Lesson Objectives:

Students should know:

• determination of the probability of a random event;
• be able to solve problems to find the probability of a random event;
• be able to apply theoretical knowledge in practice.

Lesson objectives:

Educational: to create conditions for students to master the system of knowledge, skills and abilities with the concepts of the probability of an event.

Educational: to form a scientific worldview in students

Developing: to develop students' cognitive interest, creativity, will, memory, speech, attention, imagination, perception.

Methods of organizing educational and cognitive activities:

• visual,
• practical,
• on mental activity: inductive,
• according to the assimilation of the material: partially exploratory, reproductive,
• according to the degree of independence: independent work,
• stimulating: incentives,
• types of control: verification of independently solved tasks.

Lesson plan

1. oral exercises
2. Learning new material
3. Problem solving.
4. Independent work.
5. Summing up the lesson.
6. Commenting on homework.

Equipment: multimedia projector (presentation), cards (independent work)

During the classes

I. Organizational moment.

The organization of the class throughout the lesson, the readiness of students for the lesson, order and discipline.

Setting learning goals for students, both for the entire lesson and for its individual stages.

Determine the significance of the material being studied, both in this topic and in the entire course.

II. Repetition

1. What is a probability?

Probability - the possibility of execution, the feasibility of something.

2. What definition does the founder of modern probability theory A.N. Kolmogorov?

Mathematical probability is a numerical characteristic of the degree of possibility of the occurrence of any particular event in certain conditions that can be repeated an unlimited number of times.

3. What is the classical definition of probability given by the authors of school textbooks?

The probability P(A) of an event A in a trial with equally probable elementary outcomes is the ratio of the number of outcomes m that favor event A to the number n of all outcomes of the trial.

Conclusion: in mathematics, probability is measured by a number.

Today we will continue to consider the mathematical model “dice”.

The subject of study in the theory of probability are events that appear under certain conditions and can be reproduced an unlimited number of times. Each fulfillment of these conditions is called a test.

The test is throwing a dice.

Event - six rolled or getting an even number of points.

Each side of the die has the same probability of falling out when rolling the dice multiple times (the dice is regular).

III. Oral problem solving.

1. A die (dice) was thrown once. What is the probability that a 4 is rolled?

Solution. Random experiment - throwing a die. The event is the number on the rolled edge. There are only six edges. Let's list all the events: 1, 2, 3, 4, 5, 6. So P= 6. Event A = (4 points rolled) is favored by one event: 4. Therefore t= 1. Events are equally likely, since it is assumed that the cube is fair. Therefore P(A) = t/n= 1/6 = 0,17.

2. A die (cube) was thrown once. What is the probability of getting no more than 4 points?

P= 6. Event A = (no more than 4 points fell out) is favored by 4 events: 1, 2, 3, 4. Therefore t= 4. Therefore P(A) = t/n= 4/6 = 0,67.

3. A die (cube) was thrown once. What is the probability of getting less than 4 points?

Solution. Random experiment - throwing a die. The event is the number on the rolled edge. Means P= 6. Event A = (less than 4 points fell out) is favored by 3 events: 1, 2, 3. Therefore t= 3. P(A) = t/n= 3/6 = 0,5.

4. A dice (dice) was thrown once. What is the probability of getting an odd number of points?

Solution. Random experiment - throwing a die. The event is the number on the rolled edge. Means P= 6. Event A = (an odd number of points fell out) is favored by 3 events: 1,3,5. That's why t= 3. P(A) = t/n= 3/6 = 0,5.

IV. Learning new

Today we will consider tasks when two dice are used in a random experiment or two or three throws are performed.

1. In a random experiment, two dice are thrown. Find the probability that the sum of the rolled points is 6. Round your answer to the nearest hundredth .

Solution. The outcome in this experiment is an ordered pair of numbers. The first number will fall on the first die, the second on the second. It is convenient to represent the set of outcomes in a table.

The rows correspond to the number of points on the first die, the columns correspond to the number of points on the second die. Total elementary events P= 36.

	1	2	3	4	5	6
1	2	3	4	5	6	7
2	3	4	5	6	7	8
3	4	5	6	7	8	9
4	5	6	7	8	9	10
5	6	7	8	9	10	11
6	7	8	9	10	11	12

Let's write in each cell the sum of the dropped points and paint over the cells where the sum is equal to 6.

There are 5 such cells. This means that the event A = (the sum of the dropped points is 6) is favored by 5 outcomes. Consequently, t\u003d 5. Therefore, P (A) \u003d 5/36 \u003d 0.14.

2. In a random experiment, two dice are thrown. Find the probability that the total will be 3. Round the result to hundredths .

P= 36.

The event A = (the sum is 3) is favored by 2 outcomes. Consequently, t= 2.

