Bernoulli test replication schemes presentation. Presentation on the topic "Bernoulli formula". Local and integral theorems of Moivre-Laplace

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

Let us denote these probabilities as p And q. Outcome with probability p we will call it “success”, and the outcome with probability q- “failure”.

It's obvious that

Space elementary events for each trial consists of two points. Space of elementary events for n A Bernoulli trial contains points, each of which represents one possible outcome of the 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, N, U, N, N, N)

equal to the product

Examples of Bernoulli tests.

1. Consecutive tosses of a “fair” coin. In this case p = q = 1/2 .

When tossing an asymmetrical coin, 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 from them we can select a group of outcomes that are considered “success”, calling all other outcomes “failure”.

For example, with successive throws dice“success” can be understood as rolling out a 5, and “failure” can mean rolling out 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 of points, and by “failure” - an odd number of points, then p = q = 1/2 .

4. Repeated random extraction of a ball from an urn containing a white and b black balls. If by success we mean retrieving the white ball, then , .

Feller gives the following example of a practical application of the Bernoulli test scheme. Washers produced in mass production may vary in thickness, but upon inspection they are classified as passable or defective, depending on whether the thickness is within the prescribed limits. Although products may not fully conform to Bernoulli's scheme for many reasons, this scheme sets an ideal standard for industrial product quality control, even though this standard is never quite achieved. Machines are subject to change, and therefore the probabilities do not remain the same; There is a certain consistency in the operation of the machines, with the result that long series of identical deviations are more likely than would be the case if the tests were truly independent. However, from a product quality control point of view, it is desirable that the process conform to Bernoulli's scheme, and the important thing is that this can be achieved within certain limits. The purpose of current 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.

Slide 2

Pn(k)=Cknpk(1-p)n-k If the Probability p of the occurrence of event A in each trial is constant, then the probability Pn(k) that event A will occur k times in n independent trials is equal to: T Statement of the theorem Bernoulli formula - a formula in probability theory that allows you to find the probability of the occurrence of event A during independent trials. Bernoulli's formula allows you to get rid of a large number of calculations - adding and multiplying probabilities - with a sufficiently large number of tests.

Slide 3

Historical information JACOB BERNOULLI (1654–1705) Date of birth: December 27, 1654 Place of birth: Basel Date of death: August 16, 1705 Place of death: Basel Citizenship: Switzerland Scientific field: Mathematician Place of work: University of Basel Scientific. director: Leibniz Jacob Bernoulli (German Jakob Bernoulli, December 27, 1654, Basel, - August 16, 1705, ibid.) - Swiss mathematician, brother of Johann Bernoulli; professor of mathematics at the University of Basel (since 1687). Jacob Bernoulli has significant achievements in series theory, differential calculus of variations, probability theory and number theory, where numbers with certain properties are named after him. Jacob Bernoulli also wrote works on physics, arithmetic, algebra and geometry.

Slide 4

An example of using the Bernoulli formula Every day, shares of ABC Corporation rise in price or fall in price by one point with probabilities of 0.75 and 0.25, respectively. Find the probability that the stock will return to its original price after six days. Accept the condition that changes in the stock price up and down are independent events. SOLUTION: In order for the shares to return to their original price in 6 days, they need to rise in price 3 times and fall in price three times during this time. The desired probability is calculated using the Bernoulli formula P6(3) =C36(3/4)3(1/4)3=0.13

Slide 5

Test yourself There are 20 white and 10 black balls in an urn. 4 balls are taken out in a row, and each removed ball is returned to the urn before the next one is taken out and the balls in the urn are mixed. What is the probability that out of four balls drawn, two will be white? ANSWER: SOLUTION: ANSWER: ANSWER: SOLUTION: SOLUTION: The auditor detects financial irregularities in the audited company with a probability of 0.9. Find the probability that among 4 violating firms more than half will be identified. The dice is rolled 3 times. What is the probability that 6 points will appear exactly 2 times in this series of tests? 0.01389 8/27 0.9477

