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Bernoulli tests presentation. Presentation on the topic "Bernoulli formula". III. Learning new material

A series of independent tests are carried out in
each of which has 2 possible outcomes,
which we will conditionally call Success and Failure.
For example, a student takes 4 exams, each
of which 2 outcomes are possible Success: student
passed the exam and Failure: did not pass.

The probability of success in each test is equal to
p. The probability of Failure is q=1-p.
You need to find the probability that in the series
out of n trials success will occur m times
Pn(m)

Bm Ó Ó ... Ó Í ... Í
Í Ó ... Ó Í ... Í ...
Í Í ... Í Ó ... Ó
In each case, Success occurs m times, and
Fail (n-m) times.
Number
everyone
combinations
equals
number
ways from n tests to choose those m, in
of which there was Success, i.e. Cm
n

The probability of each such combination is
theorem
about
multiplication
probabilities
will be Pmqn-m.
Since these combinations are incompatible, then
the desired probability of the event Bm will be
Pn (m) p q
m
n m
... p q
m
n m
vâñåãî C ñëàãàåì û õ C p q
m
n
m
n
m
n m

Pn (m) C p q
m
n
m
n m

It is known that if the coin lands on heads, the student
goes to the cinema if the coin lands on heads

students. What is the probability that
1) three of them will be at the lecture
2) there will be at least 3 students at the lecture
2) will at least one of the students attend the lecture?

1) In this problem, a series of n=5 is carried out
independent tests. Let's call it Success
going to a lecture (falling heads) and
Failure is a trip to the cinema (the coat of arms falls out).
p=q=1/2.
Using Bernoulli's formula we find the probability that
what will happen three times in 5 tosses of a coin?
success:
3
2
1 1
P5(3)C
2 2
5! 1 1
1
10
0,3125
3!2! 8 4
32
3
5

To find the probability that with 5 tosses
at least once the coin lands on heads,
let's move on to the possibility of the opposite
events - the coin will appear as a coat of arms all 5 times:
P5 (0).
Then the desired probability will be: P = 1 - P5 (0).
According to Bernoulli's formula:
0
5
1 1
P5(0)C
2 2
0
5
5
1
0,03125
2

Then the probability of the desired event will be
P 1 0.03125 0.96875

Bernoulli
the student is walking
in the cinema, if the coin lands heads, the student goes for it
lecture. 5 students tossed a coin. What is the most
likely number of students attending the lecture?
Probability
the winnings for 1 ticket is 0.2. What is the most
likely number of winning tickets?

Most likely number of successes in a scheme
Bernoulli

np q k np p

Most likely number of successes in a scheme
Bernoulli
Formula for most likely number of successes
np q k np p
If np-q is an integer, then this interval contains 2
integers. Both are equally likely.
If np-q is a non-integer number, then this interval contains 1
integer

Most likely number of successes in a scheme
Bernoulli
Example It is known that if a coin lands on heads,

– a student goes to a lecture. 5 tossed a coin

students going to a lecture?
np q k np p
n 5
1
p q
2

Most likely number of successes in a scheme
Bernoulli
Example It is known that if a coin lands on heads,
student goes to the cinema if the coin lands on heads
– a student goes to a lecture. 5 tossed a coin
students. What is the most likely number
students going to a lecture?
np q k np p
n 5
1
p q
2
1 1
np q 5 2
2 2
1 1
np p 5 3
2 2

Most likely number of successes in a scheme
Bernoulli
Example It is known that if a coin lands on heads,
student goes to the cinema if the coin lands on heads
– a student goes to a lecture. 5 tossed a coin
students. What is the most likely number
students going to a lecture?
2
3
3
2
5
1 1
1 10 5
P5 (2) C52 10
32 16
2 2
2
5
1 1
1 10 5
P5 (3) C53 10
32 16
2 2
2

