Coursework: Repeated and independent tests. Bernoulli's theorem on the probability frequency. Presentation on the Bernoulli formula Presentation on the Bernoulli retest scheme
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Chapter 9. Elements of mathematical statistics, combinatorics and probability theory §54. Random events and their probabilities 3. INDEPENDENT REPETITIONS OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.
Content EXAMPLE 5. Probability of hitting the 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 paragraphs a) - d) determine the values n, k, p, q and write out (without calculations) the expression for the desired probability Pn (k). Decision 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 REPETITIONS OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY. Part 3. 08.02.2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 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, teacher of mathematics 4
Solution of 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, teacher of mathematics 5
Solution of 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, teacher of mathematics 6
Solution of 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, teacher of mathematics 7
Solution of 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, teacher of mathematics 8
Note The solution given in point d) of example 5, in a particular 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 many probabilistic problems consists in the fact 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, teacher of mathematics 9
In the whole series of repetitions, it is important to know In each of these repetitions, we are interested in the question of whether this event will occur or not occur. 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 4 will come up 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 in five tosses of a coin, heads will come up exactly 4 times? 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 10
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 random event And when conducting some test, it is equal to P (A). We will consider this test as a test with only two possible outcomes: one outcome is that the event A will occur, and the other outcome is that the event A will not occur, i.e., the event Ᾱ will occur. For brevity, let's call the first outcome (the occurrence of the event A) "success", and the second outcome (the occurrence of the 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. So q = P(Ᾱ) = 1 - P(A) = 1 - p. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 11
THEOREM 3 (Bernoulli's theorem) Theorem 3 (Bernoulli's theorem). Let P n (k) be the probability of exactly k "successes" 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 trial. This theorem (we present it without proof) is of great importance for both theory and practice. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 12
EXAMPLE 6. Example 6. In each of paragraphs 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 tosses of a coin? b) Each of the 20 people independently names one of the days of the week. "Unlucky" days are Monday and Friday. What is the probability that "good luck" will be exactly half? c) Rolling the die is "successful" if it rolls 5 or 6. What is the probability that exactly 5 throws out of 25 will be "lucky"? d) The test consists of tossing three different coins at the same time. "Failure": more "tails" than "eagles". What is the probability that there will be exactly three "lucks" among 7 rolls? 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 13
Solution 6a) Example 6. In each of paragraphs 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 tosses of a coin? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 14
Solution 6b) Example 6. In each of paragraphs 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 the 20 people independently names one of the days of the week. "Unlucky" days are Monday and Friday. What is the probability that "good luck" will be exactly half? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 15
Solution 6c) Example 6. In each of paragraphs 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) Rolling the die is "successful" if it rolls 5 or 6. What is the probability that exactly 5 throws out of 25 will be "lucky"? Decision: 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 16
Solution 6d) Example 6. In each of paragraphs 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 tossing three different coins at the same time. "Failure": more "tails" than "eagles". What is the probability that there will be exactly three "lucks" among 7 rolls? Solution: d) n = 7, k = 3. “Luck” with one throw is that there are fewer “tails” than “eagles”. A total of 8 results are possible: PPP, PPO, POP, OPP, POO, ORO, OOP, OOO (P - “tails”, O - “heads”). Exactly half of them have fewer tails than heads: POO, ORO, OOP, OOO. So p = q = 0.5; P 7 (3) \u003d C 7 3 ∙ 0.5 3 ∙ 0.5 4 \u003d C 7 3 ∙ 0.5 7. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 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 of § 50 on the statistical processing of information. Consider a sequence of n independent repetitions of the same test with two outcomes - "success" and "failure". The results of these tests constitute a series of data, consisting of some sequence of two options: "success" and "failure". Simply put, there is a sequence of length n, made up of two letters U ("good 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 the Y options, 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 complicated mathematical fact is derived precisely from Bernoulli's theorem. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 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, when 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, say, 520 of them responded positively to question asked, then k/n=520/2000=0.26 and it is almost certain that for any more 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 consequences allow us to (approximately) find the probability of a random event in cases where its explicit calculation is impossible. 02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 19
For the teacher 08.02.2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 20
02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 21
02/08/2014 Tsybikova Tamara Radnazhapovna, teacher of mathematics 22
Sources Algebra and the Beginnings 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 guide for teachers, A.G. Mordkovich, P.V. Semenov, M., 2010 Tables are compiled in MS Word and MS Excel. Internet resources Tsybikova Tamara Radnazhapovna, teacher of mathematics 08.02.2014 23
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Chapter 9. Elements of mathematical statistics, combinatorics and probability theory
§54. Random events and their probabilities 3. INDEPENDENT REPETITIONS OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.
slide 2
Content
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 paragraphs 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, teacher of mathematics
*
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3. INDEPENDENT REPETITIONS OF TESTS. BERNULLI'S THEOREM AND STATISTICAL STABILITY.
