How to determine the probability of an event formula. The meaning of the word "probability"

Problems on the classical determination of probability.
Examples of solutions

In the third lesson we will look at various problems involving the direct application of the classical definition of probability. To effectively study the materials in this article, I recommend that you familiarize yourself with the basic concepts probability theory And basics of combinatorics. The task of classically determining probability with a probability tending to unity will be present in your independent/control work on terver, so we tune in to serious work. You may ask, what's so serious about this? ...just one primitive formula. I warn you against frivolity - thematic assignments are quite diverse, and many of them can easily confuse you. In this regard, in addition to working through the main lesson, try to study additional tasks on the topic that are in the piggy bank ready-made solutions for higher mathematics. Solution techniques are solution techniques, but “friends” still “need to be known by sight,” because even a rich imagination is limited and there are also enough standard tasks. Well, I'll try good quality sort out as many of them as possible.

Let's remember the classics of the genre:

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

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

- quantity elementary outcomes favorable to the event.

And immediately an immediate pit stop. Do you understand the underlined terms? This means clear, not intuitive understanding. If not, then it’s still better to return to the 1st article on probability theory and only after that move on.

Please do not skip the first examples - in them I will repeat one fundamentally important point, and will also tell you how to correctly draw up a solution and in what ways this can be done:

Problem 1

An urn contains 15 white, 5 red and 10 black balls. 1 ball is drawn at random, find the likelihood that that it will be: a) white, b) red, c) black.

Solution: The most important prerequisite for using the classical definition of probability is ability to count the total number of outcomes.

There are a total of 15 + 5 + 10 = 30 balls in the urn, and obviously the following facts are true:

– retrieving any ball is equally possible (equal opportunity outcomes), while the outcomes elementary and form full group of events (i.e., as a result of the test, one of the 30 balls will definitely be removed).

Thus, the total number of outcomes:

Consider the event: – a white ball will be drawn from the urn. This event is favored elementary outcomes, therefore, according to the classical definition:
– the probability that a white ball will be drawn from the urn.

Oddly enough, even in such a simple task one can make a serious inaccuracy, which I already focused on in the first article on probability theory. Where is the pitfall here? It is incorrect to argue here that “since half the balls are white, then the probability of drawing a white ball» . The classic definition of probability refers to ELEMENTARY outcomes, and the fraction must be written down!

With other points, similarly, consider the following events:

- will be removed from the urn red ball;
– a black ball will be drawn from the urn.

An event is favored by 5 elementary outcomes, and an event is favored by 10 elementary outcomes. So the corresponding probabilities are:

A typical check of many server tasks is carried out using theorems on the sum of probabilities of events forming a complete group. In our case, the events form a complete group, which means the sum of the corresponding probabilities must necessarily equal one: .

Let's check if this is true: that's what I wanted to make sure of.

Answer:

In principle, the answer can be written down in more detail, but personally, I’m used to putting only numbers there - for the reason that when you start “stamping out” problems in hundreds and thousands, you try to reduce the writing of the solution as much as possible. By the way, about brevity: in practice, the “high-speed” design option is common solutions:

Total: 15 + 5 + 10 = 30 balls in the urn. According to the classical definition:
– the probability that a white ball will be drawn from the urn;
– the probability that a red ball will be drawn from the urn;
– the probability that a black ball will be drawn from the urn.

Answer:

However, if there are several points in the condition, then it is often more convenient to formulate the solution in the first way, which takes a little more time, but at the same time “lays everything out on the shelves” and makes it easier to navigate the problem.

Let's warm up:

Problem 2

The store received 30 refrigerators, five of which have a manufacturing defect. One refrigerator is selected at random. What is the probability that it will be without a defect?

Select the appropriate design option and check the sample at the bottom of the page.

In the simplest examples, the number of common and the number of favorable outcomes lie on the surface, but in most cases you have to dig up the potatoes yourself. A canonical series of problems about a forgetful subscriber:

Problem 3

When dialing a phone number, the subscriber forgot the last two digits, but remembers that one of them is zero and the other is odd. Find the probability that he will dial the correct number.