Slide 6

Test yourself The coin is tossed 6 times. Find the probability that the coat of arms will appear no more than 2 times. ANSWER: SOLUTION: ANSWER: SOLUTION: Let the germination rate of wheat seeds be 90%. What is the probability that out of 7 seeds sown, 5 will sprout? 0.124 0.344

Slide 7

The probability of retrieving the white ball p=20/30=2/3 can be considered the same in all trials; 1-p=1/3 Using Bernoulli's formula, we get P4(2) = C42·p2·(1-p)2=(12/2)·(2/3)2·(1/3)2 = 8/ 27 BACK SOLUTION TO PROBLEM 1

Slide 8

BACK SOLUTION TO PROBLEM 2 The event is that out of 4 violating firms, three or four will be identified, i.e. P(A)=P4(3)+P4(4) P(A)= C340.93∙0.1+C44 0.94 = 0.93 (0.4+0.9)=0.9477

https://accounts.google.com

Slide captions:

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

Contents EXAMPLE 5. Probability of hitting the target with one shot... Solution 5a); Solution 5b); Solution 5c); Solution 5d). Note that... Throughout the 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 out (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. With a large number of independent repetitions... For the teacher. Sources. 02/08/2014 2

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

EXAMPLE 5. Probability of hitting the 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 the 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 affected; c) will be hit at least once; d) will be hit exactly once. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 4

Solution to example 5a) Example 5. The probability of hitting the 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 to example 5b) Example 5. The probability of hitting the 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 to example 5c) Example 5. The probability of hitting the 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 to example 5d) Example 5. The probability of hitting the 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 paragraph d) of Example 5, in a specific case, repeats the proof of the famous Bernoulli theorem, which refers to one of the most common probabilistic models: independent repetitions of the same test with two possible outcomes. 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 this event will happen or not. 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 is thrown ten times in a row. What is the probability that a “four” will be rolled 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 tosses of a coin, “heads” will appear exactly 4 times? 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 10

Jacob Bernoulli combined examples and questions The early 18th century Swiss mathematician Jacob Bernoulli combined examples and questions of this type into a single probabilistic scheme. Let the probability random event And when carrying out some test it is equal to P(A). Let us view this trial as a trial with only two possible outcomes: one outcome is that event A will occur, and the other outcome is that event A will not occur, i.e., event Ᾱ will occur. For brevity, let’s call the first outcome (the occurrence of event A) “success”, and the second outcome (the occurrence of event Ᾱ) “failure”. We denote the probability P(A) of “success” by p, and the probability P(Ᾱ) of “failure” by q. This means q = Р(Ᾱ) = 1 - Р(А) = 1 - р. 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 trial. 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 particular 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 points a) - d) determine the values ​​of n, k, p, q and write out (without calculations) the expression for the desired probability P n (k). a) What is the probability of getting exactly 7 heads in 10 coin tosses? b) Each of 20 people independently names one of the days of the week. Monday and Friday are considered “unlucky” days. What is the probability that the “luck” will be exactly half? c) Throwing a die is “successful” if it comes up with 5 or 6 points. What is the probability 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 “lucks” among 7 throws? 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 13