Part 3
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
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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, teacher of mathematics
*
slide 5
Solution of 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, teacher of mathematics
*
slide 6
Solution of 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, teacher of mathematics
*
Slide 7
Solution of 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, teacher of mathematics
*
Slide 8
Solution of 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, teacher of mathematics
*
Slide 9
Note
The solution given in point 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, teacher of mathematics
*
Slide 10
In the whole series of repetitions, it is important to know
In each of these repetitions, we are interested in the question of whether this event will or will not occur. 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 4 will come up 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 in five tosses of a coin, heads will come up exactly 4 times?
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
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 some test be equal to P (A). We will consider this test as a test with only two possible outcomes: one outcome is that the event A will occur, and the other outcome is that the event A will not occur, i.e., the event Ᾱ will occur. For brevity, let's call the first outcome (the occurrence of the event A) "success", and the second outcome (the occurrence of the 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. So q = P(Ᾱ) = 1 - P(A) = 1 - p.
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
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THEOREM 3 (Bernoulli's theorem)
Theorem 3 (Bernoulli's theorem). Let Pn(k) be the probability of exactly k "successes" in n independent repetitions of the same test. 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 give it without proof) is of great importance for theory, and for practice.
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
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EXAMPLE 6.
Example 6. In each of paragraphs a) - d) determine the values of n, k, p, q and write out (without calculations) the expression for the desired probability Pn(k). coins? b) Each of the 20 people independently names one of the days of the week. "Unlucky" days are Monday and Friday. What is the probability that "good luck" will be exactly half? What is the probability that exactly 5 tosses out of 25 will be "successful"? d) The test consists of tossing three different coins at the same time. "Failure": more "tails" than "eagles". What is the probability that there will be exactly three "lucks" among 7 rolls?
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
Slide 14
Solution 6a)
Example 6. In each of paragraphs a) - d) determine the values of n, k, p, q and write out (without calculations) the expression for the desired probability Pn(k). coins? Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
slide 15
Solution 6b)
Example 6. In each of paragraphs 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. "Unlucky" days are Monday and Friday. What is the probability that there will be exactly half of "good luck"? Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
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Solution 6c)
Example 6. In each of paragraphs a) - d) determine the values of n, k, p, q and write out (without calculations) the expression for the desired probability Pn(k). . What is the probability that exactly 5 throws out of 25 will be "lucky"? Solution:
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
Slide 17
Solution 6d)
Example 6. In each of paragraphs a) - d) determine the values of n, k, p, q and write out (without calculations) the expression for the desired probability Pn(k). d) The test consists in the simultaneous throwing of three different coins. "Failure": more "tails" than "eagles". What is the probability that there will be exactly three “lucks” among 7 tosses? Solution: d) n = 7, k = 3. “Luck” on one toss is that there are fewer “tails” than “eagles”. A total of 8 results are possible: PPP, PPO, POP, OPP, POO, ORO, OOP, OOO (P - “tails”, O - “heads”). Exactly half of them have fewer tails than heads: POO, ORO, OOP, OOO. So p = q = 0.5; Р7(3) = С73 ∙ 0.53 ∙ 0.54 = С73 ∙ 0.57.
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
Slide 18
Bernoulli's theorem allows...
Bernoulli's theorem makes it possible 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 of § 50 on the statistical processing of information. Consider a sequence of n independent repetitions of the same test with two outcomes - "success" and "failure". The results of these tests constitute a series of data, consisting of some sequence of two options: "success" and "failure". Simply put, there is a sequence of length n, made up of two letters Y ("good 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 the options Y, i.e. we will 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 complicated mathematical fact is derived precisely from Bernoulli's theorem.
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
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 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≈ 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, say, 520 of them gave a positive answer to the question, then k / n = 520 / 2000 = 0.26 and it is practically certain that for any larger number of respondents such a frequency will be in the range from 0.23 to 0.29. This phenomenon is called the phenomenon of statistical stability. Thus, the Bernoulli theorem and its consequences allow (approximately) finding the probability of a random event in cases where its explicit calculation is impossible.
