Probability theory: formulas and examples of problem solving. Probability theory formulas and examples of problem solving

Knowing that probability can be measured, let's try to express it in numbers. There are three possible ways.

Rice. 1.1. Measuring Probability

PROBABILITY DETERMINED BY SYMMETRY

There are situations in which possible outcomes are equally probable. For example, when tossing a coin once, if the coin is standard, the probability of “heads” or “tails” appearing is the same, i.e. P("heads") = P("tails"). Since only two outcomes are possible, then P(“heads”) + P(“tails”) = 1, therefore, P(“heads”) = P(“tails”) = 0.5.

In experiments where outcomes have equal chances of occurrence, the probability of event E, P (E) is equal to:

Example 1.1. The coin is tossed three times. What is the probability of two heads and one tail?

First, let's find all possible outcomes: To make sure that we have found all possible options, we will use a tree diagram (see Chapter 1, Section 1.3.1).

So, there are 8 equally possible outcomes, therefore, the probability of them is 1/8. Event E - two heads and tails - three occurred. That's why:

Example 1.2. A standard die is rolled twice. What is the probability that the score is 9 or more?

Let's find all possible outcomes.

Table 1.2. The total number of points obtained by rolling a die twice

So, in 10 out of 36 possible outcomes the sum of points is 9 or therefore:

EMPIRICALLY DETERMINED PROBABILITY

Example with a coin from the table. 1.1 clearly illustrates the mechanism for determining probability.

Given the total number of experiments that are successful, the probability of the required result is calculated as follows:

A ratio is the relative frequency of occurrence of a certain result over a sufficiently long experiment. The probability is calculated either based on the data of the experiment performed, based on past data.

Example 1.3. Of the five hundred electric lamps tested, 415 worked for more than 1000 hours. Based on the data from this experiment, we can conclude that the probability of normal operation of a lamp of this type for more than 1000 hours is:

Note. Testing is destructive in nature, so not all lamps can be tested. If only one lamp were tested, the probability would be 1 or 0 (i.e. whether it can last 1000 hours or not). Hence the need to repeat the experiment.

Example 1.4. In table 1.3 shows data on the length of service of men working in the company:

Table 1.3. Men's work experience

What is the probability that the next person hired by the company will work for at least two years:

Solution.

The table shows that 38 out of 100 employees have been working in the company for more than two years. The empirical probability that the next employee will remain with the company for more than two years is:

At the same time, we assume that the new employee is “typical and the working conditions are unchanged.

SUBJECTIVE PROBABILITY ASSESSMENT

In business, situations often arise in which there is no symmetry, and there is no experimental data either. Therefore, determining the likelihood of a favorable outcome under the influence of the views and experience of the researcher is subjective.

Example 1.5.

1. An investment expert estimates that the probability of making a profit in the first two years is 0.6.

2. Marketing manager's forecast: the probability of selling 1000 units of a product in the first month after its appearance on the market is 0.4.

“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 main definitions of this science will be presented in the article.

What is probability theory?

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

To make it a little clearer, let's give a small example: if you 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. They studied gambling for a long time 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 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, “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.”
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 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 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 (9th grade) mainly use the classical definition, which sounds like this:

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

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

A is actually an event. If 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 the theory of probability is, formulas and examples of solving problems that come across in the school curriculum. However, probability theory is also found in higher mathematics, which is taught in universities. Most often they operate with geometric and statistical definitions of the theory and complex formulas.

The theory of probability is very interesting. It is better to start 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 in m different ways, and a choice B can be made in n different ways, then the choice of A and B can be made by multiplication.

For example, there are 5 roads leading from city A to city B. There are 4 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 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 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. It’s difficult for an ordinary person to answer; it’s better to ask someone who has used it to win the jackpot more than once.

The probability of an event occurring in a certain test is equal to the ratio , where:

The total number of all equally possible, elementary outcomes of a given test, which form full group of events;

The number of elementary outcomes favorable to the event.

Problem 1

An urn contains 15 white, 5 red and 10 black balls. 1 ball is drawn at random, find the probability that it will be: a) white, b) red, c) black.

