Just not in an online translator.
The Golden Gate is a symbol of Kyiv, one of the oldest examples of architecture that has survived to this day. The Golden Gate of Kyiv was built under the famous Kiev prince Yaroslav the Wise in 1164. Initially they were called Southern and were part of the system of defensive fortifications of the city, practically no different from other guard gates of the city. It was the South Gate that the first Russian Metropolitan Hilarion called “Great” in his “Sermon on Law and Grace.” After the majestic Church of Hagia Sophia was built, the “Great” Gate became the main land entrance to Kyiv from the southwestern side. Realizing their significance, Yaroslav the Wise ordered the construction of a small Church of the Annunciation over the gates in order to pay tribute to the dominant Christian religion in the city and in Rus'. From that time on, all Russian chronicle sources began to call the Southern Gate of Kyiv the Golden Gate. The width of the gate was 7.5 m, the height of the passage was 12 m, and the length was about 25 m.
le sport ce n"est pas seulement des cours de gym. C"est aussi sauter toujours plus haut nager jouer au ballon danser. le sport développé ton corps et aussi ton cerveau. Quand tu prends l"escalier et non pas l"ascenseur tu fais du sport. Quand tu fais une cabane dans un arbre tu fais du sport. Quand tu te bats avec ton frere tu fais du sport. Quand tu cours, parce que tu es en retard a l"ecole, tu fais du sport.
An event is the result of a test. What is an event? One ball is taken at random from the urn. Retrieving a ball from an urn is a test. The appearance of a ball of a certain color is an event. In probability theory, an event is understood as something about which, after a certain point in time, one and only one of two things can be said. Yes, it happened. No, it didn't happen. A possible outcome of an experiment is called an elementary event, and a set of such outcomes is simply called an event.
Unpredictable events are called random. An event is called random if, under the same conditions, it may or may not occur. When rolling the dice, the result will be a six. I have a lottery ticket. After the results of the lottery are published, the event that interests me - winning a thousand rubles - either happens or does not happen. Example.
Two events that, under given conditions, can occur simultaneously are called joint, and those that cannot occur simultaneously are called incompatible. A coin is tossed. The appearance of the “coat of arms” excludes the appearance of the inscription. The events “a coat of arms appeared” and “an inscription appeared” are incompatible. Example.
An event that always occurs is called reliable. An event that cannot happen is called impossible. For example, suppose a ball is drawn from an urn containing only black balls. Then the appearance of the black ball is a reliable event; the appearance of a white ball is an impossible event. Examples. There will be no snow next year. When rolling the dice, the result will be a seven. These are impossible events. There will be snow next year. When you roll the dice, you will get a number less than seven. Daily sunrise. These are reliable events.
Problem solving For each of the described events, determine what it is: impossible, reliable or random. 1. Of the 25 students in the class, two celebrate their birthday on a) January 30; b) February 30. 2. The literature textbook randomly opens and the second word is found on the left page. This word begins: a) with the letter “K”; b) starting with the letter “Ъ”.
3. Today in Sochi the barometer shows normal atmospheric pressure. In this case: a) the water in the pan boiled at a temperature of 80º C; b) when the temperature dropped to -5º C, the water in the puddle froze. 4. Two dice are thrown: a) the first dice shows 3 points, and the second - 5 points; b) the sum of the points rolled on the two dice is 1; c) the sum of the points rolled on the two dice is 13; d) both dice got 3 points; e) the sum of points on two dice is less than 15. Problem solving
5. You opened the book to any page and read the first noun you came across. It turned out that: a) the spelling of the selected word contains a vowel; b) the spelling of the selected word contains the letter “O”; c) there are no vowels in the spelling of the selected word; d) there is a soft sign in the spelling of the selected word. Problem solving
5th grade. Introduction to Probability (4 hours)
(development of 4 lessons on this topic)
Learning goals : - introduce the definition of a random, reliable and impossible event;
Provide first ideas about solving combinatorial problems: using a tree of options and using the multiplication rule.
Educational goal: development of students' worldview.
Developmental goal : development of spatial imagination, improvement of the skill of working with a ruler.
Reliable, impossible and random events (2 hours)
Combinatorial problems (2 hours)
Reliable, impossible and random events.
First lesson
Lesson equipment: dice, coin, backgammon.
Our life largely consists of accidents. There is such a science as “Probability Theory”. Using its language, you can describe many phenomena and situations.
Even the primitive leader understood that a dozen hunters had a greater “probability” of hitting a bison with a spear than one. That's why they hunted collectively back then.
Such ancient commanders as Alexander the Great or Dmitry Donskoy, preparing for battle, relied not only on the valor and art of warriors, but also on chance.
Many people love mathematics for the eternal truths: twice two is always four, the sum of even numbers is even, the area of a rectangle is equal to the product of its adjacent sides, etc. In any problem that you solve, everyone gets the same answer - you just need to not make mistakes in the decision.
Real life is not so simple and straightforward. The outcome of many events cannot be predicted in advance. It is impossible, for example, to say for sure which side a coin thrown up will fall, when the first snow will fall next year, or how many people in the city will want to make a phone call within the next hour. Such unpredictable events are called random .
However, chance also has its own laws, which begin to manifest themselves when random phenomena are repeated many times. If you toss a coin 1000 times, it will come up heads approximately half the time, which is not the case with two or even ten tosses. "Approximately" does not mean half. This generally may or may not be the case. The law does not state anything for certain, but it does provide a certain degree of confidence that some random event will occur. Such patterns are studied by a special branch of mathematics - Probability theory . With its help, you can predict with a greater degree of confidence (but still not for sure) both the date of the first snowfall and the number of phone calls.
Probability theory is inextricably linked with our everyday life. This gives us a wonderful opportunity to establish many probabilistic laws experimentally, repeating random experiments many times. The materials for these experiments will most often be an ordinary coin, a dice, a set of dominoes, backgammon, roulette, or even a deck of cards. Each of these items is related to games in one way or another. The fact is that the case appears here in its most frequent form. And the first probabilistic tasks were related to assessing the players’ chances of winning.
Modern probability theory has moved away from gambling, but its props still remain the simplest and most reliable source of chance. After practicing with a roulette and a dice, you will learn to calculate the probability of random events in real life situations, which will allow you to evaluate your chances of success, test hypotheses, and make optimal decisions not only in games and lotteries.
When solving probabilistic problems, be very careful, try to justify every step you take, because no other area of mathematics contains so many paradoxes. Like probability theory. And perhaps the main explanation for this is its connection with the real world in which we live.
Many games use a die with a different number of dots from 1 to 6 marked on each side. The player throws the dice, looks at how many dots appear (on the side that is located on top), and makes the corresponding number of moves: 1,2,3 ,4,5, or 6. Throwing a die can be considered an experience, an experiment, a test, and the result obtained can be considered an event. People are usually very interested in guessing the occurrence of this or that event and predicting its outcome. What predictions can they make when they roll the dice? First prediction: one of the numbers 1,2,3,4,5, or 6 will appear. Do you think the predicted event will occur or not? Of course, it will definitely come. An event that is sure to occur in a given experience is called a reliable event.
