Grade 5. Introduction to Probability (4 h.)
(development of 4 lessons on this topic)
Learning objectives : - introduce the definition of a random, reliable and impossible event;
Lead the first ideas about solving combinatorial problems: using a tree of options and using the rule of multiplication.
Educational purpose: development of students' worldview.
Developing goal : development of spatial imagination, improvement of the skill of working with a ruler.
Reliable, impossible and random events (2h.)
Combinatorial tasks (2h.)
Reliable, impossible and random events.
First lesson
Lesson equipment: dice, coin, backgammon.
Our life is largely made up of accidents. There is such a science "Theory of Probabilities". Using her language, you can describe many phenomena and situations.
Even the primitive leader understood that a dozen hunters have a "probability" of hitting a bison with a spear more than one. Therefore, then they hunted collectively.
Such ancient generals as Alexander the Great or Dmitry Donskoy, preparing for battle, relied not only on the valor and skill of warriors, but also on chance.
Many people love mathematics for the eternal truths twice two are always four, the sum of even numbers is even, the area of \u200b\u200ba 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 not make mistakes in the solution.
Real life is not so simple and straightforward. The outcomes of many phenomena cannot be predicted in advance. It is impossible, for example, to say for sure which side the coin tossed up will fall on, when the first snow falls next year, or how many people in the city will want to call in the next hour. Such unpredictable phenomena are called random .
However, the case also has its own laws, which begin to manifest themselves with repeated repetition of random phenomena. If you flip a coin 1000 times, then the "heads" will fall out about half the time, which cannot be said about two or even ten tosses. "Approximately" does not mean half. This, as a rule, may or may not be so. The law does not state anything for certain at all, but gives a certain degree of certainty that some random event will occur. Such patterns are studied by a special section of mathematics - Probability theory . With its help, it is possible with a greater degree of confidence (but still not certain) to predict both the date of the first snowfall and the number of phone calls.
Probability theory is inextricably linked to our daily life. This gives us a wonderful opportunity to establish many probabilistic laws empirically, 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 somehow related to games. The fact is that the case appears here in the most frequent form. And the first probabilistic problems were associated with assessing the chances of players to win.
Modern probability theory has moved away from gambling, but its props are still the simplest and most reliable source of chance. After practicing with the roulette wheel and the dice, you will learn how to calculate the probability of random events in real life situations, which will allow you to assess 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, for no other area of \u200b\u200bmathematics contains so many paradoxes. Like probability theory. And perhaps the main explanation for this is her connection with the real world in which we live.
Many games use a dice with a different number of dots from 1 to 6 on each face. The player rolls the dice, looks at how many dots have dropped out (on the face that is on top), and makes the corresponding number of moves: 1,2,3 , 4,5, or 6. Throwing a die can be considered an experience, experiment, test, and the result obtained is an event. People are usually very interested in guessing the onset of this or that event, predicting its outcome. What predictions can they make when they roll the dice? First prediction: one of the numbers 1, 2, 3, 4, 5 will be dropped, or 6. Do you think the predicted event will come or not? Of course, it will definitely come. An event that will necessarily occur in this experience is called a reliable event.
Second prediction : number 7 will drop out. Do you think the predicted event will come or not? Of course it won't, it's just impossible. An event that cannot occur in a given experience is called impossible event.
Third prediction : number 1 will drop out. What do you think, the predicted event will come or not? We are not in a position to answer this question with full confidence, since the predicted event may or may not occur. An event that in a given experience may or may not occur is called a random event.
The task : Describe the events that are discussed in the tasks below. How reliable, impossible, or accidental.
We flip a coin. The coat of arms appeared. (random)
The hunter shot at the wolf and hit. (random)
The schoolboy goes for a walk every evening. While walking on Monday, he met three acquaintances. (random)
Let's mentally conduct the following experiment: turn the glass of water upside down. If this experiment is carried out not in space, but at home or in the classroom, then the water will pour out. (reliable)
Three shots were fired at the target. " There were five hits "(impossible)
We throw the stone up. The stone remains suspended in the air. (impossible)
We rearrange the letters of the word "antagonism" at random. The word "anachroism" will turn out. (impossible)
№959. Petya conceived natural number... The event is as follows:
a) an even number is conceived; (random) b) an odd number is conceived; (random)
c) a number is conceived that is neither even nor odd; (impossible)
d) a number that is even or odd is conceived. (reliable)
№ 961. Petya and Tolya compare their birthdays. The event is as follows:
a) their birthdays do not match; (random) b) their birthdays are the same; (random)
d) the birthdays of both fall on holidays - New Year (January 1) and Russia's Independence Day (June 12). (random)
№ 962. When playing backgammon, two dice are used. The number of moves that the participant makes is determined by adding the numbers on the two sides of the die, and if a "double" falls out (1 + 1.2 + 2.3 + 3.4 + 4.5 + 5.6 + 6), then the number of moves is doubled. You roll the dice and calculate 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 falls out, but there is no number 0 on the dice).
b) - random (if 1 + 6 or 2 + 5 is rolled).
c) - random (if a combination of 6 +6 falls).
d) - the impossible (there are no combinations of numbers from 1 to 6, the sum of which is 13; this number cannot be obtained even when the "double" appears, since it is odd).
Test yourself. (math 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 the win-win lottery (verified)
It will snow at midnight and the sun will shine in 24 hours. (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 set in a store, for which the manufacturer gives a two-year warranty. Which of the following events are impossible, which are random, which are reliable:
The TV won't break down for a year. (random)
The TV won't break down in two years. (random)
Within two years, you will not have to pay for TV repairs. (reliable)
The TV will break down in the third year. (random)
3) The bus, which carries 15 passengers, has 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 disembark at one stop. (random)
At every stop, at least someone will get out. (random)
There will be a stop where no one gets off. (random)
An even number of passengers will leave at all stops. (impossible)
An odd number of passengers will leave at all stops. (impossible)
Homework : p. 53 №960, 963, 965 (think of two reliable, random and impossible events yourself).
Second lesson.
