What is p a in probability theory. Fundamentals of probability theory and mathematical statistics

Mom washed the frame

At the end of long summer holidays it's time to slowly return to higher mathematics and solemnly open the empty Verdov file to begin creating a new section - . I admit, the first lines are not easy, but the first step is half the way, so I suggest everyone carefully study the introductory article, after which mastering the topic will be 2 times easier! I'm not exaggerating at all. …On the eve of the next September 1st, I remember first grade and the primer…. Letters form syllables, syllables form words, words form short sentences - Mom washed the frame. Mastering turver and math statistics is as easy as learning to read! However, for this you need to know key terms, concepts and designations, as well as some specific rules, which are the subject of this lesson.

But first, please accept my congratulations on the beginning (continuation, completion, note as appropriate) school year and accept the gift. The best gift is a book, and for independent work I recommend the following literature:

1) Gmurman V.E. Theory of Probability and Mathematical Statistics

Legendary tutorial, which went through more than ten reprints. It is distinguished by its intelligibility and extremely simple presentation of the material, and the first chapters are completely accessible, I think, already for students in grades 6-7.

2) Gmurman V.E. Guide to solving problems in probability theory and mathematical statistics

A solution book by the same Vladimir Efimovich with detailed examples and problems.

NECESSARILY download both books from the Internet or get their paper originals! The version from the 60s and 70s will also work, which is even better for dummies. Although the phrase “probability theory for dummies” sounds rather ridiculous, since almost everything is limited to elementary arithmetic operations. They skip, however, in places derivatives And integrals, but this is only in places.

I will try to achieve the same clarity of presentation, but I must warn that my course is aimed at problem solving and theoretical calculations are kept to a minimum. Thus, if you need a detailed theory, proofs of theorems (theorems-theorems!), please refer to the textbook. Well, who wants learn to solve problems in probability theory and mathematical statistics in the shortest possible time, follow me!

That's enough for a start =)

As you read the articles, it is advisable to become acquainted (at least briefly) with additional tasks of the types considered. On the page Ready-made solutions for higher mathematics The corresponding pdfs with examples of solutions will be posted. Significant assistance will also be provided IDZ 18.1 Ryabushko(simpler) and solved IDZ according to Chudesenko’s collection(more difficult).

1) Amount two events and the event is called which is that it will happen or event or event or both events at the same time. In the event that events incompatible, the last option disappears, that is, it may occur or event or event .

The rule also applies to a larger number of terms, for example, the event is what will happen at least one from events , A if events are incompatible – then one thing and only one thing event from this amount: or event , or event , or event , or event , or event .

There are plenty of examples:

Events (when throwing a dice, 5 points will not appear) is what will appear or 1, or 2, or 3, or 4, or 6 points.

Event (will drop no more two points) is that 1 will appear or 2points.

Event (there will be an even number of points) is what appears or 2 or 4 or 6 points.

The event is that a red card (heart) will be drawn from the deck or tambourine), and the event – that the “picture” will be extracted (jack or lady or king or ace).

A little more interesting is the case with joint events:

The event is that a club will be drawn from the deck or seven or seven of clubs According to the definition given above, at least something- or any club or any seven or their “intersection” - seven of clubs. It is easy to calculate that this event corresponds to 12 elementary outcomes (9 club cards + 3 remaining sevens).

The event is that tomorrow at 12.00 will come AT LEAST ONE of the summable joint events, namely:

– or there will be only rain / only thunderstorm / only sun;
– or only some pair of events will occur (rain + thunderstorm / rain + sun / thunderstorm + sun);
– or all three events will appear simultaneously.

That is, the event includes 7 possible outcomes.

The second pillar of the algebra of events:

2) The work two events and call an event which consists in the joint occurrence of these events, in other words, multiplication means that under some circumstances there will be And event , And event . A similar statement is true for a larger number of events, for example, a work implies that under certain conditions it will happen And event , And event , And event , …, And event .

Consider a test in which two coins are tossed and the following events:

– heads will appear on the 1st coin;
– the 1st coin will land heads;
– heads will appear on the 2nd coin;
– the 2nd coin will land heads.

Then:
And on the 2nd) heads will appear;
– the event is that on both coins (on the 1st And on the 2nd) it will be heads;
– the event is that the 1st coin will land heads And the 2nd coin is tails;
– the event is that the 1st coin will land heads And on the 2nd coin there is an eagle.