Second prediction : the number 7 will appear. Do you think the predicted event will happen or not? Of course it won’t happen, it’s simply impossible. An event that cannot occur in a given experience is called impossible event.
Third prediction : the number 1 will appear. Do you think the predicted event happened or not? We are not able to answer this question with complete certainty, since the predicted event may or may not occur. An event that may or may not occur in a given experience is called a random event.
Exercise : Describe the events discussed in the tasks below. Like certain, impossible or random.
Let's toss a coin. A coat of arms appeared. (random)
The hunter shot at the wolf and hit it. (random)
The schoolboy goes for a walk every evening. While walking on Monday, he met three acquaintances. (random)
Let's mentally carry out the following experiment: turn a glass of water upside down. If this experiment is carried out not in space, but at home or in a classroom, then water will spill out. (reliable)
Three shots were fired at the target.” There were five hits" (impossible)
Throw the stone up. The stone remains hanging in the air. (impossible)
We rearrange the letters of the word “antagonism” at random. The result is the word “anachroism.” (impossible)
№959. Petya thought of a natural number. The event is as follows:
a) an even number is intended; (random) b) an odd number is intended; (random)
c) a number is conceived that is neither even nor odd; (impossible)
d) a number is conceived that is even or odd. (reliable)
№ 961. Petya and Tolya compare their birthdays. The event is as follows:
a) their birthdays do not coincide; (random) b) their birthdays are the same; (random)
d) both of their birthdays fall on holidays - New Year (January 1) and Russian Independence Day (June 12). (random)
№ 962. When playing backgammon, two dice are used. The number of moves that a participant in the game makes is determined by adding the numbers on the two sides of the cube that fall out, and if a “double” is rolled (1 + 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6), then the number of moves doubles. You roll the dice and figure out how many moves you have to make. The event is as follows:
a) you must make one move; b) you must make 7 moves;
c) you must make 24 moves; d) you must make 13 moves.
a) – impossible (1 move can be made if the combination 1 + 0 is rolled, but there is no number 0 on the dice).
b) – random (if 1 + 6 or 2 + 5 is rolled).
c) – random (if the combination 6 +6 appears).
d) – impossible (there are no combinations of numbers from 1 to 6, the sum of which is 13; this number cannot be obtained even when a “double” is rolled, since it is odd).
Check yourself. (mathematical dictation)
1) Indicate which of the following events are impossible, which are reliable, which are random:
The football match "Spartak" - "Dynamo" will end in a draw. (random)
You will win by participating in a win-win lottery (reliable)
Snow will fall at midnight and the sun will shine 24 hours later. (impossible)
Tomorrow there will be a math test. (random)
You will be elected President of the United States. (impossible)
You will be elected president of Russia. (random)
2) You bought a TV in a store, for which the manufacturer provides a two-year warranty. Which of the following events are impossible, which are random, which are reliable:
The TV will not break for a year. (random)
The TV will not break for two years. (random)
You won't have to pay for TV repairs for two years. (reliable)
The TV will break in the third year. (random)
3) A bus carrying 15 passengers has to make 10 stops. Which of the following events are impossible, which are random, which are reliable:
All passengers will get off the bus at different stops. (impossible)
All passengers will get off at the same stop. (random)
At every stop at least someone will get off. (random)
There will be a stop where no one gets off. (random)
An even number of passengers will get off at all stops. (impossible)
An odd number of passengers will get off at all stops. (impossible)
Homework : p. 53 No. 960, 963, 965 (come up with two reliable, random and impossible events yourself).
Second lesson.
Checking homework. (orally)
a) Explain what certain, random and impossible events are.
b) Indicate which of the following events is reliable, which is impossible, which is random:
There will be no summer holidays. (impossible)
The sandwich will fall butter side down. (random)
The school year will end someday. (reliable)
They'll ask me in class tomorrow. (random)
Today I will meet a black cat. (random)
№ 960. You opened this textbook to any page and chose the first noun that came up. The event is as follows:
a) there is a vowel in the spelling of the selected word. ((reliable)
b) the spelling of the chosen word contains the letter “o”. (random)
c) there are no vowels in the spelling of the selected word. (impossible)
d) there is a soft sign in the spelling of the selected word. (random)
№ 963. You are playing backgammon again. Describe the following event:
a) the player must make no more than two moves. (impossible - with a combination of the smallest numbers 1 + 1 the player makes 4 moves; a combination of 1 + 2 gives 3 moves; all other combinations give more than 3 moves)
b) the player must make more than two moves. (reliable - any combination gives 3 or more moves)
c) the player must make no more than 24 moves. (reliable - the combination of the largest numbers 6 + 6 gives 24 moves, and all others give less than 24 moves)
d) the player must make a double-digit number of moves. (random – for example, the combination 2 + 3 gives a single-digit number of moves: 5, and rolling two fours gives a double-digit number of moves)
2. Problem solving.
№ 964. There are 10 balls in a bag: 3 blue, 3 white and 4 red. Describe the following event:
a) 4 balls were taken from the bag, and they are all blue; (impossible)
b) 4 balls were taken from the bag, and they are all red; (random)
c) 4 balls were taken out of the bag, and they all turned out to be different colors; (impossible)
d) 4 balls were taken out of the bag, and among them there was no black ball. (reliable)
Task 1. The box contains 10 red, 1 green and 2 blue pens. Two objects are drawn at random from the box. Which of the following events are impossible, which are random, which are certain:
a) two red pens are taken out (random)
b) two green handles are taken out; (impossible)
c) two blue pens are taken out; (random)
d) handles of two different colors are taken out; (random)
e) two handles are removed; (reliable)
f) two pencils are taken out. (impossible)
Task 2. Winnie the Pooh, Piglet and everyone - everyone - everyone sits down at the round table to celebrate his birthday. At what number of all - all - all is the event “Winnie the Pooh and Piglet sitting next to each other” reliable, and at what number is it random?
(if there are only 1 of all - all - all of them, then the event is reliable, if there is more than 1, then it is random).
Task 3. Among 100 charity lottery tickets, 20 are winning ones. How many tickets do you need to buy to make the “you won’t win anything” event impossible?
Task 4. There are 10 boys and 20 girls in the class. Which of the following events are impossible for this class, which are random, which are reliable
There are two people in the class who were born in different months. (random)
There are two people in the class who were born in the same month. (reliable)
There are two boys in the class who were born in the same month. (random)
There are two girls in the class who were born in the same month. (reliable)
All boys were born in different months. (reliable)
All girls were born in different months. (random)
There is a boy and a girl born in the same month. (random)
There is a boy and a girl born in different months. (random)
Task 5. There are 3 red, 3 yellow, 3 green balls in the box. We pull out 4 balls at random. Consider the event “Among the drawn balls there will be balls of exactly M colors.” For each M from 1 to 4, determine what kind of event it is - impossible, reliable or random, and fill out the table:
Independent work.
Ioption
a) your friend’s birthday number is less than 32;
c) tomorrow there will be a test in mathematics;
d) Next year the first snow in Moscow will fall on Sunday.
