Probability theory on the Unified State Examination in mathematics can be presented both in the form of simple tasks on the classical definition of probability, and in the form of quite complex ones on the application of the corresponding theorems.

In this part, we will consider problems for which it is sufficient to use the definition of probability. Sometimes here we will also use a formula to calculate the probability of the opposite event. Although you can do without this formula here, you will still need it when solving the following problems.

Theoretical part

Random is an event that may or may not occur (impossible to predict in advance) during an observation or test.

Let there be equally possible outcomes when conducting a test (throwing a coin or a dice, drawing an exam card, etc.). For example, when tossing a coin, the number of all outcomes is 2, since there cannot be any other outcomes other than heads or tails. When throwing a die, 6 outcomes are possible, since any number from 1 to 6 is equally possible to appear on the top face of the die. Let also some event A be favored by outcomes.

The probability of event A is the ratio of the number of outcomes favorable for this event to the total number of equally possible outcomes (this is the classical definition of probability). We write

For example, let event A consist of getting an odd number of points when throwing a die. There are a total of 6 possible outcomes: 1, 2, 3, 4, 5, 6 appearing on the top face of the cube. In this case, outcomes with 1, 3, 5 appearing are favorable for event A. Thus, .

Note that the double inequality is always satisfied, therefore the probability of any event A lies on the interval, that is . If your answer has a probability greater than one, it means you made a mistake somewhere and the solution needs to be double-checked.

Events A and B are called opposite each other if any outcome is favorable for exactly one of them.

For example, when throwing a die, the event “an odd number is rolled” is the opposite of the event “an even number is rolled.”

The event opposite to event A is designated. From the definition of opposite events it follows
, Means,
.

Problems about selecting objects from a set

Task 1. There are 24 teams participating in the World Championship. Using lots, they need to be divided into four groups of six teams each. There are cards with group numbers mixed in the box:

1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4.

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

The total number of outcomes is equal to the number of cards - there are 24 of them. There are 6 favorable outcomes (since number 3 is written on six cards). The required probability is equal to .

Answer: 0.25.

Task 2. There are 14 red, 9 yellow and 7 green balls in an urn. One ball is drawn at random from the urn. What is the probability that this ball will be yellow?

The total number of outcomes is equal to the number of balls: 14 + 9 + 7 = 30. The number of outcomes favorable for this event is 9. The required probability is equal to .

Task 3. There are 10 numbers on the phone keypad, from 0 to 9. What is the probability that a randomly pressed number will be even and greater than 5?

The outcome here is pressing a certain key, so there are a total of 10 equally possible outcomes. The specified event is favored by outcomes that mean pressing key 6 or 8. There are two such outcomes. The required probability is equal to .

Answer: 0.2.

Problem 4. What is the probability that a randomly selected natural number from 4 to 23 is divisible by three?

On the segment from 4 to 23 there are 23 – 4 + 1 = 20 natural numbers, which means there are a total of 20 possible outcomes. On this segment, the following numbers are multiples of three: 6, 9, 12, 15, 18, 21. There are 6 such numbers in total, so the event in question is favored by 6 outcomes. The required probability is equal to .

Answer: 0.3.

Task 5. Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will not be able to answer the ticket chosen at random?

1st method.

Since a student can answer 17 tickets, he cannot answer 3 tickets. The probability of getting one of these tickets is by definition equal to .

2nd method.

Let us denote by A the event “the student can answer the ticket.” Then . The probability of the opposite event is =1 – 0.85 = 0.15.

Answer: 0.15.

Problem 6. 20 athletes are participating in the rhythmic gymnastics championship: 6 from Russia, 5 from Germany, the rest from France. The order in which the gymnasts perform is determined by lot. Find the probability that the athlete competing seventh is from France.

There are 20 athletes in total, everyone has an equal chance to compete seventh. Therefore, there are 20 equally probable outcomes. There are 20 – 6 – 5 = 9 athletes from France, so there are 9 favorable outcomes for the specified event. The required probability is equal to .

Answer: 0.45.

