Problems on the classical determination of probability. Types of events, direct calculation of the probability of occurrence of an event Probability of an event as a percentage

“Accidents are not accidental”... It sounds like something a philosopher said, but in fact, studying randomness is the destiny of the great science of 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 possible consequences ratio 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 classical understanding studies the possibility of one of the possible events occurring 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 capital 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 written down in words.

One of the most important characteristics of events is their equal possibility. That is, if you toss a coin, all options for 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, “marked” 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.”
• And 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 range 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 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 (9th grade) mainly use the 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 theories 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 to “subtract” one card at a time from the starting point 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 number is. The formula will look like:

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

Bernoulli's formula

In probability theory, as well as in every discipline, there are works of outstanding researchers in their field who brought it to 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 the Bernoulli 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 solving problems of 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 will 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.

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 studies the constancy of random events. Of course, this concept does not really reveal the whole essence, so it is necessary to consider it in more detail.

I would like to start with the creators of the theory. As mentioned above, there were two of them, and 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 games, such as roulette, craps, 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. They, based on the work done by great geniuses, secured this item 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 quite voluminous, but without it you won’t be able to figure out 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're talking about about what “happened, although it might not have happened.”

Random events (the theory of probability pays special attention to them) 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 they 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 occur only 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 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 of a huge number of integers, as well as various permutations of both the numbers themselves and their elements, various data, etc., leading to the appearance of a number of combinations. In addition to probability theory, this branch is important for statistics, computer science and cryptography.

So, now 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 there are thirty cards in a deck of 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 necessary ways collecting deck 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 at the end multiply everything by 28. The answer is 2.4757335 ⋅〖10〗^32

Example solution. Formula for placement number

In this problem, you need to find out how many ways there are to put fifteen volumes on one shelf, but 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, given that one shelf can only hold fifteen volumes.

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 designate getting the blue ball as event A. This experiment 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 work, but also in Everyday life. With its help, you can calculate any possibility of any event.

The text also touched upon significant dates in the history of the formation of the theory of probability as a science, and the names of 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!

The probability is very easy topic, if you concentrate on the meaning of the tasks, and not on the formulas. But how to solve probability problems. First, what is probability? This is the chance that some event will happen. If we say that the probability of some event is 50%, what does that mean? That it will either happen or not happen - one of two things. Thus, calculating the probability value is very simple - you need to take the number of options that suit us and divide by the number of all possible options. For example, the chance of getting heads when tossing a coin is ½. How do we get ½? In total, we have two possible options (heads and tails), of which one suits us (tails), so we get a probability of ½.

As we have already seen, the probability can be expressed both as a percentage and in ordinary numbers. Important: on the Unified State Exam you will need to write down your answer in numbers, not as percentages. It is accepted that the probability varies from 0 (will never happen) to 1 (absolutely will happen). It can also be said that always

Probability of suitable events + probability of inappropriate events = 1

Now we understand exactly how to calculate the probability of a single event, and even such tasks are available in the FIPI Bank, but it is clear that it does not end there. To make life more fun, in probability problems, at least two events usually occur, and you need to calculate the probability taking into account each of them.

We calculate the probability of each event separately, then put signs between the fractions:

1. If you need the first AND second event, then multiply.

2. If you need the first OR second event, then add it up.

Probability problems and solutions

Task 1. Among the natural numbers from 23 to 37, one number is randomly selected. Find the probability that it is not divisible by 5.

Solution:

Probability is the ratio of favorable options to their total number.

There are 15 numbers in total in this interval. Of these, only 3 is divisible by 5, which means 12 is not divisible.

Probability then:

Answer: 0.8.

Task 2. Two students from the class are randomly selected to be on duty in the cafeteria. What is the probability that two boys will be on duty if there are 7 boys and 8 girls in the class?

Solution: Probability is the ratio of favorable options to their total number. There are 7 boys in the class, these are favorable options. And there are only 15 students.

Probability that the first boy on duty is:

Probability that the second boy on duty is:

Since both must be boys, let’s multiply the probabilities:

Answer: 0.2.