Most likely number of successes in a scheme
Bernoulli
Example It is known that if a coin lands on heads,
student goes to the cinema if the coin lands on heads
– a student goes to a lecture. 5 tossed a coin
students. What is the most likely number
students going to a lecture?
probability, Pn(k)
Probabilities of the number of students attending
lecture
0,35
0,3
0,25
0,2
0,15
0,1
0,05
0
0
1
2
3
number of students, k
4
5

Most likely number of successes in a scheme
Bernoulli
Example 10 lottery tickets were purchased.

tickets?
np q k np p
n 10
p 0.2 q 0.8

Most likely number of successes in a scheme
Bernoulli
Example 10 lottery tickets were purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
np q k np p
n 10
p 0.2 q 0.8
np q 10 0.2 0.8 1.2
np p 10 0.2 0.2 2.2

Most likely number of successes in a scheme
Bernoulli
Example 10 lottery tickets were purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
P10 (2) C 0.2 0.8
2
10
2
8
45 0, 04 0,16777216=
=0,301989888

Most likely number of successes in a scheme
Bernoulli
Example 10 lottery tickets were purchased.
The probability of winning on 1 ticket is 0.2.
What is the most likely number of winners
tickets?
Probabilities of the number of winning tickets
probability, Pn(k)
0,35
0,3
0,25
0,2
0,15
0,1
0,05
0
0
1
2
3
4
5
6
number of tickets, k
7
8
9
10

Most likely number of successes in a scheme
Bernoulli

10 contracts concluded

pay the insurance amount

one of the agreements

than under three contracts
d) find the most probable number of contracts, according to
who will have to pay the insurance amount

Most likely number of successes in a scheme
Bernoulli
Example On average, 20% of contracts are insurance
the company pays the insurance amount.
10 contracts concluded
a) Find the probability that in three
pay the insurance amount
0,201327

Most likely number of successes in a scheme
Bernoulli
Example On average, 20% of contracts are insurance
the company pays the insurance amount.
10 contracts concluded
b) The insured amount will not have to be paid either
one of the agreements
0,107374

Most likely number of successes in a scheme
Bernoulli
Example On average, 20% of contracts are insurance
the company pays the insurance amount.
10 contracts concluded
c) the insurance amount will have to be paid no more than,
than under three contracts
0,753297

If n is large, then using the formula
Pn (m) C p q
m
n
m
n m
difficult
Therefore, approximate formulas are used

Theorem: If the probability p of the occurrence of event A
in each test is close to zero,
and the number of independent tests n is quite large,
then the probability Pn(m) that in n independent trials
event A will occur m times, approximately equal to:
Pn(m)
m
m!
e
where λ=np
This formula is called Poisson's formula (law of rare events)

Pn(m)
m
m!
e, np
Usually the approximate Poisson formula is used,
when p<0,1, а npq<10.

Example Let it be known that in the manufacture of a certain drug
defective (number of packages that do not meet the standard)
is 0.2%. Approximately estimate the probability that
out of 1000 randomly selected packages there will be three packages,
not meeting the standard.
Pn(k)
k
k!
P1000 (3) ?
e,
n.p.

no more than 5 contracts are connected.

Example: On average, 1% of contracts are covered by an insurance company
pays the insurance amount. Find the probability that from
100 contracts with the occurrence of an insured event will be
no more than 5 contracts are connected.

Bernoulli's formula

Belyaeva T.Yu. GBPOU KK "AMT" Armavir Mathematics teacher

One of the founders of probability theory and mathematical analysis

Foreign member of the Paris Academy of Sciences (1699) and the Berlin Academy of Sciences (1701)

Older brother of Johann Bernoulli (the most famous representative of the Bernoulli family)

Jacob Bernoulli (1654 – 1705)

Swiss mathematician

Let it be produced P independent trials, in each of which the probability that event A will occur is equal to R , and therefore the probability that it will not happen is equal to q = 1 - p .

We need to find the probability that when P successive trials, event A will occur exactly T once.

