Prove that the events are independent. Dependent and independent events. Solve probability multiplication problems on your own and then look at the solution

The dependence of events is understood in probabilistic sense, not functional. This means that upon the appearance of one of dependent events one cannot clearly judge the appearance of another. Probabilistic dependence means that the occurrence of one of the dependent events only changes the probability of the occurrence of the other. If the probability does not change, then the events are considered independent.

Definition: Let be an arbitrary probability space, and be some random events. They say that event A does not depend on the event IN , if its conditional probability coincides with the unconditional probability:

.

If , then they say that the event A depends on the event IN.

The concept of independence is symmetrical, that is, if an event A does not depend on the event IN, then the event IN does not depend on the event A. Indeed, let . Then . Therefore they simply say that events A And IN independent.

The following symmetric definition of the independence of events follows from the rule of multiplication of probabilities.

Definition: Events A And IN, defined on the same probability space are called independent, If

If , then events A And IN are called dependent.

Note that this definition is also valid in the case when or .

Properties of independent events.

1. If events A And IN are independent, then the following pairs of events are also independent: .

▲ Let us prove, for example, the independence of events. Let's imagine an event A as: . Since the events are incompatible, then , and due to the independence of the events A And IN we get that . This is what independence means. ■

2. If the event A does not depend on events IN 1 And AT 2, which are inconsistent () , that event A does not depend on the amount.

▲ Indeed, using the axiom of additivity of probability and independence of the event A from events IN 1 And AT 2, we have:

The relationship between the concepts of independence and incompatibility.

Let A And IN- any events that have a non-zero probability: , so . If the events A And IN are inconsistent (), then equality can never take place. Thus, incompatible events are dependent.

When more than two events are considered simultaneously, their pairwise independence does not sufficiently characterize the relationship between the events of the entire group. In this case, the concept of independence in the aggregate is introduced.

Definition: Events defined on the same probability space are called collectively independent, if for any 2 £ m £ n and any combination of indices the equality is true:

At m = 2 From independence in the aggregate follows pairwise independence of events. The reverse is not true.

Example. (Bernstein S.N.)

A random experiment involves tossing a regular tetrahedron (tetrahedron). A face that has fallen down is observed. The faces of the tetrahedron are colored as follows: 1st face - white, 2nd face - black,
The 3rd side is red, the 4th side contains all colors.

Let's consider the events:

A= (white dropout); B= (black dropout);

C= (Red drop).

Then ;

Therefore, events A, IN And WITH are pairwise independent.

However, .

Therefore events A, IN And WITH are not collectively independent.

In practice, as a rule, the independence of events is not established by checking it by definition, but on the contrary: events are considered independent from any external considerations or taking into account the circumstances random experiment, and use independence to find the probabilities of events occurring.

Theorem (multiplication of probabilities for independent events).

If events defined on the same probability space are independent in the aggregate, then the probability of their product is equal to the product of the probabilities:

▲ The proof of the theorem follows from the definition of the independence of events in the aggregate or from the general theorem of multiplication of probabilities, taking into account the fact that in this case

Example 1 (typical example on finding conditional probabilities, the concept of independence, the theorem of addition of probabilities).

The electrical circuit consists of three independently operating elements. The failure probabilities of each element are respectively equal.

1) Find the probability of failure of the circuit.

2) It is known that the circuit has failed.

What is the probability that it refused:

a) 1st element; b) 3rd element?

Solution. Consider the events = (Refused k th element), and event A= (The circuit has failed). Then the event A is presented as:

.

1) Since the events are not incompatible, the axiom of additivity of probability P3) is not applicable and to find the probability one should use the general theorem of addition of probabilities, according to which

It is unlikely that many people think about whether it is possible to calculate events that are more or less random. To put it simply in simple words, is it really possible to know which side of the cube will come up next time? It was this question that two great scientists asked themselves, 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 will get the following: this is one of the branches of mathematics that deals with the study of constancy 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 they were one of the first to try to calculate the outcome of this or that event using formulas and mathematical calculations. In general, 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 the pattern and percentage of a particular number falling out. The foundation was laid in the seventeenth century by the above-mentioned scientists.

At first, their works could not be considered great achievements in this field, because all they did were simply empirical facts, and experiments were carried out visually, without using formulas. Over time, it was possible 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 Christiaan 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 pattern 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 intersect with these minds. Huygens deduced

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

• the concept of probability as the value 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 was able to present a proof of the law of large 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 observations. Russian scientists, or rather Markov, Chebyshev and Dyapunov, could not ignore this science. Based on the work done by great geniuses, they established this subject as a branch of mathematics. These figures worked already at the end of the nineteenth century, and thanks to their contribution, the following phenomena were proven:

• law of large numbers;
• Markov chain theory;
• 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 the time has come to clarify all the facts.

Basic Concepts

Before touching on laws and theorems, it is worth studying the basic concepts of probability theory. The event plays a 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 quite a few concepts of this phenomenon. Thus, 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 focuses on them Special attention) is a concept that implies absolutely any phenomenon that has the opportunity to occur. Or, conversely, this scenario may not happen if many conditions are met. It is also worth knowing that it is random events that capture the entire volume of phenomena that have occurred. The theory of probability indicates that all conditions can be repeated constantly. It is their conduct that is called “experience” or “test”.

A reliable event is a phenomenon that is one hundred percent likely to happen 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 for the described phenomenon is written as follows: C = A + B.

Incongruent events in probability theory imply that two cases are mutually exclusive. Under no circumstances can they happen at the same time. Joint events in probability theory are their antipode. What is meant here is that if A happened, then it does not prevent B in any way.

Opposite events (probability theory considers them in great detail) are easy to understand. The best way to understand them is by comparison. They are almost the same as incompatible events in probability theory. But their difference lies in the fact that one of many phenomena must happen in any case.

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

It is easier to consider an auspicious event with an example. Let's say there is an episode B and an episode A. The first is a throw dice 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 probability theory are projected only onto two or more cases and imply the independence of any action from another. For example, A is the loss of heads when tossing a coin, and B is the drawing of 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 permissible only for a set of them. They imply the dependence of one on the other, that is, phenomenon B can only occur if A has already happened or, conversely, has not happened, when this is the main condition for B.

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

Basic formulas

So, the concepts of “event” and “probability theory” were discussed above; a definition of the basic 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 complex subject as probability theory. The probability of an event plays a huge role here too.

It’s better to start with the basic ones. And before you start with them, it’s worth considering what they are.

Combinatorics is primarily a branch of mathematics; it deals with the study huge amount 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 we can move on to presenting the formulas themselves and their definition.

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

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

The equation is applied only if the elements differ only in the order of their arrangement.

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

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

A combination refers to selections that are not ordered; accordingly, this rule applies to them.

It was easy to understand the combinatorics formulas; now you 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 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 ones 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 events occurring:

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 the dependent.

The list of events will be completed by the formula of events. 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 a complete group of hypotheses.

Examples

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

Formula for the number of permutations

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

The task has been 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 formula presented above, it turns out P_30 = 30!.

Based on this rule, we 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 each other. 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 up twenty-nine places - from the first to the twenty-ninth, and the second card from the second to the thirtieth, making a total of twenty-nine places for a pair of cards. In turn, the rest can accept twenty-eight places, and in any order. That is, to rearrange 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 will be 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 to be 29 ⋅ 28! = 29!

It follows from this that there are 2 ⋅ 29 extra options!, while the necessary ways to assemble a deck are 30! - 2 ⋅ 29!. All that remains is 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, and then finally 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 provided that there are thirty volumes in total.

The solution to this problem is a little simpler than the previous one. Using the already known formula, it is necessary to calculate the total number of arrangements of 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, accordingly, will be equal to 202,843,204,931,727,360,000.

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

Before starting the solution, I would like to clarify that some problems can be solved in several ways, and this one has two methods, but both use the same formula.

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 will calculate the second shelf using the permutation formula, because fifteen books can be placed in it, while only fifteen remain. We use the formula P_15 = 15!.

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

But this problem can be solved in another 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 saw one long one in half, so we get two of fifteen. From this it turns out that there can be P_30 = 30 options for arrangement!.

Example solution. Formula for combination number

Now we will consider a version 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 ones.

To solve, 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)) ! : 15 ! = 155 117 520

That's all. Using this formula, we were able to solve this problem in the shortest possible time; the answer, accordingly, is 155,117,520.

Example solution. Classic definition of probability

Using the formula above, you can find the answer to a simple problem. But this will help to clearly see and track the progress of actions.

The problem states 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 denote the acquisition 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, six are favorable to event A. We solve using the formula:

P(A) = 6: 10 = 0.6

Applying this formula, we learned that the probability of getting the blue ball is 0.6.

Example solution. Probability of the sum of events

An option will now be presented that is solved using the sum-of-events probability formula. So, the condition is 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, they took one of them from the first and second boxes. You need to find out what is the chance that the balls you get will be gray and white.

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

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

According to the conditions of the problem, it is necessary for one of the phenomena to happen: 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 of their addition:

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

This is how you can solve similar problems using the formula.

Bottom line

The article presented information on the topic "Probability Theory", in which the probability of an event plays a vital role. Of course, not everything was taken into account, but, based on the presented text, you can theoretically familiarize yourself with this section of mathematics. The science in question can be useful not only in professional matters, 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 the people whose work was invested in it. This is how human curiosity led to the fact that people learned to calculate even random events. Once upon a time they were simply interested in this, 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!

General statement of the problem: the probabilities of some events are known, and you need to calculate the probabilities of other events that are associated with these events. In these problems, there is a need for operations with probabilities such as addition and multiplication of probabilities.

For example, while hunting, two shots are fired. Event A- hitting a duck with the first shot, event B- hit from the second shot. Then the sum of events A And B- hit with the first or second shot or with two shots.

Problems of a different type. Several events are given, for example, a coin is tossed three times. You need to find the probability that either the coat of arms will appear all three times, or that the coat of arms will appear at least once. This is a probability multiplication problem.

Addition of probabilities of incompatible events

Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.

Sum of events A And B denote A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B– an event that occurs if and only if the event occurred during observation A or event B, or simultaneously A And B.

If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.

Probability addition theorem. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:

For example, while hunting, two shots are fired. Event A– hitting a duck with the first shot, event IN– hit from the second shot, event ( A+ IN) – a hit from the first or second shot or from two shots. So, if two events A And IN– incompatible events, then A+ IN– the occurrence of at least one of these events or two events.

Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.

Solution. Let us assume that the event A- “the red ball is taken”, and the event IN- “The blue ball was taken.” Then the event is “a colored (not white) ball is taken.” Let's find the probability of the event A:

and events IN:

Events A And IN– mutually incompatible, since if one ball is taken, then it is impossible to take balls of different colors. Therefore, we use the addition of probabilities:

The theorem for adding probabilities for several incompatible events. If events constitute a complete set of events, then the sum of their probabilities is equal to 1:

The sum of the probabilities of opposite events is also equal to 1:

Opposite events form a complete set of events, and the probability of a complete set of events is 1.

Probabilities of opposite events are usually indicated in small letters p And q. In particular,

from which the following formulas for the probability of opposite events follow:

Example 2. The target in the shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the target in the first zone is 0.15, in the second zone – 0.23, in the third zone – 0.17. Find the probability that the shooter will hit the target and the probability that the shooter will miss the target.

Solution: Find the probability that the shooter will hit the target:

Let's find the probability that the shooter will miss the target:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

Addition of probabilities of mutually simultaneous events

Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a die the event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since 4 is an even number, the two events are compatible. In practice, there are problems of calculating the probabilities of the occurrence of one of the mutually simultaneous events.

Probability addition theorem for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events has the following form:

Since events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:

Event A will occur if one of two incompatible events occurs: or AB. However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:

Likewise:

Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:

When using formula (8), it should be taken into account that events A And IN can be:

• mutually independent;
• mutually dependent.

Probability formula for mutually independent events:

Probability formula for mutually dependent events:

If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is:

Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive the second car. Find:

• the probability that both cars will win;
• the probability that at least one car will win;

1) The probability that the first car will win does not depend on the result of the second car, so the events A(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:

2) Find the probability that one of the two cars will win:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

Solve the addition of probabilities problem yourself, and then look at the solution

Example 4. Two coins are tossed. Event A- loss of the coat of arms on the first coin. Event B- loss of the coat of arms on the second coin. Find the probability of an event C = A + B .

Multiplying Probabilities

Probability multiplication is used when the probability of a logical product of events must be calculated.

In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.

Probability multiplication theorem for independent events. Probability of simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:

Example 5. The coin is tossed three times in a row. Find the probability that the coat of arms will appear all three times.

Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:

Solve probability multiplication problems on your own and then look at the solution

Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games Are there any unplayed balls left in the box?

Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the letters will form the word "end".

Example 8. From full deck cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards will be of different suits.

Example 9. The same task as in example 8, but each card after being removed is returned to the deck.

More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".

The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula.

Independent events

In the practical application of probabilistic-statistical decision-making methods, the concept of independence is constantly used. For example, when applying statistical methods for product quality management, they talk about independent measurements of the values ​​of controlled parameters of the product units included in the sample, the independence of the occurrence of defects of one type from the occurrence of defects of another type, etc. The independence of random events is understood in probabilistic models in the following sense.

Definition 2. Events A And IN are called independent if P(AB) = P(A) P(B). Several events A, IN, WITH,... are called independent if the probability of their joint implementation is equal to the product of the probabilities of each of them occurring separately: R(ABC…) = R(A)R(IN)R(WITH)…

This definition corresponds to the intuition of independence: the occurrence or non-occurrence of one event should not affect the occurrence or non-occurrence of another. Sometimes the ratio R(AB) = R(A) R(IN|A) = P(B)P(A|B), valid for P(A)P(B) > 0, also called the probability multiplication theorem.

Statement 1. Let events A And IN independent. Then the events and are independent, the events and IN independent, events A and independent (here - the event opposite A, and - event opposite IN).

Indeed, from property c) in (3) it follows that for events WITH And D, whose product is empty, P(C+ D) = P(C) + P(D). Since the intersection AB And IN is empty, but there is a union IN, That P(AB) + P(B) = P(B). Since A and B are independent, then P(B) = P(B) - P(AB) = P(B) - P(A)P(B) = P(B)(1 - P(A)). Let us now note that from relations (1) and (2) it follows that P() = 1 – P(A). Means, P(B) = P()P(B).

Derivation of equality P(A) = P(A)P() differs from the previous one only by replacement everywhere A on IN, A IN on A.

To prove independence And let's take advantage of the fact that events AB, B, A, do not have pairwise common elements, but in total they constitute the entire space of elementary events. Hence, R(AB) + P(B) + P(A) + P() = 1. Using the previously proven relations, we obtain that P(B)= 1 -R(AB) - P(B)( 1 - P(A)) - P(A)( 1 - P(B))= ( 1 – R(A))( 1 – P(B)) = P()P(), which was what needed to be proven.

Example 3. Consider the experiment of throwing a die with the numbers 1, 2, 3, 4, 5,6 written on its faces. We assume that all edges have the same chance of being on top. Let us construct the corresponding probability space. Let us show that the events “at the top is a face with an even number” and “at the top is a face with a number divisible by 3” are independent.

Example analysis. The space of elementary outcomes consists of 6 elements: “at the top is the edge with 1”, “at the top is the edge with 2”, ..., “at the top is the edge with 6”. The event “on top – a face with an even number” consists of three elementary events – when 2, 4 or 6 is on top. The event “on top – a face with a number divisible by 3” consists of two elementary events – when 3 or 6 is on top. Since Since all edges have the same chance of being on top, then all elementary events must have the same probability. Since there are 6 elementary events in total, each of them has a probability of 1/6. By definition, the event “at the top is a face with an even number” has a probability of ½, and the event “at the top is a face with a number divisible by 3” has a probability of 1/3. The product of these events consists of one elementary event “at the top - edge with 6”, and therefore has a probability of 1/6. Since 1/6 = ½ x 1/3, the events in question are independent according to the definition of independence.

If, when an event occurs, the probability of the event does not change, then events And are called independent.

Theorem:Probability of co-occurrence of two independent events And (works And ) is equal to the product of the probabilities of these events.

Indeed, since events And are independent, then
. In this case, the formula for the probability of events occurring is And takes on the form.

Events
are called pairwise independent, if any two of them are independent.

Events
are called jointly independent (or simply independent), if every two of them are independent and every event and all possible products of the others are independent.

Theorem:Probability of the product of a finite number of independently independent events
is equal to the product of the probabilities of these events.

Let us illustrate the difference in the application of formulas for the probability of a product of events for dependent and independent events using examples

Example 1. The probability of the first shooter hitting the target is 0.85, the second 0.8. The guns fired one shot each. What is the probability that at least one shell hit the target?

Solution: P(A+B) =P(A) +P(B) –P(AB) Since the shots are independent, then

P(A+B) = P(A) +P(B) –P(A)*P(B) = 0.97

Example 2. The urn contains 2 red and 4 black balls. 2 balls are taken out of it in a row. What is the probability that both balls are red?

Solution: 1 case. Event A is the appearance of a red ball on the first draw, event B on the second. Event C – the appearance of two red balls.

P(C) =P(A)*P(B/A) = (2/6)*(1/5) = 1/15

Case 2. The first ball drawn is returned to the basket

P(C) =P(A)*P(B) = (2/6)*(2/6) = 1/9

Total probability formula.

Let the event can only happen with one of the incompatible events
, forming a complete group. For example, a store receives the same products from three enterprises and in different quantities. The likelihood of producing low-quality products at these enterprises varies. One of the products is randomly selected. It is required to determine the probability that this product is of poor quality (event ). Events here
– this is the selection of a product from the products of the corresponding enterprise.

In this case, the probability of the event can be considered as the sum of the products of events
.

Using the theorem for adding the probabilities of incompatible events, we obtain
. Using the probability multiplication theorem, we find

.

The resulting formula is called total probability formula.

Bayes formula

Let the event occurs simultaneously with one of incompatible events
, the probabilities of which
(
) are known before experiment ( a priori probabilities). An experiment is carried out, as a result of which the occurrence of an event is registered , and it is known that this event had certain conditional probabilities
(
). We need to find the probabilities of events
if it is known that the event happened ( a posteriori probabilities).

The problem is that, having new information (event A occurred), we need to reestimate the probabilities of events
.

Based on the theorem on the probability of the product of two events

.

The resulting formula is called Bayes formulas.

Basic concepts of combinatorics.

When solving a number of theoretical and practical problems, it is necessary to create various combinations from a finite set of elements according to given rules and count the number of all possible such combinations. Such tasks are usually called combinatorial.

When solving problems, combinatorists use the rules of sum and product.