When is the Bernoulli formula used? Numerical characteristics of a random variable distributed according to the binomial law

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1. Bogolyubov A.N. Mathematics. Mechanics: a biographical guide. - Kyiv: Naukova Dumka, 1983.

2. Gulay T.A., Dolgopolova A.F., Litvin D.B. Analysis and evaluation of the priority of sections of mathematical disciplines studied by students of economic specialties of agricultural universities // Bulletin of the APK of Stavropol. - 2013. - No. 1 (9). - P. 6-10.

3. Dolgopolova A.F., Gulay T.A., Litvin D.B. Prospects for the use of mathematical methods in economic research // Agrarian science, creativity, growth. - 2013. - S. 255-257.

In mathematics, quite often there are problems in which there are a large number of repetitions of the same condition, test or experiment. The result of each test will be considered a completely different result from the previous one. Dependence in the results will also not be observed. As a test result, several possibilities of elementary consequences can be distinguished: the occurrence of an event (A) or the occurrence of an event that complements A.

Then let's try to assume that the probability of occurrence of the event Р(А) is regular and equals р (0<р<1).

Examples of such a challenge can be a large number of tasks, such as tossing a coin, extracting black and white balls from a dark bag, or giving birth to black and white rabbits.

Such an experiment is called a repeated independent test configuration or a Bernoulli scheme.

Jacob Bernoulli was born into a family of a pharmacist. The father tried to instruct his son on the medical path, but J. Bernoulli became interested in mathematics on his own, and later it became his profession. He owns various trophies in works on topics in the theory of probability and numbers, series and differential calculus. Having studied the theory of probability from one of Huygens's works "On Calculations in Gambling", Jacob became interested in this. In this book, there was not even a clear definition of the concept of "probability". It was J. Bernoulli who introduced most of the modern concepts of probability theory into mathematics. Bernoulli was also the first to express his version of the law of large numbers. The name of Jacob is carried by various works, theorems and schemes: "Bernoulli numbers", "Bernoulli polynomial", "Bernoulli differential equation", "Bernoulli distribution" and "Bernoulli equation".

Let's get back to repetition. As already mentioned above, as a result of various tests, two outcomes are possible: either event A will appear, or the opposite of this event. The Bernoulli scheme itself denotes the production of the n-th number of typical free experiments, and in each of these experiments the event A we need may appear (the probability of this event is known: P (A) \u003d p), the probability of the opposite event to event A is indicated by q \u003d P ( A)=1-p. It is required to determine the probability that, when testing an unknown number, event A will occur exactly k times.

It is important to remember the main condition when solving problems using the Bernoulli scheme is constancy. Without it, the scheme loses all meaning.

This scheme can be used to solve problems of various levels of complexity: from simple (the same coin) to complex (interest). However, more often the Bernoulli scheme is used in solving such problems that are associated with the control of the properties of various products and confidence in a variety of mechanisms. Only to solve the problem, before starting work, all conditions and values ​​\u200b\u200bmust be known in advance.

Not all problems in probability theory are reduced to constancy under conditions. Even if we take black and white balls in a dark bag as an example: when one ball is drawn, the ratio of the number and colors of balls in the bag has changed, which means that the probability itself has changed.

However, if our conditions are constant, then we can accurately determine the required probability from us that the event A will occur exactly k times out of n possible.

This fact was compiled by Jacob Bernoulli into a theorem, which later became known as his name. "Bernoulli's theorem" is one of the main theorems in probability theory. It was first published in the work of J. Bernoulli "The Art of Assumptions". What is this theorem? “If the probability p of the occurrence of event A in each trial is constant, then the probability Pk,n that the event will occur k times in n trials that are independent of each other is equal to: , where q=1-p.”

In the proof of the effectiveness of the formula, tasks can be given.

Task #1:

Out of n glass jars per month of storage, k are broken. Randomly took m cans. Find the probability that among these jars l will not break. n=250, k=10, m=8, l=4.

Solution: We have a Bernoulli scheme with values:

p=10/250=0.04 (probability that banks will break);

n=8 (number of trials);

k=8-4=4 (number of broken jars).