We denote the desired probability R P ( T ) .

It's obvious that

p 1 (1) = p, p 1 (0) = q

R 1 (1) + p 1 (0) = p + q = 1

With two tests:

4 possible outcomes:

p 2 (2) = p 2 ; р 2 (1) = 2р·q; p 2 (0) = q 2

R 2 (2) + p 2 (1) + p 2 (0) = (p + q) 2 = 1

With three tests:

8 possible outcomes:

We get:

p 3 (2) = 3p 2 q

p 3 (1) = 3pq 2

R 3 (3) + p 3 (2) + p 3 (1) + p 3 (0) = (p + q) 3 = 1

Task 1.

The coin is tossed 8 times. What is the probability that the coat of arms will appear 4 times?

Task 2.

There are 20 balls in an urn: 15 white and 5 black. 5 balls were taken out in a row, and each ball taken out was returned to the urn before the next ball was taken out. Find the probability that out of five balls drawn there will be 2 white ones.

Formulas for finding the probability that V P tests the event will come :

A) less than t times

R P (0) + … + p P (t-1)

b) more than t times

R P (t+1) + … + p P (P)

V) no more than t times

R P (0) + … + p P (T)

G) at least t times

R P (t) + … + r P (P)

Task 3.

The probability of producing a non-standard part on an automatic machine is 0.02. Determine the probability that among six parts taken at random there will be more than 4 standard ones.

Event A - « more than 4 standard parts" (5 or 6) means

« no more than 1 defective part" (0 or 1)

Let it be produced P independent tests. In each such trial, event A may or may not occur. The probability of occurrence of event A is known.

You need to find such a number μ (0, 1, …, n), for which the probability P n (μ) will be the greatest.

Task 4.

The share of premium products at this enterprise is 31%. What is the most probable number of premium products if a batch of 75 products is selected?

According to the condition: n = 75, p = 0.31, q = 1 - 0.31 = 0.69

Task 6.

Two shooters are shooting at a target. The probability of a miss with one shot for the first shooter is 0.2, and for the second - 0.4. Find the most probable number of volleys in which there will be no hits on the target if the shooters fire 25 volleys.

According to the condition: n = 25, p = 0.2·0.4 = 0.08, q = 0.92

FEDERAL AGENCY FOR EDUCATION

State educational institution

higher professional education

"MATI" - RUSSIAN STATE TECHNOLOGICAL UNIVERSITY NAMED AFTER K.E. TSIOLKOVSKY

Department of “System Modeling and Information Technologies”

Repetition of tests. Bernoulli circuit

Guidelines for practical exercises

in the discipline "Higher Mathematics"

Compiled by: Egorova Yu.B.

Mamonov I.M.

Moscow 2006 introduction

The guidelines are intended for full-time and evening students of Faculty No. 14, specialties 150601, 160301, 230102. The guidelines highlight the basic concepts of the topic and determine the sequence of studying the material. A large number of examples discussed help in the practical development of the topic. Methodical instructions serve as a methodological basis for practical classes and individual assignments.

BERNOULLI SCHEME. BERNOULLI FORMULA

Bernoulli scheme- a scheme of repeated independent tests in which some event A can be repeated many times with constant probability R (A)= R .

Examples of tests carried out using the Bernoulli scheme: repeated tossing of a coin or dice, manufacturing a batch of parts, shooting at a target, etc.

Theorem. If the probability of an event occurring A in each test is constant and equal R, then the probability that the event A will come m once every n tests (no matter in what sequence), can be determined by Bernoulli’s formula:

Where q = 1 – p.

EXAMPLE 1. The probability that electricity consumption during one day will not exceed the established norm is equal to p= 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 electricity consumption for 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 Bernoulli’s formula is equal to:

EXAMPLE 2. The shooter fires three shots at the target. The probability of hitting the target with each shot is equal to p= 0,3. Find the probability that: a) one target is hit; b) all three targets; c) not a single target; d) at least one target; e) less than two targets.

