Solution Methods combinatorial problems
List of possible options
Simple problems are solved by an ordinary complete enumeration of possible options without compiling various tables and schemes.
Task 1.
What two-digit numbers can be formed from the numbers 1, 2, 3, 4, 5?
Answer: 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55.
Task 2.
Ivanov, Gromov and Orlov participate in the final 100m race. name possible options distribution prizes.
Answer:
Option 1: 1) Ivanov, 2) Gromov, 3) Orlov.
Option 2: 1) Ivanov, 2) Orlov, 3) Gromov.
Option 3: 1) Orlov, 2) Ivanov, 3) Gromov.
Option 4: 1) Orlov, 2) Gromov, 3) Ivanov.
Option 5: 1) Gromov, 2) Orlov, 3) Ivanov.
Option 6: 1) Gromov, 2) Ivanov, 3) Orlov.
Task 3.
Petya, Kolya, Vitya, Oleg, Tanya, Olya, Natasha, Sveta signed up for the ballroom dance club. What dance pairs of a girl and a boy can form?
Answer:
1) Tanya - Petya, 2) Tanya - Kolya, 3) Tanya - Vitya, 4) Tanya - Oleg, 5) Olya - Petya, 6) Olya - Kolya, 7) Olya - Vitya, 8) Olya - Oleg, 9) Natasha - Petya, 10) Natasha - Kolya, 11) Natasha - Vitya, 12) Natasha - Oleg, 13) Sveta - Petya, 14) Sveta - Kolya, 15) Sveta - Vitya, 16) Sveta - Oleg.
Tree of possible options
A variety of combinatorial problems are solved by drawing up special schemes. Outwardly, such a scheme resembles a tree, hence the name of the method - tree of possible options.
Task 4.
What three-digit numbers can be formed from the numbers 0, 2, 4?
Solution.Let's build a tree of possible options, given that 0 cannot be the first digit in a number.
Answer: 200, 202, 204, 220, 222, 224, 240, 242, 244, 400, 402, 404, 420, 422, 424, 440, 442, 444.
Task 5.
School tourists decided to make a trip to the mountain lake. The first stage of the journey can be overcome by train or bus. The second stage is by kayaks, bicycles or on foot. And the third stage of the journey is on foot or by cable car. What travel options do school tourists have?
Solution.Let's build a tree of possible options, designating the journey by train P, by bus - A, by kayak - B, by bicycle - C, on foot - X, by cable car - K.
Answer:The figure lists all 12 possible travel options for school tourists.
Task 6.
Write down all possible options for the schedule of five lessons per day from the subjects: mathematics, Russian, history, English language, physical education, and mathematics should be the second lesson.
Solution.Let's build a tree of possible options, denoting M - mathematics, R - Russian, I - history, A - English, F - physical education.
Answer:There are 24 possible options in total:
|
R |
R |
R |
R |
R |
R |
And |
And |
And |
And |
And |
And |
BUT |
BUT |
BUT |
BUT |
BUT |
BUT |
F |
F |
F |
F |
F |
F |
Task 7.
Sasha goes to school in trousers or jeans, and wears gray, blue, green or plaid shirts for them, interchangeable shoes takes shoes or sneakers.
a) How many days will Sasha be able to look in a new way?
b) How many days will he walk in sneakers?
c) How many days will he wear a plaid shirt and jeans?
Solution.Let's build a tree of possible options, denoting B - trousers, D - jeans, C - gray shirt, D - blue shirt, G - green shirt, P - checkered shirt, T - shoes, K - sneakers.
Answer:a) 16 days; b) 8 days; c) 2 days.
Tabulation
You can solve combinatorial problems using tables. They, like the tree of possible options, visually represent the solution of such problems.
Task 8.
How many odd two-digit numbers can be made from the numbers 1, 3, 4, 6, 7, 8, 9?
Solution.Let's make a table: on the left, the first column is the first digits of the numbers you are looking for, at the top, the first row is the second digits.
Answer: 28.
Task 9.
Masha, Olya, Vera, Ira, Andrey, Misha and Igor were preparing to become presenters at new year holiday. Name the possible options if only one girl and one boy can be the leaders.
Solution.Let's make a table: on the left, the first column is the names of the girls, at the top, the first row is the names of the boys.
Answer:All possible options are listed in the rows and columns of the table.
multiplication rule
This method of solving combinatorial problems is used when it is not required to list all possible options, but it is necessary to answer the question - how many of them exist.
Task 10.
