Attention applicants! Several tasks of the exam are analyzed here. The rest, more interesting ones, are in our free video material. Watch and act!

We will start with simple problems and basic concepts of probability theory.
Random An event is called an event that cannot be accurately predicted in advance. It can either happen or not.
You won the lottery - a random event. You invited friends to celebrate the win, and on the way to you they got stuck in the elevator - also a random event. True, the master was nearby and freed the whole company in ten minutes - and this can also be considered a happy accident ...

Our life is full random events. Each of them can be said to happen with some probability. Most likely, you are intuitively familiar with this concept. Now we will give a mathematical definition of probability.

Let's start with the simplest example. You are tossing a coin. Heads or tails?

Such an action, which can lead to one of several results, is called in probability theory test.

Heads and tails - two possible exodus tests.

The eagle will fall out in one case out of two possible. They say that probability that the coin lands heads is equal to .

Let's throw a dice. The die has six sides, so there are six possible outcomes.

For example, you guessed that three points will fall out. This is one outcome out of six possible. In probability theory, it will be called favorable outcome.

The probability of getting a triple is (one favorable outcome out of six possible).

The probability of a four is also

But the probability of the appearance of the seven is zero. After all, there is no face with seven points on the cube.

The probability of an event is equal to the ratio of the number of favorable outcomes to the total number of outcomes.

Obviously, the probability cannot be greater than one.

Here is another example. In a bag of apples, of which are red, the rest are green. The apples do not differ in shape or size. You put your hand in the bag and take out an apple at random. The probability of drawing a red apple is , and a green one is .

The probability of getting a red or green apple is .

Let's analyze the problems in the theory of probability included in the collections for preparing for the exam.

. At the taxi company this moment free cars: red, yellow and green. On a call, one of the cars left, which happened to be closest to the customer. Find the probability that a yellow taxi will arrive.

There are cars in total, that is, one out of fifteen will come to the customer. There are nine yellow ones, which means that the probability of arrival of a yellow car is , that is .

. (Demo version) In the collection of tickets on biology of all tickets, in two of them there is a question about mushrooms. At the exam, the student gets one randomly selected ticket. Find the probability that this ticket does not include the question about mushrooms.

Obviously, the probability of drawing a ticket without asking about mushrooms is , that is .

. Parental committee bought puzzles for gifts for children at the end school year, of which with pictures famous artists and pictures of animals. Gifts are distributed randomly. Find the probability that Vovochka will get the animal puzzle.

The task is solved in a similar way.

Answer: .

. Athletes participate in the gymnastics championship: from Russia, from the USA, the rest - from China. The order in which the gymnasts perform is determined by lot. Find the probability that the last athlete to compete is from China.

Let's imagine that all the athletes at the same time approached the hat and pulled out pieces of paper with numbers from it. Some of them will get the twentieth number. The probability that a Chinese athlete will pull it out is equal (since athletes are from China). Answer: .

. The student was asked to name a number from to . What is the probability that he will name a number that is a multiple of five?

Every fifth a number from the given set is divisible by . So the probability is .

A dice is thrown. Find the probability of getting an odd number of points.

Odd numbers; - even. The probability of an odd number of points is .

Answer: .

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

Note that the problem can be formulated differently: three coins are tossed at the same time. It won't affect the decision.

How many possible outcomes do you think there are?

We toss a coin. This action has two possible outcomes: heads and tails

Two coins - already four outcomes:

Three coins? That's right, outcomes, since .

Two heads and one tail come up three times out of eight.

Answer: .

. In a random experiment, two dice are thrown. Find the probability that the sum will drop points. Round the result to the nearest hundredth.

Throw the first die - six outcomes. And for each of them, six more are possible - when we roll the second die.

We get that this action- throwing two dice - the total number of possible outcomes, since .

And now for the good news:

The probability of getting eight points is .

>. The shooter hits the target with probability . Find the probability that he hits the target four times in a row.

If the probability of hitting is equal, then the probability of missing is . We argue in the same way as in the previous problem. The probability of two hits in a row is . And the probability of four hits in a row is equal to .

Probability: brute-force logic.

Here is a task from the diagnostic work, which seemed difficult to many.

Petya had ruble coins and ruble coins in his pocket. Petya, without looking, shifted some coins into another pocket. Find the probability that five-ruble coins are now in different pockets.

We know that the probability of an event is equal to the ratio of the number of favorable outcomes to the total number of outcomes. But how to calculate all these outcomes?

You can, of course, denote five-ruble coins by numbers , and ten-ruble coins by numbers - and then calculate in how many ways you can choose three elements from the set .