We use the Bernoulli formula

Got:

Answer: 0.0141

Task #2:

The probability of manufacturing a defective product in production is 0.2. Find the probability that out of 10 products manufactured at this production facility, exactly k must be in good condition. Run solution for k = 0, 1, 10.

We are interested in the event A - the production of serviceable parts, which happens once an hour with a probability p=1-0.2=0.8. We need to find the probability that the given event will occur k times. Event A is opposite to the event "not A", i.e. manufacture of a faulty product.

Therefore, we have: n=10; p=0.8; q=0.2.

As a result, we find the probability that out of 10 manufactured products all products are faulty (k=0), that one product is in good condition (k=1), that there are no faulty ones at all (k=10):

In conclusion, I would like to note that in modern times, many scientists are trying to prove that the "Bernoulli formula" does not comply with the laws of nature and that problems can be solved without applying it to use. Of course, this is possible, most problems in the theory of probability can be performed without the Bernoulli formula, the main thing is not to get confused in large volumes of numbers.

Bibliographic link

Khomutova E.A., Kalinichenko V.A. BERNULLI'S FORMULA IN PROBABILITY THEORY // International Student Scientific Bulletin. - 2015. - No. 3-4 .;
URL: http://eduherald.ru/ru/article/view?id=14141 (date of access: 03/12/2019). We bring to your attention the journals published by the publishing house "Academy of Natural History"

Brief theory

Probability theory deals with experiments that can be repeated (at least in theory) an unlimited number of times. Let some experiment be repeated once, and the results of each repetition do not depend on the outcomes of previous repetitions. Such series of repetitions are called independent trials. A special case of such tests are independent Bernoulli trials, which are characterized by two conditions:

1) the result of each test is one of two possible outcomes, called respectively "success" or "failure".

2) the probability of "success" in each subsequent test does not depend on the results of previous tests and remains constant.

Bernoulli's theorem

If a series of independent Bernoulli trials is made, in each of which "success" occurs with probability , then the probability that "success" in the trials occurs exactly once is expressed by the formula:

where is the probability of failure.

- the number of combinations of elements by (see the basic formulas of combinatorics)

This formula is called Bernoulli formula.

The Bernoulli formula allows you to get rid of a large number of calculations - addition and multiplication of probabilities - with a sufficiently large number of tests.

The Bernoulli test scheme is also called the binomial scheme, and the corresponding probabilities are called binomial, which is associated with the use of binomial coefficients.

The distribution according to the Bernoulli scheme allows, in particular, to find the most probable number of occurrence of an event .

If the number of trials n great, then enjoy:

Problem solution example

The task

The germination of seeds of a certain plant is 70%. What is the probability that out of 10 seeds sown: 8, at least 8; at least 8?

The solution of the problem

Let's use the Bernoulli formula:

In our case

Let the event - out of 10 seeds sprout 8:

Let the event - rise at least 8 (that means 8, 9 or 10)

Let the event rise at least 8 (that means 8.9 or 10)

Answer

Medium the cost of solving the control work is 700 - 1200 rubles (but not less than 300 rubles for the entire order). The price is strongly influenced by the urgency of the decision (from days to several hours). The cost of online help in the exam / test - from 1000 rubles. for the ticket solution.

The application can be left directly in the chat, having previously thrown off the condition of the tasks and informing you of the deadlines for solving it. The response time is several minutes.

Repeated independent trials are called Bernoulli trials if each trial has only two possible outcomes and the probabilities of outcomes remain the same for all trials.

Usually these two outcomes are called "success" (S) or "failure" (F) and the corresponding probabilities are denoted p and q. It's clear that p 0, q³ 0 and p+q=1.

The elementary event space of each trial consists of two events Y and H.

Space of elementary events n Bernoulli trials Ω contains 2 n elementary events, which are sequences (chains) of n symbols Y and H. Each elementary event is one of the possible outcomes of the sequence n Bernoulli trials. Since the tests are independent, then, according to the multiplication theorem, the probabilities are multiplied, that is, the probability of any particular sequence is the product obtained by replacing the symbols U and H by p and q respectively, that is, for example: R( )=(U U N U N... N U )= p p q p q ... q q p .