It is easy to see that events incompatible (because, for example, it cannot be 2 heads and 2 tails at the same time) and form full group (since taken into account All possible outcomes of tossing two coins). Let's summarize these events: . How to interpret this entry? Very simple - multiplication means a logical connective AND, and addition – OR. Thus, the amount is easy to read in understandable human language: “two heads will appear or two heads or the 1st coin will land heads And on the 2nd tails or the 1st coin will land heads And on the 2nd coin there is an eagle"

This was an example when in one test several objects are involved, in this case two coins. Another common scheme in practical problems is retesting , when, for example, the same die is rolled 3 times in a row. As a demonstration, consider the following events:

– in the 1st throw you will get 4 points;
– in the 2nd throw you will get 5 points;
– in the 3rd throw you will get 6 points.

Then the event is that in the 1st throw you will get 4 points And in the 2nd throw you will get 5 points And on the 3rd roll you will get 6 points. Obviously, in the case of a cube there will be significantly more combinations (outcomes) than if we were tossing a coin.

...I understand that perhaps they don’t understand very well interesting examples, but these are things that are often encountered in problems and there is no escape from them. In addition to a coin, a cube and a deck of cards, urns with multi-colored balls, several anonymous people shooting at a target, and a tireless worker who is constantly grinding out some details await you =)

Probability of event

Probability of event is the central concept of probability theory. ...A killer logical thing, but we had to start somewhere =) There are several approaches to its definition:

;
Geometric definition of probability ;
Statistical definition of probability .

In this article I will focus on the classical definition of probability, which is most widely used in educational tasks.

Designations. The probability of a certain event is indicated by a capital Latin letter, and the event itself is taken in brackets, acting as a kind of argument. For example:

Also, the small letter is widely used to denote probability. In particular, you can abandon the cumbersome designations of events and their probabilities in favor of the following style::

– the probability that a coin toss will result in heads;
– the probability that a dice roll will result in 5 points;
– the probability that a card of the club suit will be drawn from the deck.

This option is popular when solving practical problems, since it allows you to significantly reduce the recording of the solution. As in the first case, it is convenient to use “talking” subscripts/superscripts here.

Everyone has long guessed the numbers that I just wrote down above, and now we will find out how they turned out:

Classic definition of probability:

The probability of an event occurring in a certain test is called the ratio , where:

– total number everyone equally possible, elementary outcomes of this test, which form full group of events;

- quantity elementary outcomes, favorable event.

When tossing a coin, either heads or tails can fall out - these events form full group, thus, the total number of outcomes; at the same time, each of them elementary And equally possible. The event is favored by the outcome (heads). According to the classical definition of probability: .

Similarly, as a result of throwing a die, elementary equally possible outcomes may appear, forming a complete group, and the event is favored by a single outcome (rolling a five). That's why: THIS IS NOT ACCEPTED TO DO (although it is not forbidden to estimate percentages in your head).

It is customary to use fractions of a unit, and, obviously, the probability can vary within . Moreover, if , then the event is impossible, If - reliable, and if , then we are talking about random event.

! If, while solving any problem, you get some other probability value, look for the error!

In the classical approach to determining probability, extreme values (zero and one) are obtained through exactly the same reasoning. Let 1 ball be drawn at random from a certain urn containing 10 red balls. Consider the following events:

in a single trial a low-possibility event will not occur.

This is why you will not hit the jackpot in the lottery if the probability of this event is, say, 0.00000001. Yes, yes, it’s you – with the only ticket in a particular circulation. However, a larger number of tickets and a larger number of drawings will not help you much. ...When I tell others about this, I almost always hear in response: “but someone wins.” Okay, then let's do the following experiment: please buy a ticket for any lottery today or tomorrow (don't delay!). And if you win... well, at least more than 10 kilorubles, be sure to sign up - I will explain why this happened. For a percentage, of course =) =)

But there is no need to be sad, because there is an opposite principle: if the probability of some event is very close to one, then in a single trial it will almost certain will happen. Therefore, before jumping with a parachute, there is no need to be afraid, on the contrary, smile! After all, completely unthinkable and fantastic circumstances must arise for both parachutes to fail.

Although all this is lyricism, since depending on the content of the event, the first principle may turn out to be cheerful, and the second – sad; or even both are parallel.

Perhaps that's enough for now, in class Classical probability problems we will get the most out of the formula. In the final part of this article, we will consider one important theorem:

The sum of the probabilities of events that form a complete group is equal to one. Roughly speaking, if events form a complete group, then with 100% probability one of them will happen. In the simplest case, a complete group is formed by opposite events, for example:

– as a result of a coin toss, heads will appear;
– the result of a coin toss will be heads.