Note : zero is an even number (divisible by 2 without a remainder)

Solution: First we find the total number of outcomes. By condition, the subscriber remembers that one of the digits is zero, and the other digit is odd. Here it is more rational not to be tricky with combinatorics and use method of direct listing of outcomes . That is, when making a solution, we simply write down all the combinations:
01, 03, 05, 07, 09
10, 30, 50, 70, 90

And we count them - in total: 10 outcomes.

There is only one favorable outcome: the correct number.

According to the classical definition:
– probability that the subscriber will dial the correct number

Answer: 0,1

Decimal fractions look quite appropriate in probability theory, but you can also adhere to the traditional Vyshmatov style, operating only with ordinary fractions.

Advanced task for independent decision:

Problem 4

The subscriber has forgotten the PIN code for his SIM card, but remembers that it contains three “fives”, and one of the numbers is either a “seven” or an “eight”. What is the probability of successful authorization on the first try?

Here you can still develop the idea of ​​​​the likelihood that the subscriber will face punishment in the form of a puk code, but, unfortunately, the reasoning will already go beyond this lesson

The solution and answer are below.

Sometimes listing combinations turns out to be a very painstaking task. In particular, this is the case in the following, no less popular group problems where 2 dice are rolled (less often - larger quantities):

Problem 5

Find the probability that when throwing two dice the total number will be:

a) five points;
b) no more than four points;
c) from 3 to 9 points inclusive.

Solution: find the total number of outcomes:

Ways the side of the 1st die can fall out And in different ways the side of the 2nd cube can fall out; By rule for multiplying combinations, Total: possible combinations. In other words, each the face of the 1st cube can be ordered a couple with each the edge of the 2nd cube. Let us agree to write such a pair in the form , where is the number that appears on the 1st die, and is the number that appears on the 2nd die. For example:

– the first dice scored 3 points, the second dice scored 5 points, total points: 3 + 5 = 8;
– the first dice scored 6 points, the second dice scored 1 point, total points: 6 + 1 = 7;
– 2 points rolled on both dice, sum: 2 + 2 = 4.

Obviously, the smallest amount is given by a pair, and the largest by two “sixes”.

a) Consider the event: – when throwing two dice, 5 points will appear. Let's write down and count the number of outcomes that favor this event:

Total: 4 favorable outcomes. According to the classical definition:
– the desired probability.

b) Consider the event: – no more than 4 points will be rolled. That is, either 2, or 3, or 4 points. Again we list and count the favorable combinations, on the left I will write down the total number of points, and after the colon - suitable couples:

Total: 6 favorable combinations. Thus:
– the probability that no more than 4 points will be rolled.

c) Consider the event: – 3 to 9 points will roll, inclusive. Here you can take the straight road, but... for some reason you don’t want to. Yes, some pairs have already been listed in the previous paragraphs, but there is still a lot of work to be done.

What's the best way to proceed? IN similar cases the roundabout way turns out to be rational. Let's consider opposite event: – 2 or 10 or 11 or 12 points will be rolled.

What's the point? The opposite event is favored by a significantly smaller number of couples:

Total: 7 favorable outcomes.

According to the classical definition:
– the probability that it will appear less than three or more than 9 points.

In addition to direct listing and counting of outcomes, various combinatorial formulas. And again an epic problem about the elevator:

Problem 7

3 people entered the elevator of a 20-story building on the first floor. And let's go. Find the probability that:

a) they will exit on different floors
b) two will exit on the same floor;
c) everyone will get off on the same floor.

Is our exciting activity has come to an end, and finally, I once again strongly recommend that if not solved, then at least figure out additional problems on the classical determination of probability. As I already noted, “hand padding” matters too!

Further along the course - Geometric definition of probability And Probability addition and multiplication theorems and... luck in the main thing!

