Probability event is the ratio of the number of elementary outcomes that favor a given event to the number of all equally possible outcomes of experience in which this event may occur. The probability of an event A is denoted by P(A) (here P is the first letter of the French word probabilite - probability). According to the definition
(1.2.1)
where is the number of elementary outcomes favoring event A; - the number of all equally possible elementary outcomes of experience, forming a complete group of events.
This definition of probability is called classical. It arose at the initial stage of the development of probability theory.

The probability of an event has the following properties:
1. The probability of a certain event is equal to one. Let's designate a certain event by the letter . For a certain event, therefore
(1.2.2)
2. The probability of an impossible event is zero. We denote the impossible event by the letter . For an impossible event, therefore
(1.2.3)
3. The probability of a random event is expressed as a positive number less than one. Since the inequalities , or are satisfied for a random event, then
(1.2.4)
4. The probability of any event satisfies the inequalities
(1.2.5)
This follows from relations (1.2.2) -(1.2.4).

Example 1 An urn contains 10 balls of the same size and weight, of which 4 are red and 6 are blue. One ball is drawn from the urn. What is the probability that the drawn ball is blue?

Solution. The event "the drawn ball turned out to be blue" will be denoted by the letter A. This trial has 10 equally possible elementary outcomes, of which 6 favor event A. In accordance with formula (1.2.1), we obtain

Example 2 All natural numbers from 1 to 30 are written on identical cards and placed in an urn. After thoroughly mixing the cards, one card is removed from the urn. What is the probability that the number on the card drawn is a multiple of 5?

Solution. Denote by A the event "the number on the taken card is a multiple of 5". In this test, there are 30 equally possible elementary outcomes, of which 6 outcomes favor event A (numbers 5, 10, 15, 20, 25, 30). Consequently,

Example 3 Two dice are thrown, the sum of points on the upper faces is calculated. Find the probability of the event B, consisting in the fact that the top faces of the cubes will have a total of 9 points.

Solution. There are 6 2 = 36 equally possible elementary outcomes in this trial. Event B is favored by 4 outcomes: (3;6), (4;5), (5;4), (6;3), so

Example 4. A natural number not exceeding 10 is chosen at random. What is the probability that this number is prime?

Solution. Denote by the letter C the event "the chosen number is prime". In this case, n = 10, m = 4 (primes 2, 3, 5, 7). Therefore, the desired probability

Example 5 Two symmetrical coins are tossed. What is the probability that both coins have digits on the top sides?

Solution. Let's denote by the letter D the event "there was a number on the top side of each coin". There are 4 equally possible elementary outcomes in this test: (G, G), (G, C), (C, G), (C, C). (The notation (G, C) means that on the first coin there is a coat of arms, on the second - a number). Event D is favored by one elementary outcome (C, C). Since m = 1, n = 4, then

Example 6 What is the probability that the digits in a randomly chosen two-digit number are the same?

Solution. Two-digit numbers are numbers from 10 to 99; there are 90 such numbers in total. 9 numbers have the same digits (these are the numbers 11, 22, 33, 44, 55, 66, 77, 88, 99). Since in this case m = 9, n = 90, then
,
where A is the "number with the same digits" event.

Example 7 From the letters of the word differential one letter is chosen at random. What is the probability that this letter will be: a) a vowel b) a consonant c) a letter h?

Solution. There are 12 letters in the word differential, of which 5 are vowels and 7 are consonants. Letters h this word does not. Let's denote the events: A - "vowel", B - "consonant", C - "letter h". The number of favorable elementary outcomes: - for event A, - for event B, - for event C. Since n \u003d 12, then
, and .

Example 8 Two dice are tossed, the number of points on the top face of each dice is noted. Find the probability that both dice have the same number of points.

Solution. Let us denote this event by the letter A. Event A is favored by 6 elementary outcomes: (1;]), (2;2), (3;3), (4;4), (5;5), (6;6). In total there are equally possible elementary outcomes that form a complete group of events, in this case n=6 2 =36. So the desired probability

Example 9 The book has 300 pages. What is the probability that a randomly opened page will have a sequence number that is a multiple of 5?

Solution. It follows from the conditions of the problem that there will be n = 300 of all equally possible elementary outcomes that form a complete group of events. Of these, m = 60 favor the occurrence of the specified event. Indeed, a number that is a multiple of 5 has the form 5k, where k is a natural number, and , whence . Consequently,
, where A - the "page" event has a sequence number that is a multiple of 5".

