Open lesson in mathematics "Bernoulli scheme. Solving problems using the Bernoulli and Laplace scheme"

Didactic: the acquisition of skills and abilities to work with the Bernoulli scheme to calculate probabilities.

Developing: development of skills for applying knowledge in practice, formation and development of functional thinking of students, development of skills of comparison, analysis and synthesis, skills of working in pairs, expansion of professional vocabulary.

How to play this game:

Educational: fostering interest in the subject through the practical application of the theory, achieving a conscious assimilation of the educational material of students, the formation of the ability to work in a team, the correct use of computer terms, interest in science, respect for the future profession.

Scientific knowledge: B

Lesson type: combined lesson:

• consolidation of the material covered in previous classes;
• thematic, information-problem technology;
• generalization and consolidation of the material studied in this lesson.

Teaching method: explanatory - illustrative, problematic.

Knowledge control: frontal survey, problem solving, presentation.

Material and technical equipment of the lesson. computer, multimedia projector.

Methodological support: reference materials, presentation on the topic of the lesson, crossword puzzle.

During the classes

1. Organizational moment: 5 min.

(greeting, readiness of the group for the lesson).

2. Knowledge check:

Check questions frontally on slides: 10 min.

• definitions of the section “Probability Theory”
• the main concept of the section “Probability Theory”
• what events are studied by “Probability Theory”
• characteristic of a random event
• classical definition of probabilities

Summarizing. 5 minutes.

3. Solving problems in rows: 5 min.

Task 1. A dice is thrown. What is the probability of getting an even number less than 5?

Task 2. There are nine identical radio tubes in a box, three of which were in use. During the working day, the master had to take two radio tubes to repair the equipment. What is the probability that both lamps were used?

Task 3. There are three different films in three cinema halls. The probability that there are tickets for a certain hour at the box office of the 1st hall is 0.3, at the box office of the 2nd hall - 0.2, and at the box office of the 3rd hall - 0.4. What is the probability that at a given hour it is possible to buy a ticket for at least one movie?

4. Checking at the blackboard how to solve problems. Application 1. 5 min.

5th Conclusion on solving problems:

The probability of occurrence of an event is the same for each task: m and n - const

6. Goal-setting through the task: 5 min.

Task. Two equal chess players play chess. What is the probability of winning two games out of four?

What is the probability of winning three games out of six (draws are not taken into account)?

Question. Think and name the difference between the questions of this problem and the questions of the previous problems?

By reasoning, by comparison, achieve an answer: in questions m and n are different.

7. Lesson topic:

Calculation of the probability of occurrence of an event k times out of n experiments with p-const.

If trials are made in which the probability of occurrence of event A in each trial does not depend on the outcomes of other trials, then such trials are called independent with respect to event A. Trials, in each of which the probability of occurrence of the event is the same.

Bernoulli formula. The probability that in n independent trials, in each of which the probability of occurrence of an event is equal to p (0

or Appendix 2 Bernoulli formula, where k,n-small numbers where q = 1-p

Solution: Equal chess players are playing, so the probability of winning is p=1/2; hence the probability of losing q is also 1/2. Since the probability of winning is constant in all games and it does not matter in what order the games are won, the Bernoulli formula is applicable. 5 minutes

Find the probability that two games out of four will be won:

Find the probability that three out of six games will be won:

Since P4 (2) > P6 (3), it is more likely to win two games out of four than three out of six.

8. Task.

Find the probability that event A occurs exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.

k=70, n=243 This implies that k and n are large numbers. This means that it is difficult to calculate according to the Bernoulli formula. For such cases, the local Laplace formula is applied:

Appendix 3 for positive values ​​of x is given in Appendix 4; for negative values ​​of x use the same table and = .

9. Compose an algorithm for solving the problem: 5 min.

• find the value of x and round up to hundredths (0.01);
• according to the table of the Laplace function we will find;
• we substitute the value of the Laplace function into the Laplace formula

10. Solving the problem with analysis at the blackboard. Annex 5. 10 min.

11. Summarizing lesson information through presentations

• brief information about the section “Probability Theory”; 5 minutes.
• historical materials about the scientists Bernoulli and Laplace. 5 minutes.