According to the theorem:

It is absolutely clear that these events are equally possible and their probabilities are the same .

Due to the equality of probabilities, equally possible events are often called equally probable . And here is a tongue twister for determining the degree of intoxication =)

Example with a cube: events are opposite, therefore .

The theorem under consideration is convenient in that it allows you to quickly find the probability of the opposite event. So, if the probability that a five is rolled is known, it is easy to calculate the probability that it is not rolled:

This is much simpler than summing up the probabilities of five elementary outcomes. For elementary outcomes, by the way, this theorem is also true:
. For example, if is the probability that the shooter will hit the target, then is the probability that he will miss.

! In probability theory, it is undesirable to use letters for any other purposes.

In honor of Knowledge Day, I will not ask homework=), but it is very important that you can answer the following questions:

– What types of events exist?
– What is chance and equal possibility of an event?
– How do you understand the terms compatibility/incompatibility of events?
– What is a complete group of events, opposite events?
– What does addition and multiplication of events mean?
– What is the essence of the classical definition of probability?
– Why is the theorem for adding the probabilities of events that form a complete group useful?

No, you don’t need to cram anything, these are just the basics of probability theory - a kind of primer that will quickly fit into your head. And for this to happen as soon as possible, I suggest you familiarize yourself with the lessons

Events that happen in reality or in our imagination can be divided into 3 groups. These are certain events that will definitely happen, impossible events and random events. Probability theory studies random events, i.e. events that may or may not happen. This article will present in in brief probability theory formulas and examples of solving problems in probability theory that will be in task 4 of the Unified State Exam in mathematics (profile level).

Why do we need probability theory?

Historically, the need to study these problems arose in the 17th century in connection with the development and professionalization of gambling and the emergence of casinos. This was a real phenomenon that required its own study and research.

Playing cards, dice, and roulette created situations where any of a finite number of equally possible events could occur. There was a need to give numerical estimates of the possibility of the occurrence of a particular event.

In the 20th century, it became clear that this seemingly frivolous science plays an important role in understanding the fundamental processes occurring in the microcosm. Was created modern theory probabilities.

Basic concepts of probability theory

The object of study of probability theory is events and their probabilities. If an event is complex, then it can be broken down into simple components, the probabilities of which are easy to find.

The sum of events A and B is called event C, which consists in the fact that either event A, or event B, or events A and B occurred simultaneously.

The product of events A and B is an event C, which means that both event A and event B occurred.

Events A and B are called incompatible if they cannot occur simultaneously.

An event A is called impossible if it cannot happen. Such an event is indicated by the symbol.

An event A is called certain if it is sure to happen. Such an event is indicated by the symbol.

Let each event A be associated with a number P(A). This number P(A) is called the probability of event A if the following conditions are met with this correspondence.

An important special case is the situation when there are equally probable elementary outcomes, and arbitrary of these outcomes form events A. In this case, the probability can be entered using the formula. Probability introduced in this way is called classical probability. It can be proven that in this case properties 1-4 are satisfied.

Probability theory problems that appear on the Unified State Examination in mathematics are mainly related to classical probability. Such tasks can be very simple. Particularly simple are problems in probability theory in demo options. It is easy to calculate the number of favorable outcomes; the number of all outcomes is written right in the condition.

We get the answer using the formula.

An example of a problem from the Unified State Examination in mathematics on determining probability

There are 20 pies on the table - 5 with cabbage, 7 with apples and 8 with rice. Marina wants to take the pie. What is the probability that she will take the rice cake?

Solution.

There are 20 equally probable elementary outcomes, that is, Marina can take any of the 20 pies. But we need to estimate the probability that Marina will take the rice pie, that is, where A is the choice of the rice pie. This means that the number of favorable outcomes (choices of pies with rice) is only 8. Then the probability will be determined by the formula:

Independent, Opposite and Arbitrary Events

However, in open jar More complex tasks began to be encountered. Therefore, let us draw the reader’s attention to other issues studied in probability theory.

Events A and B are said to be independent if the probability of each does not depend on whether the other event occurs.

Event B is that event A did not happen, i.e. event B is opposite to event A. The probability of the opposite event is equal to one minus the probability of the direct event, i.e. .