Note that the outcome of the Bernoulli test is often denoted by 1 and 0, and then the elementary event in the sequence n Bernoulli tests - there is a chain consisting of zeros and ones. For example:  =(1, 0, 0, ... , 1, 1, 0).

Bernoulli trials are the most important scheme considered in probability theory. This scheme is named after the Swiss mathematician J. Bernoulli (1654-1705), who studied this model in depth in his works.

The main problem that will be of interest to us here is: what is the probability of the event that in n Bernoulli trials happened m success?

If these conditions are met, the probability that, during independent tests, an event will be observed exactly m times (no matter in which experiments), is determined by Bernoulli formula:

(21.1)

where - probability of occurrence in every test, and
is the probability that in a given experience an event Did not happen.

If we consider P n (m) as a function m, then it defines a probability distribution, which is called binomial. Let's explore this relationship P n (m) from m, 0£ m£ n.

Events B m ( m = 0, 1, ..., n) consisting of a different number of occurrences of the event BUT in n tests, are incompatible and form a complete group. Hence,
.

Consider the ratio:

=
=
=
.

Hence it follows that P n (m+1)>P n (m), if (n- m)p> (m+1)q, i.e. function P n (m) increases if m< np- q. Likewise, P n (m+1)< P n (m), if (n- m)p< (m+1)q, i.e. P n (m) decreases if m> np- q.

Thus there is a number m 0 , at which P n (m) reaches its highest value. Let's find m 0 .

According to the meaning of the number m 0 we have P n (m 0)³ P n (m 0 -1) and P n (m 0) ³ P n (m 0 +1), hence

, (21.2)

. (21.3)

Solving inequalities (21.2) and (21.3) with respect to m 0 , we get:

p/ m 0 ³ q/(n- m 0 +1) Þ m 0 £ np+ p,

q/(n- m 0 ) ³ p/(m 0 +1) Þ m 0 ³ np- q.

So the desired number m 0 satisfies the inequalities

np- q£ m 0 £ np+p. (21.4)

As p+q=1, then the length of the interval defined by inequality (21.4) is equal to one and there is at least one integer m 0 satisfying the inequalities (21.4):

1) if np - q is an integer, then there are two values m 0 , namely: m 0 = np - q and m 0 = np - q + 1 = np + p;

2) if np - q- fractional, then there is one number m 0 , namely the only integer between the fractional numbers obtained from inequality (21.4);

3) if np is an integer, then there is one number m 0 , namely m 0 = np.

Number m 0 is called the most probable or most probable value (number) of the occurrence of the event A in a series of n independent tests.

In this lesson, we will find the probability of an event occurring in independent trials when the trials are repeated. . Trials are called independent if the probability of one or another outcome of each trial does not depend on what outcomes other trials had. . Independent tests can be carried out both under the same conditions and under different conditions. In the first case, the probability of an event occurring in all trials is the same; in the second case, it varies from trial to trial.

Examples of Independent Retests :

• one of the device nodes or two or three nodes will fail, and the failure of each node does not depend on the other node, and the probability of failure of one node is constant in all tests;
• a part produced under certain constant technological conditions, or three, four, five parts, will turn out to be non-standard, and one part may turn out to be non-standard regardless of any other part, and the probability that the part will turn out to be non-standard is constant in all tests;
• out of several shots on the target, one, three or four shots hit the target regardless of the outcome of other shots and the probability of hitting the target is constant in all trials;
• when the coin is inserted, the machine will operate correctly one, two, or another number of times, regardless of what the other coin insertions have had, and the probability that the machine will operate correctly is constant in all trials.

These events can be described by one scheme. Each event occurs in each trial with the same probability, which does not change if the results of previous trials become known. Such tests are called independent, and the scheme is called Bernoulli scheme . It is assumed that such tests can be repeated as many times as desired.