Throwing a dice. Describe the event:
a) the cube, having fallen, will stand on its edge;
b) one of the numbers will appear: 1, 2, 3, 4, 5, 6;
c) the number 6 will appear;
d) a number that is a multiple of 7 will be rolled.
A box contains 3 red, 3 yellow and 3 green balls. Describe the event:
a) all the drawn balls are of the same color;
b) all the drawn balls are of different colors;
c) among the drawn balls there are balls of different colors;
c) among the drawn balls there is a red, yellow and green ball.
IIoption
Describe the event in question as reliable, impossible or accidental:
a) a sandwich that falls off the table will fall face down on the floor;
b) snow will fall in Moscow at midnight, and after 24 hours the sun will shine;
c) you will win by participating in a win-win lottery;
d) next year in May the first thunder of spring will be heard.
All two-digit numbers are written on the cards. One card is chosen at random. Describe the event:
a) there was a zero on the card;
b) there was a number on the card that was a multiple of 5;
c) there was a number on the card that was a multiple of 100;
d) there was a number on the card greater than 9 and less than 100.
The box contains 10 red, 1 green and 2 blue pens. Two objects are drawn at random from the box. Describe the event:
a) two blue pens are taken out;
b) two red pens are taken out;
c) two green handles are taken out;
d) the green and black handles are taken out.
Homework: 1). Come up with two reliable, random and impossible events.
2). Task . There are 3 red, 3 yellow, 3 green balls in the box. We draw N balls at random. Consider the event “among the drawn balls there will be balls of exactly three colors.” For each N from 1 to 9, determine what kind of event it is - impossible, reliable or random, and fill out the table:
Combinatorial problems.
First lesson
Checking homework. (orally)
a) we check the problems that the students came up with.
b) an additional task.
I’m reading an excerpt from V. Levshin’s book “Three Days in Karlikania.”
“At first, to the sounds of a smooth waltz, the numbers formed a group: 1 + 3 + 4 + 2 = 10. Then the young skaters began to change places, forming more and more new groups: 2 + 3 + 4 + 1 = 10
3 + 1 + 2 + 4 = 10
4 + 1 + 3 + 2 = 10
1 + 4 + 2 + 3 = 10, etc.
This continued until the skaters returned to their starting position.”
How many times did they change places?
Today in class we will learn how to solve such problems. They're called combinatorial.
3. Studying new material.
Task 1. How many two-digit numbers can be made from the numbers 1, 2, 3?
Solution: 11, 12, 13
31, 32, 33. 9 numbers in total.
When solving this problem, we searched through all possible options, or, as they usually say in these cases. All possible combinations. Therefore, such problems are called combinatorial. You have to calculate possible (or impossible) options in life quite often, so it’s useful to get acquainted with combinatorial problems.
№ 967. Several countries have decided to use symbols for their national flag in the form of three horizontal stripes of the same width in different colors - white, blue, red. How many countries can use such symbols, provided that each country has its own flag?
Solution. Let's assume that the first stripe is white. Then the second stripe can be blue or red, and the third stripe, respectively, red or blue. We got two options: white, blue, red or white, red, blue.
Let now the first stripe be blue, then again we get two options: white, red, blue or blue, red, white.
Let the first stripe be red, then there are two more options: red, white, blue or red, blue, white.
There were 6 possible options in total. This flag can be used by 6 countries.
So, when solving this problem, we were looking for a way to enumerate possible options. In many cases, it turns out to be useful to construct a picture - a diagram of enumerating options. This, firstly, is clear, and secondly, it allows us to take everything into account and not miss anything.
This diagram is also called a tree of possible options.
Front page
Second stripe
Third lane
The resulting combination
№ 968. How many two-digit numbers can be made from the numbers 1, 2, 4, 6, 8?
Solution. For the two-digit numbers we are interested in, the first place can be any of the given digits, except 0. If we put the number 2 in the first place, then any of the given digits can be in the second place. You will get five two-digit numbers: 2.,22, 24, 26, 28. Likewise, there will be five two-digit numbers with the first digit 4, five two-digit numbers with the first digit 6 and five two-digit numbers with the first digit 8.
Answer: There will be 20 numbers in total.
Let's build a tree of possible options to solve this problem.
Double figures
First digit
Second digit
Received numbers
20, 22, 24, 26, 28, 60, 62, 64, 66, 68,
40, 42, 44, 46, 48, 80, 82, 84, 86, 88.
Solve the following problems by constructing a tree of possible options.
№ 971. The leadership of a certain country decided to make its national flag look like this: on a single-color rectangular background, a circle of a different color is placed in one of the corners. It was decided to choose colors from three possible ones: red, yellow, green. How many variants of this flag?
exists? The figure shows some of the possible options.
Answer: 24 options.
№ 973. a) How many three-digit numbers can be made from the numbers 1,3, 5,? (27 numbers)
b) How many three-digit numbers can be made from the numbers 1,3, 5, provided that the numbers should not be repeated? (6 numbers)
№ 979. Modern pentathletes participate in competitions in five sports over two days: show jumping, fencing, swimming, shooting, and running.
a) How many options are there for the order of completing the types of competition? (120 options)
b) How many options are there for the order of the events of the competition, if it is known that the last event should be running? (24 options)
c) How many options are there for the order of the competition events if it is known that the last event should be running, and the first should be show jumping? (6 options)
№ 981. Two urns contain five balls each in five different colors: white, blue, red, yellow, green. One ball is drawn from each urn at a time.
a) how many different combinations of drawn balls are there (combinations like “white - red” and “red - white” are considered the same)?
(15 combinations)
b) How many combinations are there in which the drawn balls are of the same color?
(5 combinations)
c) how many combinations are there in which the drawn balls are of different colors?
(15 – 5 = 10 combinations)
Homework: p. 54, No. 969, 972, come up with a combinatorial problem yourself.
№ 969. Several countries have decided to use symbols for their national flag in the form of three vertical stripes of the same width in different colors: green, black, yellow. How many countries can use such symbols, provided that each country has its own flag?
№ 972. a) How many two-digit numbers can be made from the numbers 1, 3, 5, 7, 9?
b) How many two-digit numbers can be made from the numbers 1, 3, 5, 7, 9, provided that the numbers should not be repeated?
Second lesson
Checking homework. a) No. 969 and No. 972a) and No. 972b) - build a tree of possible options on the board.
b) we check the completed tasks orally.
Problem solving.
So, before this, we learned how to solve combinatorial problems using a tree of options. Is this a good way? Probably yes, but very cumbersome. Let's try to solve homework problem No. 972 differently. Who can guess how this can be done?
Answer: For each of the five colors of T-shirts there are 4 colors of panties. Total: 4 * 5 = 20 options.