Solutions and Answers:

Task 2: Solution: 30 – 5 = 25 refrigerators have no defect.

– the probability that a randomly selected refrigerator does not have a defect.
Answer :

Task 4: Solution: find the total number of outcomes:
ways you can select the place where the dubious number is located and on every Of these 4 places, 2 digits (seven or eight) can be located. According to the rule of multiplication of combinations, the total number of outcomes: .
Alternatively, the solution can simply list all the outcomes (fortunately there are few of them):
7555, 8555, 5755, 5855, 5575, 5585, 5557, 5558
There is only one favorable outcome (correct pin code).
Thus, according to the classical definition:
– probability that the subscriber logs in on the 1st attempt
Answer :

Task 6: Solution: find the total number of outcomes:
numbers on 2 dice can appear in different ways.

a) Consider the event: – when throwing two dice, the product of the points will be equal to seven. There are no favorable outcomes for a given event, according to the classical definition of probability:
, i.e. this event is impossible.

b) Consider the event: – when throwing two dice, the product of the points will be at least 20. The following outcomes are favorable for this event:

Total: 8
According to the classical definition:
– the desired probability.

c) Consider the opposite events:
– the product of points will be even;
– the product of points will be odd.
Let's list all the outcomes favorable to the event:

Total: 9 favorable outcomes.
According to the classical definition of probability:
Opposite events form a complete group, therefore:
– the desired probability.

Answer :

Problem 8: Solution: let's calculate the total number of outcomes: 10 coins can fall in different ways.
Another way: ways the 1st coin can fall And ways the 2nd coin can fall And … And ways the 10th coin can fall. According to the rule of multiplying combinations, 10 coins can fall ways.
a) Consider the event: – heads will appear on all coins. This event is favored by a single outcome, according to the classical definition of probability: .
b) Consider the event: – 9 coins will land heads, and one coin will land tails.
There are coins that can land on heads. According to the classical definition of probability: .
c) Consider the event: – heads will appear on half of the coins.
Exists unique combinations of five coins that can land heads. According to the classical definition of probability:
Answer :

Brief theory

To quantitatively compare events according to the degree of possibility of their occurrence, a numerical measure is introduced, which is called the probability of an event. The probability of a random event is a number that expresses the measure of the objective possibility of an event occurring.

The quantities that determine how significant the objective reasons are to expect the occurrence of an event are characterized by the probability of the event. It must be emphasized that probability is an objective quantity that exists independently of the knower and is conditioned by the entire set of conditions that contribute to the occurrence of an event.

The explanations we have given for the concept of probability are not a mathematical definition, since they do not quantify the concept. There are several definitions of the probability of a random event, which are widely used in solving specific problems (classical, axiomatic, statistical, etc.).

Classic definition event probability reduces this concept to the more elementary concept of equally possible events, which is no longer subject to definition and is assumed to be intuitively clear. For example, if a die is a homogeneous cube, then the loss of any of the faces of this cube will be equally possible events.

Let a reliable event be divided into equally possible cases, the sum of which gives the event. That is, the cases into which it breaks down are called favorable for the event, since the appearance of one of them ensures the occurrence.

The probability of an event will be denoted by the symbol.

The probability of an event is equal to the ratio of the number of cases favorable to it, out of the total number of uniquely possible, equally possible and incompatible cases, to the number, i.e.

This is the classic definition of probability. Thus, to find the probability of an event, it is necessary, having considered the various outcomes of the test, to find a set of uniquely possible, equally possible and incompatible cases, calculate their total number n, the number of cases m favorable for a given event, and then perform the calculation using the above formula.

The probability of an event equal to the ratio of the number of experimental outcomes favorable to the event to the total number of experimental outcomes is called classical probability random event.

The following properties of probability follow from the definition:

Property 1. The probability of a reliable event is equal to one.

Property 2. The probability of an impossible event is zero.

Property 3. The probability of a random event is a positive number between zero and one.