Probability addition and multiplication theorems, formulas

For arbitrary events A and B, the probability of the sum of these events is equal to the sum of their probabilities without the probability of their joint event, i.e. .

For independent events A and B, the probability of the occurrence of these events is equal to the product of their probabilities, i.e. in this case .

The last 2 statements are called the theorems of addition and multiplication of probabilities.

Counting the number of outcomes is not always so simple. In some cases it is necessary to use combinatorics formulas. The most important thing is to count the number of events that satisfy certain conditions. Sometimes these kinds of calculations can become independent tasks.

In how many ways can 6 students be seated in 6 empty seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways for the second student to take a place. There are 4 free places left for the third student, 3 for the fourth, 2 for the fifth, and the sixth will take the only remaining place. To find the number of all options, you need to find the product, which is denoted by the symbol 6! and reads "six factorial".

In the general case, the answer to this question is given by the formula for the number of permutations of n elements. In our case.

Let us now consider another case with our students. In how many ways can 2 students be seated in 6 empty seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways for the second student to take a place. To find the number of all options, you need to find the product.

In general, the answer to this question is given by the formula for the number of placements of n elements over k elements

In our case .

And the last case in this series. In how many ways can you choose three students out of 6? The first student can be selected in 6 ways, the second - in 5 ways, the third - in four ways. But among these options, the same three students appear 6 times. To find the number of all options, you need to calculate the value: . In general, the answer to this question is given by the formula for the number of combinations of elements by element:

In our case .

Examples of solving problems from the Unified State Exam in mathematics to determine probability

Task 1. From the collection edited by. Yashchenko.

There are 30 pies on the plate: 3 with meat, 18 with cabbage and 9 with cherries. Sasha chooses one pie at random. Find the probability that he ends up with a cherry.

Answer: 0.3.

Task 2. From the collection edited by. Yashchenko.

In each batch of 1000 light bulbs, on average, 20 are defective. Find the probability that a light bulb taken at random from a batch will be working.

Solution: The number of working light bulbs is 1000-20=980. Then the probability that a light bulb taken at random from a batch will be working:

Answer: 0.98.

The probability that student U will solve more than 9 problems correctly during a math test is 0.67. The probability that U. will correctly solve more than 8 problems is 0.73. Find the probability that U will solve exactly 9 problems correctly.

If we imagine a number line and mark points 8 and 9 on it, then we will see that the condition “U. will solve exactly 9 problems correctly” is included in the condition “U. will solve more than 8 problems correctly”, but does not apply to the condition “U. will solve more than 9 problems correctly.”

However, the condition “U. will solve more than 9 problems correctly” is contained in the condition “U. will solve more than 8 problems correctly.” Thus, if we designate events: “U. will solve exactly 9 problems correctly" - through A, "U. will solve more than 8 problems correctly" - through B, "U. will correctly solve more than 9 problems” through C. That solution will look like this:

Answer: 0.06.

In a geometry exam, a student answers one question from a list of exam questions. The probability that this is a Trigonometry question is 0.2. The probability that this is a question on External Angles is 0.15. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in the exam.

Let's think about what events we have. We are given two incompatible events. That is, either the question will relate to the topic “Trigonometry” or to the topic “External angles”. According to the probability theorem, the probability of incompatible events is equal to the sum of the probabilities of each event, we must find the sum of the probabilities of these events, that is:

Answer: 0.35.

The room is illuminated by a lantern with three lamps. The probability of one lamp burning out within a year is 0.29. Find the probability that at least one lamp will not burn out during the year.

Let's consider possible events. We have three light bulbs, each of which may or may not burn out independently of any other light bulb. These are independent events.

Then we will indicate the options for such events. Let's use the following notations: - the light bulb is on, - the light bulb is burnt out. And right next to it we will calculate the probability of the event. For example, the probability of an event in which three independent events “the light bulb is burned out”, “the light bulb is on”, “the light bulb is on” occurred: , where the probability of the event “the light bulb is on” is calculated as the probability of the event opposite to the event “the light bulb is not on”, namely: .

INTRODUCTION

Many things are incomprehensible to us not because our concepts are weak;
but because these things are not included in the range of our concepts.
Kozma Prutkov

The main goal of studying mathematics in secondary specialized educational institutions is to give students a set of mathematical knowledge and skills necessary for studying other program disciplines that use mathematics to one degree or another, for the ability to perform practical calculations, for the formation and development of logical thinking.

