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Bernoulli retest schemes presentation. Repetition of tests. Bernoulli scheme. I. Organizational moment

A series of independent trials is underway,
each of which has 2 possible outcomes,
which we will conditionally call Success and Failure.
For example, a student takes 4 exams, in each
of which 2 outcomes are possible Success: student
passed the exam and Failed: failed.

The probability of success in each trial is
p. The Failure Probability is q=1-p.
It is required to find the probability that in the series
out of n trials, success will come 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
Failed (n-m) times.
Number
all
combinations
equals
number
ways from n trials to choose those m, in
which 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
total C s lags 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 a coin falls on heads, a student
goes to the movies if the coin lands on tails

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 get to the lecture?

1) In this problem, a series of n=5
independent tests. Let's call it Success
going to a lecture (tails falling out) and
Failure - going to the cinema (falling out of the coat of arms).
p=q=1/2.
Using the Bernoulli formula, we find the probability that
What will happen 3 times after 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 after 5 tosses
at least once the coin will land tails,
let's move on to the probability of the opposite
events - the coin will drop out all 5 times with the coat of arms:
P5 (0).
Then the desired probability will be: P=1-P5(0).
According to the Bernoulli 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
P1 0.03125 0.96875

Bernoulli
student goes
in the cinema, if the coin falls tails - the student goes to
lecture. The coin was tossed by 5 students. What is the most
probable number of students going to the lecture?
Probability
winnings for 1 ticket is 0.2. What is the most
probable number of winning tickets?

The most likely number of successes in the scheme
Bernoulli

np q k np p

The most likely number of successes in the 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
whole numbers. Both are equally incredible.
If np-q is a non-integer number, then this interval contains 1
integer

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

- The student goes to the lecture. Coin tossed 5

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

The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to 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

The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to 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

The most likely number of successes in the scheme
Bernoulli
Example It is known that if a coin lands on heads,
a student goes to the cinema if the coin lands on tails
- The student goes to the lecture. Coin tossed 5
students. What is the most likely number
students going to lecture?
probability, Pn(k)
Probabilities of the number of students who attended
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

The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are purchased.

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

The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are 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

The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are 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

The most likely number of successes in the scheme
Bernoulli
Example 10 lottery tickets are 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

The most likely number of successes in the scheme
Bernoulli

10 contracts signed

pay the sum insured

one of the contracts

than three contracts
d) find the most probable number of contracts, according to
who will have to pay the sum insured

The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
a) Find the probability that three
pay the sum insured
0,201327

The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
b) The sum insured will not have to be paid out under any
one of the contracts
0,107374

The most likely number of successes in the scheme
Bernoulli
Example On average, for 20% of insurance contracts
the company pays the sum insured.
10 contracts signed
c) the amount insured will have to be paid no more than,
than 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 the event A
in each test is close to zero,
and the number of independent trials n is large enough,
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 the Poisson 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
marriage (the number of packages that do not meet the standard)
is 0.2%. Estimate the probability that
after 1000 randomly selected packages, there will be three packages,
not meeting the standard.
Pn(k)
k
k!
P1000(3) ?
e ,
np

no more than 5 contracts are connected.

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

MOU "Rudnogorsk secondary school"

Development of a lesson on probability theory

in 10th grade

on this topic

« Independent retests.

Bernoulli's theorem »

Mathematic teacher

MOU "Rudnogorsk Sosh"

Chibysheva I.A.

“... Chance is mainly

depends on our knowledge…”

Jacob Bernoulli

Subject "»

Grade:10

Lesson Objectives:

Tutorials:

Developing:

Educational:

Tasks:

Lesson type: combined.

Teaching methods: conversation, writing exercises.

Equipment: computer, multimedia projector. presentation, handout

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. Knowledge update

Let us recall the basic concepts and formulas of combinatorics.

1. What is called the factorial of the number 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 of the competition? (7 6 5 = 210. This is the number of placements of 7 elements of 3.)
4. In how many ways can 3 students out of 5 be on duty? ( this is the number of combinations of 5 elements of 3 and is equal to 10).

5. What do we call the probability of a 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 experience 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, most recently Korea hosted the Biathlon World Championships. Athletes fired a series of shots at targets, and as a rule we were not interested in the result of each individual shot, but in the total number of hits. At the same time, the results of previous experiments did not affect the subsequent ones. Such a standard scheme is often encountered in the theory of probability itself. It is called independent test 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 work "The Art of Assumptions" was published in 1713).