Task 7. The scientific conference is held over 5 days. A total of 50 reports are planned - the first three days have 12 reports each, the rest are distributed equally between the fourth and fifth days. The order of reports is determined by drawing lots. What is the probability that Professor N.’s report will be scheduled for the last day of the conference?

First, let's find how many reports are scheduled for the last day. Presentations are scheduled for the first three days. There are still 50 – 36 = 14 reports left, which are distributed equally between the remaining two days, so there are reports scheduled on the last day.

We will consider the outcome to be the serial number of Professor N.’s report. There are 50 such equally possible outcomes. There are 7 outcomes that favor the specified event (the last 7 numbers in the list of reports). The required probability is equal to .

Answer: 0.14.

Problem 8. There are 10 seats on board the aircraft next to the emergency exits and 15 seats behind the partitions separating the cabins. The remaining seats are inconvenient for tall passengers. Passenger K. is tall. Find the probability that at check-in, if a seat is randomly selected, passenger K will get a comfortable seat if there are 200 seats on the plane.

The outcome in this task is the choice of location. There are a total of 200 equally possible outcomes. The event “the chosen place is convenient” is favored by 15 + 10 = 25 outcomes. The required probability is equal to .

Answer: 0.125.

Problem 9. Out of 1000 coffee grinders assembled at the plant, 7 were defective. An expert tests one coffee grinder chosen at random from these 1000. Find the probability that the coffee grinder being tested will be defective.

When choosing a coffee grinder at random, 1000 outcomes are possible; for event A “the selected coffee grinder is defective,” 7 outcomes are favorable. By definition of probability.

Answer: 0.007.

Problem 10. The plant produces refrigerators. On average, for every 100 high-quality refrigerators, there are 15 refrigerators with hidden defects. Find the probability that the purchased refrigerator will be of high quality. Round the result to hundredths.

This task is similar to the previous one. However, the formulation “for 100 high-quality refrigerators, there are 15 with defects” indicates to us that 15 defective pieces are not included in the 100 quality ones. Therefore, the total number of outcomes is 100 + 15 = 115 (equal to the total number of refrigerators), there are 100 favorable outcomes. The required probability is equal to . To calculate the approximate value of a fraction, it is convenient to use angle division. We get 0.869... which is 0.87.

Answer: 0.87.

Problem 11. Before the start of the first round of the tennis championship, participants are randomly divided into playing pairs using lots. In total, 16 tennis players are participating in the championship, including 7 participants from Russia, including Maxim Zaitsev. Find the probability that in the first round Maxim Zaitsev will play with any tennis player from Russia.

As in the previous task, you need to carefully read the condition and understand what is an outcome and what is a favorable outcome (for example, thoughtless application of the probability formula leads to an incorrect answer).

Here the outcome is the opponent of Maxim Zaitsev. Since there are 16 tennis players in total, and Maxim cannot play against himself, there are 16 – 1 = 15 equally probable outcomes. A favorable outcome is an opponent from Russia. There are 7 – 1 = 6 such favorable outcomes (we exclude Maxim himself from the number of Russians). The required probability is equal to .

Answer: 0.4.

Problem 12. The football section is attended by 33 people, among them two brothers - Anton and Dmitry. Those attending the section are randomly divided into three teams of 11 people each. Find the probability that Anton and Dmitry will be on the same team.

We will form teams, sequentially placing players in empty seats, starting with Anton and Dmitry. First, let's place Anton on a randomly selected place from the free 33. Now we place Dmitry on the free place (we will consider the choice of a place for him to be the outcome). There are 32 free places in total (Anton has already taken one), so there are 32 possible outcomes in total. There are 10 empty spots left on the same team as Anton, so the “Anton and Dmitry on the same team” event is favored by 10 outcomes. The probability of this event is .

Answer: 0.3125.

Problem 13. A mechanical watch with a twelve-hour dial broke down at some point and stopped running. Find the probability that the hour hand is frozen, reaching 11 o'clock, but not reaching 2 o'clock.