AT football tournament several teams are involved. It turned out that they all used white, red, blue and white for shorts and T-shirts. green colors and all possible options were presented. How many teams participated in the tournament?
Solution.
Briefs can be white, red, blue or green, i.e. there are 4 options. Each of these options has 4 jersey color options.
4 x 4 = 16.
Answer: 16 teams.
Task 11.
6 students pass a test in mathematics. In how many ways can they be placed on the list?
Solution.
The first in the list can be any of the 6 students,
the second in the list can be any of the remaining 5 students,
third - any of the remaining 4 students,
fourth - any of the remaining 3 students,
fifth - any of the remaining 2 students,
sixth - the last 1 student.
6 x 5 x 4 x 3 x 2 x 1 = 720.
Answer: 720 ways.
Task 12.
How many even two-digit numbers can be made from the digits 0, 2, 3, 4, 6, 7?
Solution.
First in double digit there can be 5 digits (number 0 cannot be the first in a number), the second in a two-digit number can be 4 digits (0, 2, 4, 6, since the number must be even).
5 x 4 = 20.
Answer: 20 numbers.
Tasks for solving the consolidation of new material
Task #1. In how many ways can the 5 participants in the final
running on 5 treadmills?
Solution: R 5 \u003d 5! \u003d 1 ∙ 2 ∙ 3 ∙ 4 ∙ 5 \u003d 120 ways.
Task number 2. How three-digit numbers can be made up of the numbers 1,2,3, if each
Does the digit appear in the image of the number only once?
Solution: The number of all permutations of three elements is P 3 =3!, where 3!=1 * 2 * 3=6
This means that there are six three-digit numbers made up of the numbers 1,2,3.
Task number 3. In how many ways can four boys invite four out of six
girls to dance?
Solution: Two boys cannot invite the same girl at the same time. And
options in which the same girls dance with different boys,
considered different, so:
Task #4. How many different three-digit numbers can be formed from the numbers 1, 2, 3, 4, 5,
6, 7, 8, 9, provided that each digit is used only in the number entry
once?
Solution: In the condition of the problem, it is proposed to count the number of possible combinations from
three digits taken from the proposed nine digits, with the order
the arrangement of the numbers in the combination matters (for example, the numbers 132)
and 231 different). In other words, you need to find the number of placements out of nine
three elements.
According to the formula for the number of placements, we find:
Answer: 504 three-digit numbers.
Task #5 In how many ways can a committee of 3 be chosen from 7 people?
Solution: To consider all possible commissions, you need to consider all
possible 3-element subsets of the set consisting of 7
human. The desired number of ways is
Task number 6. 12 teams participate in the competition. How many options are there
distribution of prize (1, 2, 3) places?
Solution: A 12 3 = 12 ∙11 ∙10 = 1320 options for the distribution of prizes.
Answer: 1320 options.
Task number 7. At athletics competitions, our school was represented by a team from
10 athletes. In how many ways can the coach determine which of them
will run in the 4x100m relay in the first, second, third and fourth stages?
Solution: Choice from 10 to 4, taking into account the order: 
ways.
Answer: 5040 ways.
Task number 8. In how many ways can red, black, blue and
green balls?
Solution: In the first place, you can put any of the four balls (4 ways), on
second - any of the three remaining (3 ways), third place - any of
the remaining two (2 ways), in fourth place - the remaining last ball.
Total 4 3 2 1 = 24 ways.
P 4 = 4! \u003d 1 2 3 4 \u003d 24. Answer: 24 ways.
Task number 9. The students were given a list of 10 books to read during
vacation time. In how many ways can a student choose 6 books from them?
Solution: Choice 6 out of 10 without regard to order: 
ways.
Answer: 210 ways.
Task number 10. There are 7 students in grade 9, 9 students in grade 10, and 8 students in grade 11. For
work on the school site, it is necessary to single out two students from grade 9,
three out of 10, and one out of 11. How many ways are there to choose
students to work on the school grounds?
Solution: Choice from three sets without regard to order, each choice from
of the first set (C 7 2) can be combined with each choice from
second (C 9 3)) and with each choice of the third (C 8 1) according to the rule
multiplication we get:
Answer: 14,112 ways.
Task number 11. Ninth-graders Zhenya, Seryozha, Kolya, Natasha and Olya ran to
change to the tennis table, at which the game was already underway. How many
five ninth graders who ran up to the table can take
queue for table tennis?