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

“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 a 1 on a dice.
Group 2: At home, you had to calculate the probability of getting heads when you toss a coin. (Students name the results, conclude the reasons for the different answers, and conclude that the more trials, the better you can see what the result is striving for)
Speaking 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 a random experience or a random experiment. Note that the result of one experiment does not depend on the previous one. Several experiments are called independent, if the probability of the outcome of each of the experiments does not depend on what outcomes the other experiments had. For example, several successive tosses of a coin are independent experiments. Several successive removals of balls from the bag are independent experiments, provided that the removed ball is returned to the bag each time. Otherwise, these are dependent experiments. Jacob Bernoulli combined examples and questions of this type into a single probabilistic scheme.

Bernoulli scheme.

Consider independent repetitions of the same test with two possible outcomes, which are conditionally called "success" and "failure". It is required to find the probability that with n such repetitions exactly k "successes" will occur.

The teacher should emphasize again three conditions that the Bernoulli scheme must satisfy:

1) each trial must have two outcomes called "success" and "failure";

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

3) the results of the experiments must be independent.

1 V . Consolidation.

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

Explain why the following questions fit into the Bernoulli scheme. Indicate what "success" consists of and what equals n And k.

a) What is the probability that in 123 tosses of a coin, tails will come up exactly 45 times?

b) There are 10 white, 3 red and 7 blue balls in a black box. The balls are removed, their color is recorded and returned back. What is the probability that all of the 20 balls drawn are blue?
c) What is the probability that in 100 tosses of a coin heads will appear 73 times?
d) A pair of dice is rolled twenty times in a row. What is the probability that the sum of the points has never been equal to 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 mixed. This was repeated 4 times. What is the probability that the queen of spades was among the drawn cards each time?

TEACHER: To obtain numerical values in such problems, it is necessary to know in advance the probability of "successes" and "failures". Denoting the probability of "success" p, and the probability of "failure" 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. Number of points for the task is given in parentheses. The children 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 in ten throws of a die, 3 points will come up exactly 2 times? (2 points)

2). What is the probability that after 9 tosses of a coin, heads will come up 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 wins 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 USE in mathematics in 2005, there were 10 tasks with a choice of answers. Each of them was offered 4 possible answers, of which only one was correct. To get a positive mark on the exam, you must answer at least 6 tasks. What is the probability that a careless student will pass the exam? (4 points)

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

7). For one shot, the shooter hits the target with a probability of 0.1. Find the probability that after five shots he hits 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 verification.

v.Homework:

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

2). Sasha was given 10 tasks of the same difficulty. The probability that he will solve the problem is 0.75. Find the probability that Sasha solves: 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 which case is the expected spread of S greater if S is the number of successes?

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

Individually: presentation of material on the topic "The Law of Large Numbers", a report on the topic "The Bernoulli Family".

V1. Summarizing.

What are the key words of the lesson? Explain their meaning.

What is the key fact learned today?

What are the similarities and differences between statistics and probability?

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

Reference: SINQWINE is a technique for the development of critical thinking, at the stage of reflection.

This is a short literary work that characterizes the 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 SINQUEINE

1 line - one word - the title of the poem, the topic, 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 - an association, a synonym that repeats the essence of the topic in the 1st line, usually a noun.

Literature

V.A. Bulychev, E.A. Bunimovich. The study of probability theory and statistics in the school course of mathematics. "Mathematics in School". No. 4. 2003, p. 59. Vilenkin N. Ya. Combinatorics. – M.: Nauka, 1969.

V.N. Studinetskaya and others. "In the world of regular accidents." Volgograd: Teacher, 2007.

Gmurman V. E. Guidelines for 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. Probability theory course. – M.: Nauka, 1988.

Email textbook Abstracts and essays

Introspection of the lesson

Well: fundamentals of probability theory and mathematical statistics.

Class: 10th, physical and mathematical direction.

Lesson topic:Independent retests. Bernoulli's theorem

Lesson Objectives:

Tutorials:

Familiarize students with the Bernoulli scheme and practice its application in solving problems.