SOLUTION. The probability of hitting the target with each shot is constant and equal to R=0.75. Therefore, the probability of a miss is equal to q = 1 R= 1  0.3= 0.7. Total number of experiments performed n=3.

a) The probability of hitting one target with three shots is equal to:

b) The probability of hitting all three targets with three shots is equal to:

c) The probability of three misses with three shots is equal to:

d) The probability of hitting at least one target with three shots is equal to:

e) The probability of hitting less than two targets, that is, either one target or none:

Local and integral theorems of Moivre-Laplace

If a large number of tests are performed, then calculating probabilities using Bernoulli's formula becomes technically difficult, since the formula requires operations with huge numbers. Therefore, there are simpler approximate formulas for calculating probabilities at large n. These formulas are called asymptotic and are determined by Poisson's theorem, local and integral theorem of Laplace.

Local theorem of Moivre-Laplace. A A will happen m once every n n (n →∞ ), is approximately equal to:

where is the function
and the argument

The more n, the more accurate the calculation of probabilities. Therefore, it is advisable to apply the Moivre-Laplace theorem when npq  20.

f ( x ) special tables have been compiled (see Appendix 1). When using the table you need to keep in mind function properties f(x) :

Function f(x) is even f(  x)=f(x) .

At X ∞ function f(x) 0. In practice, we can assume that already at X>4 function f(x) ≈0.

EXAMPLE 3. Find the probability that the event A will occur 80 times in 400 trials if the probability of the event occurring A in each trial is equal p= 0,2.

SOLUTION. By condition n=400, m=80, p=0,2, q=0.8. Hence:

Using the table, we determine the value of the function f (0)=0,3989.

Integral theorem of Moivre-Laplace. If the probability of an event occurring A in each trial is constant and different from 0 and 1, then the probability that the event A comes from m 1 before m 2 once every n tests with a sufficiently large number n (n →∞ ), is approximately equal to:

Where
 integral or Laplace function,

To find the values of a function F( x ) Special tables have been compiled (for example, see Appendix 2). When using the table you need to keep in mind properties of the Laplace function Ф(x) :

Function Ф(x) is odd F(  x)=  Ф(x) .

At X ∞ function Ф(x) 0.5. In practice, we can assume that already at X>5 function Ф(x) ≈0,5.

F (0)=0.

EXAMPLE 4. The probability that the part did not pass the quality control inspection is 0.2. Find the probability that among 400 parts there will be from 70 to 100 parts untested.

SOLUTION. By condition n=400, m 1 =70, m 2 =100, p=0,2, q=0.8. Hence:

Using the table that shows the values of the Laplace function, we determine:

Ф(x 1 ) = F(  1,25 )=  F( 1,25 )=  0,3944; Ф(x 2 ) = F( 2,5 )= 0,4938.

“Elements of mathematical statistics” - Confidence interval. The science. Classification of hypotheses. Parts are made on different machines. Verification rules. Correlation dependence. Addiction. A set of criterion values. Find the confidence interval. Calculation of confidence intervals for unknown variance. Normal distribution.

“Probability and mathematical statistics” - Accuracy of the obtained values. Code for the safe. Descriptive statistics. Apple. Let's consider the events. Multiplication rule. Two shooters. Comparison curricula. Caramel. Examples of bar charts. Math grades. Multiplication rule for three. White and red roses. 9 different books. The winter vacation.

"Fundamentals of mathematical statistics" - Conditional probability. Table of standardized values. Properties of the Student distribution. Confidence interval of mathematical expectation. Sample mean. Distribution. One test can be considered as a series of one test. Quantile – to the left there should be a number of values corresponding to the quantile index.

“Probability theory and statistics” - Interval boundaries. Critical areas. Probability multiplication theorem. Distribution of a normal random variable. Derivation of Bernoulli's formula. Laws of distribution of random variables. The formulation of the LBC. The meaning and formulation of the central limit theorem. Connection of nominal features. Stochastic dependence of two random variables.