Conventionally, the dial can be divided into 12 sectors, located between the marks of adjacent numbers (between 12 and 1, 1 and 2, 2 and 3, ..., 11 and 12). We will consider the outcome to be the stop of the clock hand in one of the indicated sectors. There are a total of 12 equally possible outcomes. This event is favored by three outcomes (sectors between 11 and 12, 12 and 1, 1 and 2). The required probability is equal to .

Answer: 0.25.

Summarize

After studying the material on solving simple problems in probability theory, I recommend completing the tasks for independent solution, which we publish on our Telegram channel. You can also check whether they are completed correctly by entering your answers in the form provided.

Thank you for sharing the article on social networks.

Source “Preparation for the Unified State Exam. Mathematics. Probability Theory.” Edited by F.F. Lysenko, S.Yu. Kulabukhova

Plan for a workshop for mathematics teachers at the educational institution of the city of Tula on the topic “Solving Unified State Examination tasks in mathematics from the sections: combinatorics, probability theory. Teaching Methodology"

Time spending: 12 00 ; 15 00

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

I. Solving probability problems

1. Solving problems involving the classical determination of probability

We, as teachers, already know that the main types of problems in the Unified State Exam in probability theory are based on the classical definition of probability. Let's remember what is called the probability of an event?

Probability of the event is the ratio of the number of outcomes favorable to a given event to the total number of outcomes.

Our scientific and methodological association of mathematics teachers has developed a general scheme for solving probability problems. 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 joint discussion of problem solving, we gave this diagram. However, I want to voice it.

In our opinion, this scheme helps to quickly logically sort everything into pieces, and after that the problem can be solved much easier for both the teacher and the students.

So, I want to analyze in detail the following task.

I wanted to talk with you together to explain the methodology, how to convey to the guys such a solution, during which the children would understand this typical problem, and subsequently they would understand these problems themselves.

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

Questions about the task?

Dear colleagues, you, too, have obviously considered probability problems with dice. I think we need to analyze it, because it has its own nuances. Let's analyze this problem according to the scheme that we proposed to you. Since on each side of the cube there is a number from 1 to 6, then 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 questions will there be about the task?

How do you know that a coin is symmetrical? Let's get this straight, sometimes certain phrases cause misunderstandings. Let's understand this problem conceptually. Let's figure out with you in the experiment that is described what the elementary outcomes could be. Do you all have any idea where is heads and where is tails? What are the possible dropout options? Are there other events? What is the total number of events? According to the problem, it is known that heads came up exactly once. This means that this eventelementary events from these four ORs and ROs are favorable; this cannot happen twice. We use the formula that calculates the probability of an event. As a reminder, answers in Part B must be either a whole number or a decimal.

We show it on the interactive board. We read the problem. 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 problem there are moments 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 am giving you a ready-made solution, but during the solution it turns out that in this problem it is rational to use a solution in the form of a table. We explain what the table means. You can 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 throw - there are six of them, because the die has six sides. So are the columns. In each cell we write the sum of the points drawn. We show the completed table. Let's color the cells where the sum is equal to eight (as this is required in the condition).

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

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

(No solution) I gave this problem to the guys to solve on their own. Algorithm for solving the problem

1. Define what a random experiment consists of and what is a random event.

2. Find the total number of elementary events.

3. Find the number of events favorable to the event specified in the problem statement.

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

Students can be asked a question: if 1000 batteries go on sale, and among them 6 are faulty, then the selected battery is determined by how? What is it in our task? Next I ask the question of finding what is being used as a number hereand I suggest you find itnumber. Next I ask, what is the event here? How many accumulators contribute to the event? Next, using the formula, we calculate this probability.

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

1. What event can we consider now?

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

The guys need to be told about these formulas. They are as follows

The eighth problem can be offered to the children on their own, since it is similar to the sixth problem. 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 are involved in the tasks, 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 the Unified State Exam collection. We display the solution on the board. What questions should we ask students to understand this problem?

1. How many machines were there? If there are two machines, then there are already two events. I ask the children a question - what will the event be like?? What will be the second event?