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

Solution 6b) Example 6. In each of points a) - d) determine the values ​​of n, k, p, q and write out (without calculations) the expression for the desired probability P n (k). b) Each of 20 people independently names one of the days of the week. Monday and Friday are considered “unlucky” days. 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 points a) - d) determine the values ​​of n, k, p, q and write out (without calculations) the expression for the desired probability P n (k). c) Throwing a die is “successful” if it comes up with 5 or 6 points. What is the probability 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 points a) - d) determine the values ​​of n, k, p, q and write out (without calculations) the 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 “lucks” among 7 throws? Solution: d) n = 7, k = 3. “Luck” in one toss is that there are fewer “tails” than “heads.” There are 8 possible results in total: PPP, PPO, POP, OPP, POO, OPO, OOP, LLC (P - “tails”, O - “heads”). In exactly half of them there are fewer heads than heads: ROO, ORO, OOP, OOO. This means 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 us to establish a connection between the statistical approach to determining probability and the classical definition of the probability of a random event. To describe this connection, let us return to the terms of § 50 about statistical processing of information. Consider a sequence of n independent repetitions of the same trial with two outcomes - “success” and “failure”. The results of these tests constitute a series of data consisting of a certain sequence of two options: “success” and “failure”. Simply put, there is a sequence of length n made up of two letters U (“luck”) and H (“failure”). For example, U, U, N, N, U, N, N, N, ..., U or N, U, U, N, U, U, N, N, U, ..., N, etc. n. Let's calculate the multiplicity and frequency of variants Y, i.e. find the fraction k/n, where k is the number of “lucks” 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 complex 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 increasing accuracy is approximately equal to the probability of event A: k/n≈ P(A). For example, with n > 2000, with a probability greater than 99%, it can be stated that the absolute error | k/n - P(A)| approximate equality k/n≈ Р(А) will be less than 0.03. Therefore, in sociological surveys, it is enough to interview about 2000 randomly selected people (respondents). If, say, 520 of them answered positively to the question asked, then k/n=520/2000=0.26 and it is almost 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 make it possible (approximately) to find the probability of a random event in cases where its explicit calculation is impossible. 02/08/2014 Tsybikova Tamara Radnazhapovna, mathematics teacher 19

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

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

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

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

Preview:

To use presentation previews, create an account for yourself ( account) Google and log in: https://accounts.google.com

Slide captions:

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

Slide 2
Content
EXAMPLE 5. Probability of hitting the target with one shot...Solution 5a);Solution 5b);Solution 5c);Solution 5d).Note that...In the entire 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 out (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 TESTS. BERNOULLI'S THEOREM AND STATISTICAL STABILITY.
Part 3.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 4
EXAMPLE 5. Probability of hitting the 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 the 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 to example 5a)
Example 5. The probability of hitting the 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 to example 5b)
Example 5. The probability of hitting the 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 to example 5c)
Example 5. The probability of hitting the 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 to example 5d)
Example 5. The probability of hitting the 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 paragraph d) of Example 5, in a specific case, repeats the proof of the famous Bernoulli 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
Throughout the entire series of repetitions, it is important to know
In each of these repetitions, we are interested in the question of whether this event will happen or not. 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 is thrown ten times in a row. What is the probability that a “four” will be rolled 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 tosses of a coin, “heads” will appear exactly 4 times?
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 11
Jacob Bernoulli combined examples and questions
The Swiss mathematician of the early 18th century, Jacob Bernoulli, combined examples and questions of this type into a single probabilistic scheme. Let the probability of a random event A during a certain test be equal to P(A). Let us view this trial as a trial with only two possible outcomes: one outcome is that event A will occur, and the other outcome is that event A will not occur, i.e., event Ᾱ will occur. For brevity, let’s call the first outcome (the occurrence of event A) “success”, and the second outcome (the occurrence of event Ᾱ) “failure”. We denote the probability P(A) of “success” by p, and the probability P(Ᾱ) of “failure” by q. This means q = Р(Ᾱ) = 1 - Р(А) = 1 - р.
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 trial. Then Pn(k) = Сnk pk qn-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 theory and for 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 out (without calculations) the expression for the desired probability Pn(k).a) What is the probability of exactly 7 “heads” in 10 tosses? coins? b) Each of 20 people independently names one of the days of the week. Monday and Friday are considered “unlucky” days. What is the probability that the “success” will be exactly half? c) Throwing a die is “successful” if the result is 5 or 6 points. What is the probability that exactly 5 out of 25 tosses 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 “lucks” among 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 out (without calculations) the expression for the desired probability Pn(k).a) What is the probability of exactly 7 “heads” in 10 tosses? 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 out (without calculations) an expression for the desired probability Pn(k). b) Each of 20 people independently names one of the days of the week. Monday and Friday are considered “unlucky” days. What is the probability that the “luck” will be exactly half? Solution:
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 out (without calculations) the expression for the desired probability Pn(k).c) Throwing a die is “successful” if 5 or 6 points are rolled out . What is the probability that exactly 5 out of 25 tosses will be “successful”? Solution:
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 out (without calculations) an expression for the desired probability Pn(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 “lucks” among 7 tosses? Solution: d) n = 7, k = 3. “Luck” in one toss is that there are fewer “tails” than “heads”. There are 8 possible results in total: PPP, PPO, POP, OPP, POO, OPO, OOP, LLC (P - “tails”, O - “heads”). In exactly half of them there are fewer heads than heads: ROO, ORO, OOP, OOO. This means 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 us to establish a connection between the statistical approach to determining probability and the classical definition of the probability of a random event. To describe this connection, let us return to the terms of § 50 about statistical processing of information. Consider a sequence of n independent repetitions of the same trial with two outcomes - “success” and “failure”. The results of these tests constitute a series of data consisting of a certain sequence of two options: “success” and “failure”. Simply put, there is a sequence of length n made up of two letters U (“luck”) and H (“failure”). For example, U, U, N, N, U, N, N, N, ..., U or N, U, U, N, U, U, N, N, U, ..., N, etc. n. Let's calculate the multiplicity and frequency of variant Y, i.e., find the fraction k/n, where k is the number of “lucks” 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 complex mathematical fact is derived precisely from Bernoulli’s theorem.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 19
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 increasing accuracy is approximately equal to the probability of event A: k/n≈ P(A). For example, with n > 2000 with a probability greater than 99% , it can be argued that the absolute error |k/n- Р(А)| approximate equality k/n≈ Р(А) will be less than 0.03. Therefore, in sociological surveys, it is enough to interview about 2000 randomly selected people (respondents). If, say, 520 of them answered positively to the question asked, then k/n=520/2000=0.26 and it is almost 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 make it possible (approximately) to find the probability of a random event in cases where its explicit calculation is impossible.
08.02.2014
Tsybikova Tamara Radnazhapovna, mathematics teacher
*