"Statistical Research" - Relevance. Statistical characteristics and research. Plan. Range is the difference between the largest and smallest values of a data series. Types of statistical observation. Do you enjoy studying mathematics? Let's look at a series of numbers. Who helps you understand a difficult topic in mathematics? Do you need mathematics in your future profession?

“Basic statistical characteristics” - Basic statistical characteristics. Find the arithmetic mean. Petronius. Scope. Fashion series. The arithmetic mean of a series of numbers. Row range. Median of the series. Statistics. Median. School notebooks.

There are a total of 17 presentations in the topic

Municipal educational institution "Rudnogorsk secondary school"

Development of a lesson on probability theory

in 10th grade

on this topic

« Independent retesting.

Bernoulli's theorem »

Mathematic teacher

Municipal educational institution "Rudnogorskaya Sosh"

Chibysheva I.A.

“...Chance mainly

depends on our knowledge..."

Jacob Bernoulli

Subject "»

Class:10

Lesson objectives:

Educational:

Tasks:

Lesson type: combined.

Teaching methods: conversation, written exercises.

Equipment: computer, multimedia projector. presentation, handouts

Lesson plan:

Organizational stage -2 min

Updating basic knowledge – 3 min

Stage of learning new material – 10 min

Stage of generalization and systematization of knowledge -20 min

Homework -3 min

Summing up the lesson - 2 min

Reflection -5 min.

DURING THE CLASSES

I. Organizational moment.

II. Updating knowledge

Let us recall the basic concepts and formulas of combinatorics.

1. What is the factorial of n? (This is the product of the first n natural numbers from 1 to n.)
2. In how many ways can 4 different books be arranged on a shelf? (3! = 3 2 1. This is the number of permutations of 3 elements.)
3. In how many ways can I, II, III places be distributed among 7 participants in the competition? (7 6 5 = 210. This is the number of placements of 7 elements of 3.)
4. In how many ways can you create a duty schedule for 3 out of 5 students? ( this is the number of combinations of 5 elements of 3 and is equal to 10).

5. What we call probability random event?

6. Formulate the classical definition of probability.

III. Learning new material

In the practical application of probability theory and mathematical statistics, one often encounters problems in which the same experiment is repeated more than once. As a result of each experiment, event A may or may not appear, and we are not interested in the result of each experiment, but in the total number of occurrences of event A in a series of experiments. For example, just recently the Biathlon World Championships took place in Korea. The athletes fired a series of shots at targets, and we, as a rule, were not interested in the result of each individual shot, but in the total number of hits. Moreover, the results of previous experiments did not in any way affect subsequent ones. This standard scheme is often found in probability theory itself. It is called independent testing scheme or Bernoulli scheme . Swiss mathematician of the 17th century. Jacob Bernoulli combined examples and questions of this type into a single probabilistic problem-scheme (The Art of Conjecture, published in 1713).

Historical reference (one of the students prepares a message about the life of a scientist for the lesson):

“Jacob Bernoulli (12/27/1654, Basel, – 8/16/1705, ibid.) – professor of mathematics at the University of Basel (1687) was a native of Holland...”

Checking homework:
Group 1: At home you had to calculate the probability of rolling a 1 on a die.
Group 2: At home you had to calculate the probability of getting heads when tossing a coin. (Students name the results, a conclusion is drawn about the reasons for the different answers, and the conclusion is that the more tests, the better you can see what the result is aiming for)
When we talk about the frequency and probability of some random event A, we mean the presence of certain conditions that can be repeatedly reproduced. We call this set of conditions random experience or random experiment. Note that the result of one experiment does not depend in any way on the previous one. Several experiments are called independent, if the probability of the outcome of each experiment does not depend on what outcomes other experiments had. For example, several consecutive coin tosses are independent experiments. Several successive removals of balls from a bag are independent experiments, provided that the removed ball returns to the bag each time. Otherwise, these are dependent experiments. Jacob Bernoulli combined examples and questions of this type into a single probabilistic framework.