Check homework... (orally)
a) Explain what a certain, random, and impossible event is.
b) Indicate which of the following events is reliable, which is impossible, which is accidental:
There will be no summer holidays. (impossible)
The sandwich will fall butter down. (random)
The school year will end someday. (reliable)
They'll ask me in class tomorrow. (random)
I will meet a black cat today. (random)
№ 960. You have opened this tutorial on any page and selected the first noun that comes across. The event is as follows:
a) there is a vowel in the spelling of the selected word. ((reliable)
b) there is a letter "o" in the spelling of the selected word. (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 the combination of the smallest numbers 1 + 1, the player makes 4 moves; the combination 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 highest numbers 6 + 6 gives 24 moves, and all others - less than 24 moves)
d) the player must make a two-digit number of moves. (random - for example, a combination of 2 + 3 gives a single-digit number of moves: 5, and the fall of two fours - a two-digit number of moves)
2. Solving problems.
№ 964. The bag contains 10 balls: 3 blue, 3 white and 4 red. Describe the following event:
a) 4 balls were taken out of the bag, and they are all blue; (impossible)
b) 4 balls were taken out of the bag, and they are all red; (random)
c) 4 balls were removed from the bag, and they all turned out to be of a different color; (impossible)
d) 4 balls were taken out of the bag, and there was no black ball among them. (reliable)
Objective 1. The box contains 10 red, 1 green and 2 blue pens. Two items are taken out of the box at random. Which of the following events are impossible, which are random, which are reliable:
a) two red handles are taken out (random)
b) two green handles are taken out; (impossible)
c) two blue handles 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)
Objective 2. Winnie the Pooh, Piglet and all - all - all sit at a round table to celebrate their birthday. How many of all - all - of all the event “Winnie the Pooh and Piglet will be sitting next to each other” is reliable, and how much is it random?
(if all - all - all are only 1, then the event is reliable, if more than 1, then it is random).
Objective 3. Among the 100 charity lottery tickets, 20 winners How many tickets do you need to buy to make the “you won’t win anything” event impossible?
Problem 4. There are 10 boys and 20 girls in the class. Which of the following events are impossible for this class, which are accidental, 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 who were born in different months. (random)
Objective 5. The box contains 3 red, 3 yellow, 3 green balls. We take out 4 balls at random. Consider the event “Among the balls taken out there will be balls of exactly M colors”. For each M from 1 to 4, determine which event is impossible, reliable or accidental, and fill in the table:
Independent work.
I option
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.
Throw the dice. Describe the event:
a) the cube, having fallen, will stand on the edge;
b) one of the numbers will drop out: 1, 2, 3, 4, 5, 6;
c) the number 6 will be dropped;
d) a multiple of 7 will be dropped.
The box contains 3 red, 3 yellow and 3 green balls. Describe the event:
a) all removed balls of the same color;
b) all removed balls of different colors;
c) there are balls of different colors among the balls taken out;
c) among the balls taken out there is a red, yellow and green ball.
II option
Describe the event in question as certain, impossible or accidental:
a) a sandwich that has fallen off the table will fall butter down on the floor;
b) in Moscow it will snow at midnight, and in 24 hours the sun will shine;
c) you will win by participating in the win-win lottery;
d) next year, in May, the first spring thunder will be heard.
All two-digit numbers are written on the cards. Pick one card at random. Describe the event:
a) there was zero on the card;
b) the card contains a multiple of 5;
c) the card has a number that is a multiple of 100;
d) the card has a number greater than 9 and less than 100.
The box contains 10 red, 1 green and 2 blue pens. Two items are taken out of the box at random. Describe the event:
a) two blue handles are removed;
b) two red handles are taken out;
c) two green handles are taken out;
d) green and black handles are taken out.
Homework: 1). Come up with two reliable, random and impossible events.
2). A task . The box contains 3 red, 3 yellow, 3 green balls. Take out N balls at random. Consider the event “among the balls taken out there will be balls of exactly three colors”. For each N from 1 to 9, determine which event is impossible, reliable or accidental, and fill in the table:
Combinatorial problems.
First lesson
Homework check. (orally)
a) we check the tasks that the students came up with.
b) an additional task.
I am reading an excerpt from V. Levshin's book "Three Days in Dwarf".
“At first, to the sounds of a smooth waltz, the numbers formed a group: 1+ 3 + 4 + 2 \u003d 10. Then the young skaters began to swap places, forming more and more new groups: 2 + 3 + 4 + 1 \u003d 10
3 + 1 + 2 + 4 = 10
4 + 1 + 3 + 2 = 10
1 + 4 + 2 + 3 \u003d 10, etc.
This continued until the skaters returned to their original position. "
How many times have they changed places?
Today in the lesson we will learn how to solve such problems. They're called combinatorial.
3. Learning new material.
Objective 1. How many two-digit numbers can be made from the digits 1, 2, 3?
Decision: 11, 12, 13
31, 32, 33. There are 9 numbers in total.
When solving this problem, we carried out an enumeration of all possible options, or, as they usually say in these cases. All possible combinations. Therefore, such tasks are called combinatorial. You have to calculate possible (or impossible) options in life quite often, so it is useful to get acquainted with combinatorial problems.
№ 967. Several countries 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?
Decision. Let's assume the first stripe is white. Then the second stripe can be blue or red, and the third stripe, respectively, red or blue. It turned out two options: white, blue, red or white, red, blue.
Now let 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 are 6 possible options in total. This flag can be used by 6 countries.
So, in solving this problem, we were looking for a way to enumerate possible options. In many cases, it is useful to construct a picture - an enumeration scheme. This is, firstly, clearly, secondly, allows us to take everything into account, not to miss anything.
This scheme is also called the tree of possible options.
Front page
Second lane
Third lane
The resulting combination
№ 968. How many two-digit numbers can be made from the digits 1, 2, 4, 6, 8?
Decision. For the two-digit numbers of interest to us, 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. There will be five two-digit numbers: 2., 22, 24, 26, 28. Similarly, 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 are 20 numbers in total.
Let's build a tree of possible options for solving 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.
By building a tree of possible options, solve the following problems.
№ 971. The leadership of a country decided to make its national flag like this: a circle of a different color is placed on a one-color rectangular background in one of the corners. It was decided to choose from three colors: red, yellow, green. How many variants of such a flag
exist? The figure shows some of the possible options.
Answer: 24 options.
№ 973. a) How many three-digit numbers can be made from the digits 1,3, 5,? (27 numbers)
b) How many three-digit numbers can be made from the digits 1,3, 5, provided that the numbers should not be repeated? (6 numbers)
№ 979. Modern pentathletes for two days participate in competitions in five kinds of sports: show jumping, fencing, swimming, shooting, running.
a) How many options are there for the order of the types of competition? (120 options)
b) How many options are there for the order of passing the types of competition, if it is known that the last type should be running? (24 options)
c) How many options are there for the order of passing the types of competition, if it is known that the last type should be running, and the first - show jumping? (6 options)
№ 981. Two urns contain five balls each in five different colors: white, blue, red, yellow, green. One ball is taken out of each urn at the same time.
a) how many different combinations of balls taken out are there (combinations like "white - red" and "red - white" are considered the same)?
(15 combinations)
b) How many combinations are there in which the removed balls are of the same color?
(5 combinations)
c) how many combinations are there in which the removed balls are of different colors?
(15 - 5 \u003d 10 combinations)
Homework: p. 54, no. 969, 972, to come up with a combinatorial problem ourselves.