Solution: Any ninth-grader could be the first in line, any of them could be the second.
the remaining three, the third - any of the remaining two and the fourth -
a ninth-grader who ran up the penultimate one, and the fifth one was the last one. By
multiplication rule, five students have 5 4321=120 ways
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1 1 Basic concepts of combinatorics 1 Appendix Definition The product of all natural numbers from 1 to n inclusive is called the n-factorial and is written Example Calculate 4! 3! n! 1 3 n 4!-3!= ! 5! Example Calculate! 7! 5! 5!! Let three letters of these letters be given: 7 1! Permutations 5 3 A, B, C Let's make all possible combinations of ABC / ACB / BCA / CAB / CBA / BAC (total combinations) We see that they differ from each other only in the order of the letters Definition Combinations of n elements that differ from each other from each other only by the order of elements, are called permutations Permutations are denoted by the symbol n, where n is the number of elements included in each permutation 3 3! The number of permutations can be calculated using the formula n or using the factorial: n n 1 n 3 1 n n! Thus, the number of permutations of three elements according to the formula is, which coincides with the result of the above example 5 0 Example Calculate,! ! !- 5! 5! -fifteen! 5! 1 5 0! ! one! Example How many different five-digit numbers can be made from the numbers 1, 3, 4, 5, provided that no digit is repeated in the number?
2 5! Example Four teams participated in the competition. How many options for the distribution of places between them are possible? four! Placement Let there be four letters A, B, C, D Compose all combinations of only two letters, we get: AB, AC, AD, BA, BC, BD, CA, CB, CD, DA, DB, DC We see that all the resulting combinations differ either by letters or by their order (combinations BA and AB are considered different) Definition Combinations of m elements by n elements that differ from each other either by the elements themselves or by the order of the elements are called placements Placements are denoted by n A m n the number of elements in each combination , where m is the number of all available elements, A n m m! (m n)! Example How many options are there for the distribution of three prizes if 7 teams participate in the draw? 3 7! 7! A! four! 10 Example How many different four-digit numbers can be made from the numbers 0, 1, 8, 9? 4 10! ten! A!! Example How many schedule options can be created for one day, if there are 8 in total subjects, and only three of them can be included in the schedule for the day? 3 8! eight! A! 5! Example How many options for the distribution of three vouchers to a sanatorium of various profiles can be made for five applicants? 3 5! 5! A!!
3 Combinations Definition Combinations are all possible combinations of m elements by n, which differ from each other in at least at least one element (here m and n integers, and n 4 A random phenomenon can be characterized by the ratio of the number of its occurrences to the number of trials, in each of which, under the same conditions of all trials, it could occur or not occur. In order to record and investigate these regularities, we introduce some basic concepts and definitions. Definition Any action, phenomenon, observation with several different outcomes, realized under a given set of conditions, will be called a test. Definition The result of this action or observation will be called a random event. For example, the occurrence of a number when a coin is tossed is a random event, since it may or may not have occurred. Definition If we are interested in any specific event from all possible possible events, then we will call it the desired event (or the desired outcome) Definition All the events under consideration will be considered equally possible, those that have equal chances to occur So, when throwing a die, 1-point, 3, 4, 5 or points can appear test outcomes are equally probable In other words, equality means equality, symmetry of individual test outcomes under certain conditions. Events are usually denoted by capital letters of the Latin alphabet: A, B, C, D Definition Events are called incompatible if no two of them can occur in this experiment together Otherwise, the events are called joint. So, when a coin is tossed, the appearance of a number excludes the simultaneous appearance of a coat of arms; this is an example of incompatible events 4 5 Consider another example Suppose a circle, a diamond and a triangle are drawn on the target One shot is fired Event A hitting the circle, event B hitting the diamond, event C hitting the triangle Then events A and B, A and C, C and B are incompatible Definition Event is called reliable if it happens in this test necessarily For example, winning on a win-win lottery ticket is a reliable event Reliable events are denoted by the letter U Definition An event is called impossible if