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
Slide 20
For the teacher
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
slide 21
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
slide 22
08.02.2014
Tsybikova Tamara Radnazhapovna, teacher of mathematics
*
slide 23
Sources
Algebra and the Beginnings 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 guide for teachers, A.G. Mordkovich, P.V. Semenov, M., 2010 Tables are compiled in MS Word and MS Excel. Internet resources
Tsybikova Tamara Radnazhapovna, teacher of mathematics
08.02.2014
*
slide 1
Bernoulli's theorem
17.03.2017
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A series of n independent trials is performed. Each test has 2 outcomes: A - "success" and - "failure". The probability of "success" in each test is the same and is equal to P(A) = p Accordingly, the probability of "failure" also does not change from experience to experience and is equal.
Bernoulli scheme
What is the probability that a series of n trials will succeed k times? Find Pn(k) .
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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. We examine n products of some production, randomly selected, for quality. The shooter shoots at the target n times.
Examples
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Explain why the following questions fit into the Bernoulli scheme. Indicate what "success" consists of and what n and k are. a) What is the probability of getting a 2 three times in ten throws of a die? b) What is the probability that in 100 tosses of a coin heads will appear 73 times? c) A pair of dice is rolled twenty times in a row. What is the probability that the sum of the points has never been equal to 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 the Queen of Spades was among the drawn cards each time?
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For the number of combinations from n to k, the formula is valid
For example:
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Bernoulli's theorem
The probability Pn(k) of exactly k successes in n independent repetitions of the same test 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 the coat of arms appearing 0, 1, ...6 times? Solution. The number of experiments n=6. Event A - "success" - the loss of the coat of arms. According to the Bernoulli formula, the required probability is
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The coin is tossed 6 times. What is the probability of the coat of arms appearing 0, 1, ...6 times? Solution. The number of experiments n=6. Event A - "success" - the loss of the coat of arms.
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The coin is tossed 10 times. What is the probability that the coat of arms will appear twice? Solution. The number of experiments n=10, m=2. Event A - "success" - the loss of the coat of arms. According to the Bernoulli formula, the required probability is
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Slide 10
An urn contains 20 white and 10 black balls. 4 balls are taken out, and each ball taken out is returned to the urn before the next one is drawn and the balls in the urn are mixed. Find the probability that 2 of the 4 balls drawn are white. Solution. Event A - got white ball. Then the probabilities According to the Bernoulli formula, the required probability is
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Determine the probability that there are no girls in a family with 5 children. The probabilities of having a boy and a girl are assumed to be the same. Solution. The probability of the birth of a girl, a boy According to the Bernoulli formula, the required probability is equal to
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Find 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. The probability of the birth of a girl, a boy According to the Bernoulli formula, the required probability is equal to
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Determine the probability that a family with 5 children will have two girls. Solution. The probability of the birth of a girl, a boy According to the Bernoulli formula, the required probability is equal to
Slide 14
Find the probability that a family with 5 children will have 3 girls. Solution. The probability of the birth of a girl, a boy According to the Bernoulli formula, the required probability is equal to
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Determine the probability that a family with 5 children will have no more than 3 girls. The probabilities of having a boy and a girl are assumed to be the same. Solution. The probability of having a girl, a boy The required probability is equal to
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Among the parts processed by the worker, there are on average 4% non-standard. Find the probability that two of the 30 parts taken for testing will be non-standard. Solution. Here the experience lies in checking each of the 30 parts for quality. Event A - "appearance of a non-standard part",
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FEDERAL AGENCY FOR EDUCATION
State educational institution
higher professional education
"MATI" - RUSSIAN STATE TECHNOLOGICAL UNIVERSITY IM. K.E. TSIOLKOVSKY
Department of Systems Modeling and Information Technology
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
The guidelines are intended for students of the day and evening departments of faculty No. 14, specialties 150601, 160301, 230102. The guidelines highlight the basic concepts of the topic, determine the sequence of studying the material. A large number of considered examples helps in the practical development of the topic. Guidelines serve as a methodological basis for practical exercises and the implementation of individual tasks.
BERNULLI SCHEME. BERNULLI FORMULA
Bernoulli scheme- a scheme of repeated independent tests, in which some event BUT can be repeated many times with constant probability R (BUT)= R .