№ 980. The urns contain five balls each in five different colors: white, blue, red, yellow, green. One ball is drawn from each urn at a time. Describe the following event as certain, random, or impossible:
a) taken out balls of different colors; (random)
b) taken out balls of the same color; (random)
c) black and white balls are drawn; (impossible)
d) two balls are drawn, both of which are colored one of the following colors: white, blue, red, yellow, green. (reliable)
№ 982. A group of tourists plans to hike along the route Antonovo - Borisovo - Vlasovo - Gribovo. From Antonovo to Borisovo you can raft on the river or walk. From Borisovo to Vlasovo you can walk or ride bicycles. From Vlasovo to Gribovo you can swim along the river, ride bicycles or walk. How many trekking options can tourists choose from? How many hiking options can tourists choose, provided that they must use bicycles on at least one part of the route?
(12 route options, 8 of them using bicycles)
Independent work.
1 option
a) How many three-digit numbers can be made from the digits: 0, 1, 3, 5, 7?
b) How many three-digit numbers can be made from the digits: 0, 1, 3, 5, 7, provided that the numbers should not be repeated?
Athos, Porthos and Aramis have only a sword, a dagger and a pistol.
a) In how many ways can musketeers be armed?
b) How many weapon options are there if Aramis must wield a sword?
c) How many weapon options are there if Aramis must wield the sword and Porthos must wield the pistol?
Somewhere God sent Raven a piece of cheese, as well as feta cheese, sausage, white and black bread. Having perched on a spruce tree, the crow was just about ready to have breakfast, but she began to think: in how many ways can sandwiches be made from these products?
Option 2
a) How many three-digit numbers can be made from the digits: 0, 2, 4, 6, 8?
b) How many three-digit numbers can be made from the digits: 0, 2, 4, 6, 8, provided that the digits should not be repeated?
Count Monte Cristo decided to give Princess Hayde earrings, a necklace and a bracelet. Each piece of jewelry must contain one of the following types of gemstones: diamonds, rubies or garnets.
a) How many options are there for combining precious stone jewelry?
b) How many jewelry options are there if the earrings should be diamond?
c) How many jewelry options are there if the earrings should be diamond and the bracelet should be garnet?
For breakfast you can choose a bun, sandwich or gingerbread with coffee or kefir. How many breakfast options can you create?
Homework : No. 974, 975. (by compiling a tree of options and using the multiplication rule)
№ 974 . a) How many three-digit numbers can be made from the numbers 0, 2, 4?
b) How many three-digit numbers can be made from the numbers 0, 2, 4, provided that the numbers should not be repeated?
№ 975 . a) How many three-digit numbers can be made from the numbers 1,3, 5,7?
b) How many three-digit numbers can be made from the numbers 1,3, 5,7 under the condition. What numbers should not be repeated?
Problem numbers taken from the textbook
"Mathematics-5", I.I. Zubareva, A.G. Mordkovich, 2004.
The purpose of the lesson:
- Introduce the concept of reliable, impossible and random events.
- Develop knowledge and skills to determine the type of events.
- Develop: computing skill; attention; ability to analyze, reason, draw conclusions; group work skills.
During the classes
1) Organizational moment.
Interactive exercise: children must solve examples and decipher words; based on the results, they are divided into groups (true, impossible and random) and determine the topic of the lesson.
1 card.
| 0,5 | 1,6 | 12,6 | 5,2 | 7,5 | 8 | 5,2 | 2,08 | 0,5 | 9,54 | 1,6 |
2 card
| 0,5 | 2,1 | 14,5 | 1,9 | 2,1 | 20,4 | 14 | 1,6 | 5,08 | 8,94 | 14 |
3 card
| 5 | 2,4 | 6,7 | 4,7 | 8,1 | 18 | 40 | 9,54 | 0,78 |
2) Updating the learned knowledge.
Game “Clap”: even number - clap, odd number - stand up.
Task: from the given series of numbers 42, 35, 8, 9, 7, 10, 543, 88, 56, 13, 31, 77, ... determine even and odd.
3) Studying a new topic.
There are cubes on your tables. Let's take a closer look at them. What do you see?
Where are dice used? How?
Work in groups.
Conducting an experiment.
What predictions can you make when throwing a die?
First prediction: one of the numbers 1,2,3,4,5 or 6 will appear.
An event that is sure to occur in a given experience is called reliable.
Second prediction: the number 7 will appear.
Do you think the predicted event will happen or not?
This is impossible!
An event that cannot occur in a given experience is called impossible.
Third prediction: the number 1 will appear.
Will this event happen?
An event that may or may not occur in a given experience is called random.
4) Consolidation of the studied material.
I. Determine the type of event
-Tomorrow it will snow red.
It will snow heavily tomorrow.
Tomorrow, even though it is July, it will snow.
Tomorrow, even though it is July, there will be no snow.
Tomorrow it will snow and there will be a blizzard.
II. Add a word to this sentence in such a way that the event becomes impossible.
Kolya received an A in history.
Sasha did not complete a single task on the test.
Oksana Mikhailovna (history teacher) will explain a new topic.
III. Give examples of impossible, random and reliable events.
IV. Work from the textbook (in groups).
Describe the events discussed in the tasks below as reliable, impossible or random.
No. 959. Petya came up with a natural number. The event is as follows:
a) an even number is intended;
b) an odd number is intended;
c) a number is conceived that is neither even nor odd;
d) a number is conceived that is even or odd.
No. 960. You opened this textbook to any page and chose the first noun that came up. The event is as follows:
a) there is a vowel in the spelling of the selected word;
b) the spelling of the selected word contains the letter “o”;
c) there are no vowels in the spelling of the selected word;
d) there is a soft sign in the spelling of the selected word.
Solve No. 961, No. 964.
Discussion of solved tasks.
5) Reflection.
1. What events did you learn about in the lesson?
2. Indicate which of the following events is certain, which is impossible and which is random:
a) there will be no summer holidays;
b) the sandwich will fall butter side down;
c) the school year will end someday.
6) Homework:
Come up with two reliable, random and impossible events.
Make a drawing for one of them.
1.1. Some information from combinatorics
1.1.1. Placements
Let's consider the simplest concepts associated with the selection and arrangement of a certain set of objects.
Counting the number of ways in which these actions can be performed is often done when solving probabilistic problems.
Definition. Accommodation from n elements by k (k ≤n) is any ordered subset of k elements of a set consisting of n various elements.
Example. The following sequences of numbers are placements of 2 elements from 3 elements of the set (1;2;3): 12, 13, 23, 21, 31, 32.
Note that the placements differ in the order of the elements included in them and their composition. Placements 12 and 21 contain the same numbers, but their order is different. Therefore, these placements are considered different.
Number of different placements from n elements by k is designated and calculated by the formula:
,
Where n! = 1∙2∙...∙(n - 1)∙n(reads " n- factorial").
The number of two-digit numbers that can be made from the digits 1, 2, 3, provided that no digit is repeated equal to: .
1.1.2. Rearrangements
Definition. Permutations from n elements are called such placements of n elements that differ only in the location of the elements.
Number of permutations from n elements P n calculated by the formula: P n=n!
Example. In how many ways can 5 people line up? The number of ways is equal to the number of permutations of 5 elements, i.e.
P 5 =5!=1∙2∙3∙4∙5=120.
Definition. If among n elements k identical, then rearrangement of these n elements is called a permutation with repetitions.
Example. Let 2 of the 6 books be identical. Any arrangement of all the books on a shelf is a rearrangement with repetition.