However, there is an easier solution:

We encode coins with numbers:, (these are five-ruble), (these are ten-ruble). The condition of the problem can now be formulated as follows:

There are six chips numbered from to . In how many ways can they be distributed equally among two pockets so that the chips with numbers and do not end up together?

Let's write down what we have in the first pocket.

To do this, we will compose all possible combinations from the set . A set of three chips will be a three-digit number. It is obvious that under our conditions and are the same set of tokens. In order not to miss anything and not to repeat, we arrange the corresponding three-digit numbers in ascending order:

All! We have tried all possible combinations starting with . We continue:

total possible outcomes.

We have a condition - chips with numbers and should not be together. This means, for example, that the combination does not suit us - it means that the chips and both ended up in not the first, but the second pocket. Favorable outcomes for us are those where there is either only , or only . Here they are:

134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256 - total favorable outcomes.

Then the required probability is .

What tasks await you at the exam in mathematics?

Let's analyze one of the most difficult problems in probability theory.

In order to enter the institute for the specialty "Linguistics", the applicant Z. must score at least 70 points on the Unified State Examination in each of the three subjects - mathematics, Russian language and a foreign language. To enroll in the specialty "Commerce", you need to score at least 70 points in each of the three subjects - mathematics, Russian language and social studies.

The probability that the applicant Z. will receive at least 70 points in mathematics is 0.6, in the Russian language - 0.8, in foreign language- 0.7 and in social studies - 0.5.
Find the probability that Z. will be able to enter at least one of the two specialties mentioned.

Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at the same time and receive two diplomas. Here we need to find the probability that Z. will be able to enter at least one of these two specialties - that is, he will score the required number of points.
In order to enter at least one of the two specialties, Z. must score at least 70 points in mathematics. And in Russian. And yet - social science or foreign.
The probability of scoring 70 points in mathematics for him is 0.6.
The probability of scoring points in mathematics and Russian is 0.6 0.8.

Let's deal with foreign and social studies. The options are suitable for us when the applicant scored points in social studies, in a foreign language, or in both. The option is not suitable when he did not score points either in language or in “society”. This means that the probability of passing social studies or a foreign one by at least 70 points is equal to
1 – 0,5 0,3.
As a result, the probability of passing mathematics, Russian and social studies or a foreign one is equal to
0.6 0.8 (1 - 0.5 0.3) = 0.408. This is the answer.

Plan for a workshop for teachers of mathematics of the educational institution of the city of Tula on the topic “Solving USE tasks in mathematics from the sections: combinatorics, probability theory. Teaching Methods»

Time spending: 12 00 ; 15 00

Location: MBOU "Lyceum No. 1", room. No. 8

I. Problem solving for probability

1. Solving problems on the classical definition of probability

We, as teachers, already know that the main types of tasks in the USE in probability theory are based on the classical definition of probability. Recall what is called the probability of an event?

Probability of an event is the ratio of the number of outcomes that favor a given event to the total number of outcomes.

In our scientific and methodological association of teachers of mathematics, a general scheme for solving problems on probability has been developed. I would like to present it to your attention. By the way, we shared our work experience, and in the materials that we gave to your attention for a joint discussion of solving problems, we gave this scheme. However, I want to voice it.

In our opinion, this scheme helps to quickly logically put everything on the shelves, and after that the task can be solved much easier for both the teacher and the students.

So, I want to analyze in detail the problem of the following content.

I wanted to talk with you in order to explain the methodology of how to convey such a solution to the guys, during which the guys would understand this typical task, and later they would understand these tasks themselves.

What is a random experiment in this problem? Now we need to isolate the elementary event in this experiment. What is this elementary event? Let's list them.

Issue questions?

Dear colleagues, you, too, have obviously considered probability problems with dice. I think we need to disassemble it, because there are some nuances. Let's analyze this problem according to the scheme that we proposed to you. Since there is a number from 1 to 6 on each face of the cube, the elementary events are the numbers 1, 2, 3, 4, 5, 6. We found that the total number of elementary events is 6. Let us determine which elementary events favor the event. Only two events favor this event - 5 and 6 (since it follows from the condition that 5 and 6 points should fall out).

Explain that all elementary events are equally possible. What will be the questions on the task?

How do you understand that the coin is symmetrical? Let's get this straight, sometimes certain phrases cause misunderstandings. Let's understand this problem conceptually. Let's deal with you in that experiment, which is described, what elementary outcomes can be. Can you imagine where is the head, where is the tail? What are the fallout options? Are there other events? What is the total number of events? According to the problem, it is known that the heads fell out exactly once. So this eventelementary events from these four OR and RO favor, this cannot happen twice already. We use the formula by which the probability of an event is found. Recall that the answers in Part B must be either an integer or a decimal.