In this work, all the basic concepts of the section of mathematics “Fundamentals of Probability Theory and Mathematical Statistics”, provided for by the program and the State Educational Standards of Secondary Vocational Education (Ministry of Education of the Russian Federation. M., 2002), are consistently introduced, the main theorems are formulated, most of which are not proven . The main problems and methods for solving them and technologies for applying these methods to solving practical problems are considered. The presentation is accompanied by detailed comments and numerous examples.

Methodological instructions can be used for initial familiarization with the material being studied, when taking notes on lectures, to prepare for practical classes, to consolidate acquired knowledge, skills and abilities. In addition, the manual will also be useful for undergraduate students as a reference tool, allowing them to quickly recall what was previously studied.

At the end of the work there are examples and tasks that students can perform in self-control mode.

The guidelines are intended for part-time and full-time students.

BASIC CONCEPTS

Probability theory studies the objective patterns of mass random events. It is the theoretical basis for mathematical statistics, which deals with the development of methods for collecting, describing and processing observational results. Through observations (tests, experiments), i.e. experience in the broad sense of the word, knowledge of the phenomena of the real world occurs.

In his practical activities We often encounter phenomena the outcome of which cannot be predicted, the outcome of which depends on chance.

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.

Probability theory is a branch of mathematics in which random phenomena (events) are studied and patterns are identified when they are repeated en masse.

Mathematical statistics is a branch of mathematics that deals with the study of methods for collecting, systematizing, processing and using statistical data to obtain scientifically based conclusions and make decisions.

In this case, statistical data is understood as a set of numbers that represent the quantitative characteristics of the characteristics of the objects under study that interest us. Statistical data is obtained as a result of specially designed experiments and observations.

Statistical data by their essence depends on many random factors, therefore mathematical statistics is closely related to probability theory, which is its theoretical basis.

I. PROBABILITY. THEOREMS OF ADDITION AND MULTIPLICATION OF PROBABILITIES

1.1. Basic concepts of combinatorics

In the branch of mathematics, which is called combinatorics, some problems related to the consideration of sets and the composition of various combinations of elements of these sets are solved. For example, if we take 10 different numbers 0, 1, 2, 3,: , 9 and make combinations of them, we will get different numbers, for example 143, 431, 5671, 1207, 43, etc.

We see that some of these combinations differ only in the order of the digits (for example, 143 and 431), others - in the digits included in them (for example, 5671 and 1207), and others also differ in the number of digits (for example, 143 and 43).

Thus, the resulting combinations satisfy various conditions.

Depending on the rules of composition, three types of combinations can be distinguished: permutations, placements, combinations.

Let's first get acquainted with the concept factorial.

Product of all natural numbers from 1 to n inclusive are called n-factorial and write.

Calculate: a) ; b) ; V) .

Solution. A) .

b) Since , then we can put it out of brackets

Then we get

V) .

Rearrangements.

A combination of n elements that differ from each other only in the order of the elements is called a permutation.

Permutations are indicated by the symbol P n , where n is the number of elements included in each permutation. ( R- first letter of a French word permutation- rearrangement).

The number of permutations can be calculated using the formula

or using factorial:

Let's remember that 0!=1 and 1!=1.

Example 2. In how many ways can six different books be arranged on one shelf?

Solution. The required number of ways is equal to the number of permutations of 6 elements, i.e.

Placements.

Postings from m elements in n in each, such compounds are called that differ from each other either by the elements themselves (at least one), or by the order of their arrangement.

Placements are indicated by the symbol, where m- the number of all available elements, n- the number of elements in each combination. ( A- first letter of a French word arrangement, which means “placement, putting in order”).

At the same time, it is believed that nm.

The number of placements can be calculated using the formula

those. number of all possible placements from m elements by n equals the product n consecutive integers, of which the largest is m.

Let's write this formula in factorial form:

Example 3. How many options for distributing three vouchers to sanatoriums of various profiles can be compiled for five applicants?

Solution. The required number of options is equal to the number of placements of 5 elements of 3 elements, i.e.

Combinations.

Combinations are all possible combinations of m elements by n, which differ from each other by at least one element (here m And n- natural numbers, and n m).

Number of combinations of m elements by n are denoted by ( WITH-the first letter of a French word combination- combination).

In general, the number of m elements by n equal to the number of placements from m elements by n, divided by the number of permutations from n elements:

Using factorial formulas for the numbers of placements and permutations, we obtain:

Example 4. In a team of 25 people, you need to allocate four to work in a certain area. In how many ways can this be done?

