Lesson number 4.

Topic: Total probability formula. Bayes formula. Bernoulli scheme. Polynomial scheme. Hypergeometric scheme.

TOTAL PROBABILITY FORMULA

BAYES FORMULA

THEORY

Total Probability Formula:

Let there be a complete group of incompatible events :

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

(4.1)

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

, where are the a priori probabilities of the hypotheses

Bayes formula:

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

(4.2)

, Where posterior probabilities of hypotheses

PROBLEM SOLVING

Task 1.

Condition

In 3 batches of parts received at the warehouse, the good ones are 89 %, 92 % And 97 % respectively. The number of parts in batches is referred to as 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 turned out to be defective. Find the probabilities that it belongs to the first, second and third parties.

Solution:

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 the Bayes formula

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

Let in the first game A details. Then in the second game - 2 a details, and in the third - 3 a details. Only three games 6 a details.

(the percentage of marriage on the first line was translated into probability)

(the percentage of marriage on the second line was translated into probability)

(percentage of marriage in the third line converted to probability)

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

-answer to 1 question

The probabilities that the defective part belongs to the first, second and third batches are calculated using the Bayes formula:

Task 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 placed in the third urn. Then one ball is randomly selected from the third urn. Find the probability that the ball drawn from the third urn is white.

Solution:

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

The uncertainty lies in which balls ended up in 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=(there are 2 black balls in the third urn)

H3=( third urn contains 1 white ball and 1 black ball)

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

Task 3.

A white ball is dropped into an urn containing 2 balls of unknown color. After that, we extract 1 ball from this urn. Find the probability that the ball drawn from the urn is white. The ball taken from the urn described above 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 - to the total probability formula

2 question-on the Bayes formula

The uncertainty lies in the initial composition of the balls in the urn. Regarding the initial composition of the balls in the urn, we put forward the following hypotheses:

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

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

A = (the ball drawn from the urn after the transfer will be white)

Let's calculate the conditional probabilities:

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

Answer to 1 question

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

(decreased compared to prior probability)

(unchanged from prior probability)

(increased compared to prior probability)

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

Task 4.

Condition:

When transfusing blood, it is necessary to take into account the blood groups of the donor and the patient. The person who has fourth group blood any type of blood can be transfused, to a person with the second and third group can be poured or the blood of his group, or first. to a person with the first blood group you can transfuse blood only the first group. It is known that among the population 33,7 % have first group pu, 37,5 % have second group, 20.9% have third group And 7.9% have the 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=(in a patienti-th blood group),i=1,2,3,4

(Percentages converted to probabilities)

A=(can be transfused)

According to the total probability formula, we get:

i.e. transfusion can be performed in about 60% of cases

Bernoulli scheme (or binomial scheme)

Bernoulli trials - This independent tests 2 outcome, which we conditionally call success and failure.

p- success rate

q-probability of failure

Probability of Success does not change from experience to experience

The result of the previous test does not affect the next test.

Carrying out the tests described above is called the Bernoulli scheme or the binomial scheme.

Examples of Bernoulli tests:

Coin toss

Success - coat of arms

Failure- tails

The case of the correct coin

wrong coin case

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

Throwing a dice

Success - roll "6"

Failure - all the rest

The case of a regular dice

Case of wrong dice

p And q do not change from experience to experience, if in the process of conducting the experiment we do not change the dice

Shooting arrow at target

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 in the process of conducting the experiment we do not change the arrow

Bernoulli formula.

Let held n p. Consider events

(Vn Bernoulli trials with probability of successp 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 : probability that there will be m successes (the probabilities are multiplied, since the trials are independent, and since they are all the same, a degree appears), - the probability that n-m failures will occur (the explanation is similar as for successes), - the number of ways to implement event, that is, in how many ways can m successes be placed in n places.

Consequences of the Bernoulli formula:

Corollary 1:

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

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

(4.4)

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

Consequence 2

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

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

(4.5)

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

(all n trials will fail)

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

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

A=(coat of arms will drop exactly 5 times)

B=(coat of arms will drop no more than 5 times)

C=(coat of arms will drop at least once)

Solution:

Let us reformulate the problem in terms of Bernoulli tests:

n=10 number of trials

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 formula:

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

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

Bernoulli scheme. Calculation by approximate formulas.