Number of different permutations with repetitions (from n elements, including k identical) is calculated using the formula: .
In our example, the number of ways in which books can be arranged on a shelf is: .
1.1.3. Combinations
Definition. Combinations of n elements by k such placements are called n elements by k, which differ from one another in at least one element.
Number of different combinations of n elements by k is designated and calculated by the formula: .
By definition, 0!=1.
The following properties apply to combinations:
1.
2.
3.
4.
Example. There are 5 flowers of different colors. 3 flowers are selected for the bouquet. The number of different bouquets of 3 flowers out of 5 is equal to: .
1.2. Random Events
1.2.1. Events
Knowledge of reality in the natural sciences occurs as a result of tests (experiment, observations, experience).
Test
or experience is the implementation of a specific set of conditions that can be reproduced an arbitrarily large number of times.
Random
is an event that may or may not occur as a result of some test (experience).
Thus, the event is considered as the result of the test.
Example. Tossing a coin is a test. The appearance of an eagle during a toss is an event.
The events we observe differ in the degree of possibility of their occurrence and in the nature of their interrelation.
The event is called reliable
, if it is certain to occur as a result of this test.
Example. A student receiving a positive or negative mark on an exam is a reliable event if the exam proceeds according to the usual rules.
The event is called impossible
, if it cannot occur as a result of this test.
Example. Removing a white ball from an urn that contains only colored (non-white) balls is an impossible event. Note that under other experimental conditions the appearance of a white ball is not excluded; thus, this event is impossible only under the conditions of our experience.
In what follows, we will denote random events by capital Latin letters A, B, C... We will denote a reliable event by the letter Ω, and an impossible event by Ø.
Two or more events are called equally possible
in a given test if there is reason to believe that none of these events is more or less possible than the others.
Example. With one throw of a die, the appearance of 1, 2, 3, 4, 5 and 6 points are all equally possible events. It is assumed, of course, that the dice are made of a homogeneous material and have the correct shape.
The two events are called incompatible
in a given test, if the occurrence of one of them excludes the occurrence of the other, and joint
otherwise.
Example. The box contains standard and non-standard parts. Let's take one detail for luck. The appearance of a standard part eliminates the appearance of a non-standard part. These events are incompatible.
Several events form full group of events
in a given test, if at least one of them is sure to occur as a result of this test.
Example. The events from the example form a complete group of equally possible and pairwise incompatible events.
Two incompatible events that form a complete group of events in a given trial are called opposite events.
If one of them is designated by A, then the other is usually denoted by (read “not A»).
Example. A hit and a miss with one shot at a target are opposite events.
1.2.2. Classic definition of probability
Probability of event
– a numerical measure of the possibility of its occurrence.
Event A called favorable
event IN if whenever an event occurs A, the event comes IN.
Events A 1 , A 2 , ..., An form case diagram
, if they:
1) equally possible;
2) pairwise incompatible;
3) form a complete group.
In the scheme of cases (and only in this scheme) the classical definition of probability takes place P(A) events A. Here, a case is each of the events belonging to a selected complete group of equally possible and pairwise incompatible events.
If n is the number of all cases in the scheme, and m– number of cases favorable to the event A, That probability of an event
A is determined by the equality:
The following properties follow from the definition of probability:
1. The probability of a reliable event is equal to one.
Indeed, if an event is certain, then every case in the scheme of cases favors the event. In this case m = n and therefore 
2. The probability of an impossible event is zero.
Indeed, if an event is impossible, then no case in the pattern of cases favors the event. That's why m=0 and therefore 
The probability of a random event is a positive number between zero and one.
Indeed, only a fraction of the total number of cases in the pattern of cases is favored by a random event. Therefore 0<m<n, which means 0<m/n<1 и, следовательно, 0 < P(A) < 1.
So, the probability of any event satisfies the inequalities
0 ≤ P(A) ≤ 1.
Currently, the properties of probability are defined in the form of axioms formulated by A.N. Kolmogorov.
One of the main advantages of the classical definition of probability is the ability to calculate the probability of an event directly, i.e. without resorting to experiments, which are replaced by logical reasoning.
Problems of direct calculation of probabilities
Problem 1.1. What is the probability of an even number of points (event A) when throwing a die?
Solution. Consider the events Ai- dropped out i glasses, i= 1, 2, …,6. It is obvious that these events form a pattern of cases. Then the number of all cases n= 6. Cases favor rolling an even number of points A 2 , A 4 , A 6, i.e. m= 3. Then 
.
Problem 1.2. There are 5 white and 10 black balls in an urn. The balls are thoroughly mixed and then 1 ball is taken out at random. What is the probability that the drawn ball will be white?
Solution. There are a total of 15 cases that form a case pattern. Moreover, the expected event A– the appearance of a white ball is favored by 5 of them, therefore 
.
Problem 1.3. A child plays with six letters of the alphabet: A, A, E, K, R, T. Find the probability that he will be able to randomly form the word CARRIAGE (event A).
Solution. The solution is complicated by the fact that among the letters there are identical ones - two letters “A”. Therefore, the number of all possible cases in a given test is equal to the number of permutations with repetitions of 6 letters:
.
These cases are equally possible, pairwise inconsistent and form a complete group of events, i.e. form a diagram of cases. Only one chance favors the event A. That's why
.
Problem 1.4. Tanya and Vanya agreed to celebrate the New Year in a company of 10 people. They both really wanted to sit next to each other. What is the probability of their wish being fulfilled if it is customary to distribute places among their friends by lot?
Solution. Let us denote by A event “fulfillment of Tanya and Vanya’s wishes.” 10 people can sit at a table of 10! different ways. How many of these n= 10! equally possible ways are favorable for Tanya and Vanya? Tanya and Vanya, sitting next to each other, can take 20 different positions. At the same time, eight of their friends can sit at a table of 8! in different ways, so m= 20∙8!. Hence, 
.
Problem 1.5. A group of 5 women and 20 men selects three delegates. Assuming that each person present can be chosen with equal probability, find the probability that two women and one man will be chosen.
Solution. The total number of equally possible test outcomes is equal to the number of ways in which three delegates can be selected from 25 people, i.e. . Let us now count the number of favorable cases, i.e. the number of cases in which the event of interest occurs. A male delegate can be selected in twenty ways. At the same time, the remaining two delegates must be women, and you can choose two women out of five. Hence, . That's why
.
Problem 1.6. Four balls are randomly scattered over four holes, each ball falls into one or another hole with equal probability and independently of the others (there are no obstacles to several balls falling into the same hole). Find the probability that there will be three balls in one of the holes, one in the other, and no balls in the other two holes.
Solution. Total number of cases n=4 4 . The number of ways in which one can choose one hole where there will be three balls, . The number of ways in which you can choose a hole where there will be one ball, . The number of ways in which three of the four balls can be selected to be placed in the first hole is . Total number of favorable cases. Probability of event: 
Problem 1.7. There are 10 identical balls in the box, marked with numbers 1, 2, ..., 10. Six balls are drawn for luck. Find the probability that among the extracted balls there will be: a) ball No. 1; b) balls No. 1 and No. 2.