№ 969. Several countries 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
Homework check. a) No. 969 and No. 972a) and No. 972b) - build a tree of possible options on the board.
b) verbally check the compiled tasks.
Solving problems.
So, before that, you and I learned how to solve combinatorial problems using a variant tree. Is this a good way? Probably yes, but very cumbersome. Let's try to solve home problem No. 972 in a different way. Who can guess how this can be done?
Answer: for each of the five colors of T-shirts, there are 4 colors of the panties. Total: 4 * 5 \u003d 20 options.
№ 980. The urns have five balls each in five different colors: white, blue, red, yellow, green. One ball is taken out of each urn at the same time. Describe the following event as certain, accidental, or impossible:
a) removed balls of different colors; (random)
b) removed balls of the same color; (random)
c) black and white balls are taken out; (impossible)
d) two balls were taken out, both of which were painted in one of the following colors: white, blue, red, yellow, green. (reliable)
№ 982. A group of tourists is planning to carry out a hike along the route Antonovo - Borisovo - Vlasovo - Gribovo. From Antonovo to Borisovo you can raft along 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 hiking options can tourists choose? How many hiking options can tourists choose, provided they have to use bicycles on at least one of the sections of the route?
(12 route options, 8 of them using bicycles)
Independent work.
Option 1
a) How many three-digit numbers can be made from digits: 0, 1, 3, 5, 7?
b) How many three-digit numbers can be made from 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) How many ways can musketeers be armed?
b) How many weapon options are there if Aramis should wield a sword?
c) How many weapon options are there if Aramis should own a sword, and Porthos should own a pistol?
To the crow somewhere, God sent a piece of cheese, as well as feta cheese, sausage, white and black bread. A crow perched on a spruce tree, got ready for breakfast, but thought: how many ways can you make sandwiches from these products?
Option 2
a) How many three-digit numbers can be made from digits: 0, 2, 4, 6, 8?
b) How many three-digit numbers can be made from digits: 0, 2, 4, 6, 8, provided that the numbers should not be repeated?
Count Monte Cristo decided to present Princess Gaide with earrings, a necklace and a bracelet. Each piece of jewelry must contain one type of gem: diamonds, rubies or garnets.
a) How many options are there for combining gemstone jewelry?
b) How many jewelry options are there if the earrings are to be diamond?
c) How many jewelry options are there, if the earrings are to be diamond and the bracelet is garnet?
For breakfast, you can choose a bun, sandwich or gingerbread with coffee or kefir. How many breakfast options can you compose?
Homework : No. 974, 975. (by compiling a tree of variants and using the rule of multiplication)
№ 974 . a) How many three-digit numbers can be made from the digits 0, 2, 4?
b) How many three-digit numbers can be made from the digits 0, 2, 4, provided that the numbers should not be repeated?
№ 975 . a) How many three-digit numbers can be made up of the digits 1.3, 5.7?
b) How many three-digit numbers can be made from the digits 1.3, 5.7, provided. That numbers shouldn't be repeated?
Problem numbers are taken from the tutorial
"Mathematics-5", I.I. Zubareva, A.G. Mordkovich, 2004.
translate the text into a literal one please.Not in an online translator.
The Golden Gate is a symbol of Kiev, one of the oldest examples of architecture that has survived to our time. The Golden Gate of Kiev was built during the reign of the famous Kiev prince Yaroslav the Wise in 1164. They were originally called Southern and were part of the system defensive fortifications the city, practically not differing from other guard gates of the city. It was the South Gate that the first Russian Metropolitan Hilarion called "Great" in his "Word of Law and Grace." After the majestic Hagia Sophia was built, the “Great” gates became the main land entrance to Kiev from the southwest. Realizing their importance, Yaroslav the Wise ordered to build a small Church of the Annunciation over the gates in order to pay tribute to the dominant Christian religion in the city and in Russia. From that time on, all Russian chronicle sources began to call the Southern Gate of Kiev the Golden Gate. The gate was 7.5 m wide, the passage height 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.
Probability theory, like any branch of mathematics, operates with a certain range of concepts. Most concepts of the theory of probability are defined, but some are taken as primary, not defined, as in geometry a point, a straight line, a plane. The primary concept of the theory of probability is an event. 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 did not happen.
For example, I have a lottery ticket. After the results of the lottery draw are published, the event that interests me is that the winning of a thousand rubles either happens or does not happen. Any event occurs as a result of a test (or experience). Test (or experience) refers to the conditions that result in an event. For example, tossing a coin is a test, and the appearance of a "coat of arms" on it is an event. The event is usually designated by capital Latin letters: A, B, C,…. Events in the material world can be divided into three categories - reliable, impossible and accidental.
A credible event is an event that is known in advance that it will happen. It is denoted by the letter W. So, it is reliable to get no more than six points when throwing an ordinary dice, the appearance of a white ball when removed from an urn containing only white balls, etc.
An impossible event is an event that is known in advance that it will not happen. It is denoted by the letter E. Examples of impossible events are the removal of more than four aces from a regular deck of cards, the appearance of a red ball from an urn containing only white and black balls, etc.
A random event is an event that may or may not occur as a result of a test. Events A and B are called incompatible if the onset of one of them excludes the possibility of another. So the appearance of any possible number of points when throwing a dice (event A) is incompatible with the appearance of another number (event B). An even number of points is inconsistent with an odd number. On the contrary, the loss of even points (event A) and the number of points that is a multiple of three (event B) will not be inconsistent, because the loss of six points means the occurrence of both events A and events B, so that the occurrence of one of them does not exclude the occurrence of the other. You can perform operations with events. The union of two events C \u003d AUB is an event C that occurs if and only if at least one of these events A and B occurs. The intersection of two events D \u003d A ?? B is an event that occurs if and only when events of both A and B.
1.1. Some information from combinatorics
1.1.1. Accommodation
Consider the simplest concepts associated with the choice and location of a 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 from kelements of the set consisting of n various elements.
Example.The following sequences of numbers are 2-element placements from 3 elements of the set (1; 2; 3): 12, 13, 23, 21, 31, 32.
Note that the placements differ in the order of their elements and their composition. Placements 12 and 21 contain the same numbers, but their order is different. Therefore, these placements are considered to be different.
Number of different placements from n elements by k denoted and calculated by the formula:
,
Where n! = 1∙2∙...∙(n - 1)∙ n (reads " n - factorial ").
The number of two-digit numbers that can be formed from the digits 1, 2, 3, provided that no digit is repeated equals:.
1.1.2. Permutations
Definition... Permutations from n elements are called such placements from n elements that differ only in the arrangement of elements.
Number of permutations from n elements P n calculated by the formula: P n=n!
Example.How many ways can 5 people queue 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 rearrange these nelements is called permutation with repetitions.
Example.Let there be 2 identical books among 6 books. Any arrangement of all books on the shelf is a permutation with repetitions.