APPROXIMATE FORMULA OF MOIAVRE-LAPLACE

Local Formula

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

, (4.6)

m.

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

If , then suppose

integral formula

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

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

If , then suppose

APPROXIMATE POISSON FORMULA

Local Formula

Let the number of trials n according to the Bernoulli scheme is large, and the probability of success in one test is small, and the product 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 is large, and the probability of success in one test is small, and the product 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 over the range.

Formula	Poisson formula	Moivre-Laplace formula	Quality estimates
			estimates are rough
10			used for rough estimates calculations
			used for applied engineering calculations
100 0			used for any engineering calculations
n>1000			Very good quality ratings

You can look at the quality of examples for tasks 1.7 and 1.8 D. z.

Calculation by the Poisson formula.

Problem (Poisson's formula).

Condition:

The probability of distortion of one symbol when transmitting a message over a communication line is equal to 0.001. The message is considered accepted if there are no distortions in it. Find the probability of receiving a message consisting of 20 words 100 each characters each.

Solution:

Denote by A

-number of characters in the message

success: character is not distorted

Probability of Success

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

Probabilities for the Poisson 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 any 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:

Reformulate the problem in terms of the Bernoulli scheme

success: call received

Probability of Success

– the range within which the number of successes must lie

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

(Less than three calls will arrive) We proceed to the additional. event, since its probability is easier to calculate.

(calculation of terms see special table)

Thus,

Problem (local Mouvre-Laplace formula)

Condition

Probability of hitting the target with one shot equals 0.8. Determine the probability that at 400 shots will happen exactly 300 hits.

Solution:

Reformulate the problem in terms of the Bernoulli scheme

n=400 – number of trials

m=300 – number of successes

success - hit

(Problem question in terms of the Bernoulli scheme)

Advance paynemt:

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

p1 - ​​the probability of getting the first option in one trial

p2 - the probability of getting the second option in one trial

…………..

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

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

The sequence of tests described above is called polynomial scheme.

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

Consider the following events

А(n1,n2,….,nm)=( in n trials described above, variant 1 appeared n1 times, variant 2 appeared n2 times, ….., etc., nm times variant m appeared)

Formula for calculating probabilities using a polynomial scheme

Condition

dice throw 10 times. It is required to find the probability that "6" will fall out 2 times, and "5" will fall out 3 times.

Solution:

Denote by A the event whose probability is to be found in the problem.

n=10 - number of trials

m=3

1 option - drop 6

p1=1/6n1=2

Option 2 - Drop 5

p2=1/6n2=3

Option 3 - Drop any face except 5 and 6

p3=4/6n3=5

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

Problem for the polynomial circuit

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:

Denote by A the event whose probability is to be found in the problem.

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

n=10 - number of trials = number of people

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

Option 1 - birth in 1 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)-? (probability of the event referred to in the condition of the problem)

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

Problem for the polynomial circuit

Condition

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

Solution:

Denote by A the event whose probability is to be found in the problem.

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

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

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

Option 1 - choose a white ball

p1=1/3n1=3

Option 2 - choose the green ball

p2=1/6n2=2

3rd option - choice of the blue ball

p3=4/15n3=4

Option 4 - choose the yellow ball

p4=7/30n4=1

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

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

Let's carry out the calculation according to 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

From these n elements randomly no return choose 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 k-th type

The probability of this event is calculated by the formula

P(A(m1,…,mk))= (4.11)

Example 1

Problem for a hypergeometric scheme (sample for 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))= = can be counted to a number knowing the formula for combinations

Example 2

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

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

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

Probably, many anticipate the content of the first example =)

Wherever you spit - everywhere the urn:

Task 1

There are three identical urns. The first urn contains 4 white and 7 black balls, the second urn contains only white balls, and the third urn contains 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 drawn from a randomly selected urn. This event may or may not occur as a result of one of the following hypotheses:
– the 1st urn will be selected;
– the 2nd urn will be chosen;
– the 3rd urn will be chosen.

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

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

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

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

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

is the probability that a black ball will be drawn from a randomly selected urn.

Answer:

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

Task 2

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

Short solution and answer at the end of the lesson.

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

Task 3

There are 5 rifles in the pyramid, three of which are equipped with an optical 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 rifle taken at random.