2. is the probability of an event. We don't need to calculate it, since it is given in the condition. According to the conditions of the problem, the probability that “the coffee will run out in both machines” is 0.12. There was event A, there was event B. And a new event appears? I ask the children a question - which one? This is the event when both machines run out of coffee. In this case, in probability theory, this is a new event, which is called the intersection of two events A and B and is designated 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 be given this formula. It allows you to find the probability of a sum of events. We were asked the probability of the opposite event, the probability of which is found using 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. Problems involving the use of a tree of possible options

Based on the conditions of the problem, it is easy to draw up a diagram and find the indicated probabilities.

What theoretical material did you use to help students solve problems of this kind? Have you used a possible tree or other methods to solve such problems? Have you given the concept of graphs? In the fifth or sixth grade, children 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 possible options when solving probability problems? The fact is that not only does the Unified State Exam have such tasks, but quite complex problems have appeared that 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 us discuss the events. What events in problem 17 can be isolated?

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 is said that the first factory produces 30% of mobile phones of this brand (which one? The one they produce), which means that at the moment I am asking students, what is the probability of the first factory producing phones of this brand, the ones they produce? Since the event is the release of a phone at the first factory, the probability of this event is 30% or 0.3. The remaining phones were 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 could be produced by the first factory? With or without defect. What is the probability that a phone produced by the first factory has a defect? The condition says 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.

You need to find the probability that the phone is defective. It could be from the first factory, or maybe from the second. Then we use the formula for adding probabilities and find that the entire probability is the sum of the probabilities that the phone with a defect is from the first factory, and that the phone with a defect is from the second factory. We will find the probability that the phone has a defect and was produced at the first factory using the product of probabilities formula, which is given in the appendix.

4. One of the most difficult problems from the Unified State Exam bank on probability

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

To enter the institute for the specialty "Linguistics", applicant Z. must score at least 70 points on the Unified State Examination in each of 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 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 enroll in at least one of the two mentioned specialties.

Note that the problem does not ask whether an applicant named Z. will study both linguistics and commerce at once and receive two diplomas. Here we need to find the probability that Z. will be able to enroll in 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 also - social studies or foreign.

The probability for him to score 70 points in mathematics is 0.6.

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

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

This is the answer.

II . Solving combinatorial problems

1. Number of combinations and factorials

Let's briefly look at the theoretical material.

Expressionn ! 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, they believe that 0! = 1. Such an expression is rare, but still occurs in problems in probability theory.

Definition

Let there be objects (pencils, candies, whatever) from which you want to select exactly different objects. Then the number of options for such a choice is callednumber of combinations from elements by. This number is designated and calculated using a special formula.

Designation

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

For a better understanding, let’s look at a few simple combinatorial problems:

Task

The bartender has 6 types of green tea. To conduct a tea ceremony, you need to serve exactly 3 different types of green tea. In how many ways can the bartender fill an order?

Solution

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

Answer

Substitute into the formula. We cannot solve all problems, but we have written down typical problems and they are presented to your attention.

Task

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

Solution

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

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

Answer

190

Task

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

Solution

The problem contains quite a lot of extra data that can be confusing. The most important facts: there are onlyn = 17 servers, and the director needsk = 14 servers. We count the number of combinations:

Multipliers that are being reduced are again indicated in red. In total, there were 680 combinations. In general, the director has plenty to choose from.

Answer

680

This task is tricky because there is extra data in this task. They lead many students astray from making the right decision. 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

Law of multiplication 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 is also that these actions are independent, i.e. are not related to each other in any way. Then you can find the number of ways to perform the first and second actions using the formula:C = A · B .

Task

Petya has 4 coins of 1 ruble and 2 coins of 10 rubles. Petya, without looking, took from 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 is equal toC 2 1 = ... = 2.

Since these actions are independent, the total number of options is equal toC = 4 · 2 = 8.

Answer

Task

There are 8 white balls and 12 black balls 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, from which you must choose againk = 2 balls. The number of ways to do this isC 12 2 = ... = 66.

Since the choice of a white ball and the choice of a black ball 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 lot of options.