Number of different permutations with repetitions (from n elements, including kidentical) is calculated by the formula:.
In our example, the number of ways that you can arrange books on the shelf is:.
1.1.3. Combinations
Definition ... Combinations of n elements by k such placements from n elements by kthat differ from one another by at least one element.
Number of different combinations of n elements by k denoted and calculated by the formula:.
By definition, 0! \u003d 1.
The following properties are valid for combinations:
1.
2.
3.
4.
Example. There are 5 flowers in different colors. 3 flowers are chosen for the bouquet. The number of different bouquets of 3 flowers out of 5 is equal to:.
1.2. Random events
1.2.1. Developments
Cognition of reality in natural sciences occurs as a result of tests (experiment, observation, experience).
Test
or experience is the realization of a certain set of conditions, which can be reproduced as many times as desired.
Random
is called an event that may or may not occur as a result of some test (experience).
Thus, the event is considered as a test result.
Example. Tossing a coin is a challenge. The appearance of an eagle when thrown is an event.
The events we observe differ in the degree of possibility of their occurrence and in the nature of their relationship.
The event is called reliable
if it will necessarily occur as a result of this test.
Example. Obtaining a positive or negative mark on the exam by a student 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, which 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, random events will be denoted by large Latin letters A, B, C... A reliable event will be denoted by the letter Ω, the impossible - by Ø.
Two or more events are called equally possible
in this test, if there is reason to believe that none of these events is more possible or less possible than others.
Example.With one throw of the dice, the appearance of 1, 2, 3, 4, 5 and 6 points - all these events are equally possible. It is, of course, assumed that the dice are made of a uniform material and have the correct shape.
Two events are called inconsistent
in a given test, if the appearance of one of them excludes the appearance of the other, and joint
otherwise.
Example. The box contains standard and non-standard parts. Let's take one detail for good luck. The appearance of a standard part eliminates the appearance of a non-standard part. These events are inconsistent.
Several events form complete group of events
in this test, if at least one of them will necessarily 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 denoted by A, then the other is usually denoted by (read "not A»).
Example. Hitting and missing with one shot at a target are opposite events.
1.2.2. Classical definition of probability
Event probability
- a numerical measure of the possibility of its offensive.
Event AND called favorable
event INif whenever an event occurs AND, the event also comes IN.
Developments AND 1 , AND 2 , ..., AND n form case diagram
, if they:
1) are equally possible;
2) pairwise incompatible;
3) form a complete group.
In the case scheme (and only in this scheme) there is a classical definition of the probability P(A) developments AND... Here, a case is called each of the events belonging to the selected full group of equally possible and pairwise incompatible events.
If a n Is the number of all cases in the scheme, and m - the number of cases favorable to the event ANDthen probability of event
AND is defined 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 the event is certain, then each event in the case scheme 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 none of the events in the scheme of events favors the event. therefore m\u003d 0 and therefore 
The probability of a random event is a positive number between zero and one.
Really, random event favored only part of the total number of cases in the scheme of cases. Therefore 0<m<n, and, therefore, 0<m/n<1 и, следовательно, 0 < P (A) < 1.
So, the probability of any event satisfies the inequalities
0 ≤ P (A) ≤ 1.
At present, 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
Task 1.1... What is the probability of an even number of points (event A) occurring with one dice roll?
Decision... Consider the events AND i - dropped i points, i\u003d 1, 2, ..., 6. Obviously, these events form a case diagram. Then the number of all cases n \u003d 6. An even number of points is favored by cases AND 2 , AND 4 , AND 6, i.e. m\u003d 3. Then 
.
Task 1.2... The urn contains 5 white and 10 black balls. The balls are mixed thoroughly and then 1 ball is taken out at random. What is the likelihood that the ball you take out turns out to be white?
Decision... There are 15 cases in total, which form a case diagram. Moreover, the expected event AND - the appearance of a white ball, 5 of them favor, therefore 
.
Task 1.3... The child plays with six letters of the alphabet: A, A, E, K, P, T. Find the probability that he will be able to randomly add the word CARETA (event A).
Decision... The decision is complicated by the fact that among the letters there are the same - two letters "A". Therefore, the number of all possible cases in this test is equal to the number of permutations with repetitions of 6 letters:
.
These cases are equally possible, pairwise incompatible, and form a complete group of events, i.e. form a case diagram. Only one occasion favors an event AND... therefore
.
Task 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 the fulfillment of their desire, if it is customary to distribute places among their friends by lot?
Decision... Let us denote by AND event "fulfillment of the wishes of Tanya and Vanya". 10 people can sit at 10 table! different ways. How many of these n \u003d 10! are equally possible ways favorable for Tanya and Vanya? Tanya and Vanya, sitting side by side, can take 20 different positions. At the same time, eight of their friends can sit at table 8! in different ways, therefore m \u003d 20 ∙ 8 !. Hence, 
.
Task 1.5... A group of 5 women and 20 men selects three delegates. Assuming that each of those present with the same probability can be selected, find the probability that two women and one man will be chosen.
Decision... The total number of equally possible trial outcomes is equal to the number of ways in which three delegates out of 25 can be selected, i.e. ... Let us now count the number of favorable cases, i.e. the number of cases in which the event of interest takes place. A male delegate can be selected in twenty ways. The other two delegates must be women, and you can choose two women out of five. Hence, . therefore
.
Task 1.6. Four balls are randomly scattered over four holes, each ball hits one or another hole with the same probability and independently of the others (there are no obstacles to hitting the same hole for several balls). 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.
Decision. Total number of cases n\u003d 4 4. The number of ways you can select one hole with three balls,. The number of ways in which you can select a hole where there will be one ball,. The number of ways you can choose from the four balls is three to put them in the first hole,. The total number of favorable cases. Event probability: 
Task 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 # 1; b) balls # 1 and # 2.
Decision... a) The total number of possible elementary test outcomes is equal to the number of ways in which six balls out of ten can be extracted, i.e.
Let us find the number of outcomes favorable to the event of interest to us: among the selected six balls there is ball # 1 and, therefore, the other 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 desired probability is equal to the ratio of the number of outcomes favorable to the event under consideration to the total number of possible elementary outcomes:
b) The number of outcomes favorable to the event of interest to us (among the selected balls there are balls # 1 and # 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. Seeking 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
AND is defined by the equality:
,
Where m - the number of trials in which the event AND came n - the total number of tests performed.
J. Bernoulli proved that with an unlimited increase in the number of experiments, the relative frequency of occurrence of an event will practically differ little from a certain constant number. It turned out that this constant number is the probability of an event occurring. Therefore, naturally, the relative frequency of occurrence of an event with a sufficiently large number of tests is called the statistical probability, in contrast to the previously introduced probability.
