn will be white.

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

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

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

9. In a game consisting of X products available I defective. Selected from batch for inspection I products. Find the probability R the fact that of them exactly J items will be defective.

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

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

12. Throw two dice at the same time. Find the probabilities of the following events: A- the sum of the dropped out points is equal to 8; V- the product of the dropped out points is equal to 8; WITH- the sum of the dropped points is greater than their product.

13. Two coins are thrown. Which of the events is more likely: A - coins will fall on the same sides; V - coins will fall 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: A- balls of the same color; V - balls of different colors?

15. Three players play cards. Each of them was dealt 10 cards and two cards were left in the draw. One of the players sees that he has 6 cards of diamonds suit and 4 - not of diamonds suit. He discards two of these four cards and takes the buy for himself. Find the probability that he will buy two diamonds.

16. From an urn containing NS of the renumbered balls, take out all balls in it at random one by one. Find the probability that the numbers of the balls taken out will go in order: 1, 2, ..., NS.

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

18. A complete deck of cards (52 sheets) is divided at random into two equal packs of 26 sheets. Find the probabilities of the following events: A - each of the packs will contain two aces; V- in one of the packs there will be no aces, and in the other - all four; From one of the packs will have one ace, and the other will have three.

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

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

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

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

22. The same question as in Problem 21, but the first card after being taken out 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. From the urn, take out one by one all the balls in it and write down their colors. Find the probability that white appears before black in this list.

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

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

26. There are seven nests in the revolver drum, of which five contain cartridges, and two are left empty. The drum is set in rotation, as a result of which one of the slots is randomly opposite the barrel. After that, the trigger is pressed; if the cell was empty, no shot is fired. Find the probability R the fact that, having repeated this 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. The urn contains A; balls marked with numbers 1, 2, ..., To From the urn I one ball at a time is removed (I<к), the number of the ball is recorded and the ball is put back in the urn. Find the probability R that all recorded numbers will be different.

29. The word "book" is composed of five letters of the cut alphabet. A child who cannot read scattered these letters and then assembled them in random order. Find the probability R that he got the word "book" again.

30. The word "pineapple" is composed of the letters of the cut alphabet. A child who cannot read scattered these letters and then assembled them in random order. Find the probability R the fact that he again has the word "pineapple

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

32. N a person is randomly seated at a round table (N> 2). Find the probability R that two fixed faces A and V will be near.

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

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

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

36. Battery from M guns firing at a group consisting of N goals (M< N). The guns select their targets sequentially, randomly, provided that no two guns can shoot at the same target. Find the probability R that targets with numbers 1, 2, ..., M.

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

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

39. Four balls are randomly scattered over four holes; each ball hits one or the other hole with the same probability and independently of the others (there are no obstacles for several balls to hit 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 the same bus with him. 5 buses leave 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. The bank's IT department employs 3 analysts, 10 programmers and 20 engineers. For overtime on a public holiday, the head of the department must assign one employee. How many ways can this be done?

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

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

44. One of the belligerents captured 12 and the other 15 prisoners. 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. 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 Petya and Masha can exchange 2 action films and 7 comedies.

48. One of the belligerents captured 15 and the other 16 prisoners. 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 (only those whose spelling coincides with the Latin ones - A, B, E, K, M, H, O, P, C, T, U, X )?

50. One of the belligerents captured 14 and the other 17 prisoners. How many ways can 6 prisoners of war be exchanged?

51. How many different words can you make by rearranging the letters in the word "mom"?

52. There are 3 red and 7 green apples in the 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 the basket. They took one green apple out of it and put it aside. Then take out another 1 apple from the basket. What is the likelihood that this apple will be green?

54. In a batch of 1000 items, 4 have defects. A batch of 100 items is selected for control. What is the probability of LLP that there will be no defective ones in the control batch?

56. In the 1980s, the game "sport lotto 5 out of 36" was popular in the USSR. The player marked 5 numbers from 1 to 36 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player has not guessed a single number.

57. In the 1980s, the game "sport lotto 5 out of 36" was popular in the USSR. The player marked 5 numbers from 1 to 36 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guesses one number.

58. In the 1980s, the game "sport lotto 5 out of 36" was popular in the USSR. The player marked 5 numbers from 1 to 36 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player has guessed 3 numbers.

59. In the 1980s, the game "sport lotto 5 out of 36" was popular in the USSR. The player marked 5 numbers from 1 to 36 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player has not guessed all 5 numbers.

60. In the 1980s, the game "sport lotto 6 out of 49" was popular in the USSR. The player marked 6 numbers from 1 to 49 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player guesses 2 numbers.

61. In the 1980s, the game "sport lotto 6 out of 49" was popular in the USSR. The player marked 6 numbers from 1 to 49 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player has not guessed a single number.

62. In the 1980s, the game "sport lotto 6 out of 49" was popular in the USSR. The player marked 6 numbers from 1 to 49 on the card and received prizes of various denominations if he guessed a different number of numbers announced by the drawing commission. Find the probability that the player has guessed all 6 numbers.