If the probability p event A is constant in each trial, then the probability that in n independent test event A will come m times, located on Bernoulli formula :

(where q= 1 – p- the probability that the event will not occur)

Let's set the task - to find the probability that an event of this type in n independent trials will come m once.

Bernoulli formula: examples of problem solving

Example 1 Find the probability that among five randomly selected parts two are standard, if the probability that each part is standard is 0.9.

Decision. Event Probability BUT, consisting in the fact that a part taken at random is standard, is p=0.9 , and the probability that it is non-standard is q=1–p=0.1 . The event indicated in the condition of the problem (we denote it by AT) occurs if, for example, the first two parts are standard, and the next three are non-standard. But the event AT also occurs if the first and third parts are standard and the rest are non-standard, or if the second and fifth parts are standard and the rest are non-standard. There are other possibilities for the event to occur. AT. Any of them is characterized by the fact that out of five parts taken, two, occupying any places out of five, will turn out to be standard. Therefore, the total number of different possibilities for the occurrence of an event AT is equal to the number of possibilities for placing two standard parts in five places, i.e. is equal to the number of combinations of five elements by two, and .

The probability of each possibility, according to the probability multiplication theorem, is equal to the product of five factors, of which two, corresponding to the appearance of standard parts, are equal to 0.9, and the remaining three, corresponding to the appearance of non-standard parts, are equal to 0.1, i.e. this probability is . Since these ten possibilities are incompatible events, by the addition theorem, the probability of an event AT, which we denote

Example 2 The probability that the machine will require the attention of a worker within an hour is 0.6. Assuming that the failures on the machines are independent, find the probability that during an hour the attention of the worker will be required by any one of the four machines serviced by him.

Decision. Using Bernoulli's formula at n=4 , m=1 , p=0.6 and q=1–p=0.4 , we get

Example 3 For the normal operation of the car depot, there must be at least eight cars on the line, and there are ten of them. The probability of non-exit of each car to the line is equal to 0.1. Find the probability of normal operation of the depot in the next day.

Decision. Autobase will work fine (event F) if either or eight will enter the line (the event BUT), or nine (event AT), or all ten cars event (event C). According to the probability addition theorem,

We find each term according to the Bernoulli formula. Here n=10 , m=8; 10 and p\u003d 1-0.1 \u003d 0.9, since p should mean the probability of a car entering the line; then q=0.1 . As a result, we get

Example 4 Let the probability that a customer needs a size 41 men's shoe be 0.25. Find the probability that out of six buyers at least two need shoes of size 41.

Definition of repeated independent tests. Bernoulli formulas for calculating the probability and the most probable number. Asymptotic formulas for the Bernoulli formula (local and integral, Laplace's theorems). Using the integral theorem. Poisson's formula, for unlikely random events.

Repeated independent tests

In practice, one has to deal with such tasks that can be represented as repeatedly repeated tests, as a result of each of which the event A may or may not appear. At the same time, the outcome of interest is not the outcome of each "individual trial, but the total number of occurrences of event A as a result of a certain number of trials. In such problems, one must be able to determine the probability of any number m of occurrences of event A as a result of n trials. Consider the case when the trials are independent and the probability occurrence of event A in each trial is constant.Such trials are called repeated independents.

An example of independent testing would be testing the suitability of products taken from one of a number of batches. If these batches have the same percentage of defects, then the probability that the selected product will be defective in each case is a constant number.

Bernoulli formula

Let's use the concept difficult event, which means the combination of several elementary events, consisting in the appearance or non-appearance of event A in the i -th test. Let n independent trials be conducted, in each of which event A can either appear with probability p or not appear with probability q=1-p . Consider the event B_m, which consists in the fact that the event A in these n trials will occur exactly m times and, therefore, will not occur exactly (n-m) times. Denote A_i~(i=1,2,\ldots,(n)) occurrence of event A , a \overline(A)_i - non-occurrence of event A in the i-th trial. Due to the constancy of the test conditions, we have

Event A can appear m times in different sequences or combinations, alternating with the opposite event \overline(A) . The number of possible combinations of this kind is equal to the number of combinations of n elements by m , i.e. C_n^m . Therefore, the event B_m can be represented as a sum of complex events that are incompatible with each other, and the number of terms is equal to C_n^m :

B_m=A_1A_2\cdots(A_m)\overline(A)_(m+1)\cdots\overline(A)_n+\cdots+\overline(A)_1\overline(A)_2\cdots\overline(A)_( n-m)A_(n-m+1)\cdots(A_n),

where event A occurs in each product m times, and \overline(A) - (n-m) times.