Solution. Since the order of the four people chosen does not matter, there are ways to do this.

We find using the first formula

In addition, when solving problems, the following formulas are used, expressing the basic properties of combinations:

(by definition they assume and);

1.2. Solving combinatorial problems

Task 1. There are 16 subjects studied at the faculty. You need to put 3 subjects on your schedule for Monday. In how many ways can this be done?

Solution. There are as many ways to schedule three items out of 16 as you can arrange placements of 16 items by 3.

Task 2. Out of 15 objects, you need to select 10 objects. In how many ways can this be done?

Task 3. Four teams took part in the competition. How many options for distributing seats between them are possible?

Problem 4. In how many ways can a patrol of three soldiers and one officer be formed if there are 80 soldiers and 3 officers?

Solution. You can choose a soldier on patrol

ways, and officers in ways. Since any officer can go with each team of soldiers, there are only so many ways.

Task 5. Find , if it is known that .

Since , we get

By definition of a combination it follows that , . That. .

1.3. The concept of a random event. Types of events. Probability of event

Any action, phenomenon, observation with several different outcomes, realized under a given set of conditions, will be called test.

The result of this action or observation is called event .

If an event under given conditions can happen or not happen, then it is called random . When an event is certain to happen, it is called reliable , and in the case when it obviously cannot happen, - impossible.

The events are called incompatible , if only one of them is possible to appear each time.

The events are called joint , if, under given conditions, the occurrence of one of these events does not exclude the occurrence of another during the same test.

The events are called opposite , if under the test conditions they, being the only outcomes, are incompatible.

Events are usually denoted in capital letters of the Latin alphabet: A, B, C, D, : .

A complete system of events A 1 , A 2 , A 3 , : , A n is a set of incompatible events, the occurrence of at least one of which is obligatory during a given test.

If a complete system consists of two incompatible events, then such events are called opposite and are designated A and .

Example. The box contains 30 numbered balls. Determine which of the following events are impossible, reliable, or contrary:

took out a numbered ball (A);

got a ball with an even number (IN);

got a ball with an odd number (WITH);

got a ball without a number (D).

Which of them form a complete group?

Solution . A- reliable event; D- impossible event;

In and WITH- opposite events.

The complete group of events consists of A And D, V And WITH.

The probability of an event is considered as a measure of the objective possibility of the occurrence of a random event.

1.4. Classic definition of probability

A number that expresses the measure of the objective possibility of an event occurring is called probability this event and is indicated by the symbol R(A).

Definition. Probability of the event A is the ratio of the number of outcomes m that favor the occurrence of a given event A, to the number n all outcomes (inconsistent, only possible and equally possible), i.e. .

Therefore, to find the probability of an event, it is necessary, having considered various outcomes of the test, to calculate all possible inconsistent outcomes n, choose the number of outcomes m we are interested in and calculate the ratio m To n.

The following properties follow from this definition:

The probability of any test is a non-negative number not exceeding one.

Indeed, the number m of the required events is within . Dividing both parts into n, we get

2. The probability of a reliable event is equal to one, because .

3. The probability of an impossible event is zero, since .

Problem 1. In a lottery of 1000 tickets, there are 200 winning ones. One ticket is taken out at random. What is the probability that this ticket is a winner?

Solution. The total number of different outcomes is n=1000. The number of outcomes favorable to winning is m=200. According to the formula, we get

Problem 2. In a batch of 18 parts there are 4 defective ones. 5 parts are selected at random. Find the probability that two of these 5 parts will be defective.

Solution. Number of all equally possible independent outcomes n equal to the number of combinations of 18 by 5 i.e.

Let's count the number m that favors event A. Among 5 parts taken at random, there should be 3 good ones and 2 defective ones. The number of ways to select two defective parts from 4 existing defective ones is equal to the number of combinations of 4 by 2:

The number of ways to select three quality parts from 14 available quality parts is equal to

Any group of good parts can be combined with any group of defective parts, so the total number of combinations m amounts to

The required probability of event A is equal to the ratio of the number of outcomes m favorable to this event to the number n of all equally possible independent outcomes:

The sum of a finite number of events is an event consisting of the occurrence of at least one of them.

The sum of two events is denoted by the symbol A+B, and the sum n events with the symbol A 1 +A 2 + : +A n.

Probability addition theorem.

The probability of the sum of two incompatible events is equal to the sum of the probabilities of these events.

