Repetition of tests. Bernoulli scheme

18.12.2021

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Slide captions:

Chapter 9. Elements of mathematical statistics, combinatorics and probability theory §54. Random events and their probabilities 3. INDEPENDENT REPEAT OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.

Contents EXAMPLE 5. Probability of hitting a target with one shot ... Solution 5a); Solution 5b); Solution 5c); Solution 5d). Note that ... In the whole series of repetitions it is important to know ... Jacob Bernoulli combined examples and questions ... THEOREM 3 (Bernoulli's theorem). EXAMPLE 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability Pn (k). Solution 6 a); Solution 6 b); Solution 6 c); Solution 6 d). Bernoulli's theorem allows ... THEOREM 4. For a large number of independent repetitions ... For the teacher. Sources. 02/08/2014 2

3. INDEPENDENT REPEAT OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY. Part 3. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 3

EXAMPLE 5. Probability of hitting a target with one shot Let us slightly change the previous example: instead of two different shooters, the same shooter will shoot at the target. Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: a) will be hit three times; b) will not be amazed; c) will be struck at least once; d) will be hit exactly once. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 4

Solution of example 5a) Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: a) will be hit three times; 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 5

Solution of Example 5b) Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: b) will not be hit; Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 6

Solution of example 5c) Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: c) will be hit at least once; Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 7

Solution of example 5d) Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: d) will be hit exactly once. Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 8

Note The solution given in point d) of Example 5, in a specific case, repeats the proof of the famous Bernoulli's theorem, which refers to one of the most common probabilistic models: independent repetitions of the same test with two possible outcomes. A distinctive feature of many probabilistic problems is that the test, as a result of which the event of interest to us may occur, can be repeated many times. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 9

In the entire series of repetitions, it is important to know In each of these repetitions, we are interested in the question of whether or not this event will happen. And in the entire series of repetitions, it is important for us to know exactly how many times this event may or may not occur. For example, a dice was thrown ten times in a row. What is the probability that the “four” will be dropped exactly 3 times? 10 shots fired; what is the probability that there will be exactly 8 hits on the target? Or, what is the probability that with five coin tosses, exactly 4 times will come up heads? 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 10

Jacob Bernoulli Combines Examples and Questions An early 18th century Swiss mathematician, Jacob Bernoulli, combined examples and questions of this type into a single probabilistic scheme. Let the probability of a random event A during some test be equal to P (A). We will consider this test as a test with only two possible outcomes: one outcome is that event A will occur, and the other outcome is that event A will not happen, that is, event Ᾱ will occur. For brevity, let us call the first outcome (the onset of event A) “success”, and the second outcome (the onset of event Ᾱ) “failure”. The probability P (A) of “success” will be denoted by p, and the probability P (Ᾱ) of “failure” will be denoted by q. Hence, q = P (Ᾱ) = 1 - P (A) = 1 - p. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 11

THEOREM 3 (Bernoulli's theorem) Theorem 3 (Bernoulli's theorem). Let P n (k) be the probability of exactly k “successes” occurring in n independent repetitions of the same test. Then P n (k) = С n k  p k  q n-k, where p is the probability of “success”, and q = 1 - p is the probability of “failure” in a separate test. This theorem (we present it without proof) is of great importance for both theory and practice. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 12

EXAMPLE 6. Example 6. In each of items a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability P n (k). a) What is the probability of exactly 7 "heads" appearing with 10 coin tosses? b) Each of the 20 people independently names one of the days of the week. "Bad" days are Monday and Friday. What is the probability that the "luck" will be exactly half? c) The roll of the die is "successful" if the roll is 5 or 6 points. What is the likelihood that exactly 5 out of 25 throws will be "successful"? d) The test consists of throwing three different coins at the same time. “Failure”: there are more “tails” than “heads”. What is the probability that there will be exactly three "luck" among the 7 throws? 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 13

Solution 6a) Example 6. In each of items a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability P n (k). a) What is the probability of exactly 7 "heads" appearing with 10 coin tosses? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 14