Solution: The most important prerequisite for using the classical definition of probability is ability to count the total number of outcomes.

There are a total of 15 + 5 + 10 = 30 balls in the urn, and obviously the following facts are true:

Retrieving any ball is equally possible (equal opportunity outcomes), while the outcomes elementary and form full group of events (i.e., as a result of the test, one of the 30 balls will definitely be removed).

Thus, the total number of outcomes:

Consider the event: - a white ball will be drawn from the urn. This event is favored by elementary outcomes, therefore, according to the classical definition:
- the probability that a white ball will be drawn from the urn.

Oddly enough, even in such a simple task, serious inaccuracy can be made. Where is the pitfall here? It is incorrect to argue here that “since half the balls are white, then the probability of drawing a white ball » . The classic definition of probability refers to ELEMENTARY outcomes, and the fraction must be written down!

With other points, similarly, consider the following events:

A red ball will be drawn from the urn;
- a black ball will be drawn from the urn.

An event is favored by 5 elementary outcomes, and an event is favored by 10 elementary outcomes. So the corresponding probabilities are:

A typical check of many server tasks is carried out using theorems on the sum of probabilities of events forming a complete group. In our case, the events form a complete group, which means the sum of the corresponding probabilities must necessarily equal one: .

Let's check if this is true: that's what I wanted to make sure of.

Answer:

In practice, the “high-speed” solution design option is common:

Total: 15 + 5 + 10 = 30 balls in the urn. According to the classical definition:
- the probability that a white ball will be drawn from the urn;
- the probability that a red ball will be drawn from the urn;
- the probability that a black ball will be drawn from the urn.

Answer:

Problem 2

The store received 30 refrigerators, five of which have a manufacturing defect. One refrigerator is selected at random. What is the probability that it will be without a defect?

Problem 3

When dialing a phone number, the subscriber forgot the last two digits, but remembers that one of them is zero and the other is odd. Find the probability that he will dial the correct number.

Note: zero is an even number (divisible by 2 without a remainder)

Solution: First we find the total number of outcomes. By condition, the subscriber remembers that one of the digits is zero, and the other digit is odd. Here it is more rational not to split hairs with combinatorics and take advantage method of direct listing of outcomes . That is, when making a solution, we simply write down all the combinations:

01, 03, 05, 07, 09

10, 30, 50, 70, 90

And we count them - in total: 10 outcomes.

There is only one favorable outcome: the correct number.

According to the classical definition:
- probability that the subscriber will dial the correct number

Answer: 0,1

Advanced task for independent solution:

Problem 4

The subscriber has forgotten the PIN code for his SIM card, but remembers that it contains three “fives”, and one of the numbers is either a “seven” or an “eight”. What is the probability of successful authorization on the first try?

Here you can also develop the idea of the likelihood that the subscriber will face punishment in the form of a puk code, but, unfortunately, the reasoning will go beyond the scope of this lesson

The solution and answer are below.

Sometimes listing combinations turns out to be a very painstaking task. In particular, this is the case in the next, no less popular group of problems, where 2 dice are rolled (less often - more):

Problem 5

Find the probability that when throwing two dice the total number will be:

a) five points;

b) no more than four points;

c) from 3 to 9 points inclusive.

Solution: find the total number of outcomes:

Ways the side of the 1st die can fall out And in different ways the side of the 2nd cube can fall out; By rule for multiplying combinations, Total: possible combinations. In other words, each the face of the 1st cube can form an ordered pair with each the edge of the 2nd cube. Let’s agree to write such a pair in the form , where is the number rolled on the 1st die, is the number rolled on the 2nd die.

For example:

The first dice scored 3 points, the second dice scored 5 points, total points: 3 + 5 = 8;
- the first dice scored 6 points, the second - 1 point, the sum of points: 6 + 1 = 7;
- 2 points rolled on both dice, sum: 2 + 2 = 4.