Therefore, P(A) = 2/36 = 0.06.

3. In a random experiment, two dice are thrown. Find the probability of getting more than 10 points in total. Round the result to hundredths .

Solution. The outcome in this experiment is an ordered pair of numbers. Total events P= 36.

Event A = (more than 10 points in total) is favored by 3 outcomes.

Consequently, t

4. Luba rolls the die twice. She scored 9 points in total. Find the probability that one of the tosses rolls a 5. .

Solution The outcome in this experiment is an ordered pair of numbers. The first number will come up on the first throw, the second on the second. It is convenient to represent the set of outcomes in a table.

The rows correspond to the result of the first throw, the columns correspond to the result of the second throw.

Total events in which the sum of points 9 will be P= 4. Event A = (one of the throws scored 5 points) is favored by 2 outcomes. Consequently, t= 2.

Therefore, P(A) = 2/4 = 0.5.

5. Sveta rolls the die twice. She scored 6 points in total. Find the probability that one of the rolls rolled 1.

First toss		Second toss		Sum of points

Equivalent outcomes - 5.

Probability of the event p = 2/5 = 0.4.

6. Olya rolls the die twice. She scored 5 points in total. Find the probability of getting 3 on the first roll.

First toss		Second toss		Sum of points
	+		=
	+		=
	+		=
	+		=

Equivalent outcomes - 4.

Favorable outcomes - 1.

Event Probability R= 1/4 = 0,25.

7. Natasha and Vitya are playing dice. They roll the die once.

The one with the most points wins. If the points are equal, then there is a draw. In total, 8 points fell out. Find the probability that Natasha won.

				Sum of points
	+		=
	+		=
	+		=
	+		=
	+		=

Equivalent outcomes - 5.

Favorable outcomes - 2.

Event Probability R= 2/5 = 0,4.

8. Tanya and Natasha are playing dice. They roll the die once. The one with the most points wins. If the points are equal, then there is a draw. In total, 6 points fell out. Find the probability that Tanya lost.

Tanya		Natasha		Sum of points
	+		=
	+		=
	+		=
	+		=
	+		=

Equivalent outcomes - 5.

Favorable outcomes - 2.

Event Probability R= 2/5 = 0,4.

9. Kolya and Lena play dice. They roll the die once. The one with the most points wins. If the points are equal, then there is a draw. Kolya was the first to throw, he got 3 points. Find the probability that Lena won't win.

Kolya got 3 points.

Lena has 6 equally possible outcomes.

There are 3 favorable outcomes for losing (at 1 and at 2 and at 3).

Event Probability R= 3/6 = 0,5.

10. Masha throws a dice three times. What is the probability that even numbers will come up all three times?

Masha has 6 6 6 = 216 equally likely outcomes.

Favorable outcomes for losing - 3 3 3 = 27.

Event Probability R= 27/216 = 1/8 = 0,125.

11. In a random experiment, three dice are thrown. Find the probability that the total will be 16. Round the result to the nearest hundredth.

Solution.

		Second		Third		Sum of points
	+		+		=
	+		+		=
	+		+		=
	+		+		=
	+		+		=
	+		+		=

Equivalent outcomes - 6 6 6 = 216.

Favorable outcomes - 6.

Event Probability R\u003d 6/216 \u003d 1/36 \u003d 0.277 ... \u003d 0.28. Consequently, t\u003d 3. Therefore, P (A) \u003d 3/36 \u003d 0.08.

V. Independent work.

Option 1.

1. A die (dice) is thrown once. What is the probability of getting at least 4 points? (Answer: 0.5)
2. In a random experiment, two dice are thrown. Find the probability of getting 5 points in total. Round the result to the nearest hundredth. (Answer: 0.11)
3. Anya rolls the dice twice. She scored 3 points in total. Find the probability of getting 1 on the first roll. (Answer: 0.5)
4. Katya and Ira are playing dice. They roll the die once. The one with the most points wins. If the points are equal, then there is a draw. A total of 9 points fell out. Find the probability that Ira lost. (Answer: 0.5)
5. In a random experiment, three dice are rolled. Find the probability of getting 15 points in total. Round the result to the nearest hundredth. (Answer: 0.05)

Option 2.

1. A die (dice) is thrown once. What is the probability of getting no more than 3 points? (Answer: 0.5)
2. In a random experiment, two dice are thrown. Find the probability that the total will be 10. Round the result to the nearest hundredth. (Answer: 0.08)
3. Zhenya rolls the dice twice. She scored 5 points in total. Find the probability of getting 2 on the first throw. (Answer: 0.25)
4. Masha and Dasha are playing dice. They roll the die once. The one with the most points wins. If the points are equal, then there is a draw. In total, 11 points fell out. Find the probability that Masha won. (Answer: 0.5)
5. In a random experiment, three dice are rolled. Find the probability that the total will be 17. Round the result

VI. Homework

1. In a random experiment, three dice are rolled. In total, 12 points fell out. Find the probability of getting 5 on the first roll. Round the result to the nearest hundredth.
2. Katya throws a dice three times. What is the probability that the same number will come up all three times?