Property 4. The probability of the occurrence of events that form a complete group is equal to one.

Property 5. The probability of the occurrence of the opposite event is determined in the same way as the probability of the occurrence of event A.

The number of cases favoring the occurrence of an opposite event. Hence, the probability of the occurrence of the opposite event is equal to the difference between unity and the probability of the occurrence of event A:

An important advantage of the classical definition of the probability of an event is that with its help the probability of an event can be determined without resorting to experience, but based on logical reasoning.

When a set of conditions is met, a reliable event will definitely happen, but an impossible event will definitely not happen. Among the events that may or may not occur when a set of conditions is created, the occurrence of some can be counted on with good reason, and the occurrence of others with less reason. If, for example, there are more white balls in an urn than black balls, then there is more reason to hope for the appearance of a white ball when drawn from the urn at random than for the appearance of a black ball.

Example of problem solution

Example 1

A box contains 8 white, 4 black and 7 red balls. 3 balls are drawn at random. Find the probabilities of the following events: – at least 1 red ball is drawn, – there are at least 2 balls of the same color, – there are at least 1 red and 1 white ball.

The solution of the problem

We find the total number of test outcomes as the number of combinations of 19 (8+4+7) elements of 3:

Let's find the probability of the event– at least 1 red ball is drawn (1,2 or 3 red balls)

Required probability:

Let the event– there are at least 2 balls of the same color (2 or 3 white balls, 2 or 3 black balls and 2 or 3 red balls)

Number of outcomes favorable to the event:

Required probability:

Let the event– there is at least one red and 1 white ball

(1 red, 1 white, 1 black or 1 red, 2 white or 2 red, 1 white)

Number of outcomes favorable to the event:

Required probability:

Answer: P(A)=0.773;P(C)=0.7688; P(D)=0.6068

Example 2

Two dice are thrown. Find the probability that the sum of points is at least 5.

Solution

Let the event be a score of at least 5

Let's use the classic definition of probability:

Total number of possible test outcomes

Number of trials favoring the event of interest

On the dropped side of the first dice, one point, two points..., six points may appear. similarly, six outcomes are possible when rolling the second die. Each of the outcomes of throwing the first die can be combined with each of the outcomes of the second. Thus, the total number of possible elementary test outcomes is equal to the number of placements with repetitions (choice with placements of 2 elements from a set of volume 6):

Let's find the probability of the opposite event - the sum of points is less than 5

The following combinations of dropped points will favor the event:

The geometric definition of probability is presented and the solution is given broadly known problem about a meeting.

At When assessing the probability of the occurrence of any random event, it is very important to have a good understanding of whether the probability () of the occurrence of the event we are interested in depends on how other events develop.

In the case of the classical scheme, when all outcomes are equally probable, we can already estimate the probability values ​​of the individual event of interest to us independently. We can do this even if the event is a complex collection of several elementary outcomes. What if several random events occur simultaneously or sequentially? How does this affect the likelihood of the event we are interested in happening?

If I roll a die several times and want a six to come up, and I keep getting unlucky, does that mean I should increase my bet because, according to probability theory, I'm about to get lucky? Alas, probability theory does not state anything like this. No dice, no cards, no coins can't remember what they showed us last time. It doesn’t matter to them at all whether it’s the first time or the tenth time I’m testing my luck today. Every time I repeat the roll, I know only one thing: and this time the probability of getting a six is ​​again one sixth. Of course, this does not mean that the number I need will never come up. This only means that my loss after the first throw and after any other throw are independent events.

Events A and B are called independent, if the implementation of one of them does not in any way affect the probability of another event. For example, the probabilities of hitting a target with the first of two weapons do not depend on whether the target was hit by the other weapon, so the events “the first weapon hit the target” and “the second weapon hit the target” are independent.

If two events A and B are independent, and the probability of each of them is known, then the probability of the simultaneous occurrence of both event A and event B (denoted AB) can be calculated using the following theorem.

