How to determine the probability of a number appearing. Probability theory

Linear programming uses a graphical method to determine convex sets (solution polyhedron). If the main task linear programming has an optimal plan, then the objective function takes a value at one of the vertices of the decision polyhedron (see figure).

Purpose of the service. Using this service you can online mode solve the linear programming problem using the geometric method, and also obtain a solution to the dual problem (assess the optimal use of resources). Additionally, a solution template is created in Excel.

Instructions. Select the number of rows (number of restrictions).

The following are also used with this calculator:
Simplex method for solving ZLP

Solution of the transport problem
Solving a matrix game
Using the online service you can determine the price matrix game(lower and upper limits), check availability saddle point, find a solution mixed strategy methods: minimax, simplex method, graphical (geometric) method, Brown's method.
Extremum of a function of two variables
Calculation of limits

Solving a linear programming problem by graphical method includes the following steps:

1. Lines are constructed on the plane X 1 0X 2.
2. Half planes are determined.
3. Define a solution polygon;
4. A vector N(c 1 ,c 2) is constructed, which indicates the direction of the objective function;
5. Move forward objective function c 1 x 2 + c 2 x 2= 0 in the direction of vector N to extreme point solution polygon.
6. The coordinates of the point and the value of the objective function at this point are calculated.

Example. The company produces two types of products - P1 and P2. For the production of products, two types of raw materials are used - C1 and C2. Wholesale prices per unit of production are equal to: 5 units. for P1 and 4 units for P2. The consumption of raw materials per unit of product of type P1 and type P2 is given in the table.
Table - Consumption of raw materials for production

1. Formulate a mathematical model of a linear programming problem.
2. Solve a linear programming problem graphically(for two variables).

Resource Limits
6x 1 + 4x 2 ≤ 24
x 1 + 2x 2 ≤ 6

Demand restrictions
x 1 +1 ≥ x 2
x 2 ≤ 2

Objective function
5x 1 + 4x 2 → max

Then we get the following PLP:
6x 1 + 4x 2 ≤ 24
x 1 + 2x 2 ≤ 6
x 2 - x 1 ≤ 1
x 2 ≤ 2
x 1 , x 2 ≥ 0
5x 1 + 4x 2 → max

TOPIC 1 . Classic formula probability calculations.

Basic definitions and formulas:

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

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

Elementary outcomes events that meet the requirements are called:

1.with 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 completely describes the SE. Such a set is usually called space of elementary outcomes(PEI). The choice of PEI to describe a given SE is ambiguous and depends on the problem being solved.

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

where n – total number equally possible outcomes,

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

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

Solving typical examples

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

A– “all the drawn balls are red”;

IN– “all drawn balls are of the same color”;

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

Solution:

The elementary outcome of this SE is a triple (disordered!) of 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 drawn from five red balls, i.e. n(A)==10.

Event IN In addition to 10 red threes, black threes are also favorable, the number of which is = 1. Therefore: n (B)=10+1=11.

Event WITH Those threes of balls that contain 2 black and one non-black are favored. Each method of selecting two black balls can be combined with selecting one non-black ball (out of seven). Therefore: n (C) = = 3 * 7 = 21.

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

Example 2. In 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 extracted number is 4”;

E– “The maximum number extracted is 3.”

Solution:

To calculate n(D), we can assume that the urn has one ball with number 4, one ball with a higher number and 8 balls (3k+3h+2b) with lower numbers. Event D Those threes 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 consider: there are two balls in the urn with number 3, 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 M different particles is thrown at random into one of N cells. Find the probabilities of events:

A– all particles fell into the second cell;

IN– all particles fell into one cell;

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

D– all cells are occupied (M =N +1);

E– the second cell contains exactly To particles.

Solution:

For each particle there are N ways to get into a particular 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 (for all particles) means getting everyone into the first, or everyone into the second, or etc. everyone in Nth. But each of these N options can be implemented in one way. Therefore n (B)=1+1+…+1(N -times)=N and Р(В)=N/N M.

Event C means that each particle has one less number of placement options 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 Р(С) =

In the particular case with M =N: Р(С)=

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 reason like this: choose a cell in which there will be two particles, this can be done in =N ways; then we will select two particles for this cell, there are ways to do this. After this, we distribute the remaining (N -1) particles one at a time into the remaining (N -1) cells, for this there are (N -1)! ways.