Solution 6b) Example 6. In each of items a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability P n (k). b) Each of the 20 people independently names one of the days of the week. "Bad" days are Monday and Friday. What is the probability that the "luck" will be exactly half? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 15

Solution 6c) Example 6. In each of items a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability P n (k). c) The roll of the die is "successful" if the roll is 5 or 6 points. What is the likelihood that exactly 5 out of 25 throws will be "successful"? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 16

Solution 6d) Example 6. In each of items a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability P n (k). d) The test consists of throwing three different coins at the same time. “Failure”: there are more “tails” than “heads”. What is the probability that there will be exactly three "luck" among the 7 throws? Solution: d) n = 7, k = 3. “Luck” in one throw consists in the fact that there are fewer “tails” than “heads”. A total of 8 results are possible: PPR, PPO, POP, ORR, POO, ORO, OOP, LLC (R - "tails", O - "heads"). In exactly half of them there are fewer tails: ROO, ORO, OOP, LLC. Hence, p = q = 0.5; P 7 (3) = C 7 3 ∙ 0.5 3 ∙ 0.5 4 = C 7 3 ∙ 0.5 7. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 17

Bernoulli's theorem allows ... Bernoulli's theorem allows you to establish a connection between the statistical approach to the definition of probability and the classical definition of the probability of a random event. To describe this connection, let us return to the terms in § 50 on the statistical processing of information. Consider a sequence of n independent repetitions of the same trial with two outcomes — luck and failure. The results of these tests constitute a series of data, consisting of some sequence of two options: "luck" and "failure". Simply put, there is a sequence of length n, made up of two letters Y ("luck") and H ("bad luck"). For example, U, U, H, H, U, H, H, H, ..., U or H, U, U, H, U, U, H, H, U, ..., H, etc. n. Let's calculate the multiplicity and frequency of the variants of Y, that is, we will find the fraction k / n, where k is the number of "successes" encountered among all n repetitions. It turns out that with an unlimited increase in n, the frequency k / n of the occurrence of "successes" will be practically indistinguishable from the probability p of "success" in one trial. This rather complicated mathematical fact is derived precisely from Bernoulli's theorem. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 18

THEOREM 4. With a large number of independent repetitions THEOREM 4. With a large number of independent repetitions of the same test, the frequency of occurrence of a random event A with ever greater accuracy is approximately equal to the probability of event A: k / n ≈ P (A). For example, for n> 2000 with a probability greater than 99%, it can be argued that the absolute error | k / n - P (A) | approximate equality k / n≈ P (A) will be less than 0.03. Therefore, in sociological surveys, it is enough to interview about 2,000 randomly selected people (respondents). If, for example, 520 of them answered positively to the question asked, then k / n = 520/2000 = 0.26 and it is practically certain that for any larger number of respondents this frequency will be in the range from 0.23 to 0.29. This phenomenon is called the phenomenon of statistical stability. So, Bernoulli's theorem and its corollaries allow (approximately) to find the probability of a random event in those cases when its explicit calculation is impossible. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 19

For the teacher 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 20

02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 21

02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 22

Sources Algebra and the beginning of analysis, grades 10-11, Part 1. Textbook, 10th ed. (Basic level), A.G. Mordkovich, M., 2009 Algebra and the beginning of analysis, grades 10-11. (Basic level) Methodological manual for a teacher, A.G. Mordkovich, P.V. Semenov, M., 2010 Tables are compiled in MS Word and MS Excel. Internet resources Tsybikova Tamara Radnazhapovna, mathematics teacher 02/08/2014 23

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Slide 1
Chapter 9. Elements of mathematical statistics, combinatorics and probability theory
§54. Random events and their probabilities 3. INDEPENDENT REPEAT OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.