Example 1.8... How to estimate the approximate number of fish in the lake?
Let in the lake x fish. We throw the net and, let's say, we 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. Suppose that we find m fish in it, among which k labeled. Let the event AND - “the caught fish is marked”. Then by definition of the relative frequency.
But if in the lake x fish and we released into it n labeled, then.
Because R * (AND) » R(AND), Then.
1.2.4. Operations on events. Probability addition theorem
The sum, or a combination of several events is an event consisting in the occurrence of at least one of these events (in the same test).
Amount AND 1 + AND 2 + … + AND n denoted as follows: 
or 
.
Example... Two dice are thrown. Let the event AND consists of getting 4 points on 1 die, and the event IN - in the fall of 5 points on another die. Developments AND and IN are joint. Therefore the event AND +IN consists in dropping 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 AND - winnings for 1 loan, event IN - winnings for 2 loans. Then the event A + B - winning at least one loan (possibly two at once).
By product or the intersection of several events is an event consisting in the joint appearance of all these events (in the same test).
Composition IN events AND 1 , AND 2 , …, AND n denoted as follows:
.
Example. Developments AND and IN consist in the successful completion of the 1st and 2nd rounds, respectively, when entering the institute. Then the event AND× B consists in successfully completing both rounds.
The concepts of the sum and product of events have a clear geometric interpretation. Let the event AND there is a point hitting the area ANDand the event IN - hitting a point in the area IN... Then the event A + B there is a hit of a point in the union of these areas (Fig. 2.1), and the event ANDIN there is a hit of the point in the intersection of these areas (Fig. 2.2).
Figure: 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 be AND and Ā
- opposite events, i.e. A + Ā \u003d Ω, where Ω is a valid event. The addition theorem implies that
P (Ω) \u003d R(AND) + R(Ā
) \u003d 1, therefore
R(Ā
) = 1 – R(AND).
If events AND 1 and AND 2 are consistent, then the probability of the sum of two joint events is:
R(AND 1 + AND 2) = R(AND 1) + R(AND 2) - P ( AND 1 × AND 2).
Probability addition theorems allow you to go from direct calculation of probabilities to determining the probabilities of complex events.
Task 1.8... The shooter fires one shot at the target. The probability of knocking out 10 points (event AND), 9 points (event IN) and 8 points (event FROM) are equal to 0.11, respectively; 0.23; 0.17. Find the probability that the shooter will score less than 8 points with one shot (event D).
Decision... Let's move on to the opposite event - with one shot, the shooter will knock at least 8 points. An event occurs if it happens AND or IN, or FROM, i.e. ... Since events A, B, FROM are pairwise inconsistent, then, by the addition theorem,
from where.
Task 1.9... From the team of the brigade, which consists of 6 men and 4 women, two people are selected to the trade union conference. What is the probability that among the selected at least one woman (event AND).
Decision... If an event occurs AND, then one of the following inconsistent events will surely occur: IN - “a man and a woman were selected”; FROM - “two women were chosen”. Therefore, we can write: A \u003d B + C... Find the probability of events IN and FROM... Two people out of 10 can be selected in ways. Two women out of 4 can be chosen in ways. A man and a woman can be selected in 6 × 4 ways. Then. Since events IN and FROM are inconsistent, then, by the addition theorem,
P (A) \u003d P (B + C) \u003d P (B) + P (C) = 8/15 + 2/15 = 2/3.
Task 1.10. Fifteen textbooks are randomly arranged on a shelf in the library, five of which are bound. The librarian takes three textbooks at random. Find the probability that at least one of the textbooks taken will be bound (event AND).
Decision... First way. The requirement - at least one of the three bound textbooks taken - will be met if any of the following three inconsistent events occurs: IN - one bound textbook, FROM - two bound textbooks, D - three bound textbooks.
Event of interest to us AND can be represented as a sum of events: A \u003d B + C + D... By the addition theorem,
P (A) \u003d P (B) + P (C) + P (D). (2.1)
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 AND (at least one of the three textbooks taken is bound) and Ā
(none of the textbooks taken are bound) are opposite, so P (A) + P (Ā) \u003d 1 (the sum of the probabilities of two opposite events is 1). From here P (A) = 1 – P (Ā). Event probability Ā
(none of the textbooks taken are bound)
Seeking probability
P (A) = 1 - P (Ā) = 1 – 24/91 = 67/91.
1.2.5. Conditional probability. Probability multiplication theorem
Conditional probability P (B/AND) is the probability of event B, calculated on 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 by the conditional probability of the other, calculated under the assumption that the first event has already occurred:
P (A∙B) \u003d P (A) ∙ Р ( IN/AND). (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) \u003d P (A / B) or P (B) = P (B/AND). (2.3)
If events AND and IN are independent, then formulas (2.2) and (2.3) imply
P (A∙B) \u003d P (A)∙P (B). (2.4)
The converse is also true, i.e. if equality (2.4) holds for two events, then these events are independent. Indeed, formulas (2.4) and (2.2) imply
P (A∙B) \u003d P (A)∙P (B) = P (A) × P (B/AND), Whence P (A) = P (B/AND).
Formula (2.2) can be generalized to the case of a finite number of events AND 1 , AND 2 ,…,A n:
P (A 1 ∙AND 2 ∙…∙A n)=P (A 1)∙P (A 2 /AND 1)∙P (A 3 /AND 1 AND 2)∙…∙P (A n/AND 1 AND 2 …A n -1).
Task 1.11... From the urn, in which there are 5 white and 10 black balls, take out two balls in a row. Find the probability that both balls are white (event AND).
Decision ... Consider the events: IN - the first ball taken out is white; FROM - the second ball removed is white. Then A \u003d BC.
The experiment can be done 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 FROMindependent:
P (A) \u003d P (B)∙P (C) \u003d 5/15 × 5/15 \u003d 1/9;
2) no return: the removed ball is set aside. In this case, the events IN and FROM dependent:
P (A) \u003d P (B)∙P (C/IN).
For an event IN the conditions are the same, and for FROM the situation has changed. Happened IN, therefore, there are 14 balls left in the urn, among which 4 are white.
So, .
Task 1.12... Among the 50 light bulbs, 3 are non-standard. Find the probability that two bulbs taken at the same time are non-standard.
Decision ... Consider the events: AND - the first light is non-standard, IN - the second light is non-standard, FROM - both bulbs are non-standard. It's clear that C \u003d A∙IN... Event AND 3 cases out of 50 are favorable, i.e. P (A) \u003d 3/50. If the event AND has already arrived, then the event IN two cases out of 49 are favorable, i.e. P (B/AND) \u003d 2/49. Hence,
.
Task 1.13 ... Two athletes shoot one target independently of each other. The probability of hitting the target of the first athlete is 0.7, and the second is 0.8. What is the likelihood that the target will be hit?