Show on the interactive whiteboard. We read the task. What is the elementary outcome in this experience? Clarify that the pair is ordered - that is, the number fell on the first die, and on the second die. In any task there are times when you need to choose rational methods, forms and present the solution in the form of tables, diagrams, etc. In this problem, it is convenient to use such a table. I give you a ready-made solution, but during the solution it turns out that in this problem it is rational to use the solution in the form of a table. Explain what the table means. You understand why the columns say 1, 2, 3, 4, 5, 6.

Let's draw a square. The lines correspond to the results of the first roll - there are six of them, because the die has six faces. As are the columns. In each cell we write the sum of the dropped points. Show the completed table. Let's color the cells where the sum is equal to eight (as it is required in the condition).

I believe that the next problem, after analyzing the previous ones, can be given to the guys to solve on their own.

In the following problems, there is no need to write down all the elementary outcomes. It is enough just to count their number.

(Without solution) I gave the guys to solve this problem on their own. Algorithm for solving the problem

1. Determine what a random experiment is and what is a random event.

2. Find the total number of elementary events.

3. We find the number of events that favor the event specified in the condition of the problem.

4. Find the probability of an event using the formula.

Students can be asked a question, if 1000 batteries went on sale, and among them 6 are faulty, then the selected battery is determined as? What is it in our task? Next, I ask a question about finding what is used here as a numberand I propose to find itnumber. Then I ask, what is the event here? How many accumulators favor the completion of the event? Next, using the formula, we calculate this probability.

Here the children can be offered a second solution. Let's discuss what this method can be?

1. What event can be considered now?

2. How to find the probability of a given event?

The children need to be told about these formulas. They are next

The eighth task can be offered to the children on their own, since it is similar to the sixth task. It can be offered to them as independent work, or on a card at the board.

This problem can be solved in relation to the Olympiad, which is currently taking place. Despite the fact that different events participate in the tasks, however, the tasks are typical.

2. The simplest rules and formulas for calculating probabilities (opposite events, sum of events, product of events)

This is a task from USE collection. We put the solution on the board. What questions should we put before the students in order to analyze this problem.

1. How many machine guns were there? Once two automata, then there are already two events. I ask the children what the event will be? What will be the second event?

2. is the probability of the event. We do not need to calculate it, since it is given in the condition. According to the condition of the problem, the probability that "coffee runs out in both machines" is 0.12. There was an event A, there was an event B. And a new event appears? I ask the children the question - what? This is an event when both vending machines run out of coffee. In this case, in the theory of probability, this is a new event, which is called the intersection of two events A and B and is denoted in this way.

Let's use the probability addition formula. The formula is as follows

We give it to you in the reference material and the guys can give this formula. It allows you to find the probability of the sum of events. We were asked the probability of the opposite event, the probability of which is found by the formula.

Problem 13 uses the concept of a product of events, the formula for finding the probability of which is given in the Appendix.

3. Tasks for the use of the tree options

According to the condition of the problem, it is easy to draw up a diagram and find the indicated probabilities.

With the help of what theoretical material did you analyze the solution of problems of this kind with students? Did you use a tree of possibilities or did you use other methods for solving such problems? Did you give the concept of graphs? In the fifth or sixth grade, the guys have such problems, the analysis of which gives the concept of graphs.

I would like to ask you, have you and your students considered using a tree of possibilities when solving probability problems? The fact is that not only do the USE have such tasks, but rather complex tasks have appeared, which we will now solve.

Let's discuss with you the methodology for solving such problems - if it coincides with my methodology, as I explain to the guys, then it will be easier for me to work with you, if not, then I will help you deal with this problem.

Let's discuss the events. What events in problem 17 can be identified?

When constructing a tree on a plane, a point is designated, which is called the root of the tree. Next, we begin to consider the eventsand. We will construct a segment (in probability theory it is called a branch). According to the condition, it says that the first factory produces 30% of mobile phones of this brand (what? The one that they produce), so at the moment I am asking students what is the probability of the first factory producing phones of this brand, those that they produce? Since the event is the release of the phone at the first factory, the probability of this event is 30% or 0.3. The remaining phones are produced at the second factory - we are building the second segment, and the probability of this event is 0.7.

Students are asked the question - what type of phone can be produced by the first factory? With or without defect. What is the probability that the phone produced by the first factory has a defect? According to the condition, it is said that it is equal to 0.01. Question: What is the probability that the phone produced by the first factory does not have a defect? Since this event is opposite to the given one, its probability is equal.