Obviously, the smallest amount is given by a pair, and the largest by two “sixes”.

a) Consider the event: - when throwing two dice, 5 points will appear. Let's write down and count the number of outcomes that favor this event:

Total: 4 favorable outcomes. According to the classical definition:
- the desired probability.

b) Consider the event: - no more than 4 points will appear. That is, either 2, or 3, or 4 points. Again we list and count the favorable combinations, on the left I will write down the total number of points, and after the colon - the suitable pairs:

Total: 6 favorable combinations. Thus:
- the probability that no more than 4 points will be rolled.

c) Consider the event: - 3 to 9 points will roll, inclusive. Here you can take the straight road, but... for some reason you don’t want to. Yes, some pairs have already been listed in the previous paragraphs, but there is still a lot of work to be done.

What's the best way to proceed? In such cases, a roundabout path turns out to be rational. Let's consider opposite event: - 2 or 10 or 11 or 12 points will appear.

What's the point? The opposite event is favored by a significantly smaller number of couples:

Total: 7 favorable outcomes.

According to the classical definition:
- the probability that you will get less than three or more than 9 points.

Particularly scrupulous people can list all 29 pairs, thereby completing the check.

Answer:

In the next problem we will repeat the multiplication table:

Problem 6

Find the probability that, when throwing two dice, the product of the points is:

a) will be equal to seven;

b) there will be at least 20;

c) will be even.

A short solution and answer at the end of the lesson.

Problem 7

3 people entered the elevator of a 20-story building on the first floor. And let's go. Find the probability that:

a) they will exit on different floors;

b) two will exit on the same floor;

c) everyone will get off on the same floor.

Solution: let's calculate the total number of outcomes: ways the 1st passenger can get out of the elevator And ways - 2nd passenger And ways - the third passenger. According to the rule of multiplication of combinations: possible outcomes. That is, every 1st person exit floor can be combined with every 2nd person exit floor and with every 3rd person exit floor.

The second method is based on placements with repetitions:
- whoever understands it more clearly.

a) Consider the event: - passengers will get off on different floors. Let's calculate the number of favorable outcomes:
3 passengers on different floors can exit using these methods. Do your own reasoning based on the formula.

According to the classical definition:

c) Consider the event: - passengers will get off on the same floor. This event has favorable outcomes and, according to the classical definition, the corresponding probability: .

We come in from the back door:

b) Consider the event: - two people will get off on the same floor (and, accordingly, the third one is on the other).

Events form full group (we believe that no one will fall asleep in the elevator and the elevator will not get stuck, which means .

As a result, the desired probability is:

Thus, theorem on the addition of the probabilities of events forming a complete group, can be not only convenient, but also become a real lifesaver!

Answer:

When you get large fractions, it is good practice to indicate their approximate decimal values. Usually rounded to 2-3-4 decimal places.

Since the events of points “a”, “be”, “ve” form a complete group, it makes sense to perform a control check, and it is better with approximate values:

Which is what needed to be checked.

Sometimes, due to rounding errors, the result may be 0.9999 or 1.0001; in this case, one of the approximate values should be “adjusted” so that the total is a “pure” unit.

On one's own:

Problem 8

10 coins are tossed. Find the probability that:

a) all coins will show heads;

b) 9 coins will land heads, and one coin will land tails;

c) heads will appear on half of the coins.

Problem 9

7 people are randomly seated on a seven-seat bench. What is the probability that two certain people will be close together?

Solution: There are no problems with the total number of outcomes:
7 people can sit on a bench in different ways.

But how to calculate the number of favorable outcomes? Trivial formulas are not suitable and the only way is logical reasoning. First, let's consider the situation when Sasha and Masha were next to each other on the left edge of the bench:

Obviously, the order matters: Sasha can sit on the left, Masha on the right, and vice versa. But that is not all - for each of these two cases, the rest of the people can sit in the empty seats in other ways. Combinatorially speaking, Sasha and Masha can be rearranged in adjacent places in the following ways: And For each such permutation, other people can be rearranged in ways.

Thus, according to the rule of multiplication of combinations, favorable outcomes emerge.

But that's not all! The above facts are true for each pairs of neighboring places:

It is interesting to note that if the bench is “rounded” (connecting left and right seats), then an additional, seventh pair of adjacent places is formed. But let's not get distracted. According to the same principle of multiplying combinations, we obtain the final number of favorable outcomes:

According to the classical definition:
- the probability that two specific people will be nearby.