Section 2. Logical equivalence of formulas. Normal Forms for Propositional Algebra Formulas

Equivalence relation

With the help of truth tables, one can determine under what sets of truth values ​​of the input variables the formula will take a true or false value (as well as a statement that has the corresponding logical structure), which formulas will be tautologies or contradictions, and also establish whether two given formulas equivalent.

In logic, two sentences are said to be equivalent if they are both true or both false. The word "simultaneously" in this phrase is ambiguous. So, for the sentences "Tomorrow will be Tuesday" and "Yesterday was Sunday" this word has a literal meaning: on Monday they are both true, and on the rest of the week they are both false. For the equations " x = 2" and " 2x = 4» "simultaneously" means "with the same values ​​of the variable". The predictions “Tomorrow it will rain” and “It is not true that tomorrow it will not rain” will simultaneously be confirmed (turn out to be true) or not confirmed (turn out to be false). In essence, this is the same forecast, expressed in two different forms, which can be represented by the formulas X and . These formulas simultaneously take the value "true" or the value "false". To check, it is enough to make a truth table:

X
1	0	1
0	1	0

We see that the truth values ​​in the first and last columns are the same. Such formulas, as well as the sentences corresponding to them, are naturally considered equivalent.

Formulas F 1 and F 2 are called equivalent if their equivalent is a tautology.

The equivalence of two formulas is written as follows: (read: formula F1 is equivalent to the formula F2).

There are three ways to check whether formulas are equivalent: 1) make their equivalent and use the truth table to check if it is a tautology; 2) for each formula, make a truth table and compare the final results; if in the total columns for the same sets of variable values the truth values ​​of both formulas will be equal, then the formulas are equivalent; 3) with the help of equivalent transformations.

Example 2.1: Find out if the formulas are equivalent: 1) , ; 2) , .

1) Let us use the first method to determine equivalence, that is, find out whether the equivalence of formulas is a tautology.

Let's make an equivalence of formulas: . The resulting formula contains two different variables ( BUT and AT) and 6 operations: 1) ; 2) ; 3) ; 4) ; 5) ; 6). This means that the corresponding truth table will have 5 rows and 8 columns:

BUT	AT
1	1	0	0	0	1	0	1
1	0	0	1	1	0	1	1
0	1	1	0	1	0	1	1
0	0	1	1	1	0	1	1

From the final column of the truth table, it can be seen that the compiled equivalence is a tautology and, therefore, .

2) In order to find out whether the formulas and are equivalent, we use the second method, that is, we compile a truth table for each of the formulas and compare the final columns. ( Comment. In order to use the second method effectively, it is necessary that all compiled truth tables begin the same way, that is, the sets of variable values ​​were the same in the respective rows .)

The formula has two different variables and 2 operations, which means that the corresponding truth table has 5 rows and 4 columns:

BUT	AT
1	1	1	0
1	0	0	1
0	1	1	0
0	0	1	0

The formula has two different variables and 3 operations, which means that the corresponding truth table has 5 rows and 5 columns:

BUT	AT
1	1	0	0	1
1	0	0	1	1
0	1	1	0	0
0	0	1	1	1

Comparing the final columns of the compiled truth tables (since the tables start the same way, we can ignore the sets of variable values), we see that they do not match and, therefore, the formulas are not equivalent ().

The expression is not a formula (because the symbol " " does not refer to any logical operation). It expresses attitude between formulas (as well as equality between numbers, parallelism between lines, etc.).

The theorem on the properties of the equivalence relation is valid:

Theorem 2.1. Equivalence relation between propositional algebra formulas:

1) reflexively: ;

2) symmetrically: if , then ;

3) transitively: if and , then .

Laws of logic

The equivalences of propositional logic formulas are often called the laws of logic. We list the most important of them:

1. - the law of identity.

2. - the law of the excluded middle

3. - the law of contradiction

4. - disjunction with zero

5. - conjunction with zero

6. - disjunction with unit

7. - conjunction with unit

8. - the law of double negation

9. - commutativity of the conjunction

10. – commutativity of disjunction

11. - associativity of the conjunction

12. - disjunction associativity

13. – distributivity of the conjunction

14. – distributive disjunction

15. - laws of idempotency

16. ; - absorption laws

17. ; - De Morgan's laws

18. is the law expressing the implication through the disjunction

19. - law of counterposition

20. - laws expressing equivalence through other logical operations

The laws of logic are used to simplify complex formulas and to prove that formulas are identically true or false.