It is required to find the probability that the phone is defective. It may be from the first factory, or it may be from the second. Then we use the formula for adding probabilities and we get that the whole probability is the sum of the probabilities that the phone is defective from the first factory, and that the phone is defective from the second factory. The probability that the phone has a defect and was produced at the first factory is found by the formula for the product of probabilities, which is given in the appendix.

4. One of the most difficult tasks from the USE bank for probability

Let's analyze, for example, No. 320199 from the FIPI Task Bank. This is one of the most difficult tasks in B6.

In order to enter the institute for the specialty "Linguistics", the applicant Z. must score at least 70 points on the Unified State Examination in each of the three subjects - mathematics, Russian and a foreign language. To enter the specialty "Commerce", you need to score at least 70 points in each of the three subjects - mathematics, Russian language and social studies.

The probability that applicant Z. will receive at least 70 points in mathematics is 0.6, in Russian - 0.8, in a foreign language - 0.7 and in social studies - 0.5.

Find the probability that Z. will be able to enter at least one of the two specialties mentioned.

Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at the same time and receive two diplomas. Here we need to find the probability that Z. will be able to enter at least one of these two specialties - that is, he will score the required number of points.

In order to enter at least one of the two specialties, Z. must score at least 70 points in mathematics. And in Russian. And yet - social science or foreign.

The probability of scoring 70 points in mathematics for him is 0.6.

The probability of scoring points in mathematics and Russian is equal.

Let's deal with foreign and social studies. The options are suitable for us when the applicant scored points in social studies, in a foreign language, or in both. The option is not suitable when he did not score points either in language or in “society”. This means that the probability of passing social studies or a foreign one is at least 70 points equal. As a result, the probability of passing mathematics, Russian and social studies or a foreign one is equal to

This is the answer.

II . Solving combinatorial problems

1. Number of combinations and factorials

Let's briefly analyze the theoretical material.

Expressionn ! is read as "en-factorial" and denotes the product of all natural numbers from 1 ton inclusive:n ! = 1 2 3 ...n .

In addition, in mathematics, by definition, it is considered that 0! = 1. Such an expression is rare, but still occurs in problems in probability theory.

Definition

Let there be objects (pencils, sweets, whatever) from which it is required to choose exactly different objects. Then the number of options for such a choice is callednumber of combinations from the elements. This number is indicated and calculated according to a special formula.

Designation

What does this formula give us? In fact, almost no serious task can be solved without it.

For a better understanding, let's analyze a few simple combinatorial problems:

A task

The bartender has 6 varieties of green tea. For the tea ceremony, exactly 3 different varieties of green tea are required. In how many ways can a bartender complete an order?

Solution

Everything is simple here: there isn = 6 varieties to choose fromk = 3 varieties. The number of combinations can be found by the formula:

Answer

Substitute in the formula. We cannot solve all the tasks, but we have written out typical tasks, they are presented to your attention.

A task

In a group of 20 students, 2 representatives must be selected to speak at the conference. In how many ways can this be done?

Solution

Again, all we haven = 20 students, but you have to choosek = 2 students. Finding the number of combinations:

Please note that the factors included in different factorials are marked in red. These multipliers can be painlessly reduced and thereby significantly reduce the total amount of calculations.

Answer

190

A task

17 servers with various defects were brought to the warehouse, which cost 2 times cheaper than normal servers. The director bought 14 such servers for the school, and spent the saved money in the amount of 200,000 rubles on the purchase of other equipment. In how many ways can a director choose defective servers?

Solution

There is quite a lot of extra data in the task, which can be confusing. The most important facts: everything isn = 17 servers, and the director needsk = 14 servers. We count the number of combinations:

The red color again indicates the multipliers that are being reduced. In total, it turned out 680 combinations. In general, the director has plenty to choose from.

Answer

680

This task is capricious, as there is extra data in this task. They lead many students astray. There were 17 servers in total, and the director needed to choose 14. Substituting into the formula, we get 680 combinations.

2. Law of multiplication

Definition

multiplication law in combinatorics: the number of combinations (ways, combinations) in independent sets is multiplied.

In other words, let there beA ways to perform one action andB ways to perform another action. The path also these actions are independent, i.e. not related in any way. Then you can find the number of ways to perform the first and second action by the formula:C = A · B .

A task

Petya has 4 coins of 1 ruble each and 2 coins of 10 rubles each. Petya, without looking, took out of his pocket 1 coin with a face value of 1 ruble and another 1 coin with a face value of 10 rubles to buy a pen for 11 rubles. In how many ways can he choose these coins?

Solution

So, first Petya getsk = 1 coin fromn = 4 available coins with a face value of 1 ruble. The number of ways to do this isC 4 1 = ... = 4.