Bernoulli scheme.

Consider independent repetitions of the same trial with two possible outcomes, which are conventionally called “success” and “failure”. You need to find the probability that after n such repetitions exactly k “successes” will occur.

The teacher should emphasize once again three conditions that Bernoulli's scheme must satisfy:

1) each test must have two outcomes, called “success” and “failure”;

2) in each experiment the probability of event A must remain unchanged;

3) the results of the experiments must be independent.

1 V . Consolidation.

1. Oral work (it is possible to organize group work). The answers are discussed in groups and one representative voices them.

Explain why the following questions fit Bernoulli's scheme. Indicate what “success” consists of and what equals n And k.

a) What is the probability that in 123 tosses of a coin, the coin will land on heads exactly 45 times?

b) A black box contains 10 white, 3 red and 7 blue balls. The balls are removed, their color is recorded and returned. What is the probability that all 20 balls drawn will be blue?
c) What is the probability that in one hundred coin tosses, heads will appear 73 times?
d) They threw a pair twenty times in a row dice. What is the probability that the sum of the points never equals ten?
e) Three cards were drawn from a deck of 36 cards, the result was recorded and returned to the deck, then the cards were shuffled. This was repeated 4 times. What is the probability that the queen of spades was among the cards drawn each time?

TEACHER: To obtain numerical values in such problems, it is necessary to know in advance the probability of “success” and “failure.” Denoting the probability of “success” as p, and the probability of “failure” as q, where q = 1- p, Bernoulli proved a remarkable theorem

2. Independent work(it is possible to organize group work). Students are given 7 problems to solve. The number of points for the task is indicated in brackets. The guys discuss the solution in groups. Installation: score “5” - 17-22 points, “4” - 12- 16 points, “3” - 6-11 points.

1). What is the probability of that? that with ten throws of the dice, 3 points will appear exactly 2 times? (2 points)

2). What is the probability that with 9 tosses of a coin, “heads” will appear exactly 4 times? (2 points)

3). Ostap Bender plays 8 games against members of the chess club. Ostap plays poorly, so the probability of winning in each game is 0.01. Find the probability that Ostap will win at least one game. (3 points)

4). The probability of hitting the target with one shot is 0.125. What is the probability that out of 12 shots there will be no hits? (3 points)

5). In Part A of the Unified State Examination in mathematics in 2005 there were 10 multiple-choice questions. For each of them, 4 answer options were offered, of which only one was correct. To receive a positive mark on the exam, you must answer at least 6 questions. What is the probability that a careless student will pass the exam? (4 points)

6). We throw the dice. What is the probability that if we roll a die 8 times, we will roll a six at least 4, but not more than 6 times? (4 points)

7). In one shot, the shooter hits the target with a probability of 0.1. Find the probability that with five shots he will hit the target at least once. (4 points)

ANSWERS: 1) 0,29; 2) 0,246; 3)0,077; 4)0,2 5) 0,016; 6) 0,034; 7) 0,4095;

If there is time, then the work can be discussed, if not, then collect notebooks for checking.

V.Homework:

1). The probability of event A is 0.3. What is the probability that in a series of 6 trials event A will occur at least once? (4 points)

2). Sasha was given 10 tasks of equal difficulty. The probability that he will solve the problem is 0.75. Find the probability that Sasha will solve: a) all problems;

b) at least 8 tasks; c) at least 6 tasks.

3. A series of Bernoulli tests is carried out twice. The first time the probability of success is ½, the second time the probability of success is 1/3. In what case is the expected spread of the value of S greater if S is the number of successes that have occurred?

ANSWERS: 1). 0.882; 2) a) 0.056; b) 0.526; c) 0.922.

Individually: presentation of material on the topic “Law of Large Numbers”, report on the topic “Bernoulli Family”.