Equivalent transformations. Simplifying Formulas

If in equivalent formulas everywhere we substitute the same formula instead of some variable, then the newly obtained formulas will also turn out to be equivalent in accordance with the substitution rule. In this way, any number of new equivalences can be obtained from each equivalence.

Example 1: If in De Morgan's law instead of X substitute , instead of Y substitute , then we get a new equivalence . The validity of the obtained equivalence is easy to check using the truth table.

If any formula that is part of the formula F, be replaced by a formula equivalent to the formula , then the resulting formula will be equivalent to the formula F.

Then, for the formula from Example 2, we can make the following substitutions:

- the law of double negation;

- De Morgan's law;

- the law of double negation;

– the law of associativity;

is the law of idempotency.

By the property of transitivity of the equivalence relation, we can assert that .

The replacement of one formula by another, equivalent to it, is called equivalent transformation formulas.

Under simplification formulas that do not contain implication and equivalence signs understand an equivalent transformation that leads to a formula that does not contain negations of non-elementary formulas (in particular, double negations) or contains in total a smaller number of conjunction and disjunction signs than the original one.

Example 2.2: Let's simplify the formula .

At the first step, we applied the law that transforms the implication into a disjunction. At the second step, the commutative law was applied. At the third step, the law of idempotency was applied. On the fourth - De Morgan's law. And on the fifth - the law of double negation.

Remark 1. If a certain formula is a tautology, then any formula equivalent to it is also a tautology.

Thus, equivalent transformations can also be used to prove the identical truth of certain formulas. To do this, this formula must be reduced by equivalent transformations to one of the formulas that are tautologies.

Remark 2. Some tautologies and equivalences are combined into pairs (the law of contradiction and the law of alternative, commutative, associative laws, etc.). In these correspondences, the so-called principle of duality .

Two formulas that do not contain signs of implication and equivalence are called dual , if each of them can be obtained from the other by replacing the signs with , respectively.

The principle of duality states the following:

Theorem 2.2: If two formulas that do not contain implication and equivalence signs are equivalent, then their dual formulas are also equivalent.

normal forms

normal form is a syntactically unambiguous way of writing a formula that implements a given function.

Using the known laws of logic, any formula can be transformed into an equivalent formula of the form , where and each is either a variable, or the negation of a variable, or a conjunction of variables or their negations. In other words, any formula can be reduced to an equivalent formula of a simple standard form, which will be a disjunction of elements, each of which is a conjunction of separate different logical variables, either with or without a negation sign.

Example 2.3: In large formulas or with multiple transformations, it is customary to omit the conjunction sign (by analogy with the multiplication sign): . We see that after the transformations carried out, the formula is a disjunction of three conjunctions.

This form is called disjunctive normal form (DNF). A single element of a DNF is called elementary conjunction or constituent unit.

Similarly, any formula can be reduced to an equivalent formula, which will be a conjunction of elements, each of which will be a disjunction of logical variables with or without a negation sign. That is, each formula can be reduced to an equivalent formula of the form , where and each is either a variable, or the negation of a variable, or a disjunction of variables or their negations. This form is called conjunctive normal form (KNF).

Example 2.4:

A single element of CNF is called elementary disjunction or the constituent of zero.

Obviously, every formula has infinitely many DNFs and CNFs.

Example 2.5: Let's find several DNFs for the formula .

Perfect normal forms

SDNF (perfect DNF) is such a DNF in which each elementary conjunction contains all elementary statements, or their negations once, elementary conjunctions are not repeated.

SKNF (perfect CNF) is such a CNF in which each elementary disjunction contains all elementary propositions or their negations once, elementary disjunctions are not repeated.

Example 2.6: 1) - SDNF

2) 1 - SKNF

Let us formulate the characteristic features of SDNF (SKNF).

1) All members of the disjunction (conjunction) are different;

2) All members of each conjunction (disjunction) are different;

3) No conjunction (disjunction) contains both a variable and its negation;

4) Each conjunction (disjunction) contains all the variables included in the original formula.

As we can see, characteristics (but not forms!) satisfy the definition of duality, so it is enough to understand one form in order to learn how to get both.