Example 10. Two dice are thrown, the sum of points on the upper faces is calculated. What is more likely to get a total of 7 or 8?

Solution. Let's designate the events: A - "7 points fell out", B - "8 points fell out". Event A is favored by 6 elementary outcomes: (1; 6), (2; 5), (3; 4), (4; 3), (5; 2), (6; 1), and event B - by 5 outcomes: (2; 6), (3; 5), (4; 4), (5; 3), (6; 2). There are n = 6 2 = 36 of all equally possible elementary outcomes. Hence, and .

So, P(A)>P(B), that is, getting a total of 7 points is a more likely event than getting a total of 8 points.

Tasks

1. A natural number not exceeding 30 is chosen at random. What is the probability that this number is a multiple of 3?
2. In the urn a red and b blue balls of the same size and weight. What is the probability that a randomly drawn ball from this urn is blue?
3. A number not exceeding 30 is chosen at random. What is the probability that this number is a divisor of zo?
4. In the urn a blue and b red balls of the same size and weight. One ball is drawn from this urn and set aside. This ball is red. Then another ball is drawn from the urn. Find the probability that the second ball is also red.
5. A natural number not exceeding 50 is chosen at random. What is the probability that this number is prime?
6. Three dice are thrown, the sum of points on the upper faces is calculated. What is more likely - to get a total of 9 or 10 points?
7. Three dice are tossed, the sum of the dropped points is calculated. What is more likely to get a total of 11 (event A) or 12 points (event B)?

Answers

1. 1/3. 2 . b/(a+b). 3 . 0,2. 4 . (b-1)/(a+b-1). 5 .0,3.6 . p 1 \u003d 25/216 - the probability of getting 9 points in total; p 2 \u003d 27/216 - the probability of getting 10 points in total; p2 > p1 7 . P(A) = 27/216, P(B) = 25/216, P(A) > P(B).

Questions

1. What is called the probability of an event?
2. What is the probability of a certain event?
3. What is the probability of an impossible event?
4. What are the limits of the probability of a random event?
5. What are the limits of the probability of any event?
6. What definition of probability is called classical?

Probability shows the possibility of an event with a certain number of repetitions. This is the number of possible outcomes with one or more outcomes divided by the total number of possible events. The probability of several events is calculated by dividing the problem into separate probabilities and then multiplying these probabilities.

Steps

Probability of a single random event

1. Choose an event with mutually exclusive results. Probability can only be calculated if the event in question either occurs or does not occur. It is impossible to get any event and the opposite result at the same time. An example of such events is rolling a 5 on a game die or winning a certain horse in a race. Five will either come up or not; a certain horse will either come first or not.

   • For example, it is impossible to calculate the probability of such an event: in one roll of the die, 5 and 6 will roll at the same time.

2. Identify all possible events and outcomes that could occur. Suppose we need to determine the probability that a three of a kind will come up when a dice with 6 numbers is rolled. "Three of a kind" is an event, and since we know that any of the 6 numbers can come up, the number of possible outcomes is six. Thus, we know that in this case there are 6 possible outcomes and one event whose probability we want to determine. Below are two more examples.

   • Example 1. In this case, the event is "selecting a day that falls on the weekend", and the number of possible outcomes is equal to the number of days of the week, that is, seven.
   • Example 2. The event is "draw the red ball", and the number of possible outcomes is equal to the total number of balls, that is, twenty.

3. Divide the number of events by the number of possible outcomes. This way you determine the probability of a single event. If we consider the case of a 3 on a die roll, the number of events is 1 (three is only on one side of the die), and the total number of outcomes is 6. The result is a ratio of 1/6, 0.166, or 16.6%. The probability of an event for the two examples above is found as follows:

   • Example 1. What is the probability that you will randomly choose a day that falls on a weekend? The number of events is 2, since there are two days off in one week, and the total number of outcomes is 7. Thus, the probability is 2/7. The result obtained can also be written as 0.285 or 28.5%.
   • Example 2. A box contains 4 blue, 5 red and 11 white balls. If you draw a random ball from the box, what is the probability that it will be red? The number of events is 5, since there are 5 red balls in the box, and the total number of outcomes is 20. Find the probability: 5/20 = 1/4. The result obtained can also be written as 0.25 or 25%.