Answer:

Problem 10

Two rooks, white and black, are placed at random on a chessboard of 64 cells. How likely is it that they will not “beat” each other?

Reference: a chessboard has the size of squares; black and white rooks “beat” each other when they are located on the same rank or on the same vertical

Be sure to make a schematic drawing of the board, and even better if there is chess nearby. It's one thing to reason on paper, and quite another when you arrange the pieces with your own hands.

Problem 11

What is the probability that the four cards dealt will contain one ace and one king?

Let's calculate the total number of outcomes. In how many ways can you remove 4 cards from a deck? Probably everyone understood that we are talking about number of combinations:
using these methods you can choose 4 cards from the deck.

Now we consider favorable outcomes. According to the condition, in a selection of 4 cards there must be one ace, one king and, which is not stated in plain text - two other cards:

Ways to extract one ace;
ways you can choose one king.

We exclude aces and kings from consideration: 36 - 4 - 4 = 28

ways you can extract the other two cards.

According to the rule for multiplying combinations:
ways you can extract the desired combination of cards (1st Ace And 1st king And two other cards).

Let me comment on the combinational meaning of the notation in another way:
every ace combines with every king and with each possible pair of other cards.

According to the classical definition:
- the probability that among the four cards dealt there will be one ace and one king.

If you have time and patience, reduce large fractions as much as possible.

Answer:

A simpler task to solve on your own:

Problem 12

The box contains 15 quality and 5 defective parts. 2 parts are removed at random.

Find the probability that:

a) both parts will be of high quality;

b) one part will be of high quality, and one will be defective;

c) both parts are defective.

The events of the listed points form a complete group, so checking here suggests itself. A short solution and answer at the end of the lesson. In general, the most interesting things are just beginning!

Problem 13

The student knows the answers to 25 exam questions out of 60. What is the probability of passing the exam if you need to answer at least 2 out of 3 questions?

Solution: So, the situation is as follows: a total of 60 questions, among which 25 are “good” and, accordingly, 60 - 25 = 35 “bad”. The situation is precarious and not in favor of the student. Let's find out how good his chances are:

ways you can choose 3 questions out of 60 (total number of outcomes).

In order to pass the exam, you need to answer 2 or 3 questions. We consider favorable combinations:

Ways to choose 2 “good” questions And one is “bad”;

ways you can choose 3 “good” questions.

By rule for adding combinations:
ways you can choose a combination of 3 questions that is favorable for passing the exam (no difference with two or three “good” questions).

According to the classical definition:

Answer:

Problem 14

A poker player is dealt 5 cards. Find the probability that:

a) among these cards there will be a pair of tens and a pair of jacks;
b) the player will be dealt a flush (5 cards of the same suit);
c) the player will be dealt four of a kind (4 cards of the same value).

Which of the following combinations is most likely to be obtained?

! Attention! If the condition asks a similar question, then answer it necessary give an answer.
Reference : Poker is traditionally played with a 52-card deck, which contains cards of 4 suits ranging from deuces to aces.

Poker is the most mathematical game (those who play know it), in which you can have a noticeable advantage over less qualified opponents.

Solutions and Answers:

Task 2: Solution: 30 - 5 = 25 refrigerators have no defect.

- the probability that a randomly selected refrigerator does not have a defect.
Answer :

Task 4: Solution: find the total number of outcomes:
ways you can select the place where the dubious number is located and on every Of these 4 places, 2 digits can be located (seven or eight). According to the rule of multiplication of combinations, the total number of outcomes: .
Alternatively, the solution can simply list all the outcomes (fortunately there are few of them):

7555, 8555, 5755, 5855, 5575, 5585, 5557, 5558

There is only one favorable outcome (correct pin code).