it cannot occur in this test For example, when throwing a dice, it is impossible to get 7 points Impossible event denoted by the letter V Definition The complete system of events A 1, A, A 3, A n is a set of incompatible events, the occurrence of at least one of which is mandatory for this test So, the loss of one, two, three, four, five, six points when throwing a playing bones is a complete system of events, since all these events are incompatible and the occurrence at least one of them is required Definition If the complete system consists of two events, then such events are called opposite and are denoted by A and A Example There is one lottery ticket “6 out of 45” that he is unwinnable Are these events incompatible? Example There are 30 numbered balls in a box Determine which of the following events are impossible, certain, opposite: a numbered ball is drawn (; an even-numbered ball is drawn (an odd-numbered ball is drawn (C); a ball without a number is drawn (D) Which of them form a complete group?Example Are the events true or impossible that a single throw of a die will result in: 5 points; 7 points; from 1 to points? Which events in this trial make up a complete group? 5 6 Definition The sum of several events is an event consisting in the occurrence of at least one of them as a result of the test The sum of events A and B, denoted (A + and means that the event A, or B, or A and B together has occurred Definition The product of several events is called an event , which consists in the joint occurrence of all these events as a result of the test. The product of events A and B denotes: AB 3 Determination of the probability of an event Random events are realized with different possibilities Some occur more often, others less often To quantify the possibilities for the realization of an event, the concept of the probability of an event is introduced Definition The probability of an event A is the ratio of the number M of favorable outcomes to the total number N of equally likely outcomes forming a complete group: The probability of a certain event is 1, impossible 0, random: 0 (1 This is the classical definition of probability Relative frequency of an event A n tests: M N * (Example One letter is chosen at random from the word "polyclinic" What is the probability that this is a vowel? What is the letter K? What is a vowel or letter K? Total letters 11 Event A as a result of the experiment a vowel letter appeared Event B the letter K appeared Event A is favored by five events (5 vowels), event B is favored by two m 5 m (, n 11 n 11 m n 4 Basic theorems and formulas of probability theory Probability addition theorem Probability the occurrence of one of the incompatible events is equal to the sum of their probabilities: 7 A A A A A 1 n 1 A n The probability of the sum of two joint events A A The sum of the probabilities of opposite events (1 Definition Let A and B be two random events of the same test Designation: A B A Probability multiplication theorem The probability of the simultaneous occurrence of two independent events is equal to the product of the probabilities of these events A 7 Mathematics (BkPl-100) M.P. Kharlamov 2011/2012 academic year, 1st semester Lecture 5. Topic: Combinatorics, introduction to probability theory 1 Topic: Combinatorics Combinatorics is a branch of mathematics that studies TOPIC. THEOREMS OF ADDITION AND MULTIPLICATION OF PROBABILITIES Operations on random events. Algebra of events. The concept of the compatibility of events. Complete group of events. Dependence and independence of random events. Conditional Lecture Probability theory Basic concepts Experiment Frequency Probability Probability theory is a section of mathematics that studies the patterns of random phenomena Random events are events that, when LESSON 3 INTRODUCTION TO PROBABILITY THEORY METHODOLOGICAL RECOMMENDATIONS MISIS 2013 APPROVED: D.E. Kaputkin Chairman of the Educational and Methodological Commission for the implementation of the Agreement with the Department of Education of the mountains. 1 PART I. PROBABILITY THEORY CHAPTER 1. 1. Elements of combinatorics Definition 1. Examples: Definition. -factorial is the number denoted by !, while! = 1** * for all natural numbers 1, ; Moreover, 1) How many three-digit natural numbers are there that have only two digits less than five? There are only five digits less than 5: ( 0; 1; 2; 3; 4 ) The remaining five digits are not less than 5: ( ; ; ; ; ) 1st solution method Lecture 3 Topic Main theorems and formulas of probability theory Contents of the topic Algebra of events. Theorems of addition of probabilities. Conditional Probability. Probability multiplication theorems. Total Probability Formula. Lecture Topic: ALGEBRA OF EVENTS BASIC THEOREMS ON PROBABILITY Algebra of events The sum of events is called the event S = +, which consists in the occurrence of at least one of them The product of events and is called Department of Higher Mathematics Lectures on Probability Theory and Mathematical Statistics Section. Probability theory The subject of probability theory is the study of specific regularities in mass homogeneous CONTENTS THEME III. INTRODUCTION TO PROBABILITY THEORY... 