Probability multiplication theorem for independent events

Example.The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 =0.7; p 2 =0.8. Find the probability of a hit with one salvo by both guns simultaneously.

Solution: as we have already seen, events A (hit by the first gun) and B (hit by the second gun) are independent, i.e. P(AB)=P(A)*P(B)=p 1 *p 2 =0.56.

What will happen to our estimates if initial events are not independent? Let's change the previous example a little.

Example.Two shooters shoot at targets at a competition, and if one of them shoots accurately, the opponent begins to get nervous and his results worsen. How to turn this everyday situation into a mathematical problem and outline ways to solve it? It is intuitively clear that we need to somehow separate the two options developments, essentially create two scenarios, two different tasks. In the first case, if the opponent missed, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent has taken his chance decently, the probability of hitting the target for the second athlete decreases.

To separate possible scenarios (often called hypotheses) for the development of events, we will often use a “probability tree” diagram. This diagram is similar in meaning to the decision tree that you have probably already dealt with. Each branch represents a separate scenario for the development of events, only now it has eigenvalue so-called conditional probabilities (q 1, q 2, q 1 -1, q 2 -1).

This scheme is very convenient for analyzing sequential random events.

It remains to clarify one more important question: where do the initial values ​​of the probabilities come from? real situations ? After all, not with the same coins and dice does probability theory work? Usually these estimates are taken from statistics, and when statistical information is not available, we conduct our own research. And we often have to start it not with collecting data, but with the question of what information we actually need.

Example.Let's say we need to estimate in a city with a population of one hundred thousand inhabitants the market volume for a new product that is not an essential item, for example, for a balm for the care of colored hair. Let's consider the "probability tree" diagram. In this case, we need to approximately estimate the probability value on each “branch”. So, our estimates of market capacity:

1) of all city residents, 50% are women,

2) of all women, only 30% dye their hair often,

3) of them, only 10% use balms for colored hair,

4) of them, only 10% can muster the courage to try a new product,

5) 70% of them usually buy everything not from us, but from our competitors.

Solution: According to the law of multiplication of probabilities, we determine the probability of the event we are interested in A = (a city resident buys this new balm from us) = 0.00045.

Let's multiply this probability value by the number of city residents. As a result, we have only 45 potential customers, and considering that one bottle of this product lasts for several months, the trade is not very lively.

And yet there is some benefit from our assessments.

Firstly, we can compare forecasts of different business ideas; they will have different “forks” in the diagrams, and, of course, the probability values ​​will also be different.

Secondly, as we have already said, a random variable is not called random because it does not depend on anything at all. Just her exact the meaning is not known in advance. We know that the average number of buyers can be increased (for example, by advertising a new product). So it makes sense to focus our efforts on those “forks” where the probability distribution does not suit us particularly, on those factors that we are able to influence.

Let's look at another quantitative example of consumer behavior research.

Example. On average, 10,000 people visit the food market per day. The probability that a market visitor enters the dairy products pavilion is 1/2. It is known that this pavilion sells an average of 500 kg of various products per day.

Can we say that the average purchase in the pavilion weighs only 100 g?

Discussion. Of course not. It is clear that not everyone who entered the pavilion ended up buying something there.

As shown in the diagram, to answer the question about the average weight of a purchase, we must find an answer to the question, what is the probability that a person entering the pavilion will buy something there. If we do not have such data at our disposal, but we need it, we will have to obtain it ourselves by observing the visitors to the pavilion for some time. Let’s say our observations showed that only a fifth of pavilion visitors buy something.

Once we have obtained these estimates, the task becomes simple. Out of 10,000 people who come to the market, 5,000 will go to the dairy products pavilion; there will be only 1,000 purchases. The average purchase weight is 500 grams. It is interesting to note that to construct full picture happening, the logic of conditional “branching” must be defined at each stage of our reasoning as clearly as if we were working with a “concrete” situation, and not with probabilities.