Then Petya reaches into his pocket again and takes outk = 1 coin fromn = 2 available coins with a face value of 10 rubles. Here the number of combinations isC 2 1 = ... = 2.

Since these actions are independent, the total number of options isC = 4 2 = 8.

Answer

A task

There are 8 white balls and 12 black ones in a basket. In how many ways can you get 2 white balls and 2 black balls from this basket?

Solution

Total in cartn = 8 white balls to choose fromk = 2 balls. It can be doneC 8 2 = ... = 28 different ways.

In addition, the cart containsn = 12 black balls to choose from againk = 2 balls. The number of ways to do this isC 12 2 = ... = 66.

Since the choice of the white ball and the choice of the black one are independent events, the total number of combinations is calculated according to the multiplication law:C = 28 66 = 1848. As you can see, there can be quite a few options.

Answer

1848

The law of multiplication shows how many ways you can perform a complex action that consists of two or more simple ones - provided that they are all independent.

3. Law of addition

If the law of multiplication operates on "isolated" events that do not depend on each other, then in the law of addition the opposite is true. It deals with mutually exclusive events that never happen at the same time.

For example, “Peter took out 1 coin from his pocket” and “Peter did not take out a single coin from his pocket” are mutually exclusive events, since it is impossible to take out one coin without taking out any.

Similarly, the events "Randomly selected ball - white" and "Randomly selected ball - black" are also mutually exclusive.

Definition

Addition law in combinatorics: if two mutually exclusive actions can be performedA andB ways, respectively, these events can be combined. This will generate a new event that can be executedX = A + B ways.

In other words, when combining mutually exclusive actions (events, options), the number of their combinations is added up.

We can say that the law of addition is a logical "OR" in combinatorics, when any of the mutually exclusive options suits us. Conversely, the law of multiplication is a logical "AND", in which we are interested in the simultaneous execution of both the first and second actions.

A task

There are 9 black balls and 7 red balls in a basket. The boy takes out 2 balls of the same color. In how many ways can he do this?

Solution

If the balls are the same color, then there are few options: both of them are either black or red. Obviously, these options are mutually exclusive.

In the first case, the boy has to choosek = 2 black balls fromn = 9 available. The number of ways to do this isC 9 2 = ... = 36.

Similarly, in the second case we choosek = 2 red balls fromn = 7 possible. The number of ways isC 7 2 = ... = 21.

It remains to find the total number of ways. Since the variants with black and red balls are mutually exclusive, according to the law of addition we have:X = 36 + 21 = 57.

Answer57

A task

The stall sells 15 roses and 18 tulips. A 9th grade student wants to buy 3 flowers for his classmate, and all flowers must be the same. In how many ways can he make such a bouquet?

Solution

According to the condition, all flowers must be the same. So, we will buy either 3 roses or 3 tulips. Anyway,k = 3.

In the case of roses, you will have to choose fromn = 15 options, so the number of combinations isC 15 3 = ... = 455. For tulipsn = 18, and the number of combinations -C 18 3 = ... = 816.

Since roses and tulips are mutually exclusive options, we work according to the law of addition. Get the total number of optionsX = 455 + 816 = 1271. This is the answer.

Answer

1271

Additional terms and restrictions

Very often in the text of the problem there are additional conditions that impose significant restrictions on the combinations of interest to us. Compare two sentences:

There is a set of 5 pens in different colors. In how many ways can 3 stroke handles be selected?

There is a set of 5 pens in different colors. In how many ways can 3 stroke handles be chosen if one of them must be red?

In the first case, we have the right to take any colors that we like - there are no additional restrictions. In the second case, everything is more complicated, since we must choose a red handle (it is assumed that it is in the original set).

Obviously, any restrictions drastically reduce the total number of options. So how do you find the number of combinations in this case? Just remember the following rule:

Let there be a set ofn elements to choose fromk elements. With the introduction of additional restrictions on the numbern andk decrease by the same amount.

In other words, if you need to choose 3 out of 5 pens, and one of them must be red, then you will have to choose fromn = 5 − 1 = 4 elements byk = 3 − 1 = 2 elements. Thus, instead ofC 5 3 must be consideredC 4 2 .

Now let's see how this rule works on specific examples:

A task

In a group of 20 students, including 2 excellent students, you need to choose 4 people to participate in the conference. In how many ways can these four be chosen if the excellent students must get to the conference?

Solution

So there is a group ofn = 20 students. But you just have to choosek = 4 of them. If there were no additional restrictions, then the number of options was equal to the number of combinationsC 20 4 .