4. Add up the probabilities of all possible events and see if the total is 1. The total probability of all possible events should be 1, or 100%. If you don't get 100%, chances are you made a mistake and missed one or more possible events. Check your calculations and make sure you account for all possible outcomes.

   • For example, the probability of rolling a 3 when rolling a die is 1/6. In this case, the probability of any other number falling out of the remaining five is also equal to 1/6. As a result, we get 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6, that is, 100%.
   • If, for example, you forget about the number 4 on the die, adding the probabilities will give you only 5/6, or 83%, which is not equal to one and indicates an error.

5. Express the probability of an impossible outcome as 0. This means that the given event cannot happen and its probability is 0. This way you can account for impossible events.

   • For example, if you were calculating the probability that Easter falls on a Monday in 2020, you would get 0, since Easter is always celebrated on Sunday.

   Probability of several random events

   1. When considering independent events, calculate each probability separately. Once you determine what the probabilities of events are, they can be calculated separately. Let's say we want to know the probability of rolling a die twice in a row with a 5. We know that the probability of rolling one five is 1/6, and the probability of rolling the second five is also 1/6. The first outcome is not related to the second.

      • Several rolls of fives are called not dependent events , because what happens the first time does not affect the second event.

   2. Consider the influence of previous outcomes when calculating the probability for dependent events. If the first event affects the probability of the second outcome, it is said to calculate the probability dependent events. For example, if you choose two cards from a deck of 52 cards, after the first card is drawn, the composition of the deck changes, which affects the choice of the second card. To calculate the probability of the second of two dependent events, subtract 1 from the number of possible outcomes when calculating the probability of the second event.

      • Example 1. Consider the following event: Two cards are randomly drawn from the deck one after the other. What is the probability that both cards will have a club suit? The probability that the first card will have a club suit is 13/52, or 1/4, since there are 13 cards of the same suit in the deck.
        • After that, the probability that the second card will be a club suit is 12/51, since there is no longer one club card. This is because the first event affects the second. If you draw the 3 of clubs and don't put it back, there will be one less card in the deck (51 instead of 52).
      • Example 2. There are 4 blue, 5 red and 11 white balls in a box. If three balls are drawn at random, what is the probability that the first is red, the second is blue, and the third is white?
        • The probability that the first ball will be red is 5/20, or 1/4. The probability that the second ball will be blue is 4/19, since there is one less ball left in the box, but still 4 blue ball. Finally, the probability that the third ball is white is 11/18 since we have already drawn two balls.

   3. Multiply the probabilities of each individual event. Regardless of whether you are dealing with independent or dependent events, as well as the number of outcomes (there can be 2, 3 or even 10), you can calculate the overall probability by multiplying the probabilities of all the events in question by each other. As a result, you will get the probability of several events, the following one by one. For example, the task is Find the probability of rolling a dice twice in a row with a 5.. These are two independent events, the probability of each of which is 1/6. Thus, the probability of both events is 1/6 x 1/6 = 1/36, that is, 0.027, or 2.7%.

      • Example 1. Two cards are drawn at random from the deck, one after the other. What is the probability that both cards will have a club suit? The probability of the first event is 13/52. The probability of the second event is 12/51. We find the total probability: 13/52 x 12/51 = 12/204 = 1/17, that is, 0.058, or 5.8%.
      • Example 2. A box contains 4 blue, 5 red and 11 white balls. If three balls are randomly drawn from the box, one after the other, what is the probability that the first is red, the second is blue, and the third is white? The probability of the first event is 5/20. The probability of the second event is 4/19. The probability of the third event is 11/18. So the total probability is 5/20 x 4/19 x 11/18 = 44/1368 = 0.032, or 3.2%.