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

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

According to the total probability formula:

Answer: 0,85

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

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

By condition, – probabilities of hitting the target with the respective types of rifles.

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

Answer: 0,85

Next task for independent decision:

Task 4

The engine operates in three modes: normal, forced and idling. In idle mode, the probability of its failure is 0.05, in normal mode - 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 probability of engine failure during operation?

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

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

Problems for Bayes formulas

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

Given that that event already happened, probabilities of hypotheses overestimated according to the formulas that received the name of the English priest Thomas Bayes:

- the probability that the hypothesis took place;
- the probability that the hypothesis took place;
…
is the probability that the hypothesis was true.

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

- This a priori(estimated before tests) probabilities.

- This 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 happened.

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

Task 5

The warehouse received 2 batches of 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%. Randomly taken from the warehouse, the product 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 total probability formula. In other words, the calculations are carried out under the assumption that the test not yet produced and event "the product turned out to be standard" until it comes.

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

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

Control:

Consider the dependent event: – an item taken at random from the warehouse will standard.

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

Similarly, in the second batch 100% - 10% = 90% standard products and is the probability that a randomly selected item in the warehouse will be a standard item given that that it belongs to the 2nd party.

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

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

According to Bayes' formulas:

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

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

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

Answer:

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

You say, pure subjectivism? Partly - 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 probability that the product will be standard (0.8 and 0.9 for the 1st and 2nd shops, respectively) This preliminary(a priori) and medium estimates. But, speaking philosophically, everything flows, everything changes, including probabilities. It is quite possible that at the time of the study more successful 2nd shop increased the percentage of standard products (and/or the 1st shop reduced), and if you check more or all 10 thousand items in stock, then the overestimated values ​​will be much closer to the truth.

By the way, if Ivan Vasilyevich extracts a non-standard part, then vice versa - he will “suspect” the 1st shop more and less - the second. I suggest you check it out for yourself:

Task 6

The warehouse received 2 batches of 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%. A 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 will be 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 in order to avoid a formal overlay with Task No. 5, the event “A product taken at random from the warehouse will be non-standard” marked with .

The Bayesian scheme of re-evaluation of probabilities is found everywhere, and it is also actively exploited by various kinds of scammers. Consider a 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 trust in financial pyramid (posterior Bayesian re-evaluation 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 (“these are regular scammers”), of course, decreases and decreases. The rest, I think, is clear. It is noteworthy that the earned reputation gives the organizers time to successfully hide from Ivan Vasilyevich, who was left not only without a batch of bolts, but also without pants.

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

Task 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 lamp I bought was defective. What is the probability that it was produced by plant 2?

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

Events have 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 will will be defective.

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

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

According to the total probability formula:

- the probability that the purchased lamp will be defective.

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

According to the Bayes formula:
- the probability that the purchased defective lamp is manufactured by the second factory

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 one is better, the third one is worse). So why did it increase 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 No. 2 produced the largest number of lamps, they blame it (at least subjectively): “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 to be verified.

By the way, about underestimated and overestimated:

Task 8

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

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

Perform calculations and analyze the results of reevaluation of hypotheses.

The task is close to reality and is especially plausible for a group of part-time students, where the teacher practically does not know the abilities of this or that student. In this case, the result can cause rather unexpected consequences. (especially for exams in the 1st semester). If an ill-prepared student is lucky enough to get a ticket, then the teacher is likely to consider him a good student or even a strong student, which will bring good dividends in the future (of course, you need to “raise the bar” and maintain your image). If a student studied, crammed, 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:

Task 9

Three workshops of the plant produce parts of the same type, which are assembled in a common container for assembly. 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 workshop, the defect is 12%, in the second - 8%, in the third - 4%. For control, one part is taken from the container. What is the probability that it will be defective? What is the probability that the extracted defective part was produced by the 3rd shop?

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

Solution: in contrast to Tasks No. 5-8, a question is explicitly asked here, which is resolved using the total probability formula. But on the other hand, the condition is a little “encrypted”, and the school skill to compose the simplest equations will help us solve this rebus. For "x" it is convenient to take the smallest value:

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

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

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

Let's make and solve the equation:

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

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

For "X" it was initially possible to take the share of the 1st or the share of the 2nd shop - the probabilities will come out the same. But, one way or another, the most difficult section has been passed, and the solution is on track:

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

According to the total probability formula:
is the probability that a part randomly extracted from the container will be non-standard.