Solution. a) The total number of possible elementary outcomes of the test is equal to the number of ways in which six balls can be extracted from ten, i.e.
Let's find the number of outcomes that favor the event we are interested in: among the selected six balls there is ball No. 1 and, therefore, the remaining five balls have different numbers. The number of such outcomes is obviously equal to the number of ways in which five balls can be selected from the remaining nine, i.e.
The required probability is equal to the ratio of the number of outcomes favorable to the event in question to the total number of possible elementary outcomes:
b) The number of outcomes favorable to the event we are interested in (among the selected balls there are balls No. 1 and No. 2, therefore, four balls have different numbers) is equal to the number of ways in which four balls can be extracted from the remaining eight, i.e. Required probability 
1.2.3. Statistical probability
The statistical definition of probability is used when the outcomes of an experiment are not equally possible.
Relative event frequency
A is determined by the equality:
,
Where m– number of trials in which the event A it has arrived n– total number of tests performed.
J. Bernoulli proved that with an unlimited increase in the number of experiments, the relative frequency of the occurrence of an event will differ almost as little as desired from some constant number. It turned out that this constant number is the probability of the event occurring. Therefore, it is natural to call the relative frequency of the occurrence of an event with a sufficiently large number of trials a statistical probability, in contrast to the previously introduced probability.
Example 1.8. How to approximately determine the number of fish in the lake?
Let in the lake X fish We cast a net and, let’s say, find in it n fish We mark each of them and release them back. A few days later, in the same weather and in the same place, we cast the same net. Let us assume that we find m fish in it, among which k tagged. Let the event A- “the caught fish is marked.” Then by definition of relative frequency .
But if in the lake X fish and we released it into it n labeled, then .
Because R * (A) » R(A), That .
1.2.4. Operations on events. Probability addition theorem
Amount, or the union of several events, is an event consisting of the occurrence of at least one of these events (in the same trial).
Sum A 1 + A 2 + … + An denoted as follows: 
or 
.
Example. Two dice are thrown. Let the event A consists of rolling 4 points on 1 dice, and the event IN– when 5 points are rolled on another dice. Events A And IN joint. Therefore the event A +IN consists of rolling out 4 points on the first die, or 5 points on the second die, or 4 points on the first die and 5 points on the second at the same time.
Example. Event A– winnings for 1 loan, event IN– winnings on the 2nd loan. Then the event A+B– winning at least one loan (possibly two at once).
The work or the intersection of several events is an event consisting of the joint occurrence of all these events (in the same trial).
Work IN events A 1 , A 2 , …, An denoted as follows:
.
Example. Events A And IN consist of successfully passing the first and second rounds, respectively, upon admission to the institute. Then the event A×B consists of successfully completing both rounds.
The concepts of sum and product of events have a clear geometric interpretation. Let the event A there is a point entering the area A, and the event IN– point entering the area IN. Then the event A+B there is a point entering the union of these areas (Fig. 2.1), and the event AIN there is a point that hits the intersection of these areas (Fig. 2.2).
Rice. 2.1 Fig. 2.2
Theorem. If events A i(i = 1, 2, …, n) are pairwise inconsistent, then the probability of the sum of events is equal to the sum of the probabilities of these events: 
.
Let A And Ā
– opposite events, i.e. A + Ā= Ω, where Ω is a reliable event. From the addition theorem it follows that
Р(Ω) = R(A) + R(Ā
) = 1, therefore
R(Ā
) = 1 – R(A).
If events A 1 and A 2 are compatible, then the probability of the sum of two simultaneous events is equal to:
R(A 1 + A 2) = R(A 1) + R(A 2) – P( A 1× A 2).
Probability addition theorems allow us to move from directly calculating probabilities to determining the probabilities of the occurrence of complex events.
Problem 1.8. The shooter fires one shot at the target. Probability of scoring 10 points (event A), 9 points (event IN) and 8 points (event WITH) are equal to 0.11, respectively; 0.23; 0.17. Find the probability that with one shot the shooter will score less than 8 points (event D).
Solution. Let's move on to the opposite event - with one shot the shooter will score at least 8 points. An event occurs if it happens A or IN, or WITH, i.e. . Since events A, B, WITH are pairwise inconsistent, then, by the addition theorem,
, where .
Problem 1.9. From the team of the brigade, which consists of 6 men and 4 women, two people are selected for the trade union conference. What is the probability that among those selected at least one woman (event A).
Solution. If an event occurs A, then one of the following incompatible events will definitely occur: IN– “a man and a woman are chosen”; WITH- “two women were chosen.” Therefore we can write: A=B+C. Let's find the probability of events IN And WITH. Two out of 10 people can be chosen in different ways. Two women out of 4 can be selected in different ways. A man and a woman can be selected in 6 × 4 ways. Then . Since events IN And WITH are inconsistent, then, by the addition theorem,
P(A) = P(B + C) = P(B) + P(C) = 8/15 + 2/15 = 2/3.
Problem 1.10. There are 15 textbooks arranged randomly on a library shelf, five of them bound. The librarian takes three textbooks at random. Find the probability that at least one of the taken textbooks will be bound (event A).
Solution. First way. The requirement - at least one of the three bound textbooks taken - will be fulfilled if any of the following three incompatible events occur: IN– one bound textbook, WITH– two bound textbooks, D– three bound textbooks.
Event of interest to us A can be represented as a sum of events: A=B+C+D. According to the addition theorem,
P(A) = P(B) + P(C) + P(D). (2.1)
Let's find the probability of events B, C And D(see combinatorial schemes): 
Representing these probabilities in equality (2.1), we finally obtain
P(A)= 45/91 + 20/91 + 2/91 = 67/91.
Second way. Event A(at least one of the three textbooks taken is bound) and Ā
(none of the textbooks taken is bound) - opposite, therefore P(A) + P(Ā) = 1 (the sum of the probabilities of two opposite events is equal to 1). From here P(A) = 1 – P(Ā). Probability of event occurrence Ā
(none of the textbooks taken are bound)
Required probability
P(A) = 1 – P(Ā) = 1 – 24/91 = 67/91.
1.2.5. Conditional probability. Probability multiplication theorem
Conditional probability P(B/A) is the probability of event B, calculated under the assumption that event A has already occurred.
Theorem. The probability of the joint occurrence of two events is equal to the product of the probabilities of one of them and the conditional probability of the other, calculated under the assumption that the first event has already occurred:
P(A∙B) = P(A)∙P( IN/A). (2.2)
Two events are called independent if the occurrence of any of them does not change the probability of the occurrence of the other, i.e.
P(A) = P(A/B) or P(B) = P(B/A). (2.3)
If events A And IN are independent, then from formulas (2.2) and (2.3) it follows
P(A∙B) = P(A)∙P(B). (2.4)
The opposite statement is also true, i.e. if equality (2.4) holds for two events, then these events are independent. Indeed, from formulas (2.4) and (2.2) it follows
P(A∙B) = P(A)∙P(B) = P(A) × P(B/A), where P(A) = P(B/A).