Slide 2
Content
EXAMPLE 5. Probability of hitting a target in one shot ... Solution 5a); Solution 5b); Solution 5c); Solution 5d). Note that ... In the whole series of repetitions it is important to know ... Jacob Bernoulli combined examples and questions ... THEOREM 3 (Bernoulli's theorem ).
EXAMPLE 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) the expression for the desired probability Pn (k). Solution 6a); Solution 6b); Solution 6c); Solution 6d ). Bernoulli's theorem allows ... THEOREM 4. With a large number of independent repetitions ... For the teacher. Sources.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 3
3. INDEPENDENT REPEAT OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.
Part 3.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 4
EXAMPLE 5. Probability of hitting a target with one shot
Let's slightly change the previous example: instead of two different shooters, the same shooter will shoot at the target. Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: a) will be hit three times; b) will not be hit; c) will be hit at least once; d) will be hit exactly once.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 5
Solution of example 5a)
Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: a) will be hit three times;
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 6
Solution of example 5b)
Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: b) will not be hit; Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 7
Solution example 5c)
Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: c) will be hit at least once; Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 8
Solution example 5d)
Example 5. The probability of hitting a target with one shot is 0.8. 3 independent shots were fired. Find the probability that the target: d) will be hit exactly once. Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 9
Note
The solution given in point d) of Example 5, in a specific case, repeats the proof of the famous Bernoulli's theorem, which refers to one of the most common probabilistic models: independent repetitions of the same test with two possible outcomes. A distinctive feature of many probabilistic problems is that the test, as a result of which the event of interest to us may occur, can be repeated many times.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 10
During the entire repetition, it is important to know
In each of these repetitions, we are interested in the question of whether or not this event will happen. And in the entire series of repetitions, it is important for us to know exactly how many times this event may or may not occur. For example, a dice was thrown ten times in a row. What is the probability that the “four” will be dropped exactly 3 times? 10 shots fired; what is the probability that there will be exactly 8 hits on the target? Or, what is the probability that with five coin tosses, exactly 4 times will come up heads?
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 11
Jacob Bernoulli combined examples and questions
Jacob Bernoulli, a Swiss mathematician of the early 18th century, combined examples and questions of this type into a single probabilistic scheme: Let the probability of a random event A in some test be P (A). We will consider this test as a test with only two possible outcomes: one outcome is that event A will occur, and the other outcome is that event A will not happen, that is, event Ᾱ will occur. For brevity, let us call the first outcome (the onset of event A) “success”, and the second outcome (the onset of event Ᾱ) “failure”. The probability P (A) of “success” will be denoted by p, and the probability P (Ᾱ) of “failure” will be denoted by q. Hence, q = P (Ᾱ) = 1 - P (A) = 1 - p.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 12
THEOREM 3 (Bernoulli's theorem)
Theorem 3 (Bernoulli's theorem). Let Pn (k) be the probability of exactly k “successes” occurring in n independent repetitions of the same test. Then Pn (k) = Сnk pk qn-k, where р is the probability of "success", aq = 1-р is the probability of "failure" in a separate test. This theorem (we present it without proof) is also of great importance for theory and practice.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 13
EXAMPLE 6.
Example 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability Pn (k). coins? b) Each of the 20 people independently names one of the days of the week. "Bad" days are Monday and Friday. What is the probability that there will be exactly half of the "luck"? What is the likelihood that exactly 5 tosses out of 25 will be “successful?” D) The test consists of tossing three different coins at the same time. “Failure”: there are more “tails” than “heads”. What is the probability that there will be exactly three "luck" among the 7 throws?
