Probability theory. Probability of the event, random events (probability theory). Independent and incompatible events in probability theory

It is unlikely that many people are wondering whether it is possible to calculate events that are to some extent accidental. In simple words, is it really possible to know which side of the dice in dice will drop next time. It was this question that asked two great scientists who initiated a science such as probability theory , the probability of an event in which is studied quite extensively.

Origin

If we try to define such a concept as the probability theory, then the following result will be obtained: this is one of the branches of mathematics that deals with the study of the constancy of random events. Clearly, this concept does not really disclose the whole point, so it is necessary to consider it in more detail.

I would like to start with the founders of the theory. As mentioned above, there were two of them, this is Pierre Fermat and Blaise Pascal. They were one of the first to use the formulas and mathematical calculations to calculate the outcome of an event. In general, the beginnings of this science manifested itself in the Middle Ages. At that time, various thinkers and scientists tried to analyze gambling, such as roulette, bones and so on, thereby establishing the regularity and percentage ratio of the fall of a particular number. The foundation was laid in the seventeenth century precisely by the above-mentioned scientists.

At first, their works could not be attributed to the great achievements in this field, because everything they did was simply empirical facts, and the experiments were visualized without the use of formulas. Over time, it turned out to achieve great results, which appeared due to observation of the throwing of bones. It was this tool that helped to get the first intelligible formulas.

Like-minded people

It is impossible not to mention such a person as Christian Huygens, in the process of studying the topic called "probability theory" (the probability of the event is covered in this science). This person is very interesting. He, as well as the scientists presented above, tried to derive the laws of random events in the form of mathematical formulas. It is noteworthy that he did not do it together with Pascal and Fermat, that is, all his works did not overlap with these minds. Huygens derived the basic concepts of probability theory.

It is interesting that his work was published long before the results of the works of the discoverers, or rather, twenty years earlier. Among the designated concepts, the most famous are:

• The concept of probability as the magnitude of a chance;
• Mathematical expectation for discrete cases;
• Theorems of multiplication and addition of probabilities.

Also it is impossible not to recall Jakob Bernoulli, who also made a significant contribution to the study of the problem. Carrying out his own, no one on independent testing, he managed to present a proof of the law of large numbers. In turn, the scientists of Poisson and Laplace, who worked in the early nineteenth century, were able to prove the original theorems. It was from this moment on to use the theory of probability to analyze errors in the course of observations. The Russian scientists, or more precisely Markov, Chebyshev and Diapunov could not bypass this science either. They, based on the work done by the great geniuses, fixed this subject as a section of mathematics. These figures worked at the end of the nineteenth century, and thanks to their contribution, such phenomena as:

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

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

Basic concepts

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

The event in probability theory is Any set of outcomes of the experience. There are not so many notions of this phenomenon. So, scientist Lotman, who works in this field, said that in this case it is about what "happened, although it could not happen".

Random events (probability theory pays special attention to them) is a concept that implies absolutely any phenomenon that can occur. Or, on the contrary, this scenario may not happen when many conditions are met. It is also worthwhile to know that it is random events that capture the entire volume of the occurred phenomena. The probability theory indicates that all conditions can be repeated all the time. It was their conduct that was called "experience" or "test."

A certain event is a phenomenon that will completely happen in this trial. Accordingly, an impossible event is one that does not happen.

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

The sum of the pairs of events A and B is C, in other words, if at least one of them occurs (A or B), then the result is C. The formula for the phenomenon described is written as: C = A + B.

Non-joint events in probability theory imply that two cases mutually exclude each other. At the same time they can not happen in any case. Joint events in probability theory are their antipode. Here it is meant that if A happened, then it does not prevent V.

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

Equally possible events are those actions whose repeatability is equal. To be clearer, you can imagine throwing a coin: the fall of one of its sides is equally likely the fall of another.

A favorable event is easier to consider with an example. Suppose there is an episode B and an episode A. The first is a roll of the dice with the appearance of an odd number, and the second is the appearance of the number five on the cube. Then it turns out that A is favorable to B.

Independent events in the theory of probability are projected only in two or more cases and imply the independence of some action from the other. For example, A - dropping a tails while throwing a coin, and B - getting a jack from a deck. They are independent events in probability theory. With this moment it became clearer.

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

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

Basic Formulas

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

It is better to start with the basic formulas of combinatorics. And before you proceed 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 the numbers themselves, 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 proceed to the representation 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 is used only if the elements differ only in the order of their location.

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 placement of the element, but also to its composition.