Examples of tests carried out according to the Bernoulli scheme: multiple tossing of a coin or a dice, making a batch of parts, shooting at a target, etc.
Theorem. If the probability of an event occurring BUT in each test is constant and equal R, then the probability that the event BUT will come m once a n tests (no matter in what sequence), can be determined by the Bernoulli formula:
where q = 1 – p.
EXAMPLE 1. The probability that the consumption of electricity during one day will not exceed the established norm is equal to p= 0,75. Find the probability that in the next 6 days the electricity consumption for 4 days will not exceed the norm.
SOLUTION. The probability of normal power consumption during each of the 6 days is constant and equal to R= 0.75. Therefore, the probability of overexpenditure of electricity every day is also constant and equal to q = 1R = 1 0,75 = 0,25.
The desired probability according to the Bernoulli formula is equal to:
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) no targets; 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\u003d 1 - 0.3 \u003d 0.7. Total number of experiments 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:
c) The probability of three misses with three shots is equal to:
d) The probability of hitting at least one target with three shots is equal to:
e) Probability of hitting less than two targets, i.e. either one target or none:
Moivre-Laplace local and integral theorems
If a large number of tests are made, then the calculation of 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 for large n. These formulas are called asymptotic and are defined by Poisson's theorem, Laplace's local and integral theorems.
Local de Moivre-Laplace theorem. BUT BUT happen m once a n n (n →∞ ), is approximately equal to:
where is the function 
and the argument 
The more n, the more accurate the calculation of probabilities. Therefore, it is advisable to apply the Moivre-Laplace theorem when npq 20.
f ( x ) special tables were compiled (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 the event BUT occurs 80 times in 400 trials if the probability of occurrence of the event is BUT in each test is p= 0,2.
SOLUTION. By condition n=400, m=80, p=0,2, q=0.8. Consequently:
According to the table, we determine the value of the function f (0)=0,3989.
Moivre-Laplace integral theorem. If the probability of an event occurring BUT in each trial is constant and different from 0 and 1, then the probability that the event BUT come from m 1 before m 2 once a n tests with a sufficiently large number n (n →∞ ), is approximately equal to:
where 
- integral or Laplace function, 
To find function values F( x ) special tables have been drawn up (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, it can be considered that X>5 function Ф(x) ≈0,5.
F (0)=0.
EXAMPLE 4. The probability that the part has not passed the Quality Control Department check is 0.2. Find the probability that 70 to 100 items will be unchecked among 400 items.
SOLUTION. By condition n=400, m 1 =70, m 2 =100, p=0,2, q=0.8. Consequently:
According to the table in which the values of the Laplace function are given, we determine:
Ф(x 1 ) = F( 1,25 )= F( 1,25 )= 0,3944; Ф(x 2 ) = F( 2,5 )= 0,4938.
A series of independent trials is underway,each of which has 2 possible outcomes,
which we will conditionally call Success and Failure.
For example, a student takes 4 exams, in each
of which 2 outcomes are possible Success: student
passed the exam and Failed: failed. The probability of success in each trial is
p. The Failure Probability is q=1-p.
It is required to find the probability that in the series
out of n trials, success will come m times
Pn(m) Bm Ó Ó ... Ó Í ... Í
Í Ó ... Ó Í ... Í ...
Í Í ... Í Ó ... Ó
In each case, Success occurs m times, and
Failed (n-m) times.
Number
all
combinations
equals
number
ways from n trials to choose those m, in
which was Success, i.e. Cm
n The probability of each such combination is
theorem
about
multiplication
probabilities
will be Pmqn-m.
Since these combinations are incompatible, then
the desired probability of the event Bm will be
Pn (m) p q
m
n m
... p q
m
n m
total C s lags û õ C p q
m
n
m
n
m
n m Pn (m) C p q
m
n
m
n m It is known that if a coin falls on heads, a student
goes to the movies if the coin lands on tails
students. What is the probability that
1) three of them will be at the lecture
2) there will be at least 3 students at the lecture
2) will at least one of the students get to the lecture? 1) In this problem, a series of n=5
independent tests. Let's call it Success
going to a lecture (tails falling out) and
Failure - going to the cinema (falling out of the coat of arms).
p=q=1/2.
Using the Bernoulli formula, we find the probability that
What will happen 3 times after 5 tosses of a coin?
success:
3
2
1 1
P5(3)C
2 2
5! 1 1
1
10
0,3125
3!2! 8 4
32
3
5To find the probability that after 5 tosses
at least once the coin will land tails,
let's move on to the probability of the opposite
events - the coin will drop out all 5 times with the coat of arms:
P5 (0).