Task 3. There are 12 seats on board the aircraft next to the emergency exits and 18 seats behind the partitions separating the cabins. The remaining seats are inconvenient for tall passengers. Passenger V. is tall. Find the probability that at check-in, if a seat is randomly selected, passenger B will get a comfortable seat if there are 300 seats in total on the plane.

Solution: Passenger B has 30 comfortable seats (12 + 18 = 30), and there are 300 seats in total on the plane. Therefore, the probability that passenger B will get a comfortable seat is 30/300, i.e. 0.1.

Task 4. There are only 25 tickets in the collection of mathematics tickets, 10 of them contain a question on inequalities.

Find the probability that a student will not get a question on inequalities on a randomly selected exam ticket.

Solution: Of the 25 tickets, 15 do not contain a question on inequalities, so the probability that a student will not get a question on inequalities in a randomly selected exam ticket is 15/25, i.e. 0.6.

Problem 5. There are only 35 tickets in the collection of chemistry tickets, 7 of them contain a question on acids.

Find the probability that a student will not get a question on acids on a randomly selected exam ticket.

Solution: Of the 35 tickets, 28 do not contain a question on acids, so the probability that a student will not get a question on acids on a randomly selected exam ticket is 28/35, i.e. 0.8.

Task 6. On average, out of 500 garden pumps sold, 2 leak. Find the probability that one pump randomly selected for control does not leak.

Solution: If 2 out of 500 pumps are leaking, then 498 are not leaking. Therefore, the probability of choosing a good pump is 498/500, i.e. 0.996.

Task 7. The probability that a new vacuum cleaner will be repaired under warranty within a year is 0.065. In a certain city, out of 1,000 vacuum cleaners sold during the year, 70 units were received by the warranty workshop.

How different is the frequency of the “warranty repair” event from its probability in this city?

Solution: The frequency of the “warranty repair” event is 70/1000, i.e. 0.07. It differs from the predicted probability by 0.005 (0.07 – 0.065 = 0.005).

Task 8. 50 athletes are participating in the gymnastics championship: 18 from Russia, 14 from Ukraine, the rest from Belarus. The order in which the gymnasts perform is determined by lot.

Find the probability that the athlete competing first will be from Belarus.

Solution: There are 50 participants in the championship in total, and 18 athletes from Belarus (50 – 18 – 14 = 18).

The probability that an athlete from Belarus will compete first is 18 out of 50, i.e. 18/50, or 0.36.

Task 9. Scientific Conference carried out in 5 days. A total of 80 reports are planned - the first three days have 12 reports each, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by drawing lots.

What is the probability that Professor M.'s report will be scheduled for the last day of the conference?

Solution: In the first three days, 36 reports will be read (12 ∙ 3 ​​= 36), 44 reports are planned for the last two days. Therefore, 22 reports are planned for the last day (44: 2 = 22). This means that the probability that Professor M.’s report will be scheduled for the last day of the conference is 22/80, i.e. 0.275.

Problem 10.

Before the start of the first round of the chess championship, participants are randomly divided into playing pairs using lots. In total, 26 chess players are participating in the championship, including 14 participants from Russia, including Egor Kosov.

Find the probability that in the first round Egor Kosov will play with any chess player from Russia?

Solution: In the first round, Egor Kosov can play with 25 chess players (26 – 1 = 25), of which 13 are from Russia. This means that the probability that in the first round Egor Kosov will play with any chess player from Russia is 13/25, or 0.52.

Problem 11.

There are 16 teams participating in the World Championship. Using lots, they need to be divided into four groups of four teams each. The box contains mixed cards with group numbers: 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4.

Team captains draw one card each. What is the probability that the Russian team will be in the second group?

Solution: The probability that the Russian team will be in the second group is equal to the ratio of the number of cards with number 2 to the total number of cards, i.e. 4/16, or 0.25.

Problem 12. There are 5 people in the tourist group. Using lots, they choose two people who must go to the village to buy food. Tourist A. would like to go to the store, but he obeys the lot. What is the probability that A. will go to the store?

Solution: They choose two tourists out of five. Therefore, the probability of being selected is 2/5, i.e. 0.4.

Problem 13. There are 30 people in the group of tourists. They are dropped by helicopter into a hard-to-reach area in several stages, 6 people per flight. The order in which the helicopter transports tourists is random. Find the probability that tourist P. will take the first helicopter flight.