When assessing the probability of the occurrence of any random event, it is very important to have a good idea in advance whether the probability (probability of the event) of the occurrence of the event of interest to us 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 on our own. We can do this even if the event is a complex collection of several elementary outcomes. And if several random events occur simultaneously or sequentially? How does this affect the probability of the event of interest to us? If I roll a die a few times and I want to get a six and I'm always unlucky, does that mean I should increase my bet because, according to probability theory, I'm about to get lucky? Alas, probability theory says nothing of the sort. Neither dice, nor cards, nor coins can remember what they showed us last time. It does not matter to them at all whether for the first time or for the tenth time today I test my fate. Every time I roll again, I know only one thing: and this time the probability of rolling a "six" again is one-sixth. Of course, this does not mean that the number I need will never fall out. It only means that my loss after the first roll and after any other roll - independent events. Events A and B are called independent if the realization of one of them does not affect the probability of the other event in any way. For example, the probabilities of hitting a target with the first of two guns do not depend on whether the other gun hit the target, so the events "the first gun hit the target" and "the second gun 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 by AB) can be calculated using the following theorem.

Probability multiplication theorem for independent events

P(AB) = P(A)*P(B) the probability of the simultaneous occurrence of two independent events is equal to the product of the probabilities of these events.

Example 1. The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 = 0.7; p2 = 0.8. Find the probability of hitting with one volley by both guns simultaneously.

As we have already seen, the events A (hit by the first gun) and B (hit by the second gun) are independent, i.e. P(AB)=P(A)*P(B)=p1*p2=0.56. What happens to our estimates if the initiating events are not independent? Let's change the previous example a little.

Example 2 Two shooters in a competition shoot at targets, and if one of them shoots accurately, then the opponent starts 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 it is necessary to somehow separate the two scenarios, to compose, in fact, two scenarios, two different tasks. In the first case, if the opponent misses, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent decently realized his chance, the probability of hitting the target for the second athlete is reduced. To separate the possible scenarios (they are often called hypotheses) of the development of events, we will often use the "probability tree" scheme. This diagram is similar in meaning to the decision tree, which you have probably already had to deal with. Each branch is a separate scenario, only now it has its own value of the so-called conditional probability (q 1 , q 2 , q 1 -1, q 2 -1).

This scheme is very convenient for the analysis of successive random events. It remains to clarify one more important question: where do the initial values ​​of probabilities come from in real situations? After all, the theory of probability does not work with the same coins and dice, does it? Usually these estimates are taken from statistics, and when statistics are 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 generally need.

Example 3 In a city of 100,000 inhabitants, suppose we need to estimate the size of the market for a new non-essential product, such as a color-treated hair conditioner. Let's consider the "tree of probabilities" scheme. In this case, we need to approximately estimate the value of the probability on each "branch". So, our estimates of market capacity:

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

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

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

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

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

According to the law of multiplication of probabilities, we determine the probability of the event of interest to us A \u003d (a city resident buys this new balm from us) \u003d 0.00045. Multiply this probability value by the number of inhabitants of the city. As a result, we have only 45 potential buyers, and given that one vial of this product lasts for several months, the trade is not very lively. Still, there are benefits from our assessments. Firstly, we can compare the forecasts of different business ideas, they will have different “forks” on the diagrams, and, of course, the probability values ​​will also be different. Secondly, as we have already said, random value It is not called random because it does not depend on anything at all. It's just that its exact 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 on those "forks" where the distribution of probabilities does not particularly suit us, on those factors that we are able to influence. Consider another quantitative example of consumer behavior research.

Example 3 An average of 10,000 people visit the food market per day. The probability that a market visitor walks into a dairy pavilion is 1/2. It is known that in this pavilion, on average, 500 kg of various products are sold per day. Can it be argued 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, in order to answer the question about the average purchase weight, we must find the answer to the question, what is the probability that a person who enters the pavilion buys something there. If we do not have such data at our disposal, but we need them, we will have to obtain them ourselves, after observing the visitors of the pavilion for some time. Suppose our observations show that only a fifth of the visitors to the pavilion buy something. As soon as these estimates are obtained by us, the task becomes already simple. Of the 10,000 people who came to the market, 5,000 will go to the pavilion of dairy products, there will be only 1,000 purchases. The average purchase weight is 500 grams. It is interesting to note that in order to build a complete picture of what is 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.

Tasks for self-examination.

1. Let it eat electrical circuit, consisting of n series-connected elements, each of which works independently of the others. The probability p of non-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 of the 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, each of which operates equipment for which the probabilities of failure during the next month are, respectively, p 1 , p 2 , p 3 and p 4 . Find the probability that in a month there will be no stoppage of production due to equipment failure.