Slide 20
For the 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
Algebra and beginnings of analysis, grades 10-11, Part 1. Textbook, 10th ed. (Basic level), A.G. Mordkovich, M., 2009 Algebra and beginnings of analysis, grades 10-11. (Basic level) Methodological manual for teachers, A.G. Mordkovich, P.V. Semenov, M., 2010 Tables compiled in MS Word and MS Excel. Internet resources
Tsybikova Tamara Radnazhapovna, mathematics teacher
08.02.2014
*

“Elements of mathematical statistics” - Confidence interval. The science. Classification of hypotheses. Parts are made on different machines. Verification rules. Correlation dependence. Addiction. A set of criterion values. Find the confidence interval. Calculation of confidence intervals for unknown variance. Normal distribution.

“Probability and mathematical statistics” - Accuracy of the obtained values. Code for the safe. Descriptive statistics. Apple. Let's consider the events. Multiplication rule. Two shooters. Comparison of training programs. Caramel. Examples of bar charts. Math grades. Multiplication rule for three. White and red roses. 9 different books. The winter vacation.

"Fundamentals of mathematical statistics" - Conditional probability. Table of standardized values. Properties of the Student distribution. Confidence interval of mathematical expectation. Sample mean. Distribution. One test can be considered as a series of one test. Quantile – to the left should be the number of values ​​corresponding to the quantile index.

“Probability theory and statistics” - Interval boundaries. Critical areas. Probability multiplication theorem. Distribution of a normal random variable. Derivation of Bernoulli's formula. Laws of distribution of random variables. The formulation of the LBC. The meaning and formulation of the central limit theorem. Connection of nominal features. Stochastic dependence of two random variables.