Developing:

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

Formation of probabilistic-statistical thinking of students;

Educational:

Development of independence and self-control skills.

Motivation of students to study the topics of probability theory.

Tasks:

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

Lesson type: combined.

This material has a 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 the Bernoulli scheme and are solved according to his theorem. Rationally distributed time at all stages of the lesson. The pace of the lesson corresponded to the level of development and preparedness of students.

The lesson was conceived by me as a dialogue between the teacher and students, as the class is strong enough. The lesson contributed to the formation of basic worldview ideas, probabilistic-statistical thinking, the ability to highlight interdisciplinary connections. The children worked in groups, which allows them to develop their cognitive and communicative competence. In order for everyone to work in groups, according to their abilities and abilities, so that interest in the discipline taught is not lost, tasks are offered at different levels. Students in the lesson were active, independently came to a conclusion. 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 to make the lesson more interesting, save time for taking notes and systematizing the material.

Sinkwine example:

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

In real.

2. Oh, trials,

Independent repeated

Let's understand, understand and calculate

And help us in this, of course,

Bernoulli's theorem

The objectives of the lesson have been achieved.

Repeated independent trials are called Bernoulli trials if each trial has only two possible outcomes and the probabilities of outcomes remain the same for all trials.

We denote these probabilities as p And q. Outcome with probability p will be called “success”, and the outcome with probability q- "failure".

It's obvious that

The space of elementary events for each trial consists of two points. Space of elementary events for n The Bernoulli trial contains points, each of which represents one possible outcome of the composite experience. Since the trials are independent, the probability of a sequence of events is equal to the product of the probabilities of the corresponding outcomes. For example, the probability of a sequence of events

(U, U, N, U, N, N, N)

is equal to the product

Examples of Bernoulli tests.

1. Successive tosses of the “correct” coin. In this case p = q = 1/2 .

When tossing a biased coin, the corresponding probabilities will change their values.

2. Each result of the experiment can be considered as A or .

3. If there are several possible outcomes, then a group of outcomes can be distinguished from them, which are considered as “success”, calling all other outcomes “failure”.

For example, in successive throws of a dice, “success” can be understood as a roll of 5, and “failure” as a roll of any other number of points. In this case p = 1/6, q = 5/6.

If by “success” we mean an even number of points, and by “failure” an odd number of points, then p = q = 1/2 .

4. Repeated random extraction of the ball from the urn containing at each test a whites and b black balls. If by success we mean extraction of the white ball, then , .

Feller gives the following example of the practical application of the Bernoulli test scheme. Washers produced in mass production may vary in thickness, but when tested, they are classified into good and defective - depending on whether the thickness is within the prescribed limits. And while products may not be perfectly consistent with the Bernoulli scheme for many reasons, this scheme sets the ideal standard for industrial product quality control, even though this standard is never quite accurately achieved. Machines are subject to change, and therefore the probabilities do not remain the same; there is some constancy in the mode of operation of the machines, as a result of which long series of identical deviations are more likely than would be the case with the actual independence of the tests. However, from a product quality control point of view, it is desirable that the process conform to the Bernoulli scheme, and it is important that, within certain limits, this can be achieved. The purpose of monitoring is to detect at an early stage significant deviations from the ideal scheme and use them as indications of a threatening violation of the correct operation of the machine.

Bernoulli formula

Belyaeva T.Yu. GBPOU KK "AMT" Armavir Math 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)

Elder 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 , so the probability that it will not happen is q = 1 - p .

It is required to find the probability that P successive trials, event A will occur exactly T once.

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 outcomes are possible:

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

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

With three tests:

8 outcomes are possible:

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 drawn. Find the probability that 2 of the 5 balls drawn will be white.

Formulas for finding the probability that V P trials 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) + … + p P (P)

Task 3.

The probability of manufacturing a non-standard part on an automatic machine is 0.02. Determine the probability that among six randomly taken parts 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.

It is required to find such a number μ (0, 1, ..., n), for which the probability P n (μ) will be the largest.

Task 4.

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

By condition: n = 75, p = 0.31, q = 1 - 0.31 = 0.69

Task 6.

Two shooters shoot 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 for which there will be no hit on the target if the shooters fire 25 volleys.

By condition: n = 25, p = 0.2 0.4 = 0.08, q = 0.92

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