So n(D) =

.

The number n(E) can be calculated as follows: To particles for the second cell can be done in ways; the remaining (M – K) particles are distributed randomly over the (N -1) cell (N -1) in M-K ways. That's why:

Is it possible to win the lottery? What are the chances of matching the required number of numbers and winning the jackpot or junior category prize? The probability of winning is easy to calculate; anyone can do it themselves.

How is the probability of winning the lottery generally calculated?

Numerical lotteries are conducted according to certain formulas and the chances of each event (winning a particular category) are calculated mathematically. Moreover, this probability is calculated for any desired value, be it “5 out of 36”, “6 out of 45”, or “7 out of 49” and it does not change, since it depends only on the total number of numbers (balls, numbers) and the fact how many of them need to be guessed.

For example, for the “5 out of 36” lottery the probabilities are always as follows

• guess two numbers - 1:8
• guess three numbers - 1:81
• guess four numbers - 1: 2,432
• guess five numbers - 1: 376,992

In other words, if you mark one combination (5 numbers) on a ticket, then the chance of guessing “two” is only 1 in 8. But catching “five” numbers is much more difficult, this is already 1 chance in 376,992. This is exactly the number (376 thousand) There are all kinds of combinations in the “5 out of 36” lottery and you are guaranteed to win it if you only fill them all. True, the amount of winnings in this case will not justify the investment: if a ticket costs 80 rubles, then marking all the combinations will cost 30,159,360 rubles. The jackpot is usually much smaller.

In general, all probabilities have long been known, all that remains is to find them or calculate them yourself, using the appropriate formulas.

For those who are too lazy to look, we present the winning probabilities for the main numerical lotteries Stoloto - they are presented in this table

Necessary clarifications

The lotto widget allows you to calculate the probabilities of winning for lotteries with one lottery machine (without bonus balls) or with two lottery machines. You can also calculate the probabilities of deployed bets

Probability calculation for lotteries with one lottery machine (without bonus balls)

Only the first two fields are used, in which the numerical formula of the lottery is used, for example: - “5 out of 36”, “6 out of 45”, “7 out of 49”. In principle, you can calculate almost any world lottery. There are only two restrictions: the first value should not exceed 30, and the second - 99.

If the lottery does not use additional numbers*, then after selecting a numerical formula, all you have to do is click the calculate button and the result is ready. It doesn’t matter what probability of an event you want to know - winning a jackpot, a second/third category prize, or just finding out whether it’s difficult to guess 2-3 numbers out of the required number - the result is calculated almost instantly!

Calculation example. The chance of guessing 5 out of 36 is 1 in 376,992

Examples. Probabilities of winning the main prize for lotteries:
“5 out of 36” (Gosloto, Russia) – 1:376 922
“6 out of 45” (Gosloto, Russia; Saturday Lotto, Australia; Lotto, Austria) - 1:8 145 060
“6 out of 49” (Sportloto, Russia; La Primitiva, Spain; Lotto 6/49, Canada) - 1:13 983 816
“6 out of 52” (Super Loto, Ukraine; Illinois Lotto, USA; Mega TOTO, Malaysia) - 1:20 358 520
“7 out of 49” (Gosloto, Russia; Lotto Max, Canada) - 1:85 900 584

Lotteries with two lottery machines (+ bonus ball)

If the lottery uses two lottery machines, then all 4 fields must be filled in for calculation. In the first two - the numerical formula of the lottery (5 out of 36, 6 out of 45, etc.), in the third and fourth fields the number of bonus balls is indicated (x out of n). Important: this calculation can only be used for lotteries with two lottery machines. If the bonus ball is taken from the main lottery machine, then the probability of winning in this particular category is calculated differently.

* Since when using two lottery machines, the chance of winning is calculated by multiplying the probabilities by each other, then for the correct calculation of lotteries with one lottery machine, the choice of an additional number by default is 1 out of 1, that is, it is not taken into account.