The probability of each compound event included in formula (3.1), according to the probabilities multiplication theorem for independent events, is equal to p^(m)q^(n-m) . Since the total number of such events is equal to C_n^m , then, using the probability addition theorem for incompatible events, we obtain the probability of the event B_m (we denote it by P_(m,n) )

P_(m,n)=C_n^mp^(m)q^(n-m)\quad \text(or)\quad P_(m,n)=\frac(n{m!(n-m)!}p^{m}q^{n-m}. !}

Formula (3.2) is called Bernoulli formula, and repeated trials that satisfy the condition of independence and constancy of the probabilities of occurrence of the event A in each of them are called Bernoulli trials, or the Bernoulli scheme.

Example 1. The probability of going beyond the tolerance field when machining parts on a lathe is 0.07. Determine the probability that out of five parts randomly selected during the shift, one of the diameter dimensions does not correspond to the specified tolerance.

Decision. The condition of the problem satisfies the requirements of the Bernoulli scheme. Therefore, assuming n=5,\,m=1,\,p=0,\!07, by formula (3.2) we obtain

P_(1,5)=C_5^1(0,\!07)^(1)(0,\!93)^(5-1)\approx0,\!262.

Example 2. Observations have established that in some area in September there are 12 rainy days. What is the probability that out of 8 days randomly taken this month, 3 days will be rainy?

Decision.

P_(3;8)=C_8^3(\left(\frac(12)(30)\right)\^3{\left(1-\frac{12}{30}\right)\!}^{8-3}=\frac{8!}{3!(8-3)!}{\left(\frac{2}{5}\right)\!}^3{\left(\frac{3}{5}\right)\!}^5=56\cdot\frac{8}{125}\cdot\frac{243}{3125}=\frac{108\,864}{390\,625}\approx0,\!2787. !}

The most likely number of occurrences of the event

Most Likely Appearance event A in n independent trials is such a number m_0 for which the probability corresponding to this number is greater than or at least not less than the probability of each of the other possible numbers of occurrence of event A . To determine the most probable number, it is not necessary to calculate the probabilities of the possible number of occurrences of the event, it is enough to know the number of trials n and the probability of the occurrence of the event A in a separate trial. Let P_(m_0,n) denote the probability corresponding to the most probable number m_0 . Using formula (3.2), we write

P_(m_0,n)=C_n^(m_0)p^(m_0)q^(n-m_0)=\frac(n{m_0!(n-m_0)!}p^{m_0}q^{n-m_0}. !}

According to the definition of the most probable number, the probabilities of the event A occurring m_0+1 and m_0-1 times, respectively, should at least not exceed the probability P_(m_0,n) , i.e.

P_(m_0,n)\geqslant(P_(m_0+1,n));\quad P_(m_0,n)\geqslant(P_(m_0-1,n))

Substituting the value P_(m_0,n) and the expressions for the probabilities P_(m_0+1,n) and P_(m_0-1,n) into the inequalities, we obtain

Solving these inequalities for m_0 , we obtain

M_0\geqslant(np-q),\quad m_0\leqslant(np+p)

Combining the last inequalities, we get a double inequality, which is used to determine the most probable number:

Np-q\leqslant(m_0)\leqslant(np+p).

Since the length of the interval defined by inequality (3.4) is equal to one, i.e.

(np+p)-(np-q)=p+q=1,

and an event can occur in n trials only an integer number of times, then it should be kept in mind that:

1) if np-q is an integer, then there are two values ​​of the most probable number, namely: m_0=np-q and m"_0=np-q+1=np+p ;

2) if np-q is a fractional number, then there is one most probable number, namely: the only integer between the fractional numbers obtained from inequality (3.4);

3) if np is an integer, then there is one most probable number, namely: m_0=np .