Answer

1848

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

3. Law of addition

If the law of multiplication operates with “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, “Petya took 1 coin out of his pocket” and “Petya 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.

Likewise, the events “Ball at random is white” and “Ball at random is black” are also mutually exclusive.

Definition

Law of addition in combinatorics: if two mutually exclusive actions can be performedA AndB methods accordingly, then these events can be combined. This will create a new event that you can executeX = A + B ways.

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

We can say that the law of addition is a logical “OR” in combinatorics, when we are satisfied with any of the mutually exclusive options. 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.

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: they are both 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 is equalC 7 2 = ... = 21.

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

Answer57

Task

The stall sells 15 roses and 18 tulips. A 9th grade student wants to buy 3 flowers for his classmate, and all the 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. This means 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 isC 18 3 = ... = 816.

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

Answer

1271

Additional terms and restrictions

Very often, the text of the problem contains 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 you choose 3 pens to outline a drawing?

There is a set of 5 pens in different colors. In how many ways can you choose 3 pens for outlining a drawing if red must be among them?

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

Obviously, any restrictions sharply reduce the final number of options. Well, how can you find the number of combinations in this case? Just remember this rule:

Let there be a set ofn elements from which to choosek elements. When introducing additional restrictions on the numbern Andk decrease by the same amount.

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

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

Task

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

Solution

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

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

Answer

153

Task

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

Solution

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

Answer

III . Solving combined problems using formulas of combinatorics and probability theory

Task

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

Solution

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

In the first case, when two-ruble coins were not shifted, you will have to shift 3 ruble coins. 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, another 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 of 4 by 1:C 4 1 .

Now let's find the total number of ways to rearrange the coins. Since there are 4 + 2 = 6 coins in total, and you only need to choose 3 of them, the total number of options is equal to the number of combinations of 6 by 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 conditions of the problem, Petya, without looking, put three coins in one pocket. In answering this question, we can assume that two two-ruble coins actually remained in one pocket. Refer to the formula for adding probabilities. Show the formula again.

Task

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

Solution

To keep five-ruble coins in different pockets, you need to move 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 shifted 3 coins in total, he will have to shift 2 more coins of 10 rubles each. Petya has 4 such coins, so the number of ways is equal to the number of combinations of 4 by 2:C 4 2 .

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

We find 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, coin problems 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 coin problem, you need two numbers: 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 you count: tails or heads. The answer will be the same.

At first glance, the theorem seems too cumbersome. But once you practice a little, you will no longer want to return to the standard algorithm described above.

The coin is tossed four times. Find the probability of getting heads exactly three times.

Solution

According to the problem, the total throws weren = 4. Required number of eagles:k = 3. Substituten Andk into the formula:

You can just as easily count the number of heads:k = 4 − 3 = 1. The answer will be the same.

Answer

0,25

Task [Workbook “Unified State Exam 2012 in mathematics. Problems B6"]

The coin is tossed three times. Find the probability that you will never get heads.

Solution

Writing out the numbers againn Andk . Since the coin is tossed 3 times,n = 3. And since there shouldn’t be heads,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

Problem [Trial Unified State Exam in Mathematics 2012. Irkutsk]

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

Solution

For there to be more heads than tails, they must appear either 3 times (then there will be 1 tails) or 4 times (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 appear 3 times. Thenn = 4, k = 3. We have:

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

To get the answer, all that remains is to add up 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 State Examination, we have presented solutions to many more problems that you can choose and solve with the guys.

Materials from the State Examination Institute, Unified State Examination of various years, textbooks and websites.

IV. Reference material

Classic definition of probability

Random event – any event that may or may not occur as a result of some experience.

Probability of event R equal to the ratio of the number of favorable outcomes k to the number of possible outcomes n, i.e.

p=\frac(k)(n)

Formulas for addition and multiplication of probability theory

Event \bar(A) called opposite to event A, if event A did not occur.

Sum of probabilities of opposite events is equal to one, i.e.

P(\bar(A)) + P(A) =1

• The probability of an event cannot be greater than 1.
• If the probability of an event is 0, then it will not happen.
• If the probability of an event is 1, then it will happen.