However, we were given an additional condition: 2 excellent students must be among these four. Thus, according to the above rule, we reduce the numbersn andk by 2. We have:

Answer

153

A task

Petya has 8 coins in his pocket, of which 6 are ruble coins and 2 are 10 ruble coins. Petya shifts some three coins into another pocket. In how many ways can Petya do this if it is known that both 10-ruble coins ended up in another pocket?

Solution

So there isn = 8 coins. Petya shiftsk = 3 coins, of which 2 are ten rubles. It turns out that out of 3 coins that will be transferred, 2 are already fixed, so the numbersn andk must be reduced by 2. We have:

Answer

III . Solving combined problems on the use of formulas of combinatorics and probability theory

A task

Petya had 4 ruble coins and 2 2 ruble coins in his pocket. Petya, without looking, shifted some three coins into another pocket. Find the probability that both two-ruble coins are in the same pocket.

Solution

Suppose that both two-ruble coins really ended up in the same pocket, then 2 options are possible: either Petya did not shift them at all, or he shifted both at once.

In the first case, when two-ruble coins were not transferred, 3 ruble coins would have to be transferred. Since there are 4 such coins in total, the number of ways to do this is equal to the number of combinations of 4 by 3:C 4 3 .

In the second case, when both two-ruble coins have been transferred, one more ruble coin will have to be transferred. It must be chosen from 4 existing ones, and the number of ways to do this is equal to the number of combinations from 4 to 1:C 4 1 .

Now let's find the total number of ways to shift the coins. Since there are 4 + 2 = 6 coins in total, and only 3 of them need to be chosen, the total number of options is equal to the number of combinations from 6 to 3:C 6 3 .

It remains to find the probability:

Answer

0,4

Show on the interactive whiteboard. Pay attention to the fact that, according to the condition of the problem, Petya, without looking, shifted three coins into one pocket. In answering this question, we can assume that two two-ruble coins really remained in one pocket. Refer to the formula for adding probabilities. Show the formula again.

A task

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

Solution

In order for five-ruble coins to lie in different pockets, you need to shift only one of them. The number of ways to do this is equal to the number of combinations of 2 by 1:C 2 1 .

Since Petya transferred 3 coins in total, he will have to transfer 2 more coins of 10 rubles each. Petya has 4 such coins, so the number of ways is equal to the number of combinations from 4 to 2:C 4 2 .

It remains to find how many options there are to shift 3 coins out of 6 available. This number, as in the previous problem, is equal to the number of combinations from 6 to 3:C 6 3 .

Finding the probability:

In the last step, we multiplied the number of ways to choose two-ruble coins and the number of ways to choose ten-ruble coins, since these events are independent.

Answer

0,6

So, problems with coins have their own probability formula. It is so simple and important that it can be formulated as a theorem.

Theorem

Let the coin be tossedn once. Then the probability that heads will land exactlyk times can be found using the formula:

WhereC n k - number of combinations ofn elements byk , which is calculated by the formula:

Thus, to solve the problem with coins, two numbers are needed: the number of tosses and the number of heads. Most often, these numbers are given directly in the text of the problem. Moreover, it does not matter what exactly to count: tails or eagles. The answer will be the same.

At first glance, the theorem seems too cumbersome. But it's worth a little practice - and you no longer want to return to the standard algorithm described above.

The coin is tossed four times. Find the probability that heads will come up exactly three times.

Solution

According to the condition of the problem, the total number of throws wasn = 4. Required number of heads:k = 3. Substituten andk into the formula:

With the same success, you can count the number of tails:k = 4 − 3 = 1. The answer will be the same.

Answer

0,25

A task [ Workbook“USE 2012 in mathematics. Tasks B6»]

The coin is tossed three times. Find the probability that it never comes up tails.

Solution

Writing out the numbers againn andk . Since the coin is tossed 3 times,n = 3. And since there should be no tails,k = 0. It remains to substitute the numbersn andk into the formula:

Let me remind you that 0! = 1 by definition. That's whyC 3 0 = 1.

Answer

0,125

Task [Trial exam in mathematics 2012. Irkutsk]

In a random experiment, a symmetrical coin is tossed 4 times. Find the probability that heads will come up more times than tails.

Solution

In order for there to be more heads than tails, they must fall out either 3 times (then there will be 1 tails) or 4 (then there will be no tails at all). Let's find the probability of each of these events.