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 14
Solution 6a)
Example 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability Pn (k). coins? Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 15
Solution 6b)
Example 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability Pn (k). B) Each of the 20 people independently names one of the days of the week. "Bad" days are Monday and Friday. What is the probability that there will be exactly half the "luck"?
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 16
Solution 6c)
Example 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) an expression for the desired probability Pn (k). C) Throwing the dice is "successful" if 5 or 6 points are rolled ... What is the likelihood that exactly 5 out of 25 throws will be “successful”?
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 17
Solution 6d)
Example 6. In each of points a) - d) determine the values of n, k, p, q and write (without calculations) the expression for the desired probability Pn (k). D) The test consists in the simultaneous throwing of three different coins. “Failure”: there are more “tails” than “heads”. What is the probability that there will be exactly three "hits" among 7 throws? Solution: d) n = 7, k = 3. "Luck" in one throw is that there are fewer "tails" than "heads". A total of 8 results are possible: PPR, PPO, POP, ORR, POO, ORO, OOP, LLC (R - "tails", O - "heads"). In exactly half of them there are fewer tails: ROO, ORO, OOP, LLC. Hence, p = q = 0.5; Р7 (3) = С73 ∙ 0.53 ∙ 0.54 = С73 ∙ 0.57.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 18
Bernoulli's theorem allows ...
Bernoulli's theorem allows you to establish a connection between the statistical approach to the definition of probability and the classical definition of the probability of a random event. To describe this connection, let us return to the terms in § 50 on the statistical processing of information. Consider a sequence of n independent repetitions of the same trial with two outcomes — luck and failure. The results of these tests constitute a series of data, consisting of some sequence of two options: "luck" and "failure". Simply put, there is a sequence of length n, made up of two letters Y ("luck") and H ("bad luck"). For example, U, U, H, H, U, H, H, H, ..., U or H, U, U, H, U, U, H, H, U, ..., H, etc. n. Let's calculate the multiplicity and frequency of the variants of Y, that is, we will find the fraction k / n, where k is the number of "successes" encountered among all n repetitions. It turns out that with an unlimited increase in n, the frequency k / n of the occurrence of "successes" will be practically indistinguishable from the probability p of "success" in one trial. This rather complicated mathematical fact is derived precisely from Bernoulli's theorem.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 19
THEOREM 4. For a large number of independent repetitions
THEOREM 4. With a large number of independent repetitions of the same test, the frequency of occurrence of a random event A with ever greater accuracy is approximately equal to the probability of event A: k / n≈ P (A). For example, for n> 2000 with a probability greater than 99% , it can be argued that the absolute error | k / n- P (A) | approximate equality k / n≈ P (A) will be less than 0.03. Therefore, in sociological surveys, it is enough to interview about 2,000 randomly selected people (respondents). If, for example, 520 of them answered positively to the question asked, then k / n = 520/2000 = 0.26 and it is practically certain that for any larger number of respondents this frequency will be in the range from 0.23 to 0.29. This phenomenon is called the phenomenon of statistical stability. Thus, Bernoulli's theorem and its consequences allow (approximately) to find the probability of a random event in those cases when its explicit calculation is impossible.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 20
For teacher
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 21
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 22
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 23
Sources of
Algebra and the beginning of analysis, grades 10-11, Part 1. Textbook, 10th ed. (Basic level), A.G. Mordkovich, M., 2009 Algebra and the beginning of analysis, grades 10-11. (Basic level) Methodological manual for a teacher, A.G. Mordkovich, P.V. Semenov, M., 2010 Tables are compiled in MS Word and MS Excel. Internet resources
Tsybikova Tamara Radnazhapovna, mathematics teacher
08.02.2014
*