For large values ​​of n, it is inconvenient to use formula (3.3) to calculate the probability corresponding to the most probable number. If in equality (3.3) we substitute the Stirling formula

N!\approx(n^ne^(-n)\sqrt(2\pi(n))),

valid for sufficiently large n, and take the most probable number m_0=np , then we obtain a formula for an approximate calculation of the probability corresponding to the most probable number:

P_(m_0,n)\approx\frac(n^ne^(-n)\sqrt(2\pi(n))\,p^(np)q^(nq))((np)^(np) e^(-np)\sqrt(2\pi(np))\,(nq)^(nq)e^(-nq)\sqrt(2\pi(nq)))=\frac(1)(\ sqrt(2\pi(npq)))=\frac(1)(\sqrt(2\pi)\sqrt(npq)).

Example 2. It is known that \frac(1)(15) some of the products supplied by the factory to the trading base do not meet all the requirements of the standard. A batch of products in the amount of 250 pieces was delivered to the base. Find the most probable number of products that meet the requirements of the standard, and calculate the probability that this lot will contain the most probable number of products.

Decision. By condition n=250,\,q=\frac(1)(15),\,p=1-\frac(1)(15)=\frac(14)(15). According to inequality (3.4), we have

250\cdot\frac(14)(15)-\frac(1)(15)\leqslant(m_0)\leqslant250\cdot\frac(14)(15)+\frac(1)(15)

where 233,\!26\leqslant(m_0)\leqslant234,\!26. Therefore, the most likely number of products that meet the requirements of the standard in a batch of 250 pieces. equals 234. Substituting the data into formula (3.5), we calculate the probability of having the most probable number of items in the batch:

P_(234,250)\approx\frac(1)(\sqrt(2\pi\cdot250\cdot\frac(14)(15)\cdot\frac(1)(15)))\approx0,\!101

Local Laplace theorem

Using the Bernoulli formula for large values ​​of n is very difficult. For example, if n=50,\,m=30,\,p=0,\!1, then to find the probability P_(30,50) it is necessary to calculate the value of the expression

P_(30,50)=\frac(50{30!\cdot20!}\cdot(0,\!1)^{30}\cdot(0,\!9)^{20} !}

Naturally, the question arises: is it possible to calculate the probability of interest without using the Bernoulli formula? It turns out you can. The local Laplace theorem gives an asymptotic formula that allows you to approximately find the probability of occurrence of events exactly m times in n trials, if the number of trials is large enough.

Theorem 3.1. If the probability p of occurrence of event A in each trial is constant and different from zero and one, then the probability P_(m,n) that event A will appear in n trials exactly m times is approximately equal (the more precisely, the greater n ) to the value of the function

Y=\frac(1)(\sqrt(npq))\frac(e^(-x^2/2))(\sqrt(2\pi))=\frac(\varphi(x))(\sqrt (npq)) at .

There are tables that contain function values \varphi(x)=\frac(1)(\sqrt(2\pi))\,e^(-x^2/2)), corresponding to positive values ​​of the argument x . For negative values ​​of the argument, the same tables are used, since the \varphi(x) function is even, i.e. \varphi(-x)=\varphi(x).

So, approximately the probability that event A will appear in n trials exactly m times,

P_(m,n)\approx\frac(1)(\sqrt(npq))\,\varphi(x), where x=\frac(m-np)(\sqrt(npq)).

Example 3. Find the probability that event A occurs exactly 80 times in 400 trials if the probability of occurrence of event A in each trial is 0.2.

Decision. By condition n=400,\,m=80,\,p=0,\!2,\,q=0,\!8. We use the asymptotic Laplace formula:

P_(80,400)\approx\frac(1)(\sqrt(400\cdot0,\!2\cdot0,\!8))\,\varphi(x)=\frac(1)(8)\,\varphi (x).