VII. Lesson summary

What do you need to know to find the probability of a random event?

To calculate the classical probability, you need to know all possible outcomes of an event and favorable outcomes.

The classical definition of probability is applicable only to events with equally likely outcomes, which limits its scope.

Why do we study probability theory in school?

Many phenomena of the world around us can be described only with the help of probability theory.

Literature

1. Algebra and the beginning of mathematical analysis. 10-11 grades: textbook. for educational institutions: basic level / [Sh.A. Alimov, Yu.M. Kolyagin, M.V. Tkacheva and others]. - 16th ed., revised. – M.: Enlightenment, 2010. – 464 p.
2. Semenov A.L. USE: 3000 tasks with answers in mathematics. All tasks of group B / - 3rd ed., Revised. and additional - M .: Publishing house "Exam", 2012. - 543 p.
3. Vysotsky I.R., Yashchenko I.V. USE 2012. Mathematics. Task B10. Probability Theory. Workbook / Ed. A.L. Semenova and I.V. Yashchenko. - M.: MTsShMO, 2012. - 48 p.

Explain the principle of solving the problem. A die is thrown once. What is the probability of getting less than 4 points? and got the best answer

Answer from Divergent[guru]
50 percent
The principle is extremely simple. Total outcomes 6: 1,2,3,4,5,6
Of these, three satisfy the condition: 1,2,3, and three do not satisfy: 4,5,6. Therefore, the probability is 3/6=1/2=0.5=50%

Answer from I am superman[guru]
A total of six options can fall out (1,2,3,4,5,6)
And of these options 1, 2, and 3 are less than four
So 3 answers out of 6
To calculate the probability, we divide the favorable alignment to everything, i.e. 3 by 6 \u003d 0.5 or 50%

Answer from Yuri Dovbysh[active]
50%
divide 100% by the number of numbers on the dice,
and then multiply the percentage received by the amount you need to find out, that is, by 3)

Answer from Ivan Panin[guru]
I don’t know for sure, I’m preparing for the GIA, but the teacher told me something today, only about the probability of cars, since I understood that the ratio is shown as a fraction, from the top the number is favorable, but from the bottom, in my opinion, it’s generally general, well, we had about cars like this : The taxi company currently has 3 black, 3 yellow and 14 green cars available. One of the cars left for the customer. Find the probability that a yellow taxi will arrive. So, there are 3 yellow taxis and out of the total number of cars there are 3 of them, it turns out that we write 3 on top of the fraction, because this is a favorable number of cars, and we write 20 on the bottom, because there are 20 cars in the taxi fleet, so we get the probability 3 to 20 or 3/20 fractions, well, that's how I understood it .... As for the bones, I don't know for sure, but maybe it helped in some way ...

Answer from 3 answers[guru]

Hello! Here is a selection of topics with answers to your question: Explain the principle of solving the problem. A die is thrown once. What is the probability of getting less than 4 points?

Problem 19 ( OGE - 2015, Yashchenko I.V.)

Olya, Denis, Vitya, Artur and Rita cast lots - who should start the game. Find the probability that Rita will start the game.

Solution

In total, 5 people can start the game.

Answer: 0.2.

Problem 19 ( OGE - 2015, Yashchenko I.V.)

Misha had four sweets in his pocket - Grillage, Mask, Squirrel and Little Red Riding Hood, as well as the keys to the apartment. Taking out the keys, Misha accidentally dropped one candy. Find the probability that the candy "Mask" is lost.

Solution

There are 4 options in total.

The probability that Misha dropped the "Mask" candy is

Answer: 0.25.

Problem 19 ( OGE - 2015, Yashchenko I.V.)

A die (dice) is thrown once. What is the probability that the number rolled is not less than 3?

Solution

In total, there are 6 different options for dropping points on a die.

The number of points, not less than 3, can be: 3,4,5,6 - that is, 4 options.

So the probability is P = 4/6 = 2/3.

Answer: 2/3.

Problem 19 ( OGE - 2015, Yashchenko I.V.)

Grandmother decided to give her grandson Ilyusha some randomly selected fruit for the road. She had 3 green apples, 3 green pears and 2 yellow bananas. Find the probability that Ilyusha will receive a green fruit from his grandmother.