Thus, according to the classical definition:
- probability that the subscriber logs in on the 1st attempt
Answer :

Task 6: Solution

Task 6:Solution : find the total number of outcomes:
numbers can appear on 2 dice in different ways.

a) Consider the event: - when throwing two dice, the product of the points will be equal to seven. There are no favorable outcomes for this event,
, i.e. this event is impossible.

b) Consider the event: - when throwing two dice, the product of the points will be at least 20. The following outcomes favor this event:

Total: 8

According to the classical definition:

- the desired probability.

c) Consider the opposite events:

- the product of points will be even;

- the product of points will be odd.

Let's list all the outcomes favorable to the event :

Total: 9 favorable outcomes.

According to the classical definition of probability:

Opposite events form a complete group, therefore:

- the desired probability.

Answer :

Problem 8:Solution ways 2 coins can fall.
Another way: ways the 1st coin can fallAnd ways the 2nd coin can fallAnd … And ways the 10th coin can fall. According to the rule of multiplying combinations, 10 coins can fall ways.
a) Consider the event: - all coins will show heads. This event is favored by a single outcome, according to the classical definition of probability: .
b) Consider the event: - 9 coins will land heads, and one coin will land tails.
Exists coins that may land on heads. According to the classical definition of probability: .
c) Consider the event: - heads will appear on half of the coins.
Exists unique combinations of five coins that can land heads. According to the classical definition of probability:
Answer:

Problem 10:Solution : let's calculate the total number of outcomes:
ways to place two rooks on the board.
Another design option: ways to select two squares of a chessboardAnd ways to place a white and black rookin every out of 2016 cases. Thus, the total number of outcomes: .

Now let’s count the outcomes in which the rooks “beat” each other. Let's consider the 1st horizontal line. Obviously, the figures can be placed on it in any way, for example, like this:

In addition, the rooks can be rearranged. Let’s put the reasoning into numerical form: ways you can select two cellsAnd ways to rearrange rooksin everyout of 28 cases. Total: possible positions of figures on the horizontal.
Short version of the design: ways you can place the white and black rook on the 1st rank.

The above reasoning is correctfor each horizontal, so the number of combinations should be multiplied by eight: . Additionally, a similar story holds true for any of the eight verticals. Let's calculate the total number of formations in which the pieces “beat” each other:

Then in the remaining variants of the arrangement the rooks will not “beat” each other:
4032 - 896 = 3136

According to the classical definition of probability:
- the probability that a white and black rook placed at random on the board will not “beat” each other.

Answer :

Problem 12:Solution : total: 15 + 5 = 20 parts in a box. Let's calculate the total number of outcomes:
using these methods you can remove 2 parts from the box.
a) Consider the event: - both extracted parts will be of high quality.
using these methods you can extract 2 quality parts.
According to the classical definition of probability:
b) Consider the event: - one part will be of high quality, and one will be defective.
ways you can extract 1 quality partAnd1 defective.
According to the classical definition:
c) Consider the event: - both extracted parts are defective.
using these methods you can remove 2 defective parts.
According to the classical definition:
Examination: let's calculate the sum of the probabilities of events that form the complete group: , which was what needed to be checked.
Answer:

And now let’s take in our hands an already familiar and trouble-free learning tool - a dice with full group of events , which consist in the fact that when thrown, 1, 2, 3, 4, 5 and 6 points will appear, respectively.

Consider the event - as a result of throwing a dice, at least five points will appear. This event consists of two incompatible outcomes: (roll 5 or 6 points)
- the probability that a dice roll will result in at least five points.

Let's consider the event that no more than 4 points will be rolled and find its probability. According to the theorem of addition of probabilities of incompatible events:

Perhaps some readers have not yet fully realized essence incompatibility. Let's think about it again: a student cannot answer 2 out of 3 questions and at the same time answer all 3 questions. Thus, the events and are incompatible.

Now, using classical definition, let's find their probabilities:

The fact of successfully passing the exam is expressed by the amount (answer to 2 out of 3 questions or for all questions). According to the theorem of addition of probabilities of incompatible events:
- the probability that the student will pass the exam.

This solution is completely equivalent, choose which one you like best.

Problem 1

The store received products in boxes from four wholesale warehouses: four from the 1st, five from the 2nd, seven from the 3rd and four from the 4th. A box for sale is randomly selected. What is the probability that it will be a box from the first or third warehouse.