2 1. REFERENCE MATERIALS... 2 1.1. BASIC CONCEPTS AND DEFINITIONS... 2 1.2. ACTIONS ON RANDOM EVENTS... 4 1.3. CLASSICAL DEFINITION Lecture 2. Theorems of addition and multiplication of probabilities The sum and product of an event FEDERAL STATE BUDGETARY EDUCATIONAL INSTITUTION OF HIGHER PROFESSIONAL EDUCATION "Chelyabinsk State Academy of Culture and Art" Department of Informatics PROBABILITY THEORY PROBABILITY OF A RANDOM EVENT Kolmogorov's axioms In 1933, AN Kolmogorov in his book "Basic Concepts of Probability Theory" gave an axiomatic substantiation of the theory of probability. "This means that after NORTHERN DISTRICT DEPARTMENT OF EDUCATION WORKING PROGRAM Probability theory and statistics class Teaching aids used: Textbook: Tyurin Yu.N. etc. Theory of Probability and Statistics. M., MTsNMO: JSC Federal Agency for Education State Educational Institution of Higher Professional Education "NATIONAL RESEARCH TOMSK POLYTECHNICAL UNIVERSITY" LECTURE ON THEORY COMBINATORIAL PROBABILITY Topic 5 Lecture content 1 Introduction 2 3 4 Next paragraph 1 Introduction 2 3 4 Problem... Problem... Problem... ... and solution: Girl LECTURE OF DETERMINING THE PROBABILITY OF AN EVENT Probability of an event refers to the basic concepts of probability theory and expresses a measure of the objective possibility of an event occurring For practical activities, it is important I Definition of probability and the basic rules for its calculation Probabilistic experiment Subject of probability theory Problem book Chudesenko, probability theory, variant Two dice are thrown. Determine the probability that: a the sum of the number of points does not exceed N ; b the product of the number of points does not exceed N; in Compiled by: Associate Professor of the Department of Medical and Biological Physics Romanova N.Yu. Probability theory 1 lecture Introduction. Probability theory is a mathematical science that studies the patterns of random phenomena. MVDubatovskaya Theory of Probability and Mathematical Statistics Lecture 3 Methods for determining probabilities 0 Classical definition of probabilities We call any of the possible results of an experiment elementary Lecture 3 Topic Main theorems and formulas of probability theory Contents of the topic Algebra of events. Theorems of addition of probabilities. Conditional Probability. Probability multiplication theorems. Main categories of algebra Lecture 1. Topic: BASIC APPROACHES TO THE DETERMINATION OF PROBABILITY The subject of probability theory. Historical background M.P. Kharlamov http://vlgr.ranepa.ru/pp/hmp Synopsis Probability Theory and Mathematical Statistics Summary of the first section (questions and answers) Dr. phys.-math. Sciences Professor Mikhail Pavlovich Kharlamov Probability Theory Lecture Plan P On Probability as a Science P Basic Definitions of Probability P Frequency of a Random Event Definition of Probability P 4 Applying Combinatorics to Counting Elements of probability theory Random events Deterministic processes In science and technology, processes are considered, the outcome of which can be predicted with certainty: If a difference is applied to the ends of the conductor TOPIC 1 Combinatorics Calculation of probabilities Problem 1B 17 teams take part in the national football cup. How many ways are there to distribute gold, silver and bronze medals? Because the ( σ-algebra - field of random events - first group of Kolmogorov's axioms - second group of Kolmogorov's axioms - basic formulas of probability theory - probability addition theorem - conditional probability Fundamentals of probability theory Lecture 2 Contents 1. Conditional probability 2. Probability of the product of events 3. Probability of the sum of events 4. Total probability formula Dependent and independent events Definition N. G. TAKTAROV PROBABILITY THEORY AND MATHEMATICAL STATISTICS: A BRIEF COURSE WITH EXAMPLES AND SOLUTIONS The text has been corrected and supplemented ABSTRACT The book is a textbook in which it is concise, simple and accessible Ministry of Education and Science of the Russian Federation Federal State Budgetary Educational Institution of Higher Professional Education "Saratov State Socio-Economic University" Problems in probability theory and mathematical statistics. Random events Task. In a batch of N items, items have a hidden defect. What is the probability that out of k items taken at random, PROBABILITY THEORY. TASKS. Table of contents (by topic) 1. The formula for the classical definition of probability. Elements of combinatorics. Geometric probability 4. Operations on events. Addition and multiplication theorems Combinatorial Formulas Let there be a set consisting of n elements. Let us denote it by U