Decision ... The target will be hit if it hits either the first shooter, or the second, or both together, i.e. an event will occur A + Bwhere the event is AND is the first athlete hitting the target, and the event IN - the second. Then
P (A+IN)=P (A)+P (B)–P (A∙IN)=0, 7+0, 8–0, 7∙0,8=0,94.
Task 1.14.In the reading room there are six textbooks on probability theory, of which three are bound. The librarian took two textbooks at random. Find the probability that the two textbooks are bound.
Decision... Let us introduce the notation 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 INis this: P (B/AND) = 2/5.
Looking for the probability that both textbooks are bound, by the multiplication theorem for the probabilities of events is
P (AB) = P (A) ∙ P (B/AND) \u003d 1/2 · ∙ 2/5 \u003d 0.2.
Task 1.15. The workshop employs 7 men and 3 women. Three people were randomly selected by personnel numbers. Find the probability that all selected individuals will be men.
Decision... Let's introduce the notation of events: A - the man was selected first, IN - a man was selected second, FROM - the third is a man. The likelihood that a man will be selected first, 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, provided that two men have already been selected, i.e. conditional probability of an event FROM is this: P (C/AB) = 5/8.
Seeking the probability that all three selected individuals will be men, P (ABC) \u003d P (A) P (B/AND) P (C/AB) \u003d 7/10 2/3 5/8 \u003d 7/24.
1.2.6. Total probability formula and Bayes formula
Let be B 1 , B 2 ,…, B n - pairwise incompatible events (hypotheses) and AND - an event that can occur only in conjunction with one of them.
Let, in addition, we know P (B i) and P (A/B i) (i = 1, 2, …, n).
Under these conditions, the following formulas are valid: 
(2.5)
(2.6)
Formula (2.5) is called total probability formula
... It calculates the probability of an event AND (full probability).
Formula (2.6) is called bayes' formula
... It allows you to recalculate the probabilities of hypotheses if the event AND happened.
When compiling examples, it is convenient to assume that the hypotheses form a complete group.
Target 1.16... The basket contains apples from four trees of the same sort. From the first - 15% of all apples, from the second - 35%, from the third - 20%, from the fourth - 30%. Ripe apples account for 99%, 97%, 98%, 95%, respectively.
a) What is the probability that an apple taken at random will be ripe (event AND).
b) Provided that the random apple was ripe, calculate the probability that it is from the first tree.
Decision... a) We have 4 hypotheses:
B 1 - an apple taken at random is removed from the 1st tree;
B 2 - an apple taken at random is removed from the 2nd tree;
B 3 - an apple taken at random is removed from the 3rd tree;
B 4 - an apple taken at random is taken from the 4th tree.
Their probabilities by condition: P (B 1) = 0,15; P (B 2) = 0,35; P (B 3) = 0,2; P (B 4) = 0,3.
Conditional event probabilities AND:
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 turns out to be ripe is found by 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: 
.
Task 1.17. A white ball is dropped into an urn containing two balls, after which one ball is taken at random from it. Find the probability that the removed ball will turn out to be white if all possible assumptions about the initial composition of the balls (by color) are equally possible.
Decision... Let us denote by AND event - the white ball is removed. 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), the probability of each of the hypotheses is 1/3, i.e.
P (B 1) = P (B 2) \u003d P (B 3) = 1/3.
The conditional probability that a white ball will be drawn, provided that there were no white balls initially in the urn, P (A/B 1) \u003d 1/3. The conditional probability that a white ball will be drawn, given that there was originally one white ball in the urn, P (A/B 2) \u003d 2/3. The conditional probability that a white ball will be drawn, given that the urn originally contained two white balls P (A/B 3)=3/ 3=1.
We find the desired probability that the white ball will be drawn using the formula for the total probability:
R(AND)=P (B 1)∙P (A/B 1)+P (B 2)∙P (A/B 2)+P (B 3)∙P (A/B 3) \u003d 1/3 1/3 + 1/3 2/3 + 1/3 1 \u003d 2/3 .
Target 1.18... Two machines produce identical parts that go to a common conveyor. The productivity of the first machine is twice that of the second. The first automatic machine produces on average 60% of the 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.
Decision... Let us denote by AND the event is an excellent quality item. Two assumptions can be made: B 1 - the part is produced by the first machine, moreover (since the first machine produces twice as many parts as the second) P (A/B 1) = 2/3; B 2 - the part is 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 automatic 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 automatic machine, P (A/B 1)=0,84.
The probability that a part taken at random turns out to be of excellent quality, according to the formula of total probability, is
P (A)=P (B 1) ∙P (A/B 1)+P (B 2) ∙P (A/B 2) \u003d 2/3 0.6 + 1/3 0.84 \u003d 0.68.
Seeking the probability that the taken excellent part is produced by the first automaton, according to the Bayes formula is equal to
Target 1.19... There are three lots of parts with 20 parts each. The number of standard parts in the first, second and third lots is 20, 15, 10, respectively. A part that turned out to be standard was taken at random from the selected lot. The parts are returned to the batch and, for the second time, a part is randomly removed from the same batch, which also turns out to be standard. Find the probability that the parts were extracted from the third batch.
Decision... Let us denote by AND event - In each of the two tests (with return) a standard part was removed. 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 extracted from the third batch.
The details were taken at random from a given batch, so the probabilities of hypotheses are the same: P (B 1) = P (B 2) = P (B 3) = 1/3.
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, therefore P (A/B 1) = 1.
Find the conditional probability P (A/B 2), i.e. the probability that from the second batch two standard parts will be sequentially removed (with return): P (A/B 2)= 15/20 ∙ 15/20 = 9/16.
Find the conditional probability P (A/B 3), i.e. the probability that two standard parts will be sequentially removed from the third batch (with return): P (A/B 3) \u003d 10/20 10/20 \u003d 1/4.
The Bayesian formula for the probability that both extracted standard parts are from a third batch is 
1.2.7. Repeated tests
If several tests are performed, and the probability of an event ANDin each trial does not depend on the outcomes of other trials, then such trials are called independent regarding the event A. In various independent trials event ANDcan have either different probabilities or the same probability. We will further consider only such independent tests in which the event ANDhas the same probability.
Let it be produced pindependent tests, in each of which an event ANDmay or may not appear. Let us agree to assume that the probability of an event ANDin each test is the same, namely equal r.Therefore, the probability of non-occurrence of the event ANDin each test is also constant and equal to 1– r. Such a probabilistic scheme is called bernoulli scheme... Let us set ourselves the task of calculating the probability that for pbernoulli test event AND will come true exactly k times ( k - number of successes) and, therefore, will not be realized p- time. It is important to emphasize that it is not required that the event ANDrepeated exactly k times in a specific sequence. The required probability is denoted P n (k).