Examples. Probabilities of winning the main prize for lotteries:
“5 out of 36 + 1 out of 4” (Gosloto, Russia) – 1:1 507 978
“4 out of 20 + 4 out of 20” (Gosloto, Russia) – 1:23 474 025
“6 out of 42 + 1 out of 10” (Megalot, Ukraine) – 1:52 457 860
“5 out of 50 + 2 out of 10” (EuroJackpot) – 1:95 344 200
“5 out of 69 + 1 out of 26” (Powerball, USA) - 1: 292,201,338

Example calculation. The chance of guessing 4 out of 20 twice (in two fields) is 1 in 23,474,025

A good illustration of the complexity of playing with two lottery machines is the Gosloto 4 out of 20 lottery. The probability of guessing 4 numbers out of 20 in one field is quite fair, the chance of this is 1 in 4,845. But when you need to guess correctly and win both fields... then the probability is calculated by multiplying them. That is, in in this case 4,845 multiplied by 4,845, which gives 23,474,025. So, the simplicity of this lottery is deceptive, win it Grand Prize more difficult than "6 out of 45" or "6 out of 49"

Probability calculation (expanded bets)

In this case, the probability of winning when using expanded bets is calculated. For example, if there are 6 out of 45 in the lottery, mark 8 numbers, then the probability of winning the main prize (6 out of 45) will be 1 chance in 290,895. Whether to use expanded bets is up to you. Taking into account the fact that their cost is very high (in this case, 8 marked numbers are 28 options), it is worth knowing how this increases the chances of winning. Moreover, it is now very easy to do this!

Calculation of the probability of winning (6 out of 45) using the example of an expanded bet (8 numbers are marked)

And other possibilities

Using our widget, you can calculate the probability of winning in bingo lotteries, for example, in “ Russian lotto" The main thing that needs to be taken into account is the number of moves allocated for the onset of winning. To make it clearer: for a long time In the Russian Lotto lottery, the jackpot could be won if 15 numbers ( in one field) closed in 15 moves. The probability of such an event is absolutely fantastic, 1 chance in 45,795,673,964,460,800 (you can check and get this value yourself). This is why, by the way, for many years in the Russian Lotto lottery no one could hit the jackpot, and it was distributed forcibly.

On March 20, 2016, the rules of the Russian Lotto lottery were changed. The jackpot can now be won if 15 numbers (out of 30) were closed in 15 moves. It turns out to be an analogue of an expanded bet - after all, 15 numbers are guessed out of 30 available! And this is a completely different possibility:

Chance to win the jackpot (according to new rules) in the Russian Lotto lottery

And in conclusion, we present the probability of winning in lotteries using a bonus ball from the main lottery drum (our widget does not count such values). Of the most famous

Sportsloto “6 out of 49”(Gosloto, Russia), La Primitiva “6 out of 49” (Spain)
Category "5 + bonus ball": probability 1:2 330 636

SuperEnalotto "6 out of 90"(Italy)
Category "5 + bonus ball": probability 1:103,769,105

Oz Lotto "7 out of 45"(Australia)
Category "6 + bonus ball": probability 1:3 241 401
“5 + 1” – probability 1:29,602
“3 +1” – probability 1:87

Lotto "6 out of 59"(Great Britain)
Category "5 + 1 bonus ball": probability 1:7 509 579

A professional bettor must have a good understanding of the odds, quickly and correctly estimate the probability of an event by coefficient and, if necessary, be able to convert odds from one format to another. In this manual we will talk about what types of coefficients there are, and also use examples to show how you can calculate the probability using a known coefficient and vice versa.

What types of odds are there?

There are three main types of odds that bookmakers offer players: decimal odds, fractional odds(English) and American odds. The most common odds in Europe are decimal. IN North America American odds are popular. Fractional odds are the most traditional type; they immediately reflect information about how much you need to bet to get a certain amount.

Decimal odds

Decimal or they are also called European odds is a familiar number format, represented as a decimal fraction with an accuracy of hundredths, and sometimes even thousandths. An example of a decimal odd is 1.91. Calculating profit in the case of decimal odds is very simple; you just need to multiply the amount of your bet by this odds. For example, in the match “Manchester United” - “Arsenal”, the victory of “Manchester United” is set with a coefficient of 2.05, a draw is estimated with a coefficient of 3.9, and a victory of “Arsenal” is equal to 2.95. Let's say we're confident United will win and we bet $1,000 on them. Then our possible income is calculated as follows:

2.05 * $1000 = $2050;

It’s really not that complicated, is it?! The possible income is calculated in the same way when betting on a draw or victory for Arsenal.