In the economy, as well as in other areas of human activity or in nature, we constantly have to deal with events that cannot be accurately predicted. Thus, the volume of sales of goods depends on demand, which can vary significantly, and on a number of other factors that are almost impossible to take into account. Therefore, in the organization of production and sales, one has to predict the outcome of such activities on the basis of either one's own previous experience, or similar experience of other people, or intuition, which is also largely based on experimental data.

In order to somehow evaluate the event under consideration, it is necessary to take into account or specially organize the conditions in which this event is recorded.

The implementation of certain conditions or actions to identify the event in question is called experience or experiment.

The event is called random if, as a result of the experiment, it may or may not occur.

The event is called authentic, if it necessarily appears as a result of this experience, and impossible if it cannot appear in this experience.

For example, snowfall in Moscow on November 30th is a random event. The daily sunrise can be considered a certain event. Snowfall at the equator can be seen as an impossible event.

One of the main problems in probability theory is the problem of determining a quantitative measure of the possibility of an event occurring.

Algebra of events

Events are called incompatible if they cannot be observed together in the same experience. Thus, the presence of two and three cars in one store for sale at the same time are two incompatible events.

sum events is an event consisting in the occurrence of at least one of these events

An example of a sum of events is the presence of at least one of two products in a store.

work events is called an event consisting in the simultaneous occurrence of all these events

An event consisting in the appearance of two goods at the same time in the store is a product of events: - the appearance of one product, - the appearance of another product.

Events form a complete group of events if at least one of them necessarily occurs in the experience.

Example. The port has two berths for ships. Three events can be considered: - the absence of vessels at the berths, - the presence of one vessel at one of the berths, - the presence of two vessels at two berths. These three events form a complete group of events.

Opposite two unique possible events that form a complete group are called.

If one of the events that are opposite is denoted by , then the opposite event is usually denoted by .

Classical and statistical definitions of the probability of an event

Each of the equally possible test results (experiments) is called an elementary outcome. They are usually denoted by letters . For example, a dice is thrown. There can be six elementary outcomes according to the number of points on the sides.

From elementary outcomes, you can compose a more complex event. So, the event of an even number of points is determined by three outcomes: 2, 4, 6.

A quantitative measure of the possibility of occurrence of the event under consideration is the probability.

Most wide use received two definitions of the probability of an event: classic and statistical.

The classical definition of probability is related to the notion of a favorable outcome.

Exodus is called favorable this event, if its occurrence entails the occurrence of this event.

In the given example, the event under consideration is an even number of points on the dropped edge, has three favorable outcomes. In this case, the general
the number of possible outcomes. So, here you can use the classical definition of the probability of an event.

Classic definition equals the ratio of the number of favorable outcomes to the total number of possible outcomes

where is the probability of the event , is the number of favorable outcomes for the event, is total number possible outcomes.

In the considered example

The statistical definition of probability is associated with the concept of the relative frequency of occurrence of an event in experiments.

The relative frequency of occurrence of an event is calculated by the formula

where is the number of occurrence of an event in a series of experiments (tests).

Statistical definition. The probability of an event is the number relative to which the relative frequency is stabilized (established) with an unlimited increase in the number of experiments.

In practical problems, the relative frequency is taken as the probability of an event at a sufficiently large numbers tests.

From these definitions of the probability of an event, it can be seen that the inequality always holds

To determine the probability of an event based on formula (1.1), combinatorics formulas are often used to find the number of favorable outcomes and the total number of possible outcomes.

It is unlikely that many people think about whether it is possible to calculate events that are more or less random. Speaking in simple words, whether it is realistic to know which side of the die will fall out next time. It was this question that two great scientists asked, who laid the foundation for such a science as the theory of probability, in which the probability of an event is studied quite extensively.

Origin

If you try to define such a concept as probability theory, you get the following: this is one of the branches of mathematics that studies the constancy of random events. Of course, this concept does not really reveal the whole essence, so it is necessary to consider it in more detail.

I would like to start with the creators of the theory. As mentioned above, there were two of them, and it was they who were among the first who tried to calculate the outcome of an event using formulas and mathematical calculations. On the whole, the beginnings of this science appeared in the Middle Ages. At that time, various thinkers and scientists tried to analyze gambling, such as roulette, dice, and so on, thereby establishing a pattern and percentage of a particular number falling out. The foundation was laid in the seventeenth century by the aforementioned scientists.