Solution: total received by the store: 4 + 5 + 7 + 4 = 20 boxes.

In this task, it is more convenient to use the “fast” method of formatting without writing events in capital letters. According to the classical definition:
- the probability that a box from the 1st warehouse will be selected for sale;
- the probability that a box from the 3rd warehouse will be selected for sale.

According to the theorem of addition of incompatible events:
- the probability that a box from the first or third warehouse will be selected for sale.

Answer: 0,55

Of course, the problem is solvable and purely through classical definition of probability by directly counting the number of favorable outcomes (4 + 7 = 11), but the considered method is no worse. And even clearer.

Problem 2

The box contains 10 red and 6 blue buttons. Two buttons are removed at random. What is the probability that they will be the same color?

Similarly - here you can use combinatorial sum rule, but you never know... suddenly someone forgot it. Then the theorem for adding the probabilities of incompatible events will come to the rescue!

Probability theory is a fairly extensive independent branch of mathematics. In the school course, probability theory is discussed very superficially, but in the Unified State Examination and the State Examination Academy there are problems on this topic. However, solving school course problems is not so difficult (at least as far as arithmetic operations are concerned) - here you do not need to count derivatives, take integrals and solve complex trigonometric transformations - the main thing is to be able to handle prime numbers and fractions.

Probability theory - basic terms

The main terms of probability theory are test, outcome and random event. A test in probability theory is an experiment - tossing a coin, drawing a card, drawing lots - all these are tests. The result of the test, as you may have guessed, is called the outcome.

What is a random event? In probability theory, it is assumed that the test is carried out more than once and there are many outcomes. A random event is a set of outcomes of a trial. For example, if you toss a coin, two random events can happen - heads or tails.

Do not confuse the concepts of outcome and random event. An outcome is one result of one trial. A random event is a set of possible outcomes. By the way, there is such a term as an impossible event. For example, the event “rolling the number 8” on a standard dice is impossible.

How to find probability?

We all roughly understand what probability is, and quite often use this word in our vocabulary. In addition, we can even draw some conclusions regarding the likelihood of a particular event, for example, if there is snow outside the window, we can most likely say that it is not summer. However, how can we express this assumption numerically?

In order to introduce a formula for finding probability, we introduce one more concept - favorable outcome, that is, an outcome that is favorable for a particular event. The definition is quite ambiguous, of course, but according to the conditions of the problem it is always clear which outcome is favorable.

For example: There are 25 people in the class, three of them are Katya. The teacher assigns Olya to duty, and she needs a partner. What is the probability that Katya will become your partner?

In this example, the favorable outcome is partner Katya. We will solve this problem a little later. But first, using an additional definition, we introduce a formula for finding the probability.

P = A/N, where P is the probability, A is the number of favorable outcomes, N is the total number of outcomes.

All school problems revolve around this one formula, and the main difficulty usually lies in finding the outcomes. Sometimes they are easy to find, sometimes not so much.

How to solve probability problems?

Problem 1

So now let's solve the above problem.

The number of favorable outcomes (the teacher will choose Katya) is three, because there are three Katyas in the class, and the total outcomes are 24 (25-1, because Olya has already been chosen). Then the probability is: P = 3/24=1/8=0.125. Thus, the probability that Olya’s partner will be Katya is 12.5%. Not difficult, right? Let's look at something a little more complicated.

Problem 2

The coin was tossed twice, what is the probability of getting one head and one tail?

So, let's consider the general outcomes. How can coins land - heads/heads, tails/tails, heads/tails, tails/heads? This means that the total number of outcomes is 4. How many favorable outcomes? Two - heads/tails and tails/heads. Thus, the probability of getting a heads/tails combination is:

P = 2/4 = 0.5 or 50 percent.

Now let's look at this problem. Masha has 6 coins in her pocket: two with a face value of 5 rubles and four with a face value of 10 rubles. Masha moved 3 coins to another pocket. What is the probability that 5-ruble coins will end up in different pockets?