Formula (2.2) can be generalized to the case of a finite number of events A 1 , A 2 ,…,A n:
P(A 1 ∙A 2 ∙…∙A n)=P(A 1)∙P(A 2 /A 1)∙P(A 3 /A 1 A 2)∙…∙P(A n/A 1 A 2 …A n -1).
Problem 1.11. From an urn containing 5 white and 10 black balls, two balls are drawn in a row. Find the probability that both balls are white (event A).
Solution. Let's consider the events: IN– the first ball drawn is white; WITH– the second ball drawn is white. Then A = BC.
The experiment can be carried out in two ways:
1) with return: the removed ball, after fixing the color, is returned to the urn. In this case the events IN And WITH independent:
P(A) = P(B)∙R(S) = 5/15 ×5/15 = 1/9;
2) without returning: the removed ball is put aside. In this case the events IN And WITH dependent:
P(A) = P(B)∙R(S/IN).
For an event IN the conditions are the same, and for WITH the situation has changed. Happened IN, therefore there are 14 balls left in the urn, including 4 white ones.
So, .
Problem 1.12. Among the 50 light bulbs, 3 are non-standard. Find the probability that two light bulbs taken at the same time are non-standard.
Solution. Let's consider the events: A– the first light bulb is non-standard, IN– the second light bulb is non-standard, WITH– both bulbs are non-standard. It's clear that C = A∙IN. Event A 3 cases out of 50 possible are favorable, i.e. P(A) = 3/50. If the event A has already arrived, then the event IN two cases out of 49 possible are favorable, i.e. P(B/A) = 2/49. Hence,
.
Problem 1.13. Two athletes shoot at the same target independently of each other. The probability of the first athlete hitting the target is 0.7, and the second is 0.8. What is the probability that the target will be hit?
Solution. The target will be hit if either the first shooter, or the second, or both, hits it, i.e. an event will happen A+B, where is the event A consists of the first athlete hitting the target, and the event IN– second. Then
P(A+IN)=P(A)+P(B)–P(A∙IN)=0, 7+0, 8–0, 7∙0,8=0,94.
Problem 1.14. The reading room has six textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that two textbooks will be bound.
Solution. Let us introduce the designations of events :A– the first textbook taken is bound, IN– the second textbook is bound. The probability that the first textbook is bound is
P(A) = 3/6 = 1/2.
The probability that the second textbook is bound, provided that the first textbook taken was bound, i.e. conditional probability of an event IN, is like this: P(B/A) = 2/5.
The desired probability that both textbooks are bound, according to the theorem of multiplication of event probabilities, is equal to
P(AB) = P(A) ∙ P(B/A)= 1/2 · ∙ 2/5 = 0.2.
Problem 1.15. There are 7 men and 3 women working in the workshop. Three people were selected at random using their personnel numbers. Find the probability that all selected persons will be men.
Solution. Let us introduce event designations: A– the man is selected first, IN– the second selected is a man, WITH - The third selected was a man. The probability that a man will be selected first is P(A) = 7/10.
The probability that a man is selected second, provided that a man has already been selected first, i.e. conditional probability of an event IN next : P(B/A) = 6/9 = 2/3.
The probability that a man will be selected third, given that two men have already been selected, i.e. conditional probability of an event WITH is this: P(C/AB) = 5/8.
The desired probability that all three selected persons will be men is P(ABC) = P(A) P(B/A) P(C/AB) = 7/10 · 2/3 · 5/8 = 7/24.
1.2.6. Total Probability Formula and Bayes Formula
Let B 1 , B 2 ,…, Bn– pairwise incompatible events (hypotheses) and A– an event that can only happen together with one of them.
Let us also know P(B i) And P(A/B i) (i = 1, 2, …, n).
Under these conditions the formulas are valid: 
(2.5)
(2.6)
Formula (2.5) is called total probability formula
. It calculates the probability of an event A(total probability).
Formula (2.6) is called Bayes formula
. It allows you to recalculate the probabilities of hypotheses if the event A happened.
When compiling examples, it is convenient to assume that the hypotheses form a complete group.
Problem 1.16. The basket contains apples from four trees of the same variety. From the first - 15% of all apples, from the second - 35%, from the third - 20%, from the fourth - 30%. Ripe apples are 99%, 97%, 98%, 95%, respectively.
a) What is the probability that an apple taken at random will be ripe (event A).
b) Given that an apple taken at random turns out to be ripe, calculate the probability that it is from the first tree.
Solution. a) We have 4 hypotheses:
B 1 – an apple taken at random is taken from the 1st tree;
B 2 – an apple taken at random is taken from the 2nd tree;
B 3 – an apple taken at random is taken from the 3rd tree;
B 4 – an apple taken at random is taken from the 4th tree.
Their probabilities according to the condition: P(B 1) = 0,15; P(B 2) = 0,35; P(B 3) = 0,2; P(B 4) = 0,3.
Conditional probabilities of an event A:
P(A/B 1) = 0,99; P(A/B 2) = 0,97; P(A/B 3) = 0,98; P(A/B 4) = 0,95.
The probability that an apple taken at random will be ripe is found using the total probability formula:
P(A)=P(B 1)∙P(A/B 1)+P(B 2)∙P(A/B 2)+P(B 3)∙P(A/B 3)+P(B 4)∙P(A/B 4)=0,969.
b) Bayes formula for our case looks like: 
.
Problem 1.17. A white ball is dropped into an urn containing two balls, after which one ball is drawn at random. Find the probability that the extracted ball will be white if all possible assumptions about the initial composition of the balls (based on color) are equally possible.
Solution. Let us denote by A event – a white ball is drawn. The following assumptions (hypotheses) about the initial composition of the balls are possible: B 1– there are no white balls, AT 2– one white ball, AT 3- two white balls.
Since there are three hypotheses in total, and the sum of the probabilities of the hypotheses is 1 (since they form a complete group of events), then the probability of each of the hypotheses is 1/3, i.e.
P(B 1) = P(B 2)= P(B 3) = 1/3.
The conditional probability that a white ball will be drawn, given that there were no white balls in the urn initially, P(A/B 1)=1/3. The conditional probability that a white ball will be drawn, given that there was initially one white ball in the urn, P(A/B 2)=2/3. Conditional probability that a white ball will be drawn given that there were initially two white balls in the urn P(A/B 3)=3/ 3=1.
We find the required probability that a white ball will be drawn using the total probability formula:
R(A)=P(B 1)∙P(A/B 1)+P(B 2)∙P(A/B 2)+P(B 3)∙P(A/B 3)=1/3 1/3+1/3 2/3+1/3 1=2/3 .
Problem 1.18. Two machines produce identical parts that go onto a common conveyor. The productivity of the first machine is twice that of the second. The first machine produces on average 60% of parts of excellent quality, and the second - 84%. The part taken at random from the assembly line turned out to be of excellent quality. Find the probability that this part was produced by the first machine.
Solution. Let us denote by A event - a detail of excellent quality. Two assumptions can be made: B 1– the part was produced by the first machine, and (since the first machine produces twice as many parts as the second) P(A/B 1) = 2/3; B 2 – the part was produced by the second machine, and P(B 2) = 1/3.