Then the desired probability will be: P=1-P5(0).
According to the Bernoulli formula:
0
5
1 1
P5(0)C
2 2
0
5
5
1
0,03125
2 Then the probability of the desired event will be
P1 0.03125 0.96875
Bernoulli
student goes
in the cinema, if the coin falls tails - the student goes to
lecture. The coin was tossed by 5 students. What is the most
probable number of students going to the lecture?
Probability
winnings for 1 ticket is 0.2. What is the most
probable number of winning tickets? The most likely number of successes in the scheme
Bernoulli
np q k np p The most likely number of successes in the scheme
Bernoulli
Formula for Most Likely Number of Successes
np q k np p
If np-q is an integer, then this interval contains 2
whole numbers. Both are equally incredible.
If np-q is a non-integer number, then this interval contains 1
integer The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
- The student goes to the lecture. Coin tossed 5
students going to lecture?
np q k np p
n 5
1
p q
2The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to lecture?
np q k np p
n 5
1
p q
2
1 1
np q 5 2
2 2
1 1
np p 5 3
2 2The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to lecture?
np q k np p
n 5
1
p q
2
1 1
np q 5 2
2 2
1 1
np p 5 3
2 2
2 k 3 k 2, k 3 The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to lecture?
2
3
3
2
5
1 1
1 10 5
P5 (2) C52 10
32 16
2 2
2
5
1 1
1 10 5
P5 (3) C53 10
32 16
2 2
2 The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to lecture?
probability, Pn(k)
Probabilities of the number of students who attended
lecture
0,35
0,3
0,25
0,2
0,15
0,1
0,05
0
0
1
2
3
number of students, k
4
5The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.
tickets?
np q k np p
n 10
p 0.2 q 0.8 The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
np q k np p
n 10
p 0.2 q 0.8
np q 10 0.2 0.8 1.2
np p 10 0.2 0.2 2.2 The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
np q k np p
n 10
p 0.2 q 0.8
np q 10 0.2 0.8 1.2
1, 2 to 2, 2
np p 10 0.2 0.2 2.2
k2 The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
P10 (2) C 0.2 0.8
2
10
2
8
45 0, 04 0,16777216=
=0,301989888The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
Probabilities of the number of winning tickets
probability, Pn(k)
0,35
0,3
0,25
0,2
0,15
0,1
0,05
0
0
1
2
3
4
5
6
number of tickets, k
7
8
9
10The most likely number of successes in the scheme
Bernoulli
10 contracts signed
pay the sum insured
one of the contracts
than three contracts
d) find the most probable number of contracts, according to
who will have to pay the sum insured The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
a) Find the probability that three
pay the sum insured
0,201327The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
b) The sum insured will not have to be paid out under any
one of the contracts
0,107374The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
c) the amount insured will have to be paid no more than,
than three contracts
0,753297If n is large, then using the formula
Pn (m) C p q
m
n
m
n m
difficult
Therefore, approximate formulas are used Theorem: If the probability p of the occurrence of the event A
in each test is close to zero,
and the number of independent trials n is large enough,
then the probability Pn(m) that in n independent trials
event A will occur m times, approximately equal to:
Pn(m)
m
m!
e
where λ=np
This formula is called the Poisson formula (law of rare events) Pn(m)
m
m!
e, np
Usually the approximate Poisson formula is used,
when p<0,1, а npq<10.
Example Let it be known that in the manufacture of a certain drug
marriage (the number of packages that do not meet the standard)
is 0.2%. Estimate the probability that
after 1000 randomly selected packages, there will be three packages,
not meeting the standard.
Pn(k)
k
k!
P1000(3) ?
e ,
np Example Let it be known that in the manufacture of a certain drug
marriage (the number of packages that do not meet the standard)
is 0.2%. Estimate the probability that
after 1000 randomly selected packages, there will be three packages,
not meeting the standard.
Pn(k)
k
k!
P1000(3) ?
e, np
np 1000 0.002 2
3
2 2 8
P1000 (3) e 0.135=0.18
3!
6
no more than 5 contracts are connected. Example On average, for 1% of contracts, the insurance company
pays the sum insured. Find the probability that from
100 contracts with the occurrence of an insured event will be
no more than 5 contracts are connected.