For simplicity, let's designate the coins by numbers - 1,2 - five-ruble coins, 3,4,5,6 - ten-ruble coins. So, how can coins be in your pocket? There are 20 combinations in total:

123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456.

At first glance, it may seem that some combinations are missing, for example, 231, but in our case, combinations 123, 231 and 321 are equivalent.

Now we count how many favorable outcomes we have. For them we take those combinations that contain either the number 1 or the number 2: 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256. There are 12 of them. Thus, the probability is equal to:

P = 12/20 = 0.6 or 60%.

The probability problems presented here are quite simple, but do not think that probability is a simple branch of mathematics. If you decide to continue your education at a university (with the exception of humanities), you will definitely have classes in higher mathematics, in which you will be introduced to more complex terms of this theory, and the tasks there will be much more difficult.

Want to know the mathematical odds of your bet being successful? Then there is two good news for you. First: to calculate cross-country ability, you don’t need to carry out complex calculations and spend a lot of time. It is enough to use simple formulas, which will take a couple of minutes to work with. Second: after reading this article, you can easily calculate the probability of any of your transactions passing.

To correctly determine cross-country ability, you need to take three steps:

Calculate the percentage of probability of the outcome of an event according to the bookmaker’s office;
Calculate the probability using statistical data yourself;
Find out the value of the bet, taking into account both probabilities.

Let's look at each of the steps in detail, using not only formulas, but also examples.

Fast passage

Calculating the probability included in bookmaker odds

The first step is to find out with what probability the bookmaker himself estimates the chances of a particular outcome. It’s clear that bookmakers don’t set odds just like that. To do this we use the following formula:

PB=(1/K)*100%,

where P B is the probability of the outcome according to the bookmaker’s office;

K – bookmaker odds for the outcome.

Let’s say that the odds for London Arsenal’s victory in the match against Bayern Munich are 4. This means that the probability of their victory is assessed by the bookmaker as (1/4)*100%=25%. Or Djokovic plays against Youzhny. The multiplier for Novak's victory is 1.2, his chances are (1/1.2)*100%=83%.

This is how the bookmaker itself evaluates the chances of success of each player and team. Having completed the first step, we move on to the second.

Calculation of the probability of an event by the player

The second point of our plan is our own assessment of the probability of the event. Since we cannot mathematically take into account such parameters as motivation and game tone, we will use a simplified model and use only statistics from previous meetings. To calculate the statistical probability of an outcome, we use the formula:

PAND=(UM/M)*100%,

WherePAND– probability of an event according to the player;

UM – the number of successful matches in which such an event occurred;

M – total number of matches.

To make it clearer, let's give examples. Andy Murray and Rafael Nadal played 14 matches between themselves. In 6 of them the total was less than 21 in games, in 8 the total was more. You need to find out the probability that the next match will be played with a higher total: (8/14)*100=57%. Valencia played 74 matches against Atlético at Mestalla, in which they won 29 victories. Probability of Valencia winning: (29/74)*100%=39%.

And we learn all this only thanks to the statistics of previous games! Naturally, it will not be possible to calculate such a probability for a new team or player, so this betting strategy is only suitable for matches in which the opponents meet more than once. Now we know how to determine the bookmaker's and our own probabilities of outcomes, and we have all the knowledge to move on to the last step.

Determining the value of a bet

The value (value) of a bet and the passability have a direct connection: the higher the value, the higher the chance of passing. The value is calculated as follows:

V=PAND*K-100%,

where V is value;

P I – probability of outcome according to the bettor;

K – bookmaker odds for the outcome.

Let’s say we want to bet on Milan’s victory in the match against Roma and we calculate that the probability of the “red-blacks” winning is 45%. The bookmaker offers us odds of 2.5 for this outcome. Would such a bet be valuable? We carry out calculations: V=45%*2.5-100%=12.5%. Great, we have a valuable bet with good chances of passing.

Let's take another case. Maria Sharapova plays against Petra Kvitova. We want to make a deal for Maria to win, the probability of which, according to our calculations, is 60%. Bookmakers offer a 1.5 multiplier for this outcome. We determine the value: V=60%*1.5-100=-10%. As you can see, this bet is of no value and should be avoided.