Draw: 3.9 * $1000 = $3900;
Arsenal win: 2.95 * $1000 = $2950;

How to calculate the probability of an event using decimal odds?

Now imagine that we need to determine the probability of an event based on the decimal odds set by the bookmaker. This is also done very simply. To do this, we divide one by this coefficient.

Let's take the existing data and calculate the probability of each event:

Manchester United win: 1 / 2.05 = 0,487 = 48,7%;
Draw: 1 / 3.9 = 0,256 = 25,6%;
Arsenal win: 1 / 2.95 = 0,338 = 33,8%;

Fractional odds (English)

As the name suggests fractional coefficient represented by an ordinary fraction. An example of English odds is 5/2. The numerator of the fraction contains a number that is the potential amount of the net winnings, and the denominator contains a number indicating the amount that must be bet in order to receive this winning. Simply put, we have to bet $2 dollars to win $5. Odds of 3/2 means that in order to get $3 in net winnings, we will have to bet $2.

How to calculate the probability of an event using fractional odds?

It is also not difficult to calculate the probability of an event using fractional odds; you just need to divide the denominator by the sum of the numerator and denominator.

For the fraction 5/2 we calculate the probability: 2 / (5+2) = 2 / 7 = 0,28 = 28%;
For the fraction 3/2 we calculate the probability:

American odds

American odds unpopular in Europe, but very much so in North America. Perhaps, this type coefficients is the most complex, but this is only at first glance. In fact, there is nothing complicated in this type of coefficients. Now let's figure it all out in order.

The main feature of American odds is that they can be either positive, so negative. Example of American odds - (+150), (-120). The American odds (+150) means that in order to earn $150 we need to bet $100. In other words, a positive American coefficient reflects the potential net earnings at a bet of $100. A negative American odds reflect the amount of bet that needs to be made in order to get a net win of $100. For example, the coefficient (-120) tells us that by betting $120 we will win $100.

How to calculate the probability of an event using American odds?

The probability of an event using the American coefficient is calculated using the following formulas:

(-(M)) / ((-(M)) + 100), where M is a negative American coefficient;
100/(P+100), where P is a positive American coefficient;

For example, we have a coefficient (-120), then the probability is calculated as follows:

(-(M)) / ((-(M)) + 100); substitute the value (-120) for “M”;
(-(-120)) / ((-(-120)) + 100 = 120 / (120 + 100) = 120 / 220 = 0,545 = 54,5%;

Thus, the probability of an event with American odds (-120) is 54.5%.

For example, we have a coefficient (+150), then the probability is calculated as follows:

100/(P+100); substitute the value (+150) for “P”;
100 / (150 + 100) = 100 / 250 = 0,4 = 40%;

Thus, the probability of an event with American odds (+150) is 40%.

How, knowing the percentage of probability, convert it into a decimal coefficient?

In order to calculate the decimal coefficient based on a known percentage of probability, you need to divide 100 by the probability of the event as a percentage. For example, the probability of an event is 55%, then the decimal coefficient of this probability will be equal to 1.81.

100 / 55% = 1,81

How, knowing the percentage of probability, convert it into a fractional coefficient?

In order to calculate the fractional coefficient based on a known percentage of probability, you need to subtract one from dividing 100 by the probability of an event as a percentage. For example, if we have a probability percentage of 40%, then the fractional coefficient of this probability will be equal to 3/2.

(100 / 40%) - 1 = 2,5 - 1 = 1,5;
The fractional coefficient is 1.5/1 or 3/2.

How, knowing the percentage of probability, convert it into an American coefficient?

If the probability of an event is more than 50%, then the calculation is made using the formula:

- ((V) / (100 - V)) * 100, where V is probability;

For example, if the probability of an event is 80%, then the American coefficient of this probability will be equal to (-400).