Solution: There are 6 seats on the first flight, a total of 30 seats. Then the probability that a tourist will fly on the first flight of a helicopter is 6/30, or 0.2.

Problem 14. What is the probability that a randomly selected natural number from 10 to 19 is divisible by three?

Solution: Natural numbers from 10 to 19 ten, of which three numbers are divisible by 3: 12, 15 and 18. Therefore, the desired probability is 3/10, i.e. 0.3.

Probability of multiple events

Task 1. Before the start of a volleyball match, team captains draw fair lots to determine which team will start the game with the ball. The “Starter” team takes turns playing with the “Rotor”, “Motor” and “Strator” teams. Find the probability that the Starter will only start the second game.

Solution:

We are satisfied with the following option: “Stator” does not start the first game, starts the second game, and does not start the third game. The probability of such a development of events is equal to the product of the probabilities of each of these events. The probability of each of them is 0.5, therefore: 0.5 · 0.5 · 0.5 = 0.125.

Task 2. To advance to the next round of competition, football team you need to score at least 4 points in two games. If a team wins, it receives 3 points, if there is a draw, 1 point, and if it loses, 0 points. Find the probability that the team will advance to the next round of the competition. Consider that in each game the probabilities of winning and losing are the same and equal to 0.4.

Solution:

Question type: combination of events.

The probability of the origin of any of these 3 options is equal to the sum of the probabilities of each option: 0.08 + 0.08 + 0.16 = 0.32.

Task 3. There are 21 people in the class. Among them are two friends: Anya and Nina. The class is randomly divided into 7 groups, 3 people in each. Find the probability that Anya and Nina will be in the same group.

Solution:

Question type: group reduction.

The probability of Anya getting into one of the groups is 1. The probability of Nina getting into the same group is 2 out of 20 (2 places left in the group, and there are 20 people left). 2/20 = 1/10 = 0.1.

Task 4. Petya had 4 ruble coins and 2 two-ruble coins in his pocket. Petya, without looking, transferred some 3 coins to another pocket. Find the probability that both two-ruble coins are in the same pocket.

Solution:

Method No. 1

Task type: group reduction.

Let's imagine that six coins are divided into two groups of three coins. The probability that the first one-ruble coin will fall into one of the pockets (groups) = 1.

The probability that two two-ruble coins will fall into the same pocket = the number of remaining spaces in this pocket/the number of remaining spaces in both pockets = 2/5 = 0.4.

Method No. 2

Question type: combination of events.

The task is performed in several ways:

If Petya transferred three of the four ruble coins to another pocket (but did not transfer the two-ruble coins), or if he transferred both two-ruble coins and one ruble coin to another pocket in one of three ways: 1, 2, 2; 2, 1, 2; 2, 2, 1. You can depict this on the diagram (Petya puts it in pocket 2, so we will calculate the probabilities in the “pocket 2” column):

Problem 5. Petya had 2 coins of 5 rubles and 4 coins of 10 rubles in his pocket. Petya, without looking, transferred some 3 coins to another pocket. Find the probability that the five-ruble coins are now in different pockets.

Solution:

Task type: group reduction.

Method No. 1

Let's imagine that six coins are divided into two groups of three coins. The probability that the first two-ruble coin will fall into one of the pockets (groups) = 1. The probability that the second coin will fall into the other pocket = the number of remaining places in the other / by the number of remaining places in both pockets = 3/5 = 0.6.

Method No. 2

Question type: combination of events.

The task is performed in several ways:

In order for the five-ruble coins to end up in different pockets, Petya must take one five-ruble and two ten-ruble coins from his pocket. This can be done in three ways: 5, 10, 10; 10, 5, 10 or 10, 10, 5. You can depict this on the diagram (Petya puts it in pocket 2, so we will calculate the probabilities in the “pocket 2” column):

The probability of the origin of any of these 4 options is equal to the sum of the probabilities of each of the options:

Task 6. In a random experiment, a symmetrical coin is tossed three times. Find the probability of getting heads exactly twice.