Bernoulli's formula

Belyaeva T.Yu. GBPOU CC "AMT", Armavir Math teacher

One of the founders of the theory of probability and mathematical analysis

Foreign member of the Paris Academy of Sciences (1699) and the Berlin Academy of Sciences (1701)

Elder brother of Johann Bernoulli (the most famous member of the Bernoulli family)

Jacob Bernoulli (1654 - 1705)

Swiss mathematician

Let it be produced P independent trials, in each of which the probability that event A will occur is R , and therefore the probability that it will not happen is q = 1 - p .

It is required to find the probability that for P consecutive tests event A will occur exactly T once.

The required probability is denoted by R P ( T ) .

It's obvious that

p 1 (1) = p, p 1 (0) = q

R 1 (1) + p 1 (0) = p + q = 1

In two tests:

4 outcomes are possible:

p 2 (2) = p 2; p 2 (1) = 2p q; p 2 (0) = q 2

R 2 (2) + p 2 (1) + p 2 (0) = (p + q) 2 = 1

In three tests:

8 outcomes are possible:

We get:

p 3 (2) = 3p 2 q

p 3 (1) = 3pq 2

R 3 (3) + p 3 (2) + p 3 (1) + p 3 (0) = (p + q) 3 = 1

Objective 1.

The coin is thrown 8 times. What is the probability that the "coat of arms" will be dropped 4 times?

Objective 2.

There are 20 balls in the urn: 15 white and 5 black. They took out 5 balls in a row, with each removed ball returning to the urn before removing the next ball. Find the probability that there will be 2 white balls out of five drawn balls.

Formulas for finding the probability that v P trials event will come :

a) less than t times

R P (0) + ... + p P (t-1)

b) more than t times

R P (m + 1) + ... + p P (P)

v) no more than t times

R P (0) + ... + p P (T)

G) at least t times

R P (t) + ... + p P (P)

Objective 3.

The probability of manufacturing a non-standard part on an automatic machine is 0.02. Determine the probability that among the randomly taken six parts there will be more than 4 standard ones.

Event A - « more than 4 standard parts"(5 or 6) means

« no more than 1 defective part"(0 or 1)

Let it be produced P independent tests. With each such test, event A may or may not occur. The probability of occurrence of event A.

It is required to find such a number μ (0, 1, ..., n), for which the probability P n (μ) will be the greatest.

Task 4.

The share of premium products in this enterprise is 31%. What is the most probable number of "Extra" items if a batch of 75 items is selected?

By condition: n = 75, p = 0.31, q = 1 - 0.31 = 0.69

Task 6.

Two shooters shoot at the target. The probability of a miss in one shot for the first shooter is 0.2, and for the second - 0.4. Find the most probable number of volleys that will not hit the target if the shooters fire 25 volleys.

By condition: n = 25, p = 0.2 0.4 = 0.08, q = 0.92

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.

Let us denote these probabilities as p and q... Outcome with probability p will be called “success”, and the outcome with probability q- "failure".

It's obvious that

The elementary event space for each trial consists of two points. The space of elementary events for n Bernoulli's test contains points, each of which represents one possible outcome of a composite experiment. Since the trials are independent, the probability of a sequence of events is equal to the product of the probabilities of the corresponding outcomes. For example, the probability of a sequence of events

(U, U, H, U, H, H, H)

equal to the product

Examples of Bernoulli tests.

1. Consecutive tossing of the “correct” coin. In this case p = q = 1/2 .

When an asymmetrical coin is thrown, the corresponding probabilities will change their values.

2. Each result of the experiment can be considered as A or .

3. If there are several possible outcomes, then a group of outcomes can be distinguished from them, which are considered as “success”, calling all other outcomes “failure”.

For example, with successive throws of the dice, “success” can be understood as a roll of 5, and by “failure” - the fall of any other number of points. In this case p = 1/6, q = 5/6.

If by “success” we mean the loss of an even number, and by “failure” - an odd number of points, then p = q = 1/2 .

4. Repeated accidental withdrawal of the ball from the urn containing at each test a whites and b black balls. If by success we mean the extraction of the white ball, then,.

Feller gives the following example of the practical application of the Bernoulli test scheme. Washers made in mass production may vary in thickness, but upon inspection they are classified as good or defective, depending on whether the thickness is within the prescribed range. Although products for many reasons may not fully comply with the Bernoulli scheme, this scheme sets the ideal standard for industrial quality control of products, even though this standard is never quite accurately achieved. Machines are subject to change, and therefore the probabilities do not remain the same; there is some consistency in the operating mode of the machines, as a result of which long series of identical deviations are more likely than they would be if the tests were truly independent. However, from a product quality control point of view, it is desirable that the process follows the Bernoulli scheme, and it is important that within some limits this can be achieved. The purpose of monitoring is to detect, at an early stage, significant deviations from the ideal scheme and use them as indications of a threatening violation of the correct operation of the machine.

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higher professional education

"MATI"  RUSSIAN STATE TECHNOLOGICAL UNIVERSITY IM. K.E. TSIOLKOVSKY

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Repetition of tests. Bernoulli scheme

Methodical instructions for practical exercises

in the discipline "Higher Mathematics"

Compiled by: Egorova Yu.B.

Mamonov I.M.