Letp 1 - the probability that heads will fall out 3 times. Thenn = 4, k = 3. We have:

Now let's findp 2 - the probability that heads will fall out all 4 times. In this casen = 4, k = 4. We have:

To get the answer, it remains to add the probabilitiesp 1 andp 2 . Remember: you can only add probabilities for mutually exclusive events. We have:

p = p 1 + p 2 = 0,25 + 0,0625 = 0,3125

Answer

0,3125

In order to save your time when preparing with the guys for the Unified State Exam and the GIA, we have presented solutions to many more tasks that you can choose and solve with the guys.

Materials of the GIA, Unified State Examination of various years, textbooks and sites.

IV. Reference material

Presented to date in the open bank of USE problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is a classical definition of probability.

The easiest way to understand the formula is with examples.
Example 1 There are 9 red balls and 3 blue ones in the basket. The balls differ only in color. At random (without looking) we get one of them. What is the probability that the ball chosen in this way will be blue?

Comment. In problems in probability theory, something happens (in this case, our action of pulling the ball) that can have a different result - an outcome. It should be noted that the result can be viewed in different ways. "We pulled out a ball" is also a result. "We pulled out the blue ball" is the result. "We drew this particular ball out of all possible balls" - this least generalized view of the result is called the elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability.

Solution. Now we calculate the probability of choosing a blue ball.
Event A: "the chosen ball turned out to be blue"
Total number all possible outcomes: 9+3=12 (number of all balls we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25

Let us calculate for the same problem the probability of choosing a red ball.
The total number of possible outcomes will remain the same, 12. The number of favorable outcomes: 9. The desired probability: 9/12=3/4=0.75

The probability of any event always lies between 0 and 1.
Sometimes in everyday speech (but not in probability theory!) The probability of events is estimated as a percentage. The transition between mathematical and conversational assessment is done by multiplying (or dividing) by 100%.
So,
In this case, the probability is zero for events that cannot happen - improbable. For example, in our example, this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0 if counted according to the formula)
Probability 1 has events that will absolutely definitely happen, without options. For example, the probability that "the chosen ball will be either red or blue" is for our problem. (Number of favorable outcomes: 12, P(A)=12/12=1)

We've looked at a classic example that illustrates the definition of probability. All similar USE problems in probability theory are solved using this formula.
Instead of red and blue balls, there can be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on a certain topic (prototypes , ), defective and high-quality bags or garden pumps (prototypes , ) - the principle remains the same.

They differ slightly in the formulation of the problem of the USE probability theory, where you need to calculate the probability of an event occurring on a certain day. ( , ) As in the previous tasks, you need to determine what is an elementary outcome, and then apply the same formula.

Example 2 The conference lasts three days. On the first and second days, 15 speakers each, on the third day - 20. What is the probability that the report of Professor M. will fall on the third day, if the order of the reports is determined by lottery?

What is the elementary outcome here? - Assigning a professor's report to one of all possible serial numbers for a speech. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report can receive one of 50 numbers. This means that there are only 50 elementary outcomes.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, the probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4

The drawing of lots here is the establishment of a random correspondence between people and ordered places. In Example 2, matching was considered in terms of which of the places a particular person could take. You can approach the same situation from the other side: which of the people with what probability could get to a particular place (prototypes , , , ):

Example 3 5 Germans, 8 Frenchmen and 3 Estonians participate in the draw. What is the probability that the first (/second/seventh/last - it doesn't matter) will be a Frenchman.

The number of elementary outcomes is the number of all possible people who could get to a given place by lot. 5+8+3=16 people.
Favorable outcomes - the French. 8 people.
Desired probability: 8/16=1/2=0.5
Answer: 0.5

The prototype is slightly different. There are tasks about coins () and dice () that are somewhat more creative. Solutions to these problems can be found on the prototype pages.

Here are some examples of coin tossing or dice tossing.

Example 4 When we toss a coin, what is the probability of getting tails?
Outcomes 2 - heads or tails. (it is believed that the coin never falls on the edge) Favorable outcome - tails, 1.
Probability 1/2=0.5
Answer: 0.5.

Example 5 What if we flip a coin twice? What is the probability that it will come up heads both times?
The main thing is to determine which elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP - both times it came up tails
2) PO - first time tails, second time heads
3) OP - the first time heads, the second time tails
4) OO - heads up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one is favorable, 1.
Probability: 1/4=0.25
Answer: 0.25

What is the probability that two tosses of a coin will land on tails?
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5

In such problems, another formula may come in handy.
If at one toss of a coin we have 2 possible outcomes, then for two tosses of results there will be 2 2=2 2 =4 (as in example 5), for three tosses 2 2 2=2 3 =8, for four: 2·2·2·2=2 4 =16, … for N throws of possible outcomes there will be 2·2·...·2=2 N .