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

In this case, I had to simplify the 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 there are no “a” and “be” points in the condition, 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 extracted defective part was released by the 3rd workshop.

As you can see, the problems on the total probability formula and Bayes 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.

Additional examples are in the file with ready-made solutions for F.P.V. and Bayes formulas, in addition, there are probably those who wish to become more deeply acquainted with this topic in other sources. And the topic is really very interesting - what is it worth alone bayes paradox, which justifies the everyday advice that if a person is diagnosed with a rare disease, then it makes sense for him to conduct a second and even two repeated independent examinations. It would seem that they do it solely out of desperation ... - but no! But let's not talk about sad things.

is the probability that a randomly selected student will pass the exam.
Let the student pass the exam. According to Bayes' formulas:
A) - the probability that the student who passed the exam was prepared very well. The objective initial probability is overestimated, because almost always some "average" are lucky with questions and they answer very strongly, which gives the erroneous impression of impeccable preparation.
b) is the probability that the student who passed the exam was moderately prepared. The initial probability turns out to be slightly overestimated, because students with an average level of preparation are usually the majority, in addition, the teacher will include unsuccessfully answered “excellent students” here, and occasionally a poorly performing student who was very lucky with a ticket.
V) - the probability that the student who passed the exam was poorly prepared. The initial probability was overestimated for the worse. Not surprising.
Examination:
Answer :

Events form full group, if at least one of them will necessarily occur as a result of the experiment and are pairwise inconsistent.

Let's assume that the event A can only occur together with one of several pairwise incompatible events that form a complete group. Let's call the events i= 1, 2,…, n) hypotheses additional experience (a priori). The probability of occurrence of event A is determined by the formula full probability :

Example 16 There are three urns. The first urn contains 5 white and 3 black balls, the second urn contains 4 white and 4 black balls, and the third urn contains 8 white balls. One of the urns is chosen at random (this may mean, for example, that a selection is made from an auxiliary urn containing three balls numbered 1, 2 and 3). A ball is drawn at random from this urn. What is the probability that it will be black?

Solution. Event A– black ball is drawn. If it were known from which urn the ball is drawn, then the required probability could be calculated according to the classical definition of probability. Let us introduce assumptions (hypotheses) regarding which urn is chosen to extract the ball.

The ball can be drawn either from the first urn (hypothesis ), or from the second (hypothesis ), or from the third (hypothesis ). Since there are equal chances to choose any of the urns, then .

Hence it follows that

Example 17. Electric lamps are manufactured at three factories. The first plant produces 30% of the total number of electric lamps, the second - 25%,
and the third for the rest. The products of the first plant contain 1% of defective electric lamps, the second - 1.5%, the third - 2%. The store receives products from all three factories. What is the probability that a store-bought lamp is defective?

Solution. Assumptions must be entered as to which factory the light bulb was manufactured in. Knowing this, we can find the probability that it is defective. Let's introduce notation for events: A– the purchased electric lamp turned out to be defective, – the lamp was manufactured by the first factory, – the lamp was manufactured by the second factory,
– the lamp is manufactured by the third factory.

The desired probability is found by the total probability formula:

Bayes formula. Let be a complete group of pairwise incompatible events (hypotheses). A– random event. Then,

The last formula that allows you to overestimate the probabilities of hypotheses after the result of the test becomes known, as a result of which the event A appeared, is called Bayes formula .

Example 18. An average of 50% of patients with the disease are admitted to a specialized hospital TO, 30% with disease L, 20 % –
with disease M. The probability of a complete cure of the disease K equals 0.7 for diseases L And M these probabilities are respectively 0.8 and 0.9. The patient admitted to the hospital was discharged healthy. Find the probability that this patient had the disease K.

Solution. We introduce hypotheses: - the patient suffered from a disease TO L, the patient suffered from the disease M.

Then, by the condition of the problem, we have . Let's introduce an event A The patient admitted to the hospital was discharged healthy. By condition

According to the total probability formula, we get:

Bayes formula.

Example 19. Let there be five balls in the urn and all assumptions about the number of white balls are equally probable. A ball is taken at random from the urn and it turns out to be white. What is the most likely assumption about the initial composition of the urn?