n. A permutation of n elements is a given order in the set U n. Examples of permutations: 1) distribution CHAPTER 5 ELEMENTS OF PROBABILITY THEORY 5 Axioms of Probability Theory Various events can be classified as follows: Impossible event An event that cannot happen Certain event PRCTICUM Basic formulas of combinatorics Types of events Actions on events Classical probability Geometric probability Basic formulas of combinatorics Combinatorics studies the number of combinations, Total Probability Formula. Let there be a group of events H 1, H 2,..., H n with the following properties: 1) All events are pairwise incompatible: H i H j =; i, j=1,2,...,n; ij 2) Their union forms MINISTRY OF EDUCATION OF THE RUSSIAN FEDERATION IVANOVO STATE ENERGY UNIVERSITY DEPARTMENT OF HIGHER MATHEMATICS METHODIC INSTRUCTIONS ON PROBABILITY THEORY Compiled by: MINISTRY OF CULTURE OF THE RUSSIAN FEDERATION FEDERAL STATE BUDGET EDUCATIONAL INSTITUTION OF HIGHER VOCATIONAL EDUCATION "ST. Probability Theory and Mathematical Statistics Doctor of Phys.-Math. Sciences Professor Mikhail Pavlovich Kharlamov "Page" with methodological materials http://inter.vags.ru/hmp Volgograd branch of the RANEPA (FGOU Vorobyov V.V. "Lyceum", Kalachinsk, Omsk region Workshop on solving problems in probability theory and mathematical statistics Probability Theory and Mathematical Statistics Doctor of Phys.-Math. Sciences Professor Mikhail Pavlovich Kharlamov "Page" with methodological materials http://vlgr.ranepa.ru/pp/hmp Volgograd branch of the RANEPA PROBABILITY THEORY. DISTRIBUTION OF RANDOM VALUES Task. Choose the correct answer:. The relative frequency of a random event A is a value equal to ... a) the ratio of the number of cases that favor BASIC CONCEPTS OF PROBABILITY THEORY. 3.1. Random events. Each science, when studying the phenomena of the material world, operates with certain concepts, among which there are necessarily fundamental ones; Higher professional education Undergraduate V. S. Mkhitaryan, V. F. Shishov, A. Yu. Kozlov Probability theory and mathematical statistics Textbook Recommended by the Educational and Methodological Association for Education CONTENTS SECTION I. PROBABILITY THEORY Foreword........................................................... ......... 6 PART I. RANDOM EVENTS .................................. 7 CHAPTER 1. Elements combinatorial analysis ........................ Probability Theory and Mathematical Statistics Doctor of Phys.-Math. Sciences Professor Mikhail Pavlovich Kharlamov Internet resource with methodological materials http://www.vlgr.ranepa.ru/pp/hmp Volgograd branch Chiv through S the event, consisting in the fact that the system is not closed, can be written: S = A 1 A 2 + B = (A 1 + A 2) + B. 2.18. Similarly to the solution of Problems 2.5, 2.6, we obtain S = A(B 1 +B 2) C D; S = A + B 1 B 2 + C Ministry of Education and Science of the Russian Federation FEDERAL STATE BUDGET EDUCATIONAL INSTITUTION OF HIGHER VOCATIONAL EDUCATION "KAZAN PROBABILITY THEORY Combinatorics, product and sum rules Combinatorics as a science Combinatorics is a branch of mathematics that studies the compounds of a subset of elements extracted from Federal Agency for Education Tomsk State University of Control Systems and Radioelectronics NE Lugina WORKSHOP ON PROBABILITY THEORY Textbook Tomsk 2006 Reviewers: Cand. Lecture Random events Definition. An elementary outcome (or an elementary event) is any simple (i.e., indivisible within the framework of a given experience) outcome of an experience. The set of all elementary outcomes FEDERAL AGENCY FOR EDUCATION State Educational Institution of Higher Professional Education Ulyanovsk State Technical University S. G. Valeev S. V. Kurkina Test 4. Probability theory The test on this topic includes four tasks. Let us present the basic concepts of probability theory necessary for their implementation. To solve problems 50 50, knowledge of the topic is required Section "Probability and statistics" E.M. Udalova. Primorsky district, school 579 Probability theory is a mathematical science that allows you to find the probabilities of other random events from the probabilities of some random events Problem 1. There are 40 balls in an urn. The probability that 2 drawn balls are white is 7 60. How many white balls are in the urn? The number of ways in which k items can be chosen from n is equal to C k 4 Ministry of Education and Science of the Russian Federation State educational institution of higher professional education "Khabarovsk State Academy of Economics and Law" Department FEDERAL AGENCY FOR EDUCATION OF THE MINISTRY OF EDUCATION AND SCIENCE OF THE RUSSIAN FEDERATION GOUVPO "Perm State University" Assoc. V.V. Morozenko UDC 59. (075.8) Department of Higher Mathematics Theory FEDERAL AGENCY FOR EDUCATION State Educational Institution of Higher Professional Education "Tomsk Polytechnic University" L. I. Konstantinova PROBABILITY THEORY AND MATHEMATICAL Federal Agency of Railway Transport Branch of the Federal State Budgetary Educational Institution of Higher Professional Education "Siberian State University Definition of a matrix determinant A square matrix consists of one element A = (a). The determinant of such a matrix is A = det(a) = a. () a a Square matrix 2 2 consists of four elements A = TOMSK COLLEGE OF RAILWAY TRANSPORT BRANCH SGUPS COLLECTION OF INDIVIDUAL TASKS “Elements of combinatorics. Fundamentals of Probability Theory" disciplines Probability Theory and Mathematical Statistics