So, you can find the probability of getting 5 tails out of 5 coin tosses.
The total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR - all 5 times tails)
Probability: 1/32=0.03125

The same is true for the dice. With one throw, there are 6 possible results. So, for two throws: 6 6=36, for three 6 6 6=216, etc.

Example 6 We throw a dice. What is the probability of getting an even number?

Total outcomes: 6, according to the number of faces.
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5

Example 7 Throw two dice. What is the probability that the total rolls 10? (round to hundredths)

There are 6 possible outcomes for one die. Hence, for two, according to the above rule, 6·6=36.
What outcomes will be favorable for a total of 10 to fall out?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. So, for cubes, options are possible:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
In total, 3 options. Desired probability: 3/36=1/12=0.08
Answer: 0.08

Other types of B6 problems will be discussed in one of the following "How to Solve" articles.

The probability of an event $A$ is the ratio of the number of outcomes favorable for $A$ to the number of all equally possible outcomes

$P(A)=(m)/(n)$, where $n$ is the total number of possible outcomes and $m$ is the number of outcomes favoring $A$.

The probability of an event is a number from the segment $$

The taxi company has $50$ of cars available. $35$ of them are black, the rest are yellow. Find the probability that a yellow car will arrive at a random call.

Find the number of yellow cars:

In total, there are $50$ cars, that is, one out of fifty will come to the call. There are $15$ of yellow cars, therefore, the probability of arrival of a yellow car is $(15)/(50)=(3)/(10)=0.3$

Answer:$0.3$

Opposite events

Two events are said to be opposite if in a given trial they are incompatible and one of them necessarily occurs. The probabilities of opposite events add up to 1. An event opposite to the event $A$ is written $((A))↖(-)$.

$P(A)+P((A))↖(-)=1$

Independent events

Two events $A$ and $B$ are called independent if the probability of occurrence of each of them does not depend on whether the other event occurred or not. Otherwise, the events are called dependent.

The probability of the product of two independent events $A$ and $B$ is equal to the product of these probabilities:

$P(A B)=P(A) P(B)$

Ivan Ivanovich bought two different lottery tickets. The probability that the first lottery ticket will win is $0.15$. The probability that the second lottery ticket will win is $0.12. Ivan Ivanovich participates in both draws. Assuming that the draws are held independently of each other, find the probability that Ivan Ivanovich wins in both draws.

Probability $P(A)$ - wins the first ticket.

Probability $P(B)$ - wins the second ticket.

The events $A$ and $B$ are independent events. That is, to find the probability that both events will occur, you need to find the product of the probabilities

$P(A B)=P(A) P(B)$

$P=0.15 0.12=0.018$

Answer: $0.018

Incompatible events

Two events $A$ and $B$ are said to be incompatible if there are no outcomes favoring both event $A$ and event $B$. (Events that cannot happen at the same time)

The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:

$P(A+B)=P(A)+P(B)$

In the algebra exam, the student gets one question out of all the exams. The probability that this is a question on the topic " Quadratic equations", is equal to $0.3$. The probability that this is an Irrational Equations question is $0.18$. There are no questions related to these two topics at the same time. Find the probability that the student will get a question on one of these two topics on the exam.

These events are called incompatible, since the student will get a question EITHER on the topic “Quadricular Equations”, OR on the topic “Irrational Equations”. Topics cannot be caught at the same time. The probability of the sum of two incompatible events $A$ and $B$ is equal to the sum of the probabilities of these events:

$P(A+B)=P(A)+P(B)$

$P \u003d 0.3 + 0.18 \u003d 0.48 $

Answer: $0.48

Joint Events

Two events are said to be joint if the occurrence of one of them does not exclude the occurrence of the other in the same trial. Otherwise, the events are called incompatible.

The probability of the sum of two joint events $A$ and $B$ is equal to the sum of the probabilities of these events minus the probability of their product:

$P(A+B)=P(A)+P(B)-P(A B)$

There are two identical coffee machines in the lobby of the cinema. The probability that the machine will run out of coffee by the end of the day is $0.6$. The probability that both machines run out of coffee is $0.32$. Find the probability that at least one of the vending machines will run out of coffee by the end of the day.

Let's denote the events, let:

$A$ = coffee will end in the first machine,

$B$ = coffee will end in the second machine.

$A B =$ coffee will run out in both vending machines,

$A + B =$ coffee will run out in at least one vending machine.

By convention, $P(A) = P(B) = 0.6; P(A B) = $0.32.

The events $A$ and $B$ are joint, the probability of the sum of two joint events is equal to the sum of the probabilities of these events, reduced by the probability of their product:

$P(A + B) = P(A) + P(B) − P(A B) = 0.6 + 0.6 − 0.32 = 0.88$