At first, their work could not be attributed to the great achievements in this field, because everything they did was simply empirical facts, and the experiments were made visually, without the use of formulas. Over time, it turned out to achieve great results, which appeared as a result of observing the throwing of dice. It was this tool that helped to derive the first intelligible formulas.

Like-minded people

It is impossible not to mention such a person as Christian Huygens, in the process of studying a topic called "probability theory" (the probability of an event is covered precisely in this science). This person is very interesting. He, like the scientists presented above, tried to derive the regularity of random events in the form of mathematical formulas. It is noteworthy that he did not do this together with Pascal and Fermat, that is, all his works did not in any way intersect with these minds. Huygens brought out

An interesting fact is that his work came out long before the results of the work of the discoverers, or rather, twenty years earlier. Among the designated concepts, the most famous are:

• the concept of probability as a magnitude of chance;
• mathematical expectation for discrete cases;
• theorems of multiplication and addition of probabilities.

It is also impossible not to remember who also made a significant contribution to the study of the problem. Conducting his own tests, independent of anyone, he managed to present a proof of the law big numbers. In turn, the scientists Poisson and Laplace, who worked at the beginning of the nineteenth century, were able to prove the original theorems. It was from this moment that probability theory began to be used to analyze errors in the course of observations. side bypass this science nor could Russian scientists, or rather Markov, Chebyshev and Dyapunov. Based on the work done by the great geniuses, they fixed this subject as a branch of mathematics. These figures worked already at the end of the nineteenth century, and thanks to their contribution, phenomena such as:

• law of large numbers;
• theory of Markov chains;
• central limit theorem.

So, with the history of the birth of science and with the main people who influenced it, everything is more or less clear. Now it's time to concretize all the facts.

Basic concepts

Before touching on laws and theorems, it is worth studying the basic concepts of probability theory. The event takes the leading role in it. This topic is quite voluminous, but without it it will not be possible to understand everything else.

An event in probability theory is any set of outcomes of an experiment. There are not so many concepts of this phenomenon. So, the scientist Lotman, working in this area, said that in this case we are talking about what "happened, although it might not have happened."

Random events (probability theory pays special attention to them) is a concept that implies absolutely any phenomenon that has the ability to occur. Or, conversely, this scenario may not happen when many conditions are met. It is also worth knowing that it is random events that capture the entire volume of phenomena that have occurred. Probability theory indicates that all conditions can be repeated constantly. It was their conduct that was called "experiment" or "test".

A certain event is one that will 100% occur in a given test. Accordingly, an impossible event is one that will not happen.

The combination of a pair of actions (conditionally case A and case B) is a phenomenon that occurs simultaneously. They are designated as AB.

The sum of pairs of events A and B is C, in other words, if at least one of them happens (A or B), then C will be obtained. The formula of the described phenomenon is written as follows: C \u003d A + B.

Disjoint events in probability theory imply that the two cases are mutually exclusive. They can never happen at the same time. Joint events in probability theory are their antipode. This implies that if A happened, then it does not prevent B in any way.

Opposite events (probability theory deals with them in great detail) are easy to understand. It is best to deal with them in comparison. They are almost the same as incompatible events in probability theory. But their difference lies in the fact that one of the many phenomena in any case must occur.

Equally probable events are those actions, the possibility of repetition of which is equal. To make it clearer, we can imagine the tossing of a coin: the loss of one of its sides is equally likely to fall out of the other.

A favorable event is easier to see with an example. Let's say there is episode B and episode A. The first is the roll of the die with the appearance of an odd number, and the second is the appearance of the number five on the die. Then it turns out that A favors B.

Independent events in the theory of probability are projected only on two or more cases and imply the independence of any action from another. For example, A - dropping tails when throwing a coin, and B - getting a jack from the deck. They are independent events in probability theory. At this point, it became clearer.

Dependent events in probability theory are also admissible only for their set. They imply the dependence of one on the other, that is, the phenomenon B can occur only if A has already happened or, on the contrary, has not happened when this is the main condition for B.

The outcome of a random experiment consisting of one component is elementary events. Probability theory explains that this is a phenomenon that happened only once.

Basic Formulas

So, the concepts of "event", "probability theory" were considered above, the definition of the main terms of this science was also given. Now it's time to get acquainted directly with the important formulas. These expressions mathematically confirm all the main concepts in such a difficult subject as probability theory. The probability of an event plays a huge role here too.