Corollary 1. If the event A 1, A 2, :,A n form a complete system, then the sum of the probabilities of these events is equal to one.

Corollary 2. The sum of the probabilities of opposite events and is equal to one.

Problem 1. There are 100 lottery tickets. It is known that 5 tickets win 20,000 rubles each, 10 tickets win 15,000 rubles, 15 tickets win 10,000 rubles, 25 tickets win 2,000 rubles. and nothing for the rest. Find the probability that the purchased ticket will receive a winning of at least 10,000 rubles.

Solution. Let A, B, and C be events consisting in the fact that the purchased ticket receives a winning equal to 20,000, 15,000, and 10,000 rubles, respectively. since events A, B and C are incompatible, then

Task 2. On extramural technical school receives tests in mathematics from cities A, B And WITH. Probability of receiving a test from the city A equal to 0.6, from the city IN- 0.1. Find the probability that the next test will come from the city WITH.

The simplest example of a connection between two events is a causal relationship, when the occurrence of one of the events necessarily leads to the occurrence of the other, or vice versa, when the occurrence of one excludes the possibility of the occurrence of the other.

To characterize the dependence of some events on others, the concept is introduced conditional probability.

Definition. Let A And IN- two random events of the same test. Then the conditional probability of the event A or the probability of event A, provided that event B occurs, is called the number.

Denoting the conditional probability, we obtain the formula

, .

Task 1. Calculate the probability that in a family where there is one child, a boy, a second boy will be born.

Solution. Let the event A is that there are two boys in the family, and the event IN- that one boy.

Consider all possible outcomes: boy and boy; boy and girl; girl and boy; girl and girl.

Then, and using the formula we find

Event A called independent from the event IN, if the occurrence of an event IN does not have any effect on the probability of the event occurring 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:

The probability of occurrence of several events that are independent in the aggregate is calculated by the formula

Problem 2. 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 will be white?

A and IN there is an event AB. Hence,

b) If the first element works, then an event occurs (opposite to the event A- failure of this element); if the second element works - event IN. Let's find the probabilities of events and:

Then the event that both elements will work is and, therefore,

The classical definition of probability is based on the concept probabilistic experience, or a probability experiment. Its result is one of several possible outcomes, called elementary outcomes, and there is no reason to expect that any elementary outcome will appear more often than others when repeating a probabilistic experiment. For example, consider a probabilistic experiment involving throwing a dice. The result of this experiment is the loss of one of the 6 points drawn on the sides of the cube.

Thus, in this experiment there are 6 elementary outcomes:

and each of them is equally expected.

Event in a classical probability experiment is an arbitrary subset of the set of elementary outcomes. In the considered example of throwing a die, the event is, for example, the loss of an even number of points, which consists of elementary outcomes.

The probability of an event is the number:

where is the number of elementary outcomes that make up the event (sometimes they say that this is the number of elementary outcomes that favor the occurrence of the event), and is the number of all elementary outcomes.

In our example:

Elements of combinatorics.

When describing many probabilistic experiments, elementary outcomes can be identified with one of the following objects of combinatorics (the science of finite sets).

Rearrangement of numbers is an arbitrary ordered representation of these numbers without repetition. For example, for a set of three numbers there are 6 different permutations:

, , , , , .

For an arbitrary number of permutations is equal

(the product of consecutive numbers in the natural series, starting from 1).

A combination of is an arbitrary unordered set of any elements of the set. For example, for a set of three numbers there are 3 different combinations of 3 by 2:

For an arbitrary pair , , the number of combinations from is equal to

For example,

Hypergeometric distribution.

Consider the following probabilistic experiment. There is a black box containing white and black balls. The balls are the same size and indistinguishable to the touch. The experiment consists of drawing out balls at random. The event whose probability needs to be found is that some of these balls are white and the rest are black.

Let's renumber all the balls with numbers from 1 to . Let the numbers 1, ¼ correspond to the white balls, and the numbers , ¼, correspond to the black balls. The elementary outcome in this experiment is an unordered set of elements from the set, that is, a combination of by. Consequently, there are all elementary outcomes.

Let's find the number of elementary outcomes favorable to the occurrence of the event. The corresponding sets consist of “white” and “black” numbers. You can choose numbers from “white” numbers in three ways, and numbers from “black” numbers in ¾ ways. White and black sets can be connected arbitrarily, so there are only elementary outcomes favorable to the event.

The probability of the event is

The resulting formula is called the hypergeometric distribution.