It is easy to obtain SDNF (SKNF) from DNF (CNF) with the help of equivalent transformations. Since the rules for obtaining perfect normal forms are also dual, we will analyze in detail the rule for obtaining SMNF, and formulate the rule for obtaining SKNF independently using the definition of duality.

The general rule for reducing a formula to SDNF using equivalent transformations is:

In order to give the formula F, which is not identically false, to SDNF, it suffices:

1) bring it to some DNF;

2) remove the members of the disjunction containing the variable along with its negation (if any);

3) from the same members of the disjunction (if any), remove all but one;

4) remove all but one of the identical members of each conjunction (if any);

5) if any conjunction does not contain a variable from among the variables included in the original formula, add a term to this conjunction and apply the corresponding distributive law;

6) if the resulting disjunction contains the same terms, use prescription 3.

The resulting formula is the SDNF of this formula.

Example 2.7: Let's find SDNF and SKNF for the formula .

Since the DNF for this formula has already been found (see Example 2.5), we will start by obtaining the SDNF:

2) in the resulting disjunction there are no variables together with their negations;

3) there are no identical members in the disjunction;

4) there are no identical variables in any conjunction;

5) the first elementary conjunction contains all the variables included in the original formula, and the second elementary conjunction lacks a variable z, so let's add a term to it and apply the distributive law: ;

6) it is easy to see that the same terms appeared in the disjunction, so we remove one (prescription 3);

3) remove one of the identical disjunctions: ;

4) there are no identical terms in the remaining disjunctions;

5) none of the elementary disjunctions contains all the variables included in the original formula, so we supplement each of them with the conjunction : ;

6) there are no identical disjunctions in the resulting conjunction, so the found conjunctive form is perfect.

Since in the aggregate of SKNF and SDNF the formulas F 8 members, then most likely they are found correctly.

Each satisfiable (refutable) formula has one single SDNF and one single SKNF. A tautology has no SKNF, and a contradiction has no SDNF.

1. Two equal players play a game in which draws are excluded. What is the probability for the first player to win: a) one game out of two? b) two out of four? c) three out of six?

Answer: a) ; b) ; in)

3. Cut AB separated by a dot With in a ratio of 2:1. Four points are thrown at random on this segment. Find the probability that two of them are to the left of point C, and two are to the right.

Answer:

4. Find the probability that event A occurs exactly 70 times in 243 trials if the probability of this event occurring in each trial is 0.25.

Answer: .

5. The probability of having a boy is 0.515. Find the probability that among 100 newborns boys and girls will be equally divided.

Answer: 0,0782

6. The store received 500 bottles in glass containers. The probability that any of the bottles will be broken during transportation is 0.003. Find the probability that the store will receive broken bottles: a) exactly two; b) less than two; c) at least two; d) at least one.

Answer: a) 0.22; b) 0.20; c) 0.80; d) 0.95

7. An automobile plant produces 80% of cars without significant defects. What is the probability that among the 600 cars that came from the factory to the automotive exchange, there will be at least 500 cars without significant defects?

Answer: 0,02.

8. How many times do you need to flip a coin so that with a probability of 0.95 you can expect that the relative frequency of the coat of arms will deviate from the probability R\u003d 0.5 appearance of the coat of arms in one toss of a coin by no more than 0.02?

Answer: n ≥ 2401.

9. The probability of an event occurring in each of 100 independent events is constant and equal to p=0.8. Find the probability that the event will occur: a) at least 75 times and at most 90 times; b) at least 75 times; c) no more than 74 times.

Answer: a B C) .

10. The probability of an event occurring in each of the independent trials is 0.2. Find what deviation of the relative frequency of occurrence of an event from its probability can be expected with a probability of 0.9128 in 5000 trials.

Answer:

11. How many times should a coin be tossed so that with a probability of 0.6 it can be expected that the deviation of the relative frequency of the appearance of the coat of arms from the probability p=0.5 will be no more than 0.01 in absolute value.

Answer: n = 1764.

12. The probability of an event occurring in each of 10,000 independent trials is 0.75. Find the probability that the relative frequency of occurrence of an event deviates from its probability in absolute value by no more than 0.01.

Answer: .

13. The probability of an event occurring in each of the independent trials is 0.5. Find the number of trials n, at which with a probability of 0.7698 it can be expected that the relative frequency of the occurrence of an event deviates from its probability in absolute value by no more than 0.02.