Moscow 2006 introduction

Methodical instructions are intended for students of the day and evening departments of the faculty No. 14 of specialties 150601, 160301, 230102. The instructions highlight the basic concepts of the topic, determine the sequence of studying the material. A large number of considered examples help in the practical mastering of the topic. Methodical instructions serve as a methodological basis for practical training and individual assignments.

BERNULLY'S SCHEME. FORMULA BERNULLI

Bernoulli scheme- scheme of repeated independent tests, in which an event A can be repeated many times with constant probability R (A)= R .

Examples of tests carried out according to the Bernoulli scheme: multiple tossing of a coin or dice, making a batch of parts, shooting at a target, etc.

Theorem. If the likelihood of an event occurring A in each test is constant and equal to R, then the probability that the event A will come m once in n tests (no matter in what sequence), can be determined by the Bernoulli formula:

where q = 1 – p.

EXAMPLE 1. The probability that electricity consumption during one day will not exceed the established rate is equal to p = 0,75. Find the probability that in the next 6 days the power consumption for 4 days will not exceed the norm.

SOLUTION. The probability of normal electricity consumption during each of 6 days is constant and equal to R= 0.75. Consequently, the probability of excessive consumption of electricity per day is also constant and equal to q = 1R = 1  0,75 = 0,25.

The required probability according to the Bernoulli formula is:

EXAMPLE 2. The shooter fires three shots at the target. The probability of hitting the target with each shot is p = 0,3. Find the probability that: a) one target is hit; b) all three targets; c) not a single target; d) at least one target; e) less than two targets.

SOLUTION. The probability of hitting the target with each shot is constant and equal to R= 0.75. Therefore, the probability of a miss is q = 1 R= 1  0.3 = 0.7. Total number of experiments performed n=3.

a) The probability of hitting one target with three shots is equal to:

b) The probability of hitting all three targets with three shots is equal to:

c) The probability of three misses on three shots is equal to:

d) The probability of hitting at least one target with three shots is equal to:

e) The probability of hitting less than two targets, that is, either one target or none:

Local and integral Moivre-Laplace theorems

If a large number of tests are performed, then calculating the probabilities using the Bernoulli formula becomes technically difficult, since the formula requires operations on huge numbers. Therefore, there are simpler approximate formulas for calculating the probabilities at large n... These formulas are called asymptotic and are determined by the Poisson theorem, the local and integral Laplace theorem.

Local theorem of Moivre-Laplace. A A will happen m once in n n (n →∞ ) is approximately equal to:

where function
and the argument

The more n, the more accurate the calculation of probabilities. Therefore, it is expedient to apply the Moivre-Laplace theorem for npq  20.

f ( x ) compiled special tables (see Appendix 1). When using a table, keep in mind function properties f (x) :

Function f (x) is even f (  x) = f (x) .

At X ∞ function f (x) 0. In practice, we can assume that already at X> 4 function f (x) ≈0.

EXAMPLE 3. Find the probability that an event A occurs 80 times in 400 challenges if the probability of the occurrence of the event A in each trial is equal to p = 0,2.

SOLUTION. By condition n=400, m=80, p=0,2, q= 0.8. Hence:

Using the table, we determine the value of the function f (0)=0,3989.

Integral theorem of Moivre-Laplace. If the likelihood of an event occurring A in each test is constant and different from 0 and 1, then the probability that the event A will come from m 1 before m 2 once in n tests with a sufficiently large number n (n →∞ ) is approximately equal to:

where
 integral or Laplace function,

To find the values of the function F ( x ) compiled special tables (for example, see Appendix 2). When using a table, keep in mind properties of the Laplace function Ф (x) :

Function Ф (x) is odd F (  x) =  Ф (x) .

At X ∞ function Ф (x) 0.5. In practice, we can assume that already at X> 5 function Ф (x) ≈0,5.

F (0)=0.

EXAMPLE 4. The probability that a part has not passed the QCD inspection is 0.2. Find the probability that among the 400 parts, 70 to 100 parts will be untested.

SOLUTION. By condition n=400, m 1 =70, m 2 =100, p=0,2, q= 0.8. Hence: