How the probability is calculated. The classic formula for calculating the probability

The union (logical sum) of N events is called an event that is observed every time at least one of events ... In particular, the union of events A and B is the event A+ B(some authors
), which is observed when comesor A,or Bor both of these events at the same time(Fig. 7). The sign of intersection in the textual wording of events is the union "or".

Rice. 7. Combining events A + B

It should be borne in mind that the probability of the event P (A) corresponds to both the left part of the shaded part in Fig. 7 of the figure, and its central part, marked as
... And the outcomes corresponding to event B are located both on the right side of the shaded figure and in the marked
central part. Thus, when adding and area
will actually enter this sum twice, and the exact expression for the area of ​​the shaded figure has the form
.

So, likelihood of unification two events A and B is equal to

For a large number of events, the general calculation expression becomes extremely cumbersome due to the need to take into account the numerous options for the mutual overlap of regions. However, if the combined events are incompatible (see p. 33), then the mutual overlap of areas turns out to be impossible, and the favorable zone is determined directly by the sum of the areas corresponding to individual events.

Probability amalgamations arbitrary number inconsistent events defined by the expression

Corollary 1: A complete group of events consists of inconsistent events, one of which is necessarily realized in the experience. As a result, if events …
,form a complete group, then for them

Thus,

WITHConsequence 3 Let us take into account that the opposite statement “will occur at least one of the events …
"Is the statement" none of the events …
not implemented. " That is, in other words, “in experience, events will be observed , and , and ..., and ", Which already represents the intersection of events opposite to the original set. Hence, taking into account (2 .0), to combine an arbitrary number of events, we obtain

Corollaries 2, 3 show that in cases where direct calculation of the probability of an event is problematic, it is useful to assess the complexity of researching the event opposite to it. After all, knowing the meaning
, obtain from (2 .0) the required value
no longer presents any work.

1. Examples of calculating the probabilities of complex events

Example 1 : Two students (Ivanov and Petrov) together Irushed to the defense of laboratory work, having learned the first 8 conof the 10 available questions to this work. Checking the readiness, nthe teacher asks everyone only onen a randomly selected question. Determine the likelihood of the following events:

A= “Ivanov will defend laboratory work”;

B= “Petrov will defend laboratory work”;

C= “Both defend laboratory work”;

D= “At least one of the students will defend the job”;

E= “Only one of the students will defend the job”;

F= "None of them will defend the job."

Solution. Note that the ability to defend a job like Ivanova, tas and Petrova separately is determined only by the number of questions mastered, the poetat... (Note: in this example, the values ​​of the resulting fractions were deliberately not reduced to simplify the comparison of the calculation results.)

EventCcan be formulated differently as “both Ivanov and Petrov will defend the work”; will happenand eventA, and eventB... Thus, the eventCis the intersection of eventsAandB, and in accordance with (2 .0)

where the factor “7/9” appears due to the fact that the occurrence of the eventAmeans that Ivanov got a “good” question, which means that Petrov now has only 7 “good” questions out of the remaining 9 questions.

EventDimplies that “work will be protectedor Ivanov,or Petrov,or they are both together, ”that is, at least one of the events will occurAandB... So the eventDis a union of eventsAandB, and in accordance with (2 .0)

which is in line with expectations, since even for each student individually, the chances of success are quite high.

WITHGenesis E means that “either Ivano will defend the jobc, and Petrov "pfalls ",or Ivanov will be caught unsuccessful injust ask, and Petrov will cope with the defense. " The two alternatives are mutually exclusive (inconsistent), so

Finally, the statementFwill be true only if “and Ivanov,and Petrov with protectionnot will cope. " So,

This completes the solution of the problem, but it is useful to note the following points:

1. Each of the obtained probabilities satisfies condition (1 .0),oh if for
and
get conflictwith(1 .0) is impossible in principle, then for
try andusing (2 .0) instead of (2 .0) would lead to an explicitly uncorrectedect value
... It is important to remember that such a value of probability is fundamentally impossible, and upon obtaining such a paradoxical result, immediately start looking for an error.

2. The found probabilities satisfy the relationsm

.

NSthen it is quite expected, since developmentsC, EandFform fullgroup, and eventsDandFare opposite to each other. Consideration of theseratios on the one hand can be usedvan to double-check the calculations, and in another situation it can serve as the basis for an alternative way to solve the problem.

NS note : Don't neglect the written fixationthe exact formulation of the event, otherwise, in the course of solving the problem, you can involuntarily switch to a different interpretation of the meaning of this event, which will entail errors in reasoning.

Example 2 : In a large batch of microcircuits that have not passed the final quality control, 30% of the products are defective.If you choose any two microcircuits from this batch at random, then what isthe probability that among them:

A= “Both good”;

B= "Exactly 1 usable microcircuit";

C= “Both defective”.

Let us analyze the following version of reasoning (careful, contains a bug):

Since we are talking about a large batch of products, the removal of several microcircuits from it practically does not affect the ratio of the number of good and defective products, which means that choosing some microcircuits from this batch several times in a row, we can assume that in each of the cases there remain unchanged probabilities

= P(defective item selected) = 0.3 and

= P(good product selected) = 0.7.

For the event to occurAit is necessary thatand first,and for the second time a suitable product was chosen, and therefore (taking into account the independence from each other of the success of the choice of the first and second microcircuits) for the intersection of events we have

Similarly, for the event C to occur, both products must be defective, and to obtain B, you need to select a suitable product once, and a defective product once.

Error symptom. NSalthough all the probabilities obtained aboveand look plausible, when analyzed together, it is easy tonote that .However, casesA, BandCform a completethe group of events for which to run .This contradiction indicates the presence of some kind of error in reasoning.

WITH a lot of mistakes. We introduce into consideration two auxiliaryflax events:

= "The first microcircuit is good, the second is defective";

= "The first microcircuit is defective, the second is good."

It is obvious that, however, it is precisely this version of the calculation that was used above to obtain the probability of an eventBalthough eventsBand are not uhequivalent... Actually,
since formulationdevelopmentsBrequires that among the microcircuits exactlyone but absolutelynot necessarily the first was good (and the other was defective). Therefore, although event is not a duplicate of the event , but should take into accountindependently. Given the inconsistency of events and , the probability of their logical sum will be

After the indicated correction of the calculations, we have

which indirectly confirms the correctness of the found probabilities.

Note : Pay special attention to the difference in the wording of events such as “onlyfirst of the listed elements must ... "and" onlyone of the listed aleents should ... ”. The latter event is clearly broader and includesTthe first as one of the (possibly numerousx) options. These alternatives (even if their probabilities coincide) should be considered independently of each other.

NS note : The word "percentage" comes from "per cent”, I.e."One hundred". The representation of frequencies and probabilities in percentages allows you to operate with larger values, which sometimes makes it easier to perceive the values ​​"by ear". However, it is cumbersome and inefficient to use multiplication or division by “100%” in calculations for correct normalization. In this regard, notWhen using values, be sure to mentionexpressed as a percentage, substitute them in the calculated expressions forthe same in the form of fractions of one (for example, 35% in the calculation is writteni as “0.35”) to minimize the risk of erroneous normalization of the results.

Example 3 : The resistor set contains one resistor4 kΩ nominal, three 8 kΩ resistors and six resistorsorors with a resistance of 15 kOhm. Three resistors chosen at random are connected in parallel with each other. Determine the probability of obtaining the final resistance not exceeding 4 kOhm.

Resh enie. Parallel connection resistance cuthistory can be calculated by the formula

.

This allows events such as

A= “Three 15 kΩ resistors selected” = “
”;

B= “Inzyty two resistors of 15 kOhm and one with resistancem 8 kOhm "="
”…

The complete group of events corresponding to the condition of the problem includes a number of more options, and it is precisely those thatwhich comply with the requirement to obtain a resistance of not more than 4 kOhm. However, although the “direct” solution path, involving the calculation (and subsequent sumsration) of the probabilities characterizing all these events, and is correct, it is impractical to act in this way.

Note that to obtain a final resistance of less than 4 kΩ dit is sufficient that at least one resistor with resistanceless than 15 kOhm. Thus, only in the caseAthe requirement of the task is not met, i.e. eventAis anopposite investigated. However,

.

Thus, .

NS ri tagging : Calculating the probability of some eventA, do not forget to analyze the complexity of the definitionI the likelihood of an event opposite to it. If rassread
easy, then it is with this that you need to start solvedits tasks, completing it by applying the relation (2 .0).

NS Example 4 : The box containsnwhite,mblack andkred balls. The balls are taken out of the box one at a timeand returned back after each extraction. Determine the probabilitydevelopmentsA= “White ballwill be extracted earlier than black” .

Resh enie. Consider the following set of events

= “The white ball was removed at the first attempt”;

= “First took out the red ball and then the white one”;

= “Twice removed the red ball, and the third time - the white”…

So toAs the balls are returned, the sequence of sobout can be formally infinitely long.

These events are incompatible and together make up the set of situations in which the event occurs.A... Thus,

It is easy to see that the terms included in the sum formgeometric progression with initial element
and the denominator
... But the sumsand the elements of an infinite geometric progression is

Thus, . LIt is curious that this probability (as follows from the obtainedth expression) does not depend on the number of red balls in the box.

In economics, 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 a product depends on demand, which can vary significantly, and on a number of other factors that are practically unrealistic to take into account. Therefore, when organizing production and making sales, you have to predict the outcome of such activities based on either your 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 in question, 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 experience, it may or may not happen.

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

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

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

Algebra of events

Events are called inconsistent 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.

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

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

By product 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 a 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 receiving ships. Three events can be considered: - the absence of ships at the berths, - the presence of one vessel at one of the berths, - the presence of two ships at two berths. These three events form a complete group of events.

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

If one of the opposite events 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 rolled. There can be six elementary outcomes in total according to the number of points on the edges.

A more complex event can be composed of elementary outcomes. 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 in question is the probability.

The most widespread are two definitions of the probability of an event: classic and statistical.

The classical definition of probability is associated with the concept of a favorable outcome.

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

In the given example, the event under consideration - an even number of points on the rolled edge, has three favorable outcomes. In this case, the general
the number of possible outcomes. This means that the classical definition of the probability of an event can be used here.

Classic definition is equal to the ratio of the number of favorable outcomes to the total number of possible outcomes

where is the probability of an event, is the number of outcomes favorable to the event, is the total number of 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 occurrences 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 with a sufficiently large number of tests.

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

To determine the probability of an event on the basis of formula (1.1), combinatorial formulas are often used, according to which the number of favorable outcomes and the total number of possible outcomes are found.

THEME 1 ... The classic formula for calculating the probability.

Basic definitions and formulas:

An experiment whose outcome cannot be predicted is called random experiment(SE).

An event that in a given SE may or may not happen is called random event.

Elementary outcomes call events that meet the requirements:

1. for any implementation of SE, one and only one elementary outcome occurs;

2. every event is a certain combination, a certain set of elementary outcomes.

The set of all possible elementary outcomes fully describes the SE. Such a set is usually called space of elementary outcomes(PEI). The choice of the SEI for the description of this SE is ambiguous and depends on the problem being solved.

P (A) = n (A) / n,

where n is the total number of equally possible outcomes,

n (A) is the number of outcomes that make up event A, as they also say, favorable to event A.

The words “at random”, “at random”, “randomly” just guarantee the equal possibility of elementary outcomes.

Solution of typical examples

Example 1. From an urn containing 5 red, 3 black and 2 white balls, 3 balls are taken at random. Find the probabilities of events:

A- “all extracted balls are red”;

V- “all extracted balls are of the same color”;

WITH- “among those extracted there are exactly 2 black ones”.

Solution:

The elementary outcome of this FE is a triplet (disordered!) Balls. Therefore, the total number of outcomes is the number of combinations: n == 120 (10 = 5 + 3 + 2).

Event A consists only of those triplets that were extracted from five red balls, i.e. n (A) == 10.

Event V in addition to 10 red triplets, black triplets are also favored, the number of which is = 1. Therefore: n (B) = 10 + 1 = 11.

Event WITH those triplets of balls that contain 2 black and one non-black are favored. Each method of choosing two black balls can be combined with the choice of one non-black (out of seven). Therefore: n (C) = = 3 * 7 = 21.

So: P (A) = 10/120; P (B) = 11/120; P (C) = 21/120.

Example 2. Under the conditions of the previous problem, we will assume that the balls of each color have their own numbering, starting from 1. Find the probabilities of events:

D- “the maximum number extracted is 4”;

E- “the maximum number extracted is 3”.

Solution:

To calculate n (D), we can assume that the urn contains one ball with number 4, one ball with a higher number, and 8 balls (3k + 3h + 2b) with lower numbers. Event D those triplets of balls that necessarily contain a ball with number 4 and 2 balls with lower numbers are favored. Therefore: n (D) =

P (D) = 28/120.

To calculate n (E), we count: there are two balls with number 3 in the urn, two with large numbers and six balls with lower numbers (2k + 2h + 2b). Event E consists of triplets of two types:

1. one ball with number 3 and two with lower numbers;

2.two balls with number 3 and one with a lower number.

Therefore: n (E) =

P (E) = 36/120.

Example 3. Each of the M different particles is thrown at random into one of the N cells. Find the probabilities of events:

A- all particles hit the second cell;

V- all particles hit one cell;

WITH- each cell contains no more than one particle (M £ N);

D- all cells are busy (M = N +1);

E- the second cell contains exactly To particles.

Solution:

For each particle, there are N ways to get into this or that cell. According to the basic principle of combinatorics for M particles we have N * N * N *… * N (M-times). So, the total number of outcomes in this SE n = N M.

For each particle we have one opportunity to get into the second cell, therefore n (A) = 1 * 1 *… * 1 = 1 M = 1, and P (A) = 1 / N M.

Getting into one cell (all particles) means getting all into the first, or all into the second, or etc. everyone in the Nth. But each of these N options can be implemented in one way. Therefore, n (B) = 1 + 1 +… + 1 (N-times) = N and P (B) = N / N M.

Event C means that each particle has one less placement method than the previous particle, and the first one can fall into any of N cells. That's why:

n (C) = N * (N -1) * ... * (N + M -1) and P (C) =

In the particular case when M = N: P (C) =

Event D means that one of the cells contains two particles, and each of the (N -1) remaining cells contains one particle. To find n (D), we argue as follows: choose a cell in which there will be two particles, this can be done = N ways; then we select two particles for this cell, there are ways for this. After that we will distribute the remaining (N -1) particles one by one into the remaining (N -1) cells, for this we have (N -1)! ways.

So n (D) =

.

The number n (E) can be calculated as follows: To particles for the second cell can be ways, the remaining (M - K) particles are distributed in an arbitrary way (N -1) cell (N -1) M-K ways. That's why:

"Accidents are not accidental" ... It sounds like a philosopher said, but in fact it is the lot of the great science of mathematics to study randomness. In mathematics, chance theory deals with randomness. Formulas and examples of tasks, as well as the main definitions of this science will be presented in the article.

What is probability theory?

Probability theory is one of the mathematical disciplines that studies random events.

To make it a little clearer, let's give a small example: if you flip a coin up, it can fall "heads" or "tails". As long as the coin is in the air, both of these possibilities are possible. That is, the likelihood of possible consequences is 1: 1. If you pull one out of a deck with 36 cards, then the probability will be denoted as 1:36. It would seem that there is nothing to investigate and predict, especially with the help of mathematical formulas. Nevertheless, if you repeat a certain action many times, then you can identify a certain pattern and, on its basis, predict the outcome of events in other conditions.

To summarize all of the above, the theory of probability in the classical sense studies the possibility of the occurrence of one of the possible events in a numerical value.

From the pages of history

The theory of probability, formulas and examples of the first tasks appeared in the distant Middle Ages, when attempts were first made to predict the outcome of card games.

Initially, the theory of probability had nothing to do with mathematics. It was based on empirical facts or properties of an event that could be reproduced in practice. The first works in this area as a mathematical discipline appeared in the 17th century. The founders were Blaise Pascal and Pierre Fermat. For a long time they studied gambling and saw certain patterns, which they decided to tell the public about.

The same technique was invented by Christian Huygens, although he was not familiar with the results of the research of Pascal and Fermat. The concept of "probability theory", formulas and examples, which are considered the first in the history of the discipline, were introduced by him.

The works of Jacob Bernoulli, Laplace's and Poisson's theorems are also important. They made probability theory more like a mathematical discipline. The theory of probability, formulas and examples of basic tasks received their present form thanks to Kolmogorov's axioms. As a result of all the changes, the theory of probability has become one of the mathematical branches.

Basic concepts of probability theory. Developments

The main concept of this discipline is "event". There are three types of events:

• Credible. Those that will happen anyway (the coin will fall).
• Impossible. Events that will not happen under any scenario (the coin will remain hanging in the air).
• Random. Those that will or will not happen. They can be influenced by various factors, which are very difficult to predict. If we talk about a coin, then random factors that can affect the result: the physical characteristics of the coin, its shape, initial position, power of the throw, etc.

All events in the examples are designated in capital Latin letters, with the exception of P, which has a different role. For example:

• A = "students came to the lecture."
• Ā = "students did not come to the lecture."

In practical exercises, it is customary to write down events in words.

One of the most important characteristics of events is their equality of opportunity. That is, if you flip a coin, all variants of the initial fall are possible until it falls. But also events are not equally possible. This happens when someone specifically influences the outcome. For example, "marked" playing cards or dice in which the center of gravity is shifted.

Also, events are compatible and incompatible. Compatible events do not exclude each other from occurring. For example:

• A = "a student came to the lecture."
• B = "student came to the lecture."

These events are independent of each other, and the appearance of one of them does not affect the appearance of the other. Incompatible events are determined by the fact that the appearance of one excludes the appearance of the other. If we talk about the same coin, then the “tails” falling out makes it impossible for the “heads” to appear in the same experiment.

Actions on events

Events can be multiplied and added, respectively, logical connectives "AND" and "OR" are introduced in the discipline.

The amount is determined by the fact that either event A, or B, or two can occur at the same time. In the case when they are incompatible, the last option is impossible, either A or B.

The multiplication of events consists in the appearance of A and B at the same time.

Now you can give a few examples to better remember the basics, probability theory and formulas. Examples of problem solving further.

Exercise 1: The firm is participating in a competition for contracts for three types of work. Possible events that can occur:

• A = "the firm will receive the first contract."
• A 1 = "the firm will not receive the first contract."
• B = "the firm will receive a second contract."
• B 1 = "the firm will not receive a second contract"
• C = "the firm will receive a third contract."
• C 1 = "the firm will not receive a third contract."

Let's try to express the following situations using actions on events:

• K = "the firm will receive all contracts."

In mathematical form, the equation will look like this: K = ABC.

• M = "the firm will not receive a single contract."

M = A 1 B 1 C 1.

Complicating the task: H = "the firm will receive one contract." Since it is not known which contract the firm will receive (first, second or third), it is necessary to record the entire series of possible events:

Н = А 1 ВС 1 υ AB 1 С 1 υ А 1 В 1 С.

A 1 BC 1 is a series of events where the firm does not receive the first and third contracts, but receives the second. Other possible events were recorded by the corresponding method. The symbol υ in the discipline denotes the "OR" link. If we translate the given example into human language, then the company will receive either a third contract, or a second, or first. Similarly, you can write down other conditions in the discipline "Theory of Probability". The formulas and examples of problem solving presented above will help you do it yourself.

Actually, the probability

Perhaps, in this mathematical discipline, the probability of an event is the central concept. There are 3 definitions of probability:

• classic;
• statistical;
• geometric.

Each has its place in the study of probabilities. Probability theory, formulas and examples (grade 9) mainly use the classical definition, which sounds like this:

• The probability of situation A is equal to the ratio of the number of outcomes that favor its occurrence to the number of all possible outcomes.

The formula looks like this: P (A) = m / n.

A is actually an event. If there is a case opposite to A, it can be written as Ā or A 1.

m is the number of possible favorable cases.

n - all events that can happen.

For example, A = "draw a card of the heart suit." There are 36 cards in a standard deck, 9 of them are hearts. Accordingly, the formula for solving the problem will look like:

P (A) = 9/36 = 0.25.

As a result, the probability that a heart-suit card is drawn from the deck is 0.25.

Towards higher mathematics

Now it has become a little known what the theory of probability is, formulas and examples of solving tasks that come across in the school curriculum. However, probability theory is also found in higher mathematics, which is taught in universities. Most often, they operate with geometric and statistical definitions of the theory and complex formulas.

The theory of probability is very interesting. It is better to start learning formulas and examples (higher mathematics) small - with a statistical (or frequency) definition of probability.

The statistical approach does not contradict the classical one, but slightly expands it. If in the first case it was necessary to determine with what degree of probability an event will occur, then in this method it is necessary to indicate how often it will occur. Here we introduce a new concept "relative frequency", which can be denoted by W n (A). The formula is no different from the classic one:

If the classical formula is calculated for forecasting, then the statistical one - according to the results of the experiment. Take a small assignment, for example.

The technological control department checks the products for quality. Among 100 products, 3 were found to be of poor quality. How do you find the probability of the frequency of a quality product?

A = "the appearance of a quality product."

W n (A) = 97/100 = 0.97

Thus, the frequency of a quality product is 0.97. Where did you get 97 from? Of the 100 items that were checked, 3 were found to be of poor quality. We subtract 3 from 100, we get 97, this is the amount of quality goods.

A little about combinatorics

Another method of probability theory is called combinatorics. Its basic principle is that if a certain choice of A can be made in m different ways, and the choice of B in n different ways, then the choice of A and B can be made by multiplication.

For example, there are 5 roads leading from city A to city B. There are 4 ways from city B to city C. How many ways can you get from city A to city C?

It's simple: 5x4 = 20, that is, you can get from point A to point C in twenty different ways.

Let's complicate the task. How many ways are there to play cards in solitaire? There are 36 cards in the deck - this is the starting point. To find out the number of ways, you need to "subtract" one card from the starting point and multiply.

That is, 36x35x34x33x32 ... x2x1 = the result does not fit on the calculator screen, so you can simply designate it as 36 !. Sign "!" next to a number indicates that the entire series of numbers is multiplied among themselves.

In combinatorics, there are concepts such as permutation, placement, and combination. Each of them has its own formula.

An ordered set of elements in a set is called an arrangement. Placements can be repetitive, that is, one element can be used multiple times. And no repetition, when the elements are not repeated. n are all elements, m are elements that participate in the placement. The formula for placement without repetitions would be:

A n m = n! / (N-m)!

Connections of n elements that differ only in the order of placement are called permutations. In mathematics, this is: P n = n!

Combinations of n elements by m are such compounds in which it is important what elements they were and what their total number was. The formula will look like:

A n m = n! / M! (N-m)!

Bernoulli's formula

The theory of probability, as well as in every discipline, has the works of outstanding researchers in their field who have taken it to a new level. One of these works is the Bernoulli formula, which allows you to determine the probability of a certain event occurring under independent conditions. This suggests that the appearance of A in an experiment does not depend on the appearance or non-appearance of the same event in earlier or subsequent tests.

Bernoulli's equation:

P n (m) = C n m × p m × q n-m.

The probability (p) of the occurrence of event (A) is unchanged for each trial. The probability that the situation will occur exactly m times in n number of experiments will be calculated by the formula presented above. Accordingly, the question arises of how to find out the number q.

If event A occurs p number of times, respectively, it may not occur. One is a number that is used to designate all outcomes of a situation in a discipline. Therefore, q is a number that denotes the possibility of the event not occurring.

Now you know Bernoulli's formula (probability theory). We will consider examples of solving problems (the first level) further.

Assignment 2: The store visitor will make a purchase with a probability of 0.2. 6 visitors entered the store independently. What is the likelihood that a visitor will make a purchase?

Solution: Since it is not known how many visitors should make a purchase, one or all six, it is necessary to calculate all possible probabilities using Bernoulli's formula.

A = "the visitor makes a purchase."

In this case: p = 0.2 (as indicated in the task). Accordingly, q = 1-0.2 = 0.8.

n = 6 (since there are 6 customers in the store). The number m will change from 0 (no customer will make a purchase) to 6 (all visitors to the store will purchase something). As a result, we get the solution:

P 6 (0) = C 0 6 × p 0 × q 6 = q 6 = (0.8) 6 = 0.2621.

None of the buyers will make a purchase with a probability of 0.2621.

How else is Bernoulli's formula (probability theory) used? Examples of problem solving (second level) below.

After the above example, questions arise about where C and p have gone. With respect to p, the number to the power of 0 will be equal to one. As for C, it can be found by the formula:

C n m = n! / m! (n-m)!

Since in the first example m = 0, respectively, C = 1, which, in principle, does not affect the result. Using the new formula, let's try to find out what is the probability of two visitors buying goods.

P 6 (2) = C 6 2 × p 2 × q 4 = (6 × 5 × 4 × 3 × 2 × 1) / (2 × 1 × 4 × 3 × 2 × 1) × (0.2) 2 × (0.8) 4 = 15 × 0.04 × 0.4096 = 0.246.

The theory of probability is not that complicated. Bernoulli's formula, examples of which are presented above, is a direct proof of this.

Poisson's formula

Poisson's equation is used to calculate unlikely random situations.

Basic formula:

P n (m) = λ m / m! × e (-λ).

Moreover, λ = n x p. Here is such a simple Poisson formula (probability theory). We will consider examples of solving problems further.

Assignment 3: The factory produced parts in the amount of 100,000 pieces. Defective part appearance = 0.0001. What is the probability that there will be 5 defective parts in a batch?

As you can see, marriage is an unlikely event, and therefore Poisson's formula (probability theory) is used for the calculation. Examples of solving problems of this kind are no different from other tasks of the discipline, we substitute the necessary data in the given formula:

A = "a randomly selected part will be defective."

p = 0.0001 (according to the condition of the task).

n = 100000 (number of parts).

m = 5 (defective parts). We substitute the data into the formula and get:

P 100000 (5) = 10 5/5! X e -10 = 0.0375.

Just like Bernoulli's formula (probability theory), examples of solutions with which are written above, Poisson's equation has an unknown e. In fact, it can be found by the formula:

е -λ = lim n -> ∞ (1-λ / n) n.

However, there are special tables that contain almost all the values ​​of e.

Moivre-Laplace theorem

If the number of tests in the Bernoulli scheme is large enough, and the probability of occurrence of event A in all schemes is the same, then the probability of occurrence of event A a certain number of times in a series of tests can be found by the Laplace formula:

Р n (m) = 1 / √npq x ϕ (X m).

X m = m-np / √npq.

To better remember the Laplace formula (probability theory), examples of problems to help you below.

First, we find X m, substitute the data (they are all indicated above) into the formula and get 0.025. Using the tables, we find the number ϕ (0.025), the value of which is 0.3988. Now you can substitute all the data in the formula:

R 800 (267) = 1 / √ (800 x 1/3 x 2/3) x 0.3988 = 3/40 x 0.3988 = 0.03.

So the probability that the flyer will fire exactly 267 times is 0.03.

Bayes formula

Bayes' formula (probability theory), examples of solving problems with the help of which will be given below, is an equation that describes the probability of an event, based on the circumstances that could be associated with it. The basic formula looks like this:

P (A | B) = P (B | A) x P (A) / P (B).

A and B are specific events.

P (A | B) - conditional probability, that is, event A can occur provided that event B is true.

P (B | A) - conditional probability of event B.

So, the final part of the short course "Theory of Probability" is the Bayes formula, examples of solutions to problems with which are below.

Assignment 5: Phones from three companies were brought to the warehouse. At the same time, part of the phones that are manufactured at the first plant is 25%, at the second - 60%, at the third - 15%. It is also known that the average percentage of defective products in the first factory is 2%, in the second - 4%, and in the third - 1%. It is necessary to find the probability that a randomly selected phone will turn out to be defective.

A = "randomly picked phone."

B 1 - the phone that was made by the first factory. Accordingly, there will be input B 2 and B 3 (for the second and third factories).

As a result, we get:

P (B 1) = 25% / 100% = 0.25; P (B 2) = 0.6; P (B 3) = 0.15 - thus we found the probability of each option.

Now you need to find the conditional probabilities of the desired event, that is, the probability of defective products in firms:

P (A / B 1) = 2% / 100% = 0.02;

P (A / B 2) = 0.04;

P (A / B 3) = 0.01.

Now we plug the data into the Bayes formula and get:

P (A) = 0.25 x 0.2 + 0.6 x 0.4 + 0.15 x 0.01 = 0.0305.

The article presents the theory of probability, formulas and examples of problem solving, but this is only the tip of the iceberg of a vast discipline. And after all that has been written, it will be logical to ask the question of whether the theory of probability is needed in life. It is difficult for an ordinary person to answer, it is better to ask about this from the one who has hit the jackpot more than once with its help.

There will also be tasks for an independent solution, to which you can see the answers.

The general formulation of the problem: the probabilities of some events are known, but the probabilities of other events that are associated with these events need to be calculated. In these tasks, there is a need for such actions on probabilities as addition and multiplication of probabilities.

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

Tasks of a different type. Several events are given, for example, the coin is tossed three times. It is required to find the probability that either the coat of arms will be dropped all three times, or that the coat of arms will be drawn at least once. This is a problem of multiplying probabilities.

Adding the probabilities of inconsistent events

Probability addition is used when you need to calculate the probability of a union 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 when at least one of the events occurs. It means that A + B- an event that occurs if and only if an event occurred during observation A or event B, or at the same time 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 test is calculated using the addition of the probabilities.

The addition theorem for probabilities. 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 from the first shot, event V- hit from the second shot, event ( A+ V) - hit from the first or second shot or from two shots. So if two events A and V- incompatible events, then A+ V- the onset of at least one of these events or two events.

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

Solution. Let's assume that the event A- "the red ball is taken", and the event V- "a blue ball is taken." Then the event is “a colored (not white) ball is taken”. Find the probability of an event A:

and events V:

Developments A and V- mutually incompatible, since if one ball is taken, then you cannot take balls of different colors. Therefore, we use the addition of probabilities:

The theorem of addition of probabilities for several inconsistent events. If events make up the complete set of events, then the sum of their probabilities is 1:

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

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

The probabilities of opposite events are usually denoted 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: Let's find the probability that the shooter will hit the target:

Let's find the probability that the shooter misses the target:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various problems on addition and multiplication of probabilities".

Addition of probabilities of mutually compatible 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 dice, the event A the fall of the number 4 is considered, and the event V- an even number dropped out. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of one of the mutually joint 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 is as follows:

Since events A and V compatible, event A+ V 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 inconsistent 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:

Similarly:

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

When using formula (8), it should be borne in mind that events A and V may be:

• mutually independent;
• mutually dependent.

Probability formula for mutually independent events:

Probability formula for mutually dependent events:

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

Example 3. In a car race, when driving in the first car, there is a chance of winning, when driving in a second car. Find:

• the likelihood 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, therefore the events A(the first car wins) and V(second car wins) - independent events. Let's find the probability that both cars will win:

2) Let's find the probability that one of the two cars will win:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various problems on addition and multiplication of probabilities".

Solve the probability addition problem yourself, and then see the solution

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

Multiplication of probabilities

Probability multiplication is used when calculating the probability of the logical product of events.

Moreover, random events must be independent. Two events are called mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.

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

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

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

Solve probability multiplication problems yourself, and then see the solution

Example 6. There is a box with nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, played and unplayed are not distinguished. What is the probability that after three games there will be no balls left in the box?

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

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

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

More difficult tasks in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events - on the page "Various problems on 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.