Solution. Let be the hypothesis that the urn contains white balls , i.e., it is possible to make six assumptions. Then, by the condition of the problem, we have .

Let's introduce an event A A randomly drawn white ball. Let's calculate . Since , then according to the Bayes formula we have:

Thus, the hypothesis is the most probable, since .

Example 20. Two out of three independently operating elements of the computing device failed. Find the probability that the first and second elements failed if the failure probabilities of the first, second and third elements are respectively equal to 0.2; 0.4 and 0.3.

Solution. Denote by A event - two elements failed. The following hypotheses can be made:

- the first and second elements failed, and the third element is serviceable. Since the elements work independently, the multiplication theorem applies: .

Since, under hypotheses, the event A reliably, then the corresponding conditional probabilities are equal to one: .

According to the total probability formula:

According to the Bayes formula, the desired probability that the first and second elements failed.

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

or, using the total probability formula, we get

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

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

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

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

H 1 - a man is selected,

H 2 - a woman is selected.

These events can be chosen as hypotheses.

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

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

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

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

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

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

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

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

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

Probability of drawing a white ball from the second box

Probability of drawing a white ball from the third box

We find the desired probability using the Bayes formula:

Repetition of tests. Bernoulli formula.

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

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

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

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

Bernoulli formula.

The following formulas are obvious:

P n (m less k times in n trials.

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

If the event A can only happen when one of the events that form 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 complete group of incompatible events , whose probabilities of occurrence are . Event A can only occur together with any of the events that we will call hypotheses . Then according to the total probability formula

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

According to the probability multiplication theorem

.

Similarly, for other hypotheses

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

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

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

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

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

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

Solution. Three hypotheses are possible:

The first shooter is called to the line of fire,

The second shooter is called to the line of fire,

A third shooter was called to the line of fire.

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

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

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

Example. On three automatic machines, parts of the same type are processed, which arrive after processing on a common conveyor. The first machine gives 2% rejects, the second - 7%, the third - 10%. The productivity of the first machine is 3 times greater than the productivity of the second, and the third is 2 times less than the second.

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

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

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

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

The dependencies between machine performances mean the following:

And since the hypotheses form a complete group, then .

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

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

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

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

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

Practical tasks

Exercise 1

Solving problems in the main sections of probability theory

The goal is to gain practical skills in solving problems on

sections of probability theory

Preparation for the practical task

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

Task execution order

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

Initial data options

Table 1

	task number

The composition of the report for task 1

5 solved problems according to the variant number.

Tasks for independent solution

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

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

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

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

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

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

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

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

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

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

11. A die is thrown twice. Find Probability R that the same number of points will appear both times.

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

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

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

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

16. From an urn containing P numbered balls, randomly take out one by one all the balls in it. Find the probability that the numbers of the drawn balls will be in order: 1, 2,..., P.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

36. Battery out M guns firing at a group consisting of N goals (M< N). The guns choose their targets sequentially, randomly, provided that no two guns can fire at the same target. Find Probability R the fact that targets with numbers 1, 2, ..., will be fired upon M.

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

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

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

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

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

42. The head of the security service of the bank must daily place 10 guards in 10 posts. In how many ways can this be done?

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

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

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

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

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

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

49. How many cars can be registered in 1 city if the number has 3 digits and 3 letters )?

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

75. From a basket containing 3 red and 7 green apples, take out all the apples in turn. What is the probability that the 2nd apple is red?

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

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

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

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

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

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

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

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

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

85. In urn A whites and B

86. In urn A whites and B

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

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

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

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

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

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

93. Two shooters, independently of one another, fire two shots (each at their own target). Probability of hitting the target with one shot for the first shooter p1 for the second p2. The winner of the competition is the shooter, in the target of which there will be more holes. Find Probability Rx what the first shooter wins.

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

95. 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the word "end" will be obtained.

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

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

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

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

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

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

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

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

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

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

106. 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability of getting the word "book".

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

108. Two balls are scattered randomly and independently of each other over four cells located one after the other in a straight line. Each ball with the same probability 1/4 hits each cell. Find the probability that 2 balls fall into the same cell

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

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

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

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

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

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

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

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

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

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

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

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