"Statistical Research" - Relevance. Statistical characteristics and research. Plan. Range is the difference between the largest and smallest values ​​of a data series. Types of statistical observation. Do you enjoy studying mathematics? Let's look at a series of numbers. Who helps you understand a difficult topic in mathematics? Do you need mathematics in your future profession?

“Basic statistical characteristics” - Basic statistical characteristics. Find the arithmetic mean. Petronius. Scope. Fashion series. The arithmetic mean of a series of numbers. Row range. Median of the series. Statistics. Median. School notebooks.

There are a total of 17 presentations in the topic

Slide 1

Bernoulli's theorem
17.03.2017

Slide 2

A series of n independent tests is carried out. Each test has 2 outcomes: A - “success” and - “failure”. The probability of “success” in each trial is the same and is equal to P(A) = p. Accordingly, the probability of “failure” also does not change from experiment to experiment and is equal.
Bernoulli scheme
What is the probability that in a series of n experiments there will be success k times? Find Pn(k) .

Slide 3

The coin is tossed n times. A card is drawn from the deck n times, and each time the card is returned, the deck is shuffled. n products of a certain production, selected at random, are examined for quality. The shooter shoots at the target n times.
Examples

Slide 4

Explain why the following questions fit Bernoulli's scheme. State what “success” is and what n and k are equal to. a) What is the probability of getting a “2” three times when throwing a die ten times? b) What is the probability that in one hundred coin tosses, heads will appear 73 times? c) A pair of dice were thrown twenty times in a row. What is the probability that the sum of the points never equals ten? d) Three cards were drawn from a deck of 36 cards, the result was recorded and returned to the deck, then the cards were shuffled. This was repeated 4 times. What is the probability that each time the queen of spades was among the cards drawn?

Slide 5

For the number of combinations from n to k, the following formula is valid:
For example:

Slide 6

Bernoulli's theorem
The probability Pn(k) of the occurrence of exactly k successes in n independent repetitions of the same trial is found by the formula, where p is the probability of “success”, q = 1- p is the probability of “failure” in a separate experiment.

Slide 7

The coin is tossed 6 times. What is the probability of getting the coat of arms 0, 1, ...6 times? Solution. Number of experiments n=6. Event A – “success” – loss of the coat of arms. According to Bernoulli's formula, the required probability is equal to
;
;
;
;
;
;

Slide 8

The coin is tossed 6 times. What is the probability of getting the coat of arms 0, 1, ...6 times? Solution. Number of experiments n=6. Event A – “success” – loss of the coat of arms.
;
;
;
;
;
;

Slide 9

The coin is tossed 10 times. What is the probability of the coat of arms appearing twice? Solution. Number of experiments n=10, m=2. Event A – “success” – loss of the coat of arms. According to Bernoulli's formula, the required probability is equal to
;
;
;
;
;
;

Slide 10

There are 20 white and 10 black balls in an urn. 4 balls are taken out, and each removed ball is returned to the urn before the next one is taken out and the balls in the urn are mixed. Find the probability that out of four drawn balls there will be 2 white ones. Solution. Event A – a white ball is taken out. Then the probabilities According to Bernoulli's formula, the required probability is equal to

Slide 11

Determine the probability that a family with 5 children has no girls. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 12

Determine the probability that a family with 5 children will have one girl. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 13

Determine the probability that a family with 5 children will have two girls. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 14

Determine the probability that a family with 5 children will have three girls. Solution. Probability of having a girl or a boy According to Bernoulli’s formula, the required probability is equal to

Slide 15

Determine the probability that in a family with 5 children there will be no more than three girls. The probabilities of having a boy and a girl are assumed to be the same. Solution. Probability of having a girl or a boy The required probability is equal to
.

Slide 16

Among the parts processed by a worker, on average 4% are non-standard. Find the probability that among 30 parts taken for testing, two will be non-standard. Solution. Here the experience consists of checking each of the 30 parts for quality. Event A - “appearance of a non-standard part”,