- (80 / (100 - 80)) * 100 = - (80 / 20) * 100 = - 4 * 100 = (-400);

If the probability of an event is less than 50%, then the calculation is made using the formula:

((100 - V) / V) * 100, where V is probability;

For example, if we have a percentage probability of an event of 20%, then the American coefficient of this probability will be equal to (+400).

((100 - 20) / 20) * 100 = (80 / 20) * 100 = 4 * 100 = 400;

How to convert the coefficient to another format?

There are times when it is necessary to convert odds from one format to another. For example, we have a fractional odds of 3/2 and we need to convert it to decimal. To convert a fractional odds to a decimal odds, we first determine the probability of an event with a fractional odds, and then convert this probability into a decimal odds.

The probability of an event with a fractional odds of 3/2 is 40%.

2 / (3+2) = 2 / 5 = 0,4 = 40%;

Now let’s convert the probability of an event into a decimal coefficient; to do this, divide 100 by the probability of the event as a percentage:

100 / 40% = 2.5;

Thus, the fractional odds of 3/2 are equal to the decimal odds of 2.5. In a similar way, for example, American odds are converted to fractional, decimal to American, etc. The most difficult thing in all this is just the calculations.

“Accidents are not accidental”... It sounds like something a philosopher said, but in fact, studying accidents is the destiny great science mathematics. In mathematics, chance is dealt with by probability theory. Formulas and examples of tasks, as well as the basic 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 throw a coin up, it can land on heads or tails. While the coin is in the air, both of these probabilities are possible. That is, the probability of possible consequences is 1:1. If one is drawn from a deck of 36 cards, then the probability will be indicated as 1:36. It would seem that there is nothing to explore and predict here, especially with the help of mathematical formulas. However, if you repeat a certain action many times, you can identify a certain pattern and, based on it, predict the outcome of events in other conditions.

To summarize all of the above, probability theory 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 to predict the outcome of card games first arose.

Initially, probability theory had nothing to do with mathematics. It was justified by 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. Long time they studied gambling and saw certain patterns, which they decided to tell the public about.

The same technique was invented by Christiaan 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 of no small importance. They made probability theory more like a mathematical discipline. Probability theory, formulas and examples of basic tasks received their current form thanks to Kolmogorov’s axioms. As a result of all the changes, probability theory became one of the mathematical branches.

Basic concepts of probability theory. Events

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

• Reliable. Those that will happen anyway (the coin will fall).
• Impossible. Events that will not happen under any circumstances (the coin will remain hanging in the air).
• Random. The ones that will happen or won't happen. They can be influenced by various factors that are very difficult to predict. If we talk about a coin, then there are random factors that can affect the result: the physical characteristics of the coin, its shape, its original position, the force of the throw, etc.

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

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

IN practical tasks Events are usually recorded in words.

One of the most important characteristics events - their equal possibility. That is, if you toss a coin, all variants of the initial fall are possible until it falls. But events are also not equally possible. This happens when someone deliberately influences an outcome. For example, "labeled" playing cards or dice in which the center of gravity is shifted.

Events can also be compatible and incompatible. Compatible events do not exclude each other's occurrence. For example:

• A = “the student came to the lecture.”
• B = “the student came to the lecture.”

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

Actions on events

Events can be multiplied and added; accordingly, 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 simultaneously. If they are incompatible, the last option is impossible; either A or B will be rolled.

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

Now we can give several examples to better remember the basics, probability theory and formulas. Examples of problem solving below.

Exercise 1: The company takes part in a competition to receive contracts for three types of work. Possible events that may 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.”

Using actions on events, we will try to express the following situations:

• K = “the company will receive all contracts.”

In mathematical form, the equation will have the following form: K = ABC.

• M = “the company will not receive a single contract.”

M = A 1 B 1 C 1.

Let’s complicate the task: H = “the company will receive one contract.” Since it is not known which contract the company will receive (first, second or third), it is necessary to record the entire series of possible events:

H = A 1 BC 1 υ AB 1 C 1 υ A 1 B 1 C.

And 1 BC 1 is a series of events where the firm does not receive the first and third contract, but receives the second. Other possible events were recorded using the appropriate method. The symbol υ in the discipline denotes the connective “OR”. If we translate the above example into human language, the company will receive either the third contract, or the second, or the first. In a similar way, you can write down other conditions in the discipline “Probability Theory”. The formulas and examples of problem solving presented above will help you do this 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 probability. Probability theory, formulas and examples (grade 9) mainly use classic 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 a case opposite to A appears, 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 of hearts. Accordingly, the formula for solving the problem will look like:

P(A)=9/36=0.25.