V1. Summarizing.

Which keywords lesson can be highlighted? Explain their meaning.

What key fact has been learned today?

What are the similarities and differences between statistics and probability?

V11. Reflection. At the reflection stage, students are asked to compose a syncwine and express their attitude to the material studied in poetic form.

Help: SINQWAIN is a technique for developing critical thinking at the stage of reflection.

This is a short literary work characterizing a subject (topic), consisting of five lines, which is written according to a certain plan. The word "cinquain" comes from the French word for "five".

RULES FOR WRITING SINQWAIN

1 line – one word – title of the poem, theme, usually a noun.

Line 2 – two words (adjectives or participles). Description of the topic, words can be connected by conjunctions and prepositions.

Line 3 – three words (verbs). Actions related to the topic.

Line 4 – four words – a sentence. A phrase that shows the author’s attitude to the topic in the 1st line.

Line 5 – one word – association, synonym that repeats the essence of the topic in line 1, usually a noun.

Literature

V.A. Bulychev, E.A. Bunimovich. Studying probability theory and statistics in a school mathematics course. “Mathematics at school.” No. 4. 2003 p. 59. Vilenkin N. Ya. Combinatorics. – M.: Nauka, 1969.

V.N. Studinetskaya et al. “In the world of natural accidents.” Volgograd: Teacher, 2007.

Gmurman V. E. Guide to solving problems in probability theory and mathematical statistics. – M.: Higher School, 1975.

Gmurman V. E. Probability theory and mathematical statistics. – M.: Higher School, 1977.

Gnedenko B.V. Course in probability theory. – M.: Nauka, 1988.

Email textbook Abstracts and essays

Self-analysis of the lesson

Well: fundamentals of probability theory and mathematical statistics.

Class: 10th, physics and mathematics.

Lesson topic:Independent retests. Bernoulli's theorem

Lesson objectives:

Educational:

Introducing students to Bernoulli's scheme and practicing its application in solving problems.

Educational:

Formation of a unified scientific picture of the world and elements of a scientific worldview among students through the study of interdisciplinary connections between probability theory and various sciences;

Formation of probabilistic and statistical thinking of students;

Educational:

Development of independence and self-control skills.

Motivating students to study topics in probability theory.

Tasks:

consolidate knowledge and skills in solving combinatorial problems;
develop skills in using Bernoulli's scheme when solving problems,
develop problem solving skills using Bernoulli’s formula,
develop the basic mental operations of students: the ability to compare, analyze.

Lesson type: combined.

This material has practical application, as it allows solve problems in which the same experience is repeated repeatedly. As a result of each experiment, event A may or may not appear, and we are not interested in the result of each experiment, but in the total number of occurrences of event A in a series of experiments. In this lesson, the children learned the formula for solving such problems, learned to identify problems that fit Bernoulli’s scheme and are solved using his theorem. Time is rationally distributed at all stages of the lesson. The pace of the lesson corresponded to the level of development and preparedness of the students.

The lesson was conceived by me as a dialogue between the teacher and students, since the class is quite strong. The lesson contributed to the formation of basic ideological ideas, probabilistic-statistical thinking, and the ability to identify interdisciplinary connections. The children worked in groups, which allowed them to develop their cognitive and communicative competence. In order for everyone to work in groups according to their capabilities and abilities, so that they do not lose interest in the taught discipline, tasks are proposed at different levels. Students in the lesson were active and independently came to conclusions. The content of the lesson contributed to the development of interest in learning, as evidenced by the reflective stage of the lesson. The presentation helped make the lesson more interesting, save time for taking notes and systematizing the material.

Example of syncwine:

1. Bernoulli's theorem
New, interesting.
We met, understood, became interested.
Allows you to find the probability

In real.

2. Oh, trials,

Independent repeat

Let's analyze, understand and calculate

And it will help us with this, of course.

Bernoulli's theorem

The goals set during the lesson were achieved.

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