Solution: Type of question: finding the desired and the actual \ combining events We are satisfied with three options:

Heads - tails - heads;

Eagle - eagle - tails;

Tails - heads - heads;

The probability of each case is 1/2, and of each option is 1/8 (1/2 ∙ 1/2 ∙ 1/2 = 1/8)

We will be satisfied with either the first, second, or third option. Therefore, we add up their probabilities and get 3/8 (1/8 + 1/8 + 1/8 = 3/8), i.e. 0.375.

Task 7. If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.5. If A. plays black, then A. wins against B. with probability 0.34. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.

Solution:

Question type: combination of events.

In any case, A. will play both white and black, so we are satisfied with the option when grandmaster A. wins playing white (probability - 0.5) and also playing black (probability - 0.34). Therefore, we need to multiply the probabilities of these two events: 0.5 ∙ 0.34 = 0.17.

Task 8. The probability that the battery is defective is 0.02. A buyer in a store chooses a random package containing two of these batteries. Find the probability that both batteries are good.

Solution:

Question type: combination of events.

The probability that the battery is good is 0.98. The buyer needs both the first and second batteries to be in good working order: 0.98 · 0.98 = 0.9604.

Task 9. Bands perform at the rock festival - one from each of the declared countries. The order of performance is determined by lot. What is the probability that a group from the USA will perform after a group from Canada and after a group from China? Round the result to the nearest hundredth.

Solution:

Question type: combination of events.

The total number of groups performing at the festival is not important to answer the question. No matter how many there are, there are 6 ways for these countries relative position among the speakers (KIT - China, CAN = Canada):

... USA, CAN, KIT ...

...USA, KIT, CAN...

... KIT, USA, CAN ...

... CAN, USA, KIT ...

... KAN, KIT, USA ...

...KIT, CAN, USA...

The US is behind China and Canada in the last two cases. Therefore, the probability that the groups will be randomly distributed in this way is equal to:

Complementary Probability

Task 1.

An automatic line produces batteries. The probability that a finished battery is faulty is 0.02. Before packaging, each battery goes through a control system. The probability that the system will reject a faulty battery is 0.97. The probability that the system will mistakenly reject a working battery is 0.05.

Find the probability that a randomly selected battery will be rejected.

Solution:

There are 2 options that suit us:

Option A: the battery is rejected, it is faulty;

Option B: the battery is defective, it is working.

Probability of option A: 0.02 ∙ 0.97 = 0.0194;

Probability of option B: 0.05 ∙ 0.98 = 0.049;

We will be satisfied with either the first or second option: 0.0194 + 0.049 = 0.0684.

Task 2. Two factories produce identical glasses for car headlights. The first factory produces 60% of these glasses, the second - 40%. The first factory produces 3% of defective glass, and the second - 5%. Find the probability that glass accidentally purchased in a store will be defective.

Solution:

The probability that the glass was purchased at the first factory and is defective: 0.6 · 0.03 = 0.018.

The probability that the glass was purchased from a second factory and is defective: 0.4 · 0.05 = 0.02.

The probability that glass accidentally purchased in a store will be defective is 0.018 + 0.02 = 0.038.

Task 3. At a ceramic tableware factory, 10% of the plates produced are defective. During product quality control, 80% of defective plates are identified. The remaining plates are on sale. Find the probability that a plate randomly selected upon purchase has no defects. Round the result to the nearest thousand.

Solution:

Suppose we have x plates initially (after all, we constantly deal with percentages, so nothing prevents us from operating with specific quantities).

Then 0.1x are defective plates, and 0.9x are normal plates, which will arrive in the store immediately. Of the defective ones, 80% is removed, that is, 0.08x, and 0.02x remains, which will also go to the store. Thus, the total number of plates on the shelves in the store will be: 0.9x + 0.02x = 0.92x. Of these, 0.9x will be normal. Accordingly, according to the formula, the probability will be 0.9x/0.92x ≈ 0.978.

Task 4. Based on customer reviews, Igor Igorevich assessed the reliability of the two online stores. The probability that the desired product will be delivered from store A is 0.91. The probability that this product will be delivered from store B is 0.89. Igor Igorevich ordered goods from both stores at once. Assuming that online stores operate independently of each other, find the probability that no store will deliver the product.