The third equation of combinatorics, and it is the latter, is called the formula for the number of combinations:

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

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

With the combinatorics formulas it was possible to sort out without difficulty, now we can proceed to the classical definition of probabilities. This expression looks like this:

P (A) = m: n.

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

There are a lot of expressions, the article will not cover everything, but the most important ones will be affected, such as the probability of the sum of events:

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

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

Probability of the event:

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

(P (A ⋅ B) = P (A) ⋅ P (B|A); P (A ⋅ B) = P (A) ⋅ P (A; B)), and this for the dependent ones.

End the list of event formulas. The theory of probability tells us about the theorem Bayes, which looks like this:

P (H_m|A) = (P (H_m) P (A|H_m)): (Σ_ (k = 1) n n P (H_k) P (AHH_k)), m = 1, N

In this formula, H ₁ , H ₂ , ..., H _n is a complete set of hypotheses.

We will dwell on this, we will further consider examples of applying formulas to solve specific problems from practice.

Examples

If you carefully study any section of mathematics, it does not do without exercises and sample solutions. So the theory of probability: events, examples here are an integral component, confirming scientific calculations.

The formula for the number of permutations

Let's say that there are thirty cards in a card deck, starting with a face value of one. Next question. How many ways there are ways to fold the deck so that cards with face value one and two are not located side by side?

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

Based on this rule, we learn 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 will be next. To do this, we 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 30th, you get only twenty-nine places for a pair of cards. In turn, the rest can take twenty-eight places, and in an arbitrary order. That is, for the interchange of twenty-eight cards there are twenty-eight variants P_28 = 28!

In the end, it turns out that if we consider the solution, when the first card is over the second, the extra possibilities will turn out to be 29 ⋅ 28! = 29!

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

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

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

Now we need to multiply all the numbers from one to twenty-nine, then multiply everything by 28. At the end, we get 2,4757335 ⋅ 〖10〗 ^ 32

Solution of the example. The formula for the number of placements

In this task it is necessary to find out how many ways there are to put fifteen volumes on one shelf, but on the condition that there are thirty volumes altogether.

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 to 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 202 843 204 931 727 360 000.

Now let's take the task a little more complicated. It is necessary to find out how many ways there are to place thirty books on two bookshelves, provided that only fifteen volumes can be on one shelf.

Before the beginning of the solution, I would like to clarify that some problems are solved in several ways, so there are two ways in this, but both use the same formula.

In this task, we can take the answer from the previous one, because there we calculated how many times it is possible to fill the shelf for fifteen books in different ways. It turned out that A_30 ^ 15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16.

The second shelf will be calculated according to the permutation formula, since fifteen books are placed in it, while there are only fifteen books left. We use the formula P_15 = 15 !.

It turns out that the sum will be A_30 ^ 15 ⋅ P_15 ways, but besides this, the product of all numbers from thirty to sixteen will have to be multiplied by the product of numbers from one to fifteen, in the end we get the product of all numbers from one to thirty, that is the answer Is equal to 30!

But this task can be solved in a different way - it's 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 are two shelves, then we cut one long saw in half, we get two by fifteen. From this it turns out that the variants of the arrangement can be P_30 = 30 !.

Solution of the example. The formula for the combination number

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

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 it is necessary to find out the total number of combinations of thirty books by fifteen.

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

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

Solution of the example. The classical definition of probability

Using the formula above, you can find the answer in a simple task. But this will help visually see and follow the course of action.

In the problem it 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 know the probability of getting blue.

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

P (A) = 6: 10 = 0.6

Applying this formula, we learned that the ability to get a blue ball is 0.6.

Solution of the example. Probability of the sum of events

Now a variant will be presented, which is solved using the probability formula of the sum of events. So, in the condition it is given that there are two boxes, in the first there is one gray and five white balls, and in the second - 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 received balls will be gray and white.

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

• So, A - took the gray ball from the first drawer: P (A) = 1/6.
• A '- took a white ball also from the first drawer: P (A') = 5/6.
• B - extracted the gray ball from the second box: P (B) = 2/3.
• B '- took a gray ball from the second box: P (B') = 1/3.

By the condition of the problem, it is necessary that one of the events 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 was used. Next, to find out the answer, you need to apply the equation of their addition:

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

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

The result

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

The text also touched upon significant dates in the history of the emergence of probability theory as a science, and the names of people whose works were invested in it. That's how human curiosity has led to the fact that people have learned to count even random events. Once they just became interested in it, but today everyone knows about it. And no one will say what awaits us in the future, what other ingenious discoveries connected with the theory under consideration will be committed. But one thing is for sure - research on the spot is not worth it!