As a result, the probability that a card of the heart suit will be drawn from the deck will be 0.25.

Towards higher mathematics

Now it has become a little known what probability theory is, formulas and examples of solving problems that come across in 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 studying formulas and examples (higher mathematics) small - with the 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 probability an event will occur, then in this method it is necessary to indicate how often it will occur. Here a new concept of “relative frequency” is introduced, which can be denoted by W n (A). The formula is no different from the classic one:

If the classical formula is calculated for prediction, then the statistical one is calculated according to the results of the experiment. Let's take a small task for example.

The technological control department checks products for quality. Among 100 products, 3 were found to be of poor quality. How to find the frequency probability 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? Out of 100 products that were checked, 3 were found to be of poor quality. We subtract 3 from 100 and 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 A can be made m different ways, and the choice of B is in n different ways, then the choice of A and B can be done by multiplication.

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

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

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

That is, 36x35x34x33x32...x2x1= the result does not fit on the calculator screen, so it can simply be designated 36!. Sign "!" next to the number indicates that the entire series of numbers is multiplied together.

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

An ordered set of elements of a set is called an arrangement. Placements can be repeated, that is, one element can be used several times. And without repetition, when elements are not repeated. n are all elements, m are elements that participate in the placement. The formula for placement without repetition will look like:

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

Connections of n elements that differ only in the order of placement are called permutations. In mathematics it looks like: P n = n!

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

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

Bernoulli's formula

In probability theory, as in every discipline, there are 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 occurrence of A in an experiment does not depend on the occurrence or non-occurrence of the same event in earlier or subsequent trials.

Bernoulli's equation:

P n (m) = C n m ×p m ×q n-m.

The probability (p) of the occurrence of event (A) is constant 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, accordingly, it may not occur. Unit 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 an event not occurring.

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

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

Solution: Since it is unknown 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 will make 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 vary from 0 (not a single 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 probability 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 r went. Relative to p, a 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 purchasing 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 direct proof of this.

Poisson's formula

Poisson's equation is used to calculate low probability random situations.

Basic formula:

P n (m)=λ m /m! × e (-λ) .

In this case λ = n x p. Here is a simple Poisson formula (probability theory). We will consider examples of problem solving below.

Task 3: The factory produced 100,000 parts. Occurrence of a defective part = 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 the Poisson formula (probability theory) is used for calculation. Examples of problem solving this kind are no different from other tasks in the discipline; we substitute the necessary data into the given formula:

A = “a randomly selected part will be defective.”

p = 0.0001 (according to the task conditions).

n = 100000 (number of parts).

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

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

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

e -λ = lim n ->∞ (1-λ/n) n .

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

De Moivre-Laplace theorem

If in the Bernoulli scheme the number of trials is sufficiently large, 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 Laplace’s formula:

Р n (m)= 1/√npq x ϕ(X m).

X m = m-np/√npq.

To better remember Laplace’s formula (probability theory), examples of problems are below to help.

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

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

Thus, the probability that the flyer will work exactly 267 times is 0.03.

Bayes formula

The 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 is as follows:

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

A and B are definite events.

P(A|B) is a 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 “Probability Theory” is the Bayes formula, examples of solutions to problems with which are below.

Task 5: Phones from three companies were brought to the warehouse. At the same time, the share of 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 at the first factory is 2%, at the second - 4%, and at the third - 1%. You need to find the probability that a randomly selected phone will be defective.

A = “randomly picked phone.”

B 1 - the phone that the first factory produced. Accordingly, introductory B 2 and B 3 will appear (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 companies:

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

P(A/B 2) = 0.04;

P (A/B 3) = 0.01.

Now let’s substitute 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 probability theory, formulas and examples of problem solving, but this is only the tip of the iceberg of a vast discipline. And after everything that has been written, it would be logical to ask the question of whether the theory of probability is needed in life. To the common man It’s difficult to answer, it’s better to ask someone who has used it to win the jackpot more than once.