Problem 5.1. The box contains 55 standard and 6 defective parts of the same type. What is the probability that among three randomly selected parts, at least one will be defective?

Solution. There are 61 parts in total, we take 3. An elementary outcome is a combination of 61 by 3. The number of all elementary outcomes is equal to . Favorable outcomes are divided into three groups: 1) these are those outcomes in which 1 part is defective and 2 are good; 2) 2 parts are defective, and 1 is good; 3) all 3 parts are defective. The number of sets of the first type is equal to , the number of sets of the second type is equal to , and the number of sets of the third type is equal to . Consequently, the occurrence of an event is favored by elementary outcomes. The probability of the event is

Algebra of events

The space of elementary events is the set of all elementary outcomes related to a given experience.

Amount two events is called an event that consists of elementary outcomes belonging to the event or event.

The work two events is called an event consisting of elementary outcomes that simultaneously belong to the events and .

Events and are called incompatible if .

The event is called opposite event, if the event is favored by all those elementary outcomes that do not belong to the event. In particular, , .

SUM THEOREM.

In particular, .

Conditional probability event, provided that the event occurred, is called the ratio of the number of elementary outcomes belonging to the intersection to the number of elementary outcomes belonging to . In other words, conditional probability event is determined by the classical probability formula, in which the new probability space is . The conditional probability of an event is denoted by .

Product THEOREM. .

The events are called independent, If . For independent events, the product theorem gives the relation .

A consequence of the sum and product theorems is the following two formulas.

Total probability formula. A complete group of hypotheses is an arbitrary set of incompatible events , , ¼, , which together make up the entire probability space:

In this situation, for an arbitrary event, a formula called the total probability formula is valid,

where is the function Laplace. . . The Laplace function is tabulated, and its values, given a given value, can be found in any textbook on probability theory and mathematical statistics.

Problem 5.3. It is known that in a large batch of parts there are 11% defective. 100 parts are selected for testing. What is the probability that among them there are no more than 14 defective ones? Estimate the answer using the Moivre-Laplace theorem.

Solution. We're dealing with a test Bernoulli, Where , , . A success is considered to be the discovery of a defective part, and the number of successes satisfies the inequality . Hence,

Direct calculation gives:

, , , , , , , , , , , , , , .

Hence, . Now let's apply the Moivre-Laplace integral theorem. We get:

Using the table of function values, taking into account the oddness of the function, we obtain

The error of the approximate calculation does not exceed .

Random variables

A random variable is a numerical characteristic of a probabilistic experiment, which is a function of elementary outcomes. If , , ¼, is a set of elementary outcomes, then the random variable is a function of . It is more convenient, however, to characterize a random variable by listing all its possible values and the probabilities with which it takes this value.

Such a table is called the distribution law of a random variable. Since the events form a complete group, the law of probabilistic normalization is satisfied

The mathematical expectation, or average value, of a random variable is a number equal to the sum of the products of the values of the random variable and the corresponding probabilities.

The dispersion (the degree of spread of values around the mathematical expectation) of a random variable is expected value random variable,

It can be shown that

Magnitude

is called the mean square deviation of the random variable.

The distribution function for a random variable is the probability of falling into the set, that is

It is a non-negative, non-decreasing function taking values from 0 to 1. For a random variable that has a finite set of values, it is a piecewise constant function that has discontinuities of the second kind at state points. Moreover, and is continuous on the left.

Problem 5.4. Two dice are thrown in succession. If one, three or five points appear on one dice, the player loses 5 rubles. If two or four points are rolled, the player receives 7 rubles. If six points are rolled, the player loses 12 rubles. Random value x is the player's payoff for two dice rolls. Find the distribution law x, plot the distribution function, find the mathematical expectation and variance x.

Solution. Let's first consider what the player's winnings are equal to when throwing a die. Let the event be that 1, 3 or 5 points are rolled. Then, the winnings will be rubles. Let the event be that 2 or 4 points are rolled. Then, the winnings will be rubles. Finally, let the event mean rolling a 6. Then the winnings are equal to rubles.

Now let's consider all possible combinations of events , and with two dice throws, and determine the winning values for each such combination.

If an event occurred, then, at the same time.

Similarly, when we get , .

We write all the found states and the total probabilities of these states in the table:

We check the fulfillment of the law of probabilistic normalization: on the real line you need to be able to determine the probability of a random variable falling into this interval 1) and rapidly decreasing at, ¼,