Self-test tasks

1. Let there be an electrical circuit consisting of n elements connected in series, each of which operates independently of the others.

The probability p of failure of each element is known. Determine the probability of proper operation of the entire section of the circuit (event A).

2. The student knows 20 out of 25 exam questions. Find the probability that the student knows the three questions given to him by the examiner.

3. Production consists of four successive stages, at each of which equipment operates, for which the probabilities of failure over the next month are equal to p 1, p 2, p 3 and p 4, respectively. Find the probability that there will be no production stoppages due to equipment failure in a month.

1. Formulas for calculating the probability of events

1.3.1. Sequence of independent tests (Bernoulli circuit)

Suppose that some experiment can be carried out repeatedly under the same conditions. Let this experience be made n times, i.e., a sequence of n tests.

Definition. Subsequence n tests are called mutually independent , if any event related to a given test is independent of any events related to other tests.

Let's assume that some event A likely to happen p as a result of one test or not likely to happen q= 1- p.

Definition . Sequence of n tests forms a Bernoulli scheme if the following conditions are met:

subsequence n tests are mutually independent,

2) probability of an event A does not change from trial to trial and does not depend on the result in other trials.

Event A is called the “success” of the test, and the opposite event is called “failure.” Consider the event

=( in n tests happened exactly m“success”).

To calculate the probability of this event, the Bernoulli formula is valid

p() =
, m = 1, 2, …, n , (1.6)

Where - number of combinations of n elements by m :

=
=
.

Example 1.16. The die is tossed three times. Find:

a) the probability that 6 points will appear twice;

b) the probability that the number of sixes will not appear more than twice.

Solution . We will consider the “success” of the test to be when the side with the image of 6 points appears on the die.

a) Total number of tests – n=3, number of “successes” – m = 2. Probability of “success” - p=, and the probability of “failure” is q= 1 - =. Then, according to Bernoulli's formula, the probability that, as a result of throwing a die three times, the side with six points will appear twice, will be equal to

.

b) Let us denote by A an event that means that a side with a score of 6 will appear no more than twice. Then the event can be represented as the sum of three incompatible events A=
,

Where IN 3 0 – an event when the edge of interest never appears,

IN 3 1 - event when the edge of interest appears once,

IN 3 2 - event when the edge of interest appears twice.

Using the Bernoulli formula (1.6) we find

p(A) = p (
) = p(
)=
+
+
=

=
.

1.3.2. Conditional probability of an event

Conditional probability reflects the influence of one event on the probability of another. Changing the conditions under which the experiment is carried out also affects

on the probability of occurrence of the event of interest.

Definition. Let A And B– some events, and probability p(B)> 0.

Conditional probability events A provided that the “event Balready happened” is the ratio of the probability of the occurrence of these events to the probability of an event that occurred earlier than the event whose probability is required to be found. Conditional probability is denoted as p(A B). Then by definition

p (A  B) =
. (1.7)

Example 1.17. Two dice are tossed. The space of elementary events consists of ordered pairs of numbers

(1,1) (1,2) (1,3) (1,4) (1,5) (1,6)

(2,1) (2,2) (2,3) (2,4) (2,5) (2,6)

(3,1) (3,2) (3,3) (3,4) (3,5) (3,6)

(4,1) (4,2) (4,3) (4,4) (4,5) (4,6)

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6).

In Example 1.16 it was determined that the event A=(number of points on the first die > 4) and event C=(sum of points is 8) dependent. Let's make a relation

.

This relationship can be interpreted as follows. Let us assume that the result of the first roll is known to be that the number of points on the first die is > 4. It follows that throwing the second die can lead to one of the 12 outcomes that make up the event A:

(5,1) (5,2) (5,3) (5,4) (5,5) (5,6)

(6,1) (6,2) (6,3) (6,4) (6,5) (6,6) .

At this event C only two of them can match (5,3) (6,2). In this case, the probability of the event C will be equal
. Thus, information about the occurrence of an event A influenced the likelihood of an event C.