Probability addition theorem:

“The probability of the sum of two incompatible events is equal to the sum of the probabilities of these events.”

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

Probability amounts two joint events equal to the sum of the probabilities of these events without taking into account their joint occurrence:

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

Probability multiplication theorem

“The probability of two events occurring is equal to the product of the probabilities of one of them and the conditional probability of the other, calculated under the condition that the first occurred.”

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

Events are called incompatible, if the appearance of one of them excludes the appearance of others. That is, only one specific event or another can happen.

Events are called joint, if the occurrence of one of them does not exclude the occurrence of the other.

Two random events A and B are called independent, if the occurrence of one of them does not change the probability of the occurrence of the other. Otherwise, events A and B are called dependent.

At a ceramic tile factory, 5% of the tiles produced have a defect. During product quality control, only 40% of defective tiles are detected. The remaining tiles are sent for sale. Find the probability that a tile chosen at random upon purchase will have no defects. Round your answer to the nearest hundredth.

Solution

During product quality control, 40% of defective tiles are identified, which account for 5% of the tiles produced, and they do not go on sale. This means that 0.4 · 5% = 2% of the tiles produced do not go on sale. The rest of the tiles produced - 100% - 2% = 98% - go on sale.

100% - 95% of produced tiles are free from defects. The probability that the purchased tile does not have a defect is 95%: 98% = \frac(95)(98)\approx 0.97

Answer

Condition

The probability that the battery is not charged is 0.15. A customer in a store purchases a random package that contains two of these batteries. Find the probability that both batteries in this package will be charged.

Solution

The probability that the battery is charged is 1-0.15 = 0.85. Let’s find the probability of the event “both batteries are charged.” Let us denote by A and B the events “the first battery is charged” and “the second battery is charged”. We got P(A) = P(B) = 0.85. The event “both batteries are charged” is the intersection of events A \cap B, its probability is equal to P(A\cap B) = P(A)\cdot P(B) = 0.85\cdot 0.85 = 0,7225.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The probability that a new washing machine will be repaired under warranty within a year is 0.065. In a certain city, 1,200 washing machines were sold during the year, of which 72 were delivered to the warranty workshop. Determine how different the relative frequency of the occurrence of the “warranty repair” event is from its probability in this city?

Solution

The frequency of the event “the washing machine will be repaired under warranty within a year” is equal to \frac(72)(1200) = 0.06. It differs from probability by 0.065-0.06=0.005.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The probability that the pen is defective is 0.05. A customer in a store purchases a random package that contains two pens. Find the probability that both pens in this package will be good.

Solution

The probability that the handle is working is 1-0.05 = 0.95. Let's find the probability of the event “both handles are working.” Let us denote by A and B the events “the first handle is working” and “the second handle is working”. We got P(A) = P(B) = 0.95. The event “both handles are working” is the intersection of events A\cap B, its probability is equal to P(A\cap B) = P(A)\cdot P(B) = 0.95\cdot 0.95 = 0,9025.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

The picture shows a labyrinth. The beetle crawls into the maze at the “Entrance” point. The beetle cannot turn around and crawl in the opposite direction, so at each fork it chooses one of the paths it has not been on yet. With what probability is the beetle coming to exit D if the choice of the further path is random?

Solution

Let's place arrows at intersections in the directions in which the beetle can move (see figure).

At each intersection we will choose one direction out of two possible ones and assume that when it gets to the intersection the beetle will move in the direction we have chosen.

In order for the beetle to reach exit D, it is necessary that at each intersection the direction indicated by the solid red line is chosen. In total, the choice of direction is made 4 times, each time regardless of the previous choice. The probability that the solid red arrow is selected each time is \frac12\cdot\frac12\cdot\frac12\cdot\frac12= 0,5^4= 0,0625.

Answer

Source: “Mathematics. Preparation for the Unified State Exam 2017. Profile level." Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Condition

There are 16 athletes in the section, among them two friends - Olya and Masha. Athletes are randomly assigned to 4 equal groups. Find the probability that Olya and Masha will end up in the same group.