For example, the symbol R 5 (3) means the probability that in five tests the event will occur exactly 3 times and, therefore, will not occur 2 times.
The problem can be solved using the so-called bernoulli formulas, which looks like: 
.
Task 1.20.The probability that the power consumption during one day will not exceed the established rate is r\u003d 0.75. Find the probability that in the next 6 days the power consumption for 4 days will not exceed the norm.
Decision. The probability of normal consumption of electricity during each of 6 days is constant and equal to r\u003d 0.75. Therefore, the probability of excessive consumption of electricity per day is also constant and equal to q \u003d1–r=1–0,75=0,25.
The sought probability by the Bernoulli formula is
.
Task 1.21... Two equivalent chess players play chess. Which is more likely: to win two games out of four or three games out of six (draws are not taken into account)?
Decision... Equivalent chess players play, so the probability of winning r \u003d 1/2, therefore, the probability of losing q is also 1/2. Because in all games the probability of winning is constant and it does not matter in what sequence the games will be won, the Bernoulli formula is applicable.
Let's find the probability that two games out of four will be won:
Let us find the probability that three games out of six will be won:
Because P 4 (2) > P 6 (3), it is more likely to win two games out of four than three out of six.
However, one can see that using the Bernoulli formula for large values n it is quite difficult, since the formula requires performing actions on huge numbers and therefore in the process of calculations errors accumulate; 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 carrying out a large number of tests according to the Bernoulli scheme (with n \u003d\u003e ∞) and with a small number of favorable outcomes k (it is assumed that the probability of success p is small), Bernoulli's formula approaches Poisson's formula 
.
Example 1.22. The probability of a marriage when the enterprise releases a unit of production is p\u003d 0.001. What is the probability that with the release of 5000 units of products there will be less than 4 defective ones ( AND Decision... Because n is large, we use the local Laplace theorem: 
We calculate x:
Function 
- even, therefore φ (–1.67) \u003d φ (1.67).
According to the table in Appendix A.1, we find φ (1.67) \u003d 0.0989.
Seeking probability P 2400 (1400) = 0,0989.
3. Laplace's integral theorem
If the probability r occurrence of an event A in each test according to the Bernoulli scheme is constant and different from zero and one, then with a large number of tests n , probability P n (k 1 , k 2) occurrence of the event A in these tests from k 1 to k 2 times is approximately equal to
R p(k 1 , k 2) \u003d Φ ( x "") – Φ ( x "), where 
- Laplace function,
The definite integral in the Laplace function is not calculated in the class of analytic functions, therefore, to calculate it, Table 1 is used. A.2 given in the appendix.
Example 1.24.The probability of occurrence of an event in each of one hundred independent tests is constant and equal to p \u003d 0.8. Find the probability that the event will occur: a) at least 75 times and no more than 90 times; b) at least 75 times; c) no more than 74 times.
Decision... We will use Laplace's integral theorem:
R p(k 1 , k 2) \u003d Φ ( x "") – Φ( x "), where Ф ( x) Is the Laplace function,
a) By condition, n = 100, p = 0,8, q = 0,2, k 1 = 75, k 2 \u003d 90. Calculate x "" and x " :
Considering that the Laplace function is odd, i.e. Ф (- x) \u003d - Ф ( x), we get
P 100 (75; 90) \u003d Ф (2.5) - Ф (–1.25) \u003d Ф (2.5) + Ф (1.25).
According to the table. A.2. we will find applications:
F (2.5) \u003d 0.4938; Ф (1.25) \u003d 0.3944.
Seeking probability
P 100 (75; 90) = 0,4938 + 0,3944 = 0,8882.
b) The requirement that the event appears at least 75 times means that the number of occurrences of the event can be equal to 75, or 76, ..., or 100. Thus, in this case, one should accept k 1 = 75, k 2 \u003d 100. Then 
.
According to the table. A.2. applications, we find Ф (1.25) \u003d 0.3944; Ф (5) \u003d 0.5.
Seeking probability
P 100 (75;100) = (5) – (–1,25) = (5) + (1,25) = 0,5 + 0,3944 = 0,8944.
c) Event - " AND appeared at least 75 times "and" AND appeared no more than 74 times "are opposite, so the sum of the probabilities of these events is 1. Therefore, the desired probability
P 100 (0;74) = 1 – P 100 (75; 100) = 1 – 0,8944 = 0,1056.
Lesson topic: "Random, reliable and impossible events"
Place of the lesson in the curriculum: “Combinatorics. Random events "lesson 5/8
Lesson type: Lesson in the formation of new knowledge
Lesson objectives:
Educational:
o introduce the definition of a random, reliable and impossible event;
o teach in the process of a real situation to define the terms of the theory of probability: reliable, impossible, equiprobable events;
Developing:
o promote the development of logical thinking,
o cognitive interest of students,
o the ability to compare and analyze,
Educational:
o fostering interest in the study of mathematics,
o development of students' outlook.
o possession of intellectual skills and mental operations;
Teaching methods: explanatory and illustrative, reproductive, mathematical dictation.
UMK: Mathematics: textbook for 6th grade. under the editorship of, and others, publishing house "Education", 2008, Mathematics, 5-6: book. for teacher / [, [ ,]. - M.: Education, 2006.
Didactic material: posters on the board.
Literature:
1. Mathematics: textbook. for 6 cl. general education. institutions /, etc.]; ed. ,; Grew up. acad. Sciences, Ros. acad. education, publishing house "Education". - 10th ed. - M.: Education, 2008.-302 p .: ill. - (Academic school textbook).
2. Mathematics, 5-b: book. for teacher / [,]. - M.: Education, 2006 .-- 191 p. : ill.
4. Solving problems in statistics, combinatorics and probability theory. 7-9 grades. / author - comp. ... Ed. 2nd, rev. - Volgograd: Teacher, 2006.-428 p.
5. Lessons in mathematics using information technology. 5-10 grades. Methodical - a manual with an electronic attachment / etc. 2nd ed., Stereotype. - M .: Publishing house "Globus", 2010. - 266 p. (Modern school).
6. Teaching mathematics in modern schools. Guidelines. Vladivostok: PIPPKRO Publishing House, 2003.
LESSON PLAN
I. Organizational moment.
II. Oral work.
III. Learning new material.
IV. Formation of skills and abilities.
V. Lesson summary.
V. Homework.
DURING THE CLASSES
1. Organizational moment
2. Updating knowledge
15*(-100) |
Oral work:
3. Explanation of the new material
Teacher: Our life is a lot of chance. There is such a science "Theory of Probabilities". Using her language, you can describe many phenomena and situations.
Such ancient generals as Alexander the Great or Dmitry Donskoy, preparing for battle, relied not only on the valor and skill of warriors, but also on chance.