It is better to start with the main ones. And before proceeding to them, it is worth considering what it is.

Combinatorics is primarily a branch of mathematics, it deals with the study of a huge number of integers, as well as various permutations of both the numbers themselves and their elements, various data, etc., leading to the appearance of a number of combinations. In addition to probability theory, this branch is important for statistics, computer science, and cryptography.

So, now you can move on to the presentation of the formulas themselves and their definition.

The first of these will be an expression for the number of permutations, it looks like this:

P_n = n ⋅ (n - 1) ⋅ (n - 2)…3 ⋅ 2 ⋅ 1 = n!

The equation applies only if the elements differ only in their order.

Now the placement formula will be considered, it looks like this:

A_n^m = n ⋅ (n - 1) ⋅ (n-2) ⋅ ... ⋅ (n - m + 1) = n! : (n - m)!

This expression is applicable not only to the order of the element, but also to its composition.

The third equation from combinatorics, and it is also the last one, is called the formula for the number of combinations:

C_n^m = n ! : ((n - m))! :m!

A combination is called a selection that is not ordered, respectively, and this rule applies to them.

It turned out to be easy to figure out the formulas of combinatorics, now we can move on to the classical definition of probabilities. This expression looks like this:

In this formula, m is the number of conditions favorable to the event A, and n is the number of absolutely all equally possible and elementary outcomes.

There are a large number of expressions, the article will not cover all of them, but the most important of them will be touched upon, such as, for example, the probability of the sum of events:

P(A + B) = P(A) + P(B) - this theorem is for adding only incompatible events;

P(A + B) = P(A) + P(B) - P(AB) - and this one is for adding only compatible ones.

Probability of producing events:

P(A ⋅ B) = P(A) ⋅ P(B) - this theorem is for independent events;

(P(A ⋅ B) = P(A) ⋅ P(B∣A); P(A ⋅ B) = P(A) ⋅ P(A∣B)) - and this one is for dependents.

The event formula will end the list. Probability theory tells us about Bayes' theorem, which looks like this:

P(H_m∣A) = (P(H_m)P(A∣H_m)) : (∑_(k=1)^n P(H_k)P(A∣H_k)),m = 1,..., n

In this formula, H 1 , H 2 , …, H n is the full group of hypotheses.

Examples

If you carefully study any branch of mathematics, it is not complete without exercises and sample solutions. So is the theory of probability: events, examples here are an integral component that confirms scientific calculations.

Formula for number of permutations

Let's say there are thirty cards in a deck of cards, starting with face value one. Next question. How many ways are there to stack the deck so that cards with a face value of one and two are not next to each other?

The task is set, now let's move on to solving it. First you need to determine the number of permutations of thirty elements, for this we take the above formula, it turns out P_30 = 30!.

Based on this rule, we will find out how many options there are to fold the deck in different ways, but we need to subtract from them those in which the first and second cards are next. To do this, let's start with the option when the first is above the second. It turns out that the first card can take twenty-nine places - from the first to the twenty-ninth, and the second card from the second to the thirtieth, it turns out only twenty-nine places for a pair of cards. In turn, the rest can take twenty-eight places, and in any order. That is, for a permutation of twenty-eight cards, there are twenty-eight options P_28 = 28!

As a result, it turns out that if we consider the solution when the first card is above the second, there are 29 ⋅ 28 extra possibilities! = 29!

Using the same method, you need to calculate the number of redundant options for the case when the first card is under the second. It also turns out 29 ⋅ 28! = 29!

From this it follows that there are 2 ⋅ 29! extra options, while there are 30 necessary ways to build the deck! - 2 ⋅ 29!. It remains only to count.

30! = 29! ⋅ 30; 30!- 2 ⋅ 29! = 29! ⋅ (30 - 2) = 29! ⋅ 28

Now you need to multiply all the numbers from one to twenty-nine among themselves, and then at the end multiply everything by 28. The answer is 2.4757335 ⋅〖10〗^32

Example solution. Formula for Placement Number

In this problem, you need to find out how many ways there are to put fifteen volumes on one shelf, but on the condition that there are thirty volumes in total.

In this problem, the solution is slightly simpler than in the previous one. Using the already known formula, it is necessary to calculate the total number of arrangements from thirty volumes of fifteen.