À. Ì. Ïîïîâ, Â. Í. Ñîòíèêîâ ÒÅÎÐÈß ÂÅÐÎßÒÍÎÑÒÅÉ È ÌÀÒÅÌÀÒÈ ÅÑÊÀß ÑÒÀÒÈÑÒÈÊÀ Âûñøàÿ ìàòåìàòèêà äëÿ ýêîíîìèñòîâ Ó ÅÁÍÈÊ ÄËß ÁÀÊÀËÀÂÐÎÂ Ïîä ðåäàêöèåé À. Ì. Ïîïîâà Ðåêîìåíäîâàíî Ó åáíî-ìåòîäè åñêèì öåíòðîì MINISTRY OF EDUCATION AND SCIENCE OF RUSSIA Federal State Budgetary Educational Institution of Higher Professional Education "Ukhta State Technical University" (USTU) Workshop on discipline MVDubatov Probability Theory and Mathematical Statistics Lecture 4 Addition and Multiplication Theorems Total Probability Formula Bayes Formula Let and B be incompatible events and probabilities TASKS: 1. Using curly brackets, write down the set of natural numbers located on the ray between the numbers 10 and 15. Which of the numbers 0; ten; eleven; 12; fifteen; 50 belong to this set? 2. Write down the set FEDERAL EDUCATIONAL AGENCY State Educational Institution of Higher Professional Education "NATIONAL RESEARCH TOMSK POLYTECHNICAL UNIVERSITY" L.I. KONSTANTINOV Lecture 5 Topic Bernoulli scheme. Topic content Bernoulli scheme. Bernoulli formula. The most probable number of successes in the Bernoulli scheme. Binomial random variable. The main categories of Newton's binomial, scheme As a result of studying the section, the student must: know: ¾ basic concepts of combinatorics; ¾ classical definition of probability; ¾ definition of a random variable; ¾ mathematical characteristics of a random variable: mathematical expectation and variance; be able to: ¾ solve problems to find the probability of an event; ¾ solve problems for finding the mathematical expectation and variance of a random variable. Basic concepts of combinatorics In the section of mathematics called combinatorics, some problems are solved related to the consideration of sets and the compilation of various combinations of elements of these sets. For example, if we take 10 different numbers 0, 1, 2, ..., 9 and make combinations of them, we will get different numbers, for example, 345, 534, 1036, 5671, 45, etc. We see that some of these combinations differ only in the order of the digits (345 and 534), others in the numbers included in them (1036, 5671), others also differ in the number of digits (345 and 45). Thus, the obtained combinations satisfy various conditions. Depending on the composition rules, three types of combinations can be distinguished: placements, permutations and combinations. However, let's first get acquainted with the concept of factorial. The product of all natural numbers from 1 to n inclusive is called n-factorial. 1. Accommodations
. Arrangements of n elements, m each, are such connections that differ from each other either by the elements themselves or by the order of their arrangement. Example. How many two-digit numbers can be formed from the five digits 1, 2, 3, 4, 5, provided that none of them repeats? Solution. Since two-digit numbers differ from each other either by the numbers themselves or by their order, the required number is equal to the number of placements of five elements by two: Exercise. In how many ways can three persons be chosen from eight candidates for three positions? Answer: 336. 2. Permutations
. Permutations of n elements are such compounds of all n elements that differ from each other in the order of the elements. Example. Let three letters A, B, C be given. How many combinations of these letters can be made? Solution. The number of permutations of three elements can be calculated using the formula: 3! == 6. Exercise. In how many ways can 7 people be seated in 7 places? Solution. ________________________________________________________________________________________________________________________________________________________________ Answer: 5040. 3. Example.In how many ways can three attendants be chosen if there are 30 students in the class? Solution. Since you need to choose 3 out of 30 students, you can make combinations that differ from each other by at least one element, i.e. combinations of 30 to 3: Answer: 4060. Exercise. In how many ways can teams of 5 people be formed out of 15 workers? ______________________________________________________________________________________________________________________________________________________