Solution. The probability that the first store will not deliver the goods is 1 − 0.91 = 0.09. The probability that the second store will not deliver the goods is 1 − 0.89 = 0.11. The probability of these two events occurring simultaneously is equal to the product of the probabilities of each of them: 0.09 · 0.11 = 0.0099.

Task 5. When manufacturing bearings with a diameter of 70 mm, the probability that the diameter will differ from the specified one by less than 0.01 mm is 0.961. Find the probability that a random bearing will have a diameter less than 69.99 mm or greater than 70.01 mm.

Solution: We are given the probability of an event in which the diameter will be between 69.99 mm and 70.01 mm, and it is equal to 0.961. We can find the probability of all other options using the principle of complementary probability: 1 − 0.961 = 0.039.

Task 6. The probability that a student will solve more than 9 problems correctly on a history test is 0.68. The probability of solving more than 8 problems correctly is 0.78. Find the probability of solving exactly 9 problems correctly.

Solution: The probability that T. will correctly solve more than 8 problems includes the probability of solving exactly 9 problems. At the same time, events in which O. solves more than 9 problems are not suitable for us. Therefore, subtracting from the probability of solving more than 9 problems the probability of solving more than 8 problems, we will find the probability of solving only 9 problems: 0.78 – 0.68 = 0.1.

Task 7. A bus runs daily from the district center to the village. The probability that there will be fewer than 21 passengers on the bus on Monday is 0.88. The probability that there will be fewer than 12 passengers is 0.66. Find the probability that the number of passengers will be from 12 to 20.

Solution. The probability that a bus will have fewer than 21 passengers includes the probability that it will have between 12 and 20 passengers. At the same time, events in which there will be less than 12 passengers are not suitable for us. Therefore, subtracting the second probability (less than 12) from the first probability (less than 21), we find the probability that there will be from 12 to 20 passengers: 0.88 – 0.66 = 0.22.

Task 8. In the Magic Land there are two types of weather: good and excellent, and the weather, once established in the morning, remains unchanged all day. It is known that with probability 0.9 the weather tomorrow will be the same as today. On April 10, the weather in Magic Land is good. Find the probability that the weather will be great in Fairyland on April 13th.

Solution:

The task is performed in several options (“X” - good weather, “O” - excellent weather):

The probability of the origin of any of these 4 options is equal to the sum of the probabilities of each option: 0.081 + 0.081 + 0.081 + 0.001 = 0.244.

Task 9. In the Magic Land there are two types of weather: good and excellent, and the weather, once established in the morning, remains unchanged all day. It is known that with probability 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in the Magic Land is good. Find the probability that the weather will be great in Fairyland on July 6th.

Solution:

The task is performed in several options (“X” - good weather, “O” - excellent weather):

The probability of the origin of any of these 4 options is equal to the sum of the probabilities of each option: 0.128 + 0.128 + 0.128 + 0.008 = 0.392.

probability- a number between 0 and 1 that reflects the chances that a random event will occur, where 0 is the complete absence of probability of the event occurring, and 1 means that the event in question will definitely occur.

The probability of event E is a number from to 1.
The sum of the probabilities of mutually exclusive events is equal to 1.

empirical probability- probability, which is calculated as the relative frequency of an event in the past, extracted from the analysis of historical data.

The probability of very rare events cannot be calculated empirically.

subjective probability- probability based on a personal subjective assessment of an event without regard to historical data. Investors who make decisions to buy and sell shares often act based on considerations of subjective probability.

prior probability -

The chance is 1 in... (odds) that an event will occur through the concept of probability. The chance of an event occurring is expressed through probability as follows: P/(1-P).

For example, if the probability of an event is 0.5, then the chance of the event is 1 out of 2 because 0.5/(1-0.5).

The chance that an event will not occur is calculated using the formula (1-P)/P

Inconsistent probability- for example, the price of shares of company A takes into account possible event E by 85%, and the price of shares of company B only takes into account 50%. This is called inconsistent probability. According to the Dutch Betting Theorem, inconsistent probability creates profit opportunities.

Unconditional probability is the answer to the question “What is the probability that the event will occur?”

Conditional probability- this is the answer to the question: “What is the probability of event A if event B occurs.” Conditional probability is denoted as P(A|B).