Probability of events happening

Multiplication theorem

Probability of events happeningA 1 A 2  A n is determined by the formula

p(A 1 A 2  A n)= p(A 1)p(A 2  A 1)) p(A n  A 1 A 2  A n- 1). (1.8)

For the product of two events it follows that

p(AB)= p(A B)p{B)= p(B A)p{A). (1.9)

Example 1.18. In a batch of 25 products, 5 products are defective. 3 items are selected at random in succession. Determine the probability that all selected products are defective.

Solution. Let's denote the events:

A 1 = (first product is defective),

A 2 = (second product is defective),

A 3 = (third product is defective),

A = (all products are defective).

Event A is the product of three events A = A 1 A 2 A 3 .

From the multiplication theorem (1.6) we get

p(A)= p( A 1 A 2 A 3 ) = p(A 1) p(A 2  A 1))p(A 3  A 1 A 2).

The classical definition of probability allows us to find p(A 1) is the ratio of the number of defective products to total number products:

p(A 1)= ;

p(A 2) – This the ratio of the number of defective products remaining after the removal of one, to total number remaining products:

p(A 2  A 1))= ;

p(A 3) – this is the ratio of the number of defective products remaining after the removal of two defective ones to the total number of remaining products:

p(A 3  A 1 A 2)=.

Then the probability of the event A will be equal

p(A) ==
.

• Probability - degree (relative measure, quantification) the possibility of some event occurring. When the reasons for some possible event to actually occur outweigh the opposite reasons, then this event is called probable, otherwise - unlikely or improbable. The preponderance of positive reasons over negative ones, and vice versa, can be to varying degrees, as a result of which the probability (and improbability) can be greater or lesser. Therefore, probability is often assessed at a qualitative level, especially in cases where a more or less accurate quantitative assessment is impossible or extremely difficult. Various gradations of “levels” of probability are possible.

  The study of probability from a mathematical point of view constitutes a special discipline - probability theory. In probability theory and mathematical statistics, the concept of probability is formalized as a numerical characteristic of an event - a probability measure (or its value) - a measure on a set of events (subsets of a set of elementary events), taking values ​​from

  (\displaystyle 0)

  (\displaystyle 1)

  Meaning

  (\displaystyle 1)

  Corresponds to a reliable event. An impossible event has a probability of 0 (the converse is generally not always true). If the probability of an event occurring is

  (\displaystyle p)

  Then the probability of its non-occurrence is equal to

  (\displaystyle 1-p)

  In particular, the probability

  (\displaystyle 1/2)

  Means equal probability of occurrence and non-occurrence of an event.

  The classic definition of probability is based on the concept of equal probability of outcomes. The probability is the ratio of the number of outcomes favorable for a given event to the total number of equally possible outcomes. For example, the probability of getting heads or tails in a random coin toss is 1/2 if it is assumed that only these two possibilities occur and that they are equally possible. This classic "definition" of probability can be generalized to the case of an infinite number of possible values ​​- for example, if some event can occur with equal probability at any point (the number of points is infinite) of some limited region of space (plane), then the probability that it will occur in some part of this admissible region is equal to the ratio of the volume (area) of this part to the volume (area) of the region of all possible points.

  The empirical “definition” of probability is related to the frequency of occurrence of an event based on the fact that with sufficient large number testing frequency should tend to the objective degree of possibility of this event. In the modern presentation of probability theory, probability is defined axiomatically, as a special case abstract theory measures of a set. However, the connecting link between the abstract measure and the probability, which expresses the degree of possibility of the occurrence of an event, is precisely the frequency of its observation.

  The probabilistic description of certain phenomena has become widespread in modern science, in particular in econometrics, statistical physics of macroscopic (thermodynamic) systems, where even in the case of a classical deterministic description of the movement of particles, a deterministic description of the entire system of particles does not seem practically possible and appropriate. IN quantum physics The described processes themselves are of a probabilistic nature.