Answer: 3003. Questions for self-control 1. List the main tasks of combinatorics. 2. What are called permutations? 3. Write down the formula for permutations of n elements. 4. What is called placements? 5. Write down the formula for the number of placements of n elements by m. 6. What are called combinations? 7. Write down the formula for the number of combinations of n elements by m. Control task PRACTICAL TASKS FOR SELF-CONTROL How many options are there for the distribution of three prizes if 7 teams participate in the draw? In how many ways can two students be selected for the conference if there are 33 people in the group? Solve Equations From a group of 15 people, a foreman and 4 members of the brigade should be selected. In how many ways can this be done? Morse code letters are made up of symbols (dots and dashes). How many letters can be represented if each letter is required to contain no more than five characters? In how many ways can four-color ribbons be made up of seven ribbons of different colors? In how many ways can four persons be chosen for four different positions from nine candidates? How many ways can you choose 3 out of 6 cards? Before graduation, a group of 30 students exchanged photographs. How many photographs were given out. In how many ways can 10 guests be seated in ten places at the festive table? How many games must 20 football teams play in a one-round championship? In how many ways can 12 people be divided into teams if there are 6 people in each team? Probability theory Nine different books are arranged at random on one shelf. Find the probability that four certain books will be placed side by side (event C). Out of 10 tickets, 2 are winning. Determine the probability that among 5 tickets taken at random, one is winning. 3 cards are drawn at random from a deck of cards (52 cards). Find the probability that it is a three, seven, ace. The child plays with the five letters of the split alphabet A, K, R, W, Y. What is the probability that with a random arrangement of letters in a row, he will receive the word "Roof". A box contains 6 white and 4 red balls. Two balls are taken at random. What is the probability that they are the same color? The first urn contains 6 black and 4 white balls, the second urn contains 5 black and 7 white balls. One ball is drawn from each urn. What is the probability that both balls are white? Random variable, mathematical expectation and variance of a random variable The probability that a student will find the book he needs in the library is 0.3. Draw up a distribution law for the number of libraries that he will visit if there are four libraries in the city. The hunter shoots at the game before the first hit, but manages to make no more than four shots. Find the variance of the number of misses if the probability of hitting the target with one shot is 0.7. Find the mathematical expectation of a random variable X, if the law of its distribution is given by the table: X R The plant has four automatic lines. The probability that during the working shift the first line will not require adjustment is 0.9, the second - 0.8, the third - 0.75, the fourth - 0.7. find the mathematical expectation of the number of lines that do not require adjustment during the work shift. X R V. ANSWERS Combinatorics Probability theory Random variable, mathematical expectation and variance of a random variable. 0,046656 2. 0,3 3. 13 EMBED Equation.3 1415 4. 13 EMBED Equation.3 1415 5.13 EMBED Equation.3 1415 6.13 EMBED Equation.3 1415. Root EntryEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation NativeEquation Native
Combinations
. Combinations of n elements, m each, are such compounds that differ from each other by at least one element.
Combinatorics
How many different five-digit numbers can be made from the digits 1, 3, 5, 7, 9, provided that no digit repeats in the number?
a) 13 EMBED Equation.3 1415. b) 13 EMBED Equation.3 1415.
How many four-digit numbers divisible by 5 can be made from the digits 0, 1, 2, 5, 7 if each number must not contain the same digits?
An urn contains 7 red and 6 blue balls. Two balls are taken out of the urn at the same time. What is the probability that both balls are red (event A)?
Write down the distribution law for the number of hits on the target with six shots, if the probability of hitting with one shot is 0.4.
1
2
3
4
0,3
0,1
0,2
0,4
Find the variance of a random variable X, knowing the law of its distribution:
0
1
2
3
4
0,2
0,4
0,3
0,08
0,02
1. 13 EMBED Equation.3 1415. 2. 13 EMBED Equation.3 1415. 3. 13 EMBED Equation.3 1415. 4. a) 13 EMBED Equation.3 1415, 5; b) 13 EMBED Equation.3 1415. 5. 13 EMBED Equation.3 1415. 6.13 EMBED Equation.3 1415. 7. 13 EMBED Equation.3 1415. 8. 13 EMBED Equation.3 1415. 9.13 EMBED Equation.3 1415. 10.13 EMBED Equation.3 1415. 11. 13 EMBED Equation.3 1415. 12. 13 EMBED Equation.3 1415. 13.190.14.924.
1. 13 EMBED Equation.3 1415 2.13 EMBED Equation.3 1415 3. 13 EMBED Equation.3 1415 4. 13 EMBED Equation.3 14155. 13 EMBED Equation.3 14156.13 EMBED Equation.3 1415 7. 13 EMBED Equation.3 1415
1.
0
1
2
3
4
5
6
0,186624
0,311040
0,276480
0,138240
0,036864
0,004096
1
2
3
4
0,21
0,147
0,343