Joint probability- the probability that events A and B will occur simultaneously. Denoted as P(AB).

P(A|B) = P(AB)/P(B) (1)

P(AB) = P(A|B)*P(B)

Rule for summing up probabilities:

The probability that either event A or event B will happen is

P (A or B) = P(A) + P(B) - P(AB) (2)

If events A and B are mutually exclusive, then

P (A or B) = P(A) + P(B)

Independent events- events A and B are independent if

P(A|B) = P(A), P(B|A) = P(B)

That is, it is a sequence of results where the probability value is constant from one event to the next.
A coin toss is an example of such an event - the result of each subsequent toss does not depend on the result of the previous one.

Dependent Events- these are events where the probability of the occurrence of one depends on the probability of the occurrence of another.

The rule for multiplying the probabilities of independent events:
If events A and B are independent, then

P(AB) = P(A) * P(B) (3)

Total probability rule:

P(A) = P(AS) + P(AS") = P(A|S")P(S) + P (A|S")P(S") (4)

S and S" are mutually exclusive events

expected value random variable is the average of possible outcomes random variable. For event X, the expectation is denoted as E(X).

Let’s say we have 5 values ​​of mutually exclusive events with a certain probability (for example, a company’s income was such and such an amount with such a probability). The expected value is the sum of all outcomes multiplied by their probability:

Dispersion of a random variable is the expectation of square deviations of a random variable from its expectation:

s 2 = E( 2 ) (6)

Conditional expected value is the expected value of a random variable X, provided that the event S has already occurred.

Every person encounters the concept of probability every day. People calculate the chance of catching the bus, the probability that they will receive a salary today, and they come up with various combinations for winning the lottery. The theory of probability in computer programs and artificial intelligence is seriously affected, and it is also closely intertwined with financial exchanges and the like. There are elementary examples of how to find probability.

The classic case is with a coin. It is thrown up, and there are two different options for its landing: falling on the obverse and falling on the reverse. The possibility of falling on an edge is excluded in advance, that is, there are two probable outcomes. Since there are only two of them, and they happen with the same frequency, the probability of getting, for example, heads is 1/2. This is the basic law of how to find probability in mathematics.

Where did this 1/2 come from? The principle is that the probability of one (1) event out of two (2) possible events is calculated. Their ratio is resolved by the division operation, which gives 1/2. Similarly, you can calculate the probability of a certain number falling out on a dice. As you know, the surface of a cube has 6 faces, therefore any number from 1 to 6 can appear - six different options. How to find the probability of rolling, for example, a four?

Four can only come out in the only way (1) out of six in every possible way, therefore, the probability will be equal to 1: 6 = 1/6. One sixth can be converted to decimal by dividing on a calculator: 1/6 = 0.6(6). By multiplying the value by 100 and adding the “%” sign, you can get an estimate of the probability of the event as a percentage. It is extremely important to know that the probability of an event is estimated as a number from 0 to 1, which in percentage ranges from 0% to 100%.

All other probability values ​​are absurd. A specific example should be considered: a random card is drawn from a classic deck of cards (36 cards). What is the probability that the card will be red and its number will be odd? A red odd card can only be a seven or nine of diamonds or hearts. There are 4 such cards in total. This means that the probability of such a card appearing is 4 / 36 = 1 / 9 = 0.1(1). The probability should be calculated as a percentage, this is equal to 1.1%.

Very often in problems you should use the complex probability formula. For example, there are 10 balls in an urn, 3 black and 7 white. What is the probability that two balls drawn at random in a row will be black? This problem should be solved as two separate ones. First, you should calculate the probability of drawing a black ball from all of them. There are 3 such balls, and there are 10 of them in total, which means the probability will be equal to 3/10. Next, we need to move on to the second part of the problem, where probability theory allows us to harmonize the results.

After extraction, 9 balls will remain in the urn, 2 of which will be black. In this case, the chance of getting a black ball is 2/9. Next, you should multiply the obtained probabilities for the final result: 3/10 * 2/9 = 6/90 = 1/15 = 0.6(6), which is approximately equal to 6.7%. This means that the probability of this event is quite low.