Let's calculate the value x defined by the problem data:

X=\frac(m-np)(\sqrt(npq))=\frac(80-400\cdot0,\!2)(8)=0.

According to the table adj, 1 we find \varphi(0)=0,\!3989. Desired probability

P_(80,100)=\frac(1)(8)\cdot0,\!3989=0,\!04986.

The Bernoulli formula leads to approximately the same result (calculations are omitted due to their cumbersomeness):

P_(80,100)=0,\!0498.

Laplace integral theorem

Assume that there are n independent trials, in each of which the probability of occurrence of the event A is constant and equal to p . It is necessary to calculate the probability P_((m_1,m_2),n) that event A will appear in n trials at least m_1 and at most m_2 times (for brevity, we will say "from m_1 to m_2 times"). This can be done using Laplace's integral theorem.

Theorem 3.2. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then approximately the probability P_((m_1,m_2),n) that event A will appear in trials from m_1 to m_2 times,

P_((m_1,m_2),n)\approx\frac(1)(\sqrt(2\pi))\int\limits_(x")^(x"")e^(-x^2/2) \,dx, where .

When solving problems that require the application of the Laplace integral theorem, special tables are used, since the indefinite integral \int(e^(-x^2/2)\,dx) is not expressed in terms of elementary functions. Integral table \Phi(x)=\frac(1)(\sqrt(2\pi))\int\limits_(0)^(x)e^(-z^2/2)\,dz given in app. 2, where the values ​​of the function \Phi(x) are given for positive values ​​of x , for x<0 используют ту же таблицу (функция \Phi(x) нечетна, т. е. \Phi(-x)=-\Phi(x) ). Таблица содержит значения функции \Phi(x) лишь для x\in ; для x>5 can take \Phi(x)=0,\!5 .

So, approximately the probability that event A will appear in n independent trials from m_1 to m_2 times,

P_((m_1,m_2),n)\approx\Phi(x"")-\Phi(x"), where x"=\frac(m_1-np)(\sqrt(npq));~x""=\frac(m_2-np)(\sqrt(npq)).

Example 4. Probability that a part is manufactured in violation of standards, p=0,\!2 . Find the probability that among 400 randomly selected non-standard parts there will be from 70 to 100 parts.

Decision. By condition p=0,\!2,\,q=0,\!8,\,n=400,\,m_1=70,\,m_2=100. Let's use Laplace's integral theorem:

P_((70,100),400)\approx\Phi(x"")-\Phi(x").

Let us calculate the limits of integration:

lower

X"=\frac(m_1-np)(\sqrt(npq))=\frac(70-400\cdot0,\!2)(\sqrt(400\cdot0,\!2\cdot0,\!8)) =-1,\!25,

upper

X""=\frac(m_2-np)(\sqrt(npq))=\frac(100-400\cdot0,\!2)(\sqrt(400\cdot0,\!2\cdot0,\!8) )=2,\!5,

Thus

P_((70,100),400)\approx\Phi(2,\!5)-\Phi(-1,\!25)=\Phi(2,\!5)+\Phi(1,\!25) .

According to the table app. 2 find

\Phi(2,\!5)=0,\!4938;~~~~~\Phi(1,\!25)=0,\!3944.

Desired probability

P_((70,100),400)=0,\!4938+0,\!3944=0,\!8882.

Application of Laplace's integral theorem

If the number m (the number of occurrences of event A in n independent trials) will change from m_1 to m_2, then the fraction \frac(m-np)(\sqrt(npq)) will change from \frac(m_1-np)(\sqrt(npq))=x" before \frac(m_2-np)(\sqrt(npq))=x"". Therefore, Laplace's integral theorem can also be written as follows:

P\left\(x"\leqslant\frac(m-np)(\sqrt(npq))\leqslant(x"")\right\)=\frac(1)(\sqrt(2\pi))\ int\limits_(x")^(x"")e^(-x^2/2)\,dx.