Solution

3+3+2 = 8 - total fruits. Of these, green - 6 (3 apples and 3 pears).

Then the probability that Ilyusha will receive a green fruit from his grandmother is

P=6/8=3/4=0.75.

Answer: 0.75.

Problem 19 ( OGE - 2015, Yashchenko I.V.)

A die is thrown twice. Find the probability that a number greater than 3 is rolled both times.

Solution

6 * 6 = 36 - total number of numbers falling out during two throws of a dice.

We have options for:

There are 9 options in total.

So the probability of getting a number greater than 3 both times is

P = 9/36 = 1/4 = 0.25.

Answer: 0.25.

Problem 19 ( OGE - 2015, Yashchenko I.V.)

A die (dice) is thrown 2 times. Find the probability that a number greater than 3 is rolled once and a number less than 3 is rolled another time.

Solution

Total options: 6 * 6 = 36.

We have the following outcomes:

Tasks for dice probability no less popular than coin tossing problems. The condition of such a problem usually sounds like this: when throwing one or more dice (2 or 3), what is the probability that the sum of the points will be 10, or the number of points is 4, or the product of the number of points, or divisible by 2 the product of the number of points and etc.

The application of the classical probability formula is the main method for solving problems of this type.

One die, probability.

The situation is quite simple with one dice. is determined by the formula: P=m/n, where m is the number of favorable outcomes for the event, and n is the number of all elementary equally possible outcomes of the experiment with tossing a die or a die.

Problem 1. A die is thrown once. What is the probability of getting an even number of points?

Since the dice is a cube (or it is also called a regular dice, the cube will fall on all faces with the same probability, since it is balanced), the die has 6 faces (the number of points from 1 to 6, which are usually indicated by dots), which means , that in the task the total number of outcomes: n=6. The event is favored only by outcomes in which a face with even points 2,4 and 6 falls out, for a cube of such faces: m=3. Now we can determine the desired probability of a dice: P=3/6=1/2=0.5.

Task 2. A dice is thrown once. What is the probability of getting at least 5 points?

Such a problem is solved by analogy with the example indicated above. When throwing a dice, the total number of equally possible outcomes is: n=6, and satisfy the condition of the problem (at least 5 points fell out, that is, 5 or 6 points fell out) only 2 outcomes, which means m=2. Next, we find the desired probability: P=2/6=1/3=0.333.

Two dice, probability.

When solving problems with throwing 2 dice, it is very convenient to use a special score table. On it, the number of points that fell on the first dice is plotted horizontally, and the number of points that fell on the second dice is plotted vertically. The workpiece looks like this:

But the question arises, what will be in the empty cells of the table? It depends on the task to be solved. If the problem is about the sum of points, then the sum is written there, and if it is about the difference, then the difference is written, and so on.

Problem 3. 2 dice are thrown at the same time. What is the probability of getting a sum less than 5 points?

First you need to figure out what will be the total number of outcomes of the experiment. Everything was obvious when throwing one die 6 faces of the die - 6 outcomes of the experiment. But when there are already two dice, then the possible outcomes can be represented as ordered pairs of numbers of the form (x, y), where x shows how many points fell on the first dice (from 1 to 6), and y - how many points fell on the second dice (from 1 until 6). In total there will be such numerical pairs: n=6*6=36 (36 cells exactly correspond to them in the table of outcomes).

Now you can fill in the table, for this, the number of the sum of points that fell on the first and second dice is entered in each cell. The completed table looks like this:

Thanks to the table, we will determine the number of outcomes that favor the event "drops in total less than 5 points". Let's count the number of cells, the value of the sum in which will be less than the number 5 (these are 2, 3 and 4). For convenience, we paint over such cells, they will be m = 6:

Given the table data, dice probability equals: P=6/36=1/6.

Problem 4. Two dice were thrown. Determine the probability that the product of the number of points will be divisible by 3.

To solve the problem, we will make a table of the products of points that fell on the first and second dice. In it, we immediately select numbers that are multiples of 3:

We write down the total number of outcomes of the experiment n=36 (the reasoning is the same as in the previous problem) and the number of favorable outcomes (the number of cells that are shaded in the table) m=20. The probability of an event is: P=20/36=5/9.

Problem 5. A dice is thrown twice. What is the probability that the difference between the number of points on the first and second dice will be between 2 and 5?

To determine dice probability Let's write down the table of score differences and select those cells in it, the value of the difference in which will be between 2 and 5:

The number of favorable outcomes (the number of cells shaded in the table) is equal to m=10, the total number of equally possible elementary outcomes will be n=36. Determines the probability of an event: P=10/36=5/18.

In the case of a simple event and when throwing 2 dice, you need to build a table, then select the necessary cells in it and divide their number by 36, this will be considered a probability.