63. In a batch of 1000 items, 4 have defects. A batch of 100 items is selected for control. What is the probability of a limited liability partnership that there will be only 1 defective batch in the control batch?

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

65. How many different words can you make by rearranging the letters in the word "pineapple"?

66. 6 people entered the elevator, and the hostel has 7 floors. What is the probability that all 6 people will 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, a wire break occurred in the section between km 40 and 79 of the power transmission line. Assuming that the cliff is equally possible at any point, find the probability that the cliff occurred between the 40th and 45th kilometers.

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

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

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

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

73. Petya and Masha agreed to meet at the monument to Pushkin from 12:00 to 13:00, but no one could indicate the exact time of arrival. They agreed to wait 15 minutes for each other. What is the likelihood of their meeting?

74. The fishermen caught 120 fish in the pond, of which 10 were ringed. What is the chance of catching a ringed fish?

75. Take out all the apples in turn from the basket containing 3 red and 7 green apples. what is the probability that the 2nd apple turns out to be red?

76. Take out all the apples in turn from the basket containing 3 red and 7 green apples. what is the likelihood that the last apple will be green?

77. Students consider 10 out of 50 tickets to be “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 they both got a “good” ticket?

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

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

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

82. Statistics of requests for loans in the bank is as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of non-repayment of loans is 0.01, 0.05 and 0.2, respectively. What proportion of loans are not refunded?

83. The probability that the weekly turnover of the ice cream merchant will exceed 2000 rubles. is 80% in clear weather, 50% in variable cloudiness 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 variable cloudiness and rains - 40% each.

84. In urn A white (b) and B black (h) balls. Two balls are taken out of the urn (simultaneously or sequentially). Find the probability that both balls will be 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 removed from the urn, its color is noted 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. Includes a box of nine new tennis balls. Three balls are taken for the game; after the game they are put back. When choosing balls, played and unplayed are not distinguished. What is the probability that after three games there will be no balls left 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 wear galoshes in the dark. Each of them can distinguish the right galosh from the left, but cannot distinguish his own from another. 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 they come across.

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

93. Two shooters, independently of one another, fire two shots (each at its own target). Probability of hitting a target in one shot for the first shooter p1 for the second p2. The winner of the competition is the shooter with the most holes in the target. Find the probability Px what the first shooter will win.

94. behind a space object, the object is detected with a probability R. The detection of an object in each cycle occurs independently of the others. Find the probability that for NS cycles the object will be detected.

95. 32 letters of the Russian alphabet are written on the cards of the split alphabet. Five cards are taken out at random one after the other and placed on the table in the order of appearance. Find the probability that you get the word "end".

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

97. Incendiary rounds are fired at the aircraft. The fuel on the aircraft is concentrated in four tanks located in the fuselage one after the other. The areas of the tanks are the same. In order to ignite the plane, it is enough to hit with two shells either into the same tank or into adjacent tanks. It is known that two shells hit the area of ​​the tanks. 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 different suits.

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

100. When the ignition is turned on, the engine starts to run 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 - 20%. The likelihood of failure of the device over time t in normal mode it is 0.1; in the abnormal - 0.7. Find the full 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 marriage rate is 5, 6 and 8 percent, respectively. What is the likelihood that the purchased defective product came from a second supplier.

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

104. The flow of cars past the gas station consists of 60% trucks and 40% cars. What is the probability of finding a truck at a gas station if the probability of refueling it is 0.1, and that of a passenger 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 marriage rate 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 the cards of the split alphabet. Five cards are taken out at random one after the other and placed on the table in the order of appearance. Find the probability that you get 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 marriage rate 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 in four cells located one after the other in a straight line. Each ball with the same probability of 1/4 hits each cell. Find the probability that 2 balls will fall into one 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 when the ignition is switched on a second time;

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

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

112. In urn A whites and B black balls. One ball is removed from the urn, its color is noted and the ball is returned to the urn. After that, another ball is taken from the urn. Find the probability that both balls taken out will be 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 in four cells located one after the other in a straight line. Each ball with the same probability of 1/4 hits each cell. Find the probability that the balls will fall into adjacent cells.

115. Masha came to the exam knowing the answers to 20 of the 25 questions of the program. 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. What is the likelihood that they both got a “good” ticket?

117. Statistics of requests for loans in the bank is as follows: 10% - state. authorities, 20% - other banks, the rest - individuals. The probability of non-repayment of loans is 0.01, 0.05 and 0.2, respectively. What proportion of loans are not refunded?

118. 32 letters of the Russian alphabet are written on the cards of the split alphabet. Five cards are taken out at random one after the other and placed on the table in the order of appearance. Find the probability that you get the word "end".

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

120. The probability that the weekly turnover of the ice cream merchant will exceed 2000 rubles. is 80% in clear weather, 50% in variable cloudiness 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 variable cloudiness and rains - 40% each.