Let's set the task to find the probability that the absolute value of the deviation of the relative frequency \frac(m)(n) from the constant probability p does not exceed the given number \varepsilon>0 . In other words, we find the probability of the inequality \left|\frac(m)(n)-p\right|\leqslant\varepsilon, which is the same -\varepsilon\leqslant\frac(m)(n)-p\leqslant\varepsilon. This probability will be denoted as follows: P\left\(\left|\frac(m)(n)-p\right|\leqslant\varepsilon\right\). Taking into account formula (3.6), for this probability, we obtain

P\left\(\left|\frac(m)(n)-p\right|\leqslant\varepsilon\right\)\approx2\Phi\left(\varepsilon\,\sqrt(\frac(n)(pq ))\right).

Example 5. The probability that the part is non-standard, p=0,\!1 . Find the probability that among randomly selected 400 parts, the relative frequency of appearance of non-standard parts deviates from the probability p=0,\!1 in absolute value by no more than 0.03.

Decision. By condition n=400,\,p=0,\!1,\,q=0,\!9,\,\varepsilon=0,\!03. We need to find the probability P\left\(\left|\frac(m)(400)-0,\!1\right|\leqslant0,\!03\right\). Using formula (3.7), we obtain

P\left\(\left|\frac(m)(400)-0,\!1\right|\leqslant0,\!03\right\)\approx2\Phi\left(0,\!03\sqrt( \frac(400)(0,\!1\cdot0,\!9))\right)=2\Phi(2)

According to the table app. 2 we find \Phi(2)=0,\!4772 , therefore 2\Phi(2)=0,\!9544 . So, the desired probability is approximately equal to 0.9544. The meaning of the result obtained is as follows: if we take a sufficiently large number of samples of 400 parts each, then approximately in 95.44% of these samples the deviation of the relative frequency from the constant probability p=0,\!1 in absolute value will not exceed 0.03.

Poisson formula for unlikely events

If the probability p of the occurrence of an event in a separate trial is close to zero, then even with a large number of trials n, but with a small value of the product np, the probabilities P_(m,n) obtained by the Laplace formula are not accurate enough and there is a need for another approximate formula.

Theorem 3.3. If the probability p of the occurrence of event A in each trial is constant but small, the number of independent trials n is large enough, but the value of the product np=\lambda remains small (no more than ten), then the probability that event A occurs m times in these trials,

P_(m,n)\approx\frac(\lambda^m)(m\,e^{-\lambda}. !}

To simplify calculations using the Poisson formula, a table of Poisson function values ​​has been compiled \frac(\lambda^m)(m\,e^{-\lambda} !}(see appendix 3).

Example 6. Let the probability of manufacturing a non-standard part be 0.004. Find the probability that among 1000 parts there will be 5 non-standard ones.

Decision. Here n=1000,p=0.004,~\lambda=np=1000\cdot0,\!004=4. All three numbers satisfy the requirements of Theorem 3.3, so to find the probability of the desired event P_(5,1000) we use the Poisson formula. According to the table of values ​​of the Poisson function (app. 3) with \lambda=4;m=5 we get P_(5,1000)\approx0,\!1563.

Let's find the probability of the same event using Laplace's formula. To do this, we first calculate the x value corresponding to m=5 :

X=\frac(5-1000\cdot0,\!004)(\sqrt(1000\cdot0,\!004\cdot0,\!996))\approx\frac(1)(1,\!996)\approx0 ,\!501.

Therefore, according to the Laplace formula, the desired probability

P_(5,1000)\approx\frac(\varphi(0,\!501))(1,\!996)\approx\frac(0,\!3519)(1,\!996)\approx0,\ !1763

and according to the Bernoulli formula, its exact value

P_(5,1000)=C_(1000)^(5)\cdot0,\!004^5\cdot0,\!996^(995)\approx0,\!1552.

Thus, the relative error in calculating the probabilities P_(5,1000) using the approximate Laplace formula is

\frac(0,\!1763-0,\!1552)(0,\!1552)\approx0,\!196, or 13,\!6\%

and according to the Poisson formula -

\frac(0,\!1563-0,\!1552)(0,\!1552)\approx0,\!007, or 0,\!7\%

That is, many times less.
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