A_30^15 = 30 ⋅ 29 ⋅ 28⋅... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16 = 202 843 204 931 727 360 000

The answer, respectively, will be equal to 202,843,204,931,727,360,000.

Now let's take the task a little more difficult. You need to find out how many ways there are to arrange thirty books on two bookshelves, provided that only fifteen volumes can be on one shelf.

Before starting the solution, I would like to clarify that some problems are solved in several ways, so there are two ways in this one, but the same formula is used in both.

In this problem, you can take the answer from the previous one, because there we calculated how many times you can fill a shelf with fifteen books in different ways. It turned out A_30^15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ...⋅ 16.

We calculate the second shelf according to the permutation formula, because fifteen books are placed in it, while only fifteen remain. We use the formula P_15 = 15!.

It turns out that in total there will be A_30^15 ⋅ P_15 ways, but, in addition, the product of all numbers from thirty to sixteen will need to be multiplied by the product of numbers from one to fifteen, as a result, the product of all numbers from one to thirty will be obtained, that is, the answer equals 30!

But this problem can be solved in a different way - easier. To do this, you can imagine that there is one shelf for thirty books. All of them are placed on this plane, but since the condition requires that there be two shelves, we cut one long one in half, it turns out two fifteen each. From this it turns out that the placement options can be P_30 = 30!.

Example solution. Formula for combination number

Now we will consider a variant of the third problem from combinatorics. You need to find out how many ways there are to arrange fifteen books, provided that you need to choose from thirty absolutely identical ones.

For the solution, of course, the formula for the number of combinations will be applied. From the condition it becomes clear that the order of the identical fifteen books is not important. Therefore, initially you need to find out the total number of combinations of thirty books of fifteen.

C_30^15 = 30 ! : ((30-15)) ! : fifteen ! = 155 117 520

That's all. Using this formula, in the shortest possible time it was possible to solve such a problem, the answer, respectively, is 155 117 520.

Example solution. The classical definition of probability

Using the formula above, you can find the answer in a simple problem. But it will help to visually see and trace the course of actions.

The problem is given that there are ten absolutely identical balls in the urn. Of these, four are yellow and six are blue. One ball is taken from the urn. You need to find out the probability of getting blue.

To solve the problem, it is necessary to designate getting the blue ball as event A. This experience can have ten outcomes, which, in turn, are elementary and equally probable. At the same time, six out of ten are favorable for event A. We solve using the formula:

P(A) = 6: 10 = 0.6

By applying this formula, we found out that the probability of getting a blue ball is 0.6.

Example solution. Probability of the sum of events

Now a variant will be presented, which is solved using the formula for the probability of the sum of events. So, in the condition given that there are two boxes, the first contains one gray and five white balls, and the second contains eight gray and four white balls. As a result, one of them was taken from the first and second boxes. It is necessary to find out what is the chance that the balls taken out will be gray and white.

To solve this problem, it is necessary to designate events.

• So, A - take a gray ball from the first box: P(A) = 1/6.
• A '- they took a white ball also from the first box: P (A ") \u003d 5/6.
• B - a gray ball was taken out already from the second box: P(B) = 2/3.
• B' - they took a gray ball from the second box: P(B") = 1/3.

According to the condition of the problem, it is necessary that one of the phenomena occur: AB 'or A'B. Using the formula, we get: P(AB") = 1/18, P(A"B) = 10/18.

Now the formula for multiplying the probability has been used. Next, to find out the answer, you need to apply the equation for their addition:

P = P(AB" + A"B) = P(AB") + P(A"B) = 11/18.

So, using the formula, you can solve similar problems.

Outcome

The article provided information on the topic "Probability Theory", in which the probability of an event plays a crucial role. Of course, not everything was taken into account, but, based on the text presented, one can theoretically get acquainted with this section of mathematics. The science in question can be useful not only in professional work, but also in Everyday life. With its help, you can calculate any possibility of any event.

The text also touched upon significant dates in the history of the formation of the theory of probability as a science, and the names of people whose works were invested in it. This is how human curiosity led to the fact that people learned to calculate even random events. Once they were just interested in it, but today everyone already knows about it. And no one will say what awaits us in the future, what other brilliant discoveries related to the theory under consideration will be made. But one thing is for sure - research does not stand still!