The conditional probability that the part will be of excellent quality if it is produced by the first machine, P(A/B 1)=0,6.
The conditional probability that the part will be of excellent quality if it is produced by the second machine is P(A/B 1)=0,84.
The probability that a part taken at random will be of excellent quality, according to the total probability formula, is equal to
P(A)=P(B 1) ∙P(A/B 1)+P(B 2) ∙P(A/B 2)=2/3·0.6+1/3·0.84 = 0.68.
The required probability that the selected excellent part was produced by the first machine, according to Bayes’ formula, is equal to
Problem 1.19. There are three batches of parts, each containing 20 parts. The number of standard parts in the first, second and third batches are respectively 20, 15, 10. A part that turned out to be standard was randomly removed from the selected batch. The parts are returned to the batch and a part is randomly removed from the same batch, which also turns out to be standard. Find the probability that the parts were removed from the third batch.
Solution. Let us denote by A event - in each of the two trials (with return), a standard part was retrieved. Three assumptions (hypotheses) can be made: B 1 – parts are removed from the first batch, IN 2
– parts are removed from the second batch, IN 3 – parts are removed from the third batch.
The parts were extracted at random from a given batch, so the probabilities of the hypotheses are the same: P(B 1) = P(B 2) = P(B 3) = 1/3.
Let's find the conditional probability P(A/B 1), i.e. the probability that two standard parts will be sequentially removed from the first batch. This event is reliable, because in the first batch all parts are standard, so P(A/B 1) = 1.
Let's find the conditional probability P(A/B 2), i.e. the probability that two standard parts will be sequentially removed (and returned) from the second batch: P(A/B 2)= 15/20 ∙ 15/20 = 9/16.
Let's find the conditional probability P(A/B 3), i.e. the probability that two standard parts will be sequentially removed (and returned) from the third batch: P(A/B 3) = 10/20 · 10/20 = 1/4.
The desired probability that both extracted standard parts are taken from the third batch, according to Bayes’ formula, is equal to 
1.2.7. Repeated tests
If several tests are performed, and the probability of the event A in each test does not depend on the outcomes of other tests, then such tests are called independent with respect to event A. In different independent trials the event A may have either different probabilities or the same probability. We will further consider only such independent tests in which the event A has the same probability.
Let it be produced P independent trials, in each of which the event A may or may not appear. Let us agree to assume that the probability of an event A in each trial is the same, namely equal R. Therefore, the probability of the event not occurring A in each trial is also constant and equal to 1– R. This probabilistic scheme is called Bernoulli scheme. Let us set ourselves the task of calculating the probability that when P Bernoulli test event A will come true k once ( k– number of successes) and, therefore, will not come true P- once. It is important to emphasize that it is not required that the event A repeated exactly k times in a certain sequence. We denote the desired probability R p (k).
For example, the symbol R 5(3) means the probability that in five trials the event will appear exactly 3 times and therefore not occur 2 times.
The problem posed can be solved using the so-called Bernoulli formulas, which looks like: 
.
Problem 1.20. The probability that electricity consumption during one day will not exceed the established norm is equal to R=0.75. Find the probability that in the next 6 days, electricity consumption for 4 days will not exceed the norm.
Solution. The probability of normal energy consumption during each of 6 days is constant and equal to R=0.75. Consequently, the probability of excessive energy consumption every day is also constant and equal to q= 1–R=1–0,75=0,25.
The required probability according to the Bernoulli formula is equal to
.
Problem 1.21. Two equal chess players play chess. What is more likely: winning two games out of four or three games out of six (draws are not taken into account)?
Solution. Equal chess players are playing, so the probability of winning R= 1/2, therefore, the probability of losing q is also equal to 1/2. Because in all games the probability of winning is constant and it does not matter in what sequence the games are won, then Bernoulli’s formula is applicable.
Let's find the probability that two games out of four will be won:
Let's find the probability that three games out of six will be won:
Because P 4 (2) > P 6 (3), then it is more likely to win two games out of four than three out of six.
However, it can be seen that using Bernoulli’s formula for large values n quite difficult, since the formula requires operations on huge numbers and therefore errors accumulate during the calculation process; As a result, the final result may differ significantly from the true one.
To solve this problem, there are several limit theorems that are used for the case of a large number of tests.
1. Poisson's theorem
When conducting a large number of tests using the Bernoulli scheme (with n=> ∞) and with a small number of favorable outcomes k(it is assumed that the probability of success p small), Bernoulli's formula approaches Poisson's formula 
.
Example 1.22. The probability of defects when an enterprise produces a unit of product is equal to p=0.001. What is the probability that when producing 5000 units of product, less than 4 of them will be defective (event A Solution. Because n is large, we use Laplace’s local theorem: 
Let's calculate x:
Function 
– even, so φ(–1.67) = φ(1.67).
Using the table in Appendix A.1, we find φ(1.67) = 0.0989.
Required probability P 2400 (1400) = 0,0989.
3. Laplace's integral theorem
If the probability R occurrence of an event A in each trial according to the Bernoulli scheme is constant and different from zero and one, then with a large number of trials n, probability R p (k 1 , k 2) occurrence of the event A in these tests from k 1 to k 2 times approximately equal
R p(k 1 , k 2) = Φ ( x"") – Φ ( x"), Where 
– Laplace function,
The definite integral in the Laplace function cannot be calculated on the class of analytic functions, so the table is used to calculate it. Clause 2, given in the appendix.
Example 1.24. The probability of an event occurring in each of one hundred independent trials is constant and equal to p= 0.8. Find the probability that the event will appear: a) at least 75 times and not more than 90 times; b) at least 75 times; c) no more than 74 times.
Solution. Let's use Laplace's integral theorem:
R p(k 1 , k 2) = Φ ( x"") – Φ( x"), where Ф( x) – Laplace function,
a) According to the condition, n = 100, p = 0,8, q = 0,2, k 1 = 75, k 2 = 90. Let's calculate x"" And x" :
Considering that the Laplace function is odd, i.e. F(- x) = – Ф( x), we get
P 100 (75;90) = Ф (2.5) – Ф(–1.25) = Ф(2.5) + Ф(1.25).
According to the table P.2. we will find applications:
F(2,5) = 0.4938; F(1.25) = 0.3944.
Required probability
P 100 (75; 90) = 0,4938 + 0,3944 = 0,8882.
b) The requirement that an event appear at least 75 times means that the number of occurrences of the event can be 75, or 76, ..., or 100. Thus, in the case under consideration, it should be accepted k 1 = 75, k 2 = 100. Then 
.
According to the table P.2. application we find Ф(1.25) = 0.3944; Ф(5) = 0.5.
Required probability
P 100 (75;100) = (5) – (–1,25) = (5) + (1,25) = 0,5 + 0,3944 = 0,8944.
c) Event – “ A appeared at least 75 times" and " A appeared no more than 74 times" are opposite, so the sum of the probabilities of these events is equal to 1. Therefore, the desired probability
P 100 (0;74) = 1 – P 100 (75; 100) = 1 – 0,8944 = 0,1056.
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