Many people love mathematics for the eternal truths twice two are always four, the sum of even numbers is even, the area of \u200b\u200ba rectangle is equal to the product of its adjacent sides, etc. In any problems that you solved, everyone gets the same answer - you just need not make mistakes in the solution.
Real life is not so simple and straightforward. The outcomes of many phenomena cannot be predicted in advance. It is impossible, for example, to say for sure which side the coin tossed up will fall, when the first snow falls next year, or how many people in the city will want to make a phone call in the next hour. Such unpredictable phenomena are called random .
However, the case also has its own laws, which begin to manifest themselves with repeated repetition of random phenomena. If you flip a coin 1000 times, then "heads" will fall out about half of the time, which cannot be said about two or even ten tosses. "Approximately" does not mean half. This, as a rule, may or may not be so. The law does not state anything for certain at all, but gives a certain degree of certainty that some random event will occur.
Such patterns are studied by a special section of mathematics - Probability theory . With its help, it is possible with a greater degree of confidence (but still not certain) to predict both the date of the first snowfall and the number of phone calls.
Probability theory is inextricably linked to our daily life. This gives us a wonderful opportunity to establish many probabilistic laws empirically, 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 somehow related to games. The fact is that the case appears here in the most frequent form. And the first probabilistic problems were associated with assessing the chances of players to win.
Modern probability theory has moved away from gambling, but its props are still the simplest and most reliable source of chance. After practicing with a roulette wheel and a dice, you will learn how to calculate the probability of random events in real life situations, which will allow you to assess 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, for no other area of \u200b\u200bmathematics contains so many paradoxes. Like probability theory. And, perhaps, the main explanation for this is her connection with the real world in which we live.
Many games use a dice with a different number of dots from 1 to 6 on each face. The player throws a dice, looks at how many dots have dropped out (on the face that is on top), and makes the corresponding number of moves: 1,2,3 , 4,5, or 6. Throwing a die can be considered an experience, experiment, test, and the result obtained is an event. People are usually very interested in guessing the onset of this or that event, predicting its outcome. What predictions can they make when they roll the dice?
First prediction: one of the numbers 1, 2, 3, 4, 5 will be dropped, or 6. Do you think the predicted event will come or not? Of course, it will definitely come.
An event that will necessarily occur in this experience is called reliableevent.
Second prediction : number 7 will drop out. Do you think the predicted event will come or not? Of course it won't, it's just impossible.
An event that cannot occur in a given experience is called impossible event.
Third prediction : number 1 will drop out. Do you think the predicted event will come or not? We are not in a position to answer this question with full confidence, since the predicted event may or may not occur.
Events that, under the same conditions, may or may not happen are called random.
Example. The box contains 5 candies in a blue wrapper and one in a white one. Without looking into the box, they take out one candy at random. Can you tell in advance what color it will be?
The task : Describe the events that are discussed in the tasks below. How reliable, impossible, or accidental.
1. Flip a coin. The coat of arms appeared. (random)
2. The hunter shot at the wolf and hit. (random)
3. The student goes for a walk every evening. During a walk on Monday, he met three acquaintances. (random)
4. Let's mentally carry out the following experiment: turn the glass of water upside down. If this experiment is carried out not in space, but at home or in the classroom, then the water will pour out. (reliable)
5. Three shots were fired at the target. " There were five hits " (impossible)
6. Throw the stone up. The stone remains suspended in the air. (impossible)
ExamplePetya conceived a natural number. The event is as follows:
a) an even number is conceived; (random)
b) an odd number is conceived; (random)
c) a number is conceived that is neither even nor odd; (impossible)
d) a number that is even or odd is conceived. (reliable)
Events that under these conditions have equal chances are called equiprobable.
Random events that have an equal chance are called equally possible or equiprobable .
Place a poster on the board.
In the oral exam, the student takes one of the tickets laid out in front of him. The chances of taking any of the exam tickets are equal. It is equally probable to get any number of points from 1 to 6 when throwing a dice, as well as "heads" or "tails" when throwing a coin.
But not all events are equally possible... The alarm may not ring, the light bulb may burn out, the bus may break down, but under normal conditions such events unlikely. It is more likely that the alarm will ring, the light will come on, and the bus will move.
Some events chancesmore, it means they are more likely - closer to reliable. Others have less chances, they are less likely - closer to impossible.
Impossible events have no chances to happen, and reliable events have all the chances to happen, under certain conditions they will definitely happen.
ExamplePetya and Kolya compare their birthdays. The event is as follows:
a) their birthdays do not match; (random)
b) their birthdays are the same; (random)
d) the birthdays of both fall on holidays - New Year (January 1) and Russia's Independence Day (June 12). (random)
3. Formation of skills and abilities
Problem from the textbook number 000. Which of the following random events are reliable, possible:
a) the turtle will learn to speak;
b) the water in the kettle on the stove will boil;
d) you win by participating in the lottery;
e) you will not win by participating in a win-win lottery;
f) you will lose a game of chess;
g) tomorrow you will meet an alien;
h) the weather will deteriorate next week; i) you pressed the call, but it did not ring; j) today is Thursday;
k) after Thursday it will be Friday; l) Will it be Thursday after Friday?
The boxes contain 2 red, I yellow and 4 green balls. Three balls are taken out of the box at random. Which of the following events are impossible, random, reliable:
A: three green balls will be drawn;
Q: three red balls will be drawn;
C: balls of two colors will be drawn;
D: balls of the same color will be drawn;
E: there is blue among the elongated balls;
F: there are balls of three colors among the elongated ones;
G: Are there two yellow balls among the outstretched ones?
Test yourself. (math dictation)
1) Indicate which of the following events are impossible, which are reliable, which are random:
· Football match "Spartak" - "Dynamo" will end in a draw (random)
You will win by participating in the win-win lottery ( reliable)
Snow will fall at midnight and the sun will shine in 24 hours (impossible)
· Tomorrow there will be a test in mathematics. (random)
· You will be elected President of the United States. (impossible)
· You will be elected President of Russia. (random)
2) You bought a TV set in a store, for which the manufacturer gives a two-year warranty. Which of the following events are impossible, which are random, which are reliable:
· The TV will not break down within a year. (random)
TV will not break down for two years ... (random)
· Within two years, you will not have to pay for TV repairs. (reliable)
· The TV will break down in the third year. (random)
3) The bus with 15 passengers will have 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 disembark at one stop. (random)
· At every stop at least someone gets off. (random)
· There will be a stop where no one gets off. (random)
· An even number of passengers will leave at all stops. (impossible)
· An odd number of passengers will leave at all stops. (impossible)
Lesson summary
Questions to students:
What events are called random?
What events are called equiprobable?
What events are called credible? impossible?
What events are called more likely? less likely?
Homework : {!LANG-3a3bd07cf949bda7c9e88e192a9da2a3!}
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