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Report on the parent meeting "At the round table

Tatyana Alferova

IN middle group"Rose" prepared and held the first in the middle group parent meeting in the form of a round table.


Beginnings meeting with statements about the upbringing of children.

Raising children is the most important area of ​​our lives.

Our children are future fathers and mothers, they will also be educators of their children. Our children should grow up to be good fathers and mothers. But this is not All: our children are our old age. Proper upbringing- this is our happy old age, bad Education- this is our future grief, these are our tears, this is our guilt before other people.


Then attention parents The game "Club" was presented. Spinning a ball of yarn, each parent called affectionate words relating to the nature of the child. And vice versa, they twisted the ball, naming the words when the child upsets parents.


Parents answered a number of questions related to education. Every question was discussed.



The problem to which the most serious attention must be paid is the question of the purpose of education. In some families, one can observe complete thoughtlessness in this question: just live nearby parents and children, And parents are hopeful that everything will work itself out. At parents have no clear goal, no specific program. Of course, in this case, the results will always be random, and often such parents are surprised why did they grow up bad kids. Nothing can be done well if you do not know what you want to achieve.

Carried out a musical workout with parents.




Also parents 5 pedagogical situations were presented for discussion.

We chose ourselves parents counting who should read the situation and this parent chose it who should answer. and of course collectively discussed its content.

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I present to your attention the report from the parent meeting

Round table with 26 chairs

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Round table with 26 chairs

Placements of tourists are random, so the events are equally probable. To visit a small museum, the group is randomly divided into three subgroups of equal size. All decisions are the property of the site. Advanced training 36 courses from 1500 rubles. True, she also has a table twice as large, oval, but very wide. Find the probability that they both fail this test. There are 17 people in each subgroup. On individual chairs, much will not fit, but here you are. This site uses cookies. Determine through which longest time after starting work, turn off the device. I really want a round table, but there are big suspicions that it will take up a lot of space, taking into account the chairs.

In hepatitis patients, the analysis gives positive result with a probability of 0.9. Start date new group: 24 May. The probability that both machines run out of coffee is 0.16. Uncomfortable and not turn around in the kitchen. Based on the results of training, students are issued printed diplomas of the established sample. This leaves us with 1 girl and 16 chairs.

Two servicemen on exercises independently pass an obstacle course. Mechanical watches with a twelve-hour dial at some point broke and stopped walking. With one shot, the probability of a miss is 0.6. About the project All rights to the materials posted on the site are protected by copyright and related rights and cannot be reproduced or used in any way without written permission the copyright holder and putting down an active link to the main page of the Eva portal. The probability that both machines run out of coffee is 0.16. The dependence of temperature in degrees Kelvin on time for the heating element of some device was obtained experimentally. Since the table is round, it doesn't matter where the boys sit.

Round table with 26 chairs

For the first, the probability of passing it is 0.8, and for the second, 0.5. At a round table on 9 chairs, 7 boys and 2 girls are randomly seated. In the evening there were 4 chairs, and in the afternoon they folded 2, which significantly saved space. We bought a set of wooden table and a velvet corner with stools about 7 years ago a budget option was, no firm. The shots are repeated until the target is destroyed. For the second Frenchman, there are 50 places left in the subgroup -16 places.

You can specify the storage conditions and access to cookies in your browser. Let's color the cells where the sum is 7, there are six of them.

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  • 00:01

    The fourth task of the One state exam This is a task in the theory of probability.
    The vast majority of tasks are tasks for events that have a finite number of outcomes.
    In this case, the probability of the event A is called the fraction ,
    where is the numerator m- the number of outcomes that favor the occurrence of event A,
    and the denominator n is the number of all possible outcomes of a particular experiment.

  • 00:33

    Task. Suppose our experience is that a coin is tossed three times.
    Find the probability that heads come up exactly once.

    Such a problem is solved by enumeration. Let the letter O denote the event corresponding to the loss of an eagle,
    and the letter P - tails. Let's write down all the possible outcomes of our experience.
    If we tossed a coin three times, then it could fall out
    all three times tails (RRR),
    or tails, tails, heads (RPO),
    or tails, heads, tails (ROP),
    or tails, eagle, eagle (ROO).
    And so on. Total options we have 8.
    Three out of eight options are required.
    That is, they correspond to the condition that heads appear exactly once.
    Then the probability of an event that heads up only once is equal to

    Problem solved.

  • 01:32

    Another example of a task.
    20 athletes participate in the gymnastics championship:
    8 from Russia, 7 from the USA, and the rest from Denmark.
    The order in which the athletes compete is determined by lot.
    Find the probability that the athlete who competes last is a Danish athlete.
    So our event is that the last competing athlete is a Dane.
    How many Danes are taking part in the competition?
    Of the 20 athletes, 8+7 are not Danes. So the remaining 5 are Danes.
    So, from the Danes, all 5 people can speak last, and in total 20 people can speak last,
    that is, any athlete who takes part in the competition.
    Then the desired probability is the ratio of the number of outcomes favoring the event "last Dane to speak" to all possible outcomes.

    Problem solved.

  • 02:37

    A lot of people make mistakes in the following problem.
    26 athletes take part in badminton competitions, 10 of them are from Russia,
    including Ruslan Orlov.
    What is the probability of the event that Ruslan Orlov will play with an opponent from the Russian team?
    As usual, we must determine how many people can play with Ruslan Orlov
    and how many Russians can play with Ruslan Orlov.
    According to the condition of the problem, 26 athletes take part in the competition, and one of them is Ruslan Orlov. Therefore, 25 opponents can potentially face Ruslan Orlov.
    There are 10 people in the Russian team, and one of them is Ruslan Orlov.
    This means that Ruslan Orlov's potential rivals from the Russian team are 9 people.
    Then the probability of the event that Ruslan Orlov will play with an opponent from the Russian team is equal to:

    Problem solved.

  • 03:42

    Another problem that causes difficulties.
    There are on average 8 defective bags per 100 bags. The question is, what is the probability that the purchased bag will be of high quality? Round the result to the nearest hundredth.
    Let's make a small digression and solve another problem.
    Out of 1,000 pumps, an average of 8 can leak.
    What is the probability that the store bought a working pump?
    If out of 1,000 pumps 8 leak, then 992 pumps are serviceable.
    And, therefore, the probability of buying a quality pump is equal to:

    As for the bags, the task was formulated as follows:
    There are 8 defective bags per 100 bags.
    This means that 8 defective bags are not out of 100 bags, but in total 108 bags. And there are 100 quality bags.
    Therefore, the probability of buying quality bag is equal to:

    Rounding to hundredths, we get 0.93. Recall the rounding rules: if we discard values ​​that start with 5, 6, 7, 8, or 9, then we increase the number in the previous digit by 1, but if we discard smaller numbers - 0, 1, 2, 3, or 4, then we the previous digit number is not changed.

    Problem solved.

  • 05:14

    The next thing to pay attention to when preparing is the probabilities that are calculated for the placements in a circle.
    Task. At a round table designed for 9 chairs, 7 boys and 2 girls are randomly seated. Find the probability that the girls will sit next to each other.

    In order to solve this problem, it is convenient to translate it into a graphic language.
    Let's draw a table, mark 9 places, and this drawing will be the key to solving the problem.
    Let the first girl sit on some chair. If we want the second girl to sit next to her, then either the chair on the right or the chair on the left suits her.
    So, favorable to our event (the girls are sitting next to each other) cases - 2.
    And in total free places- 8, because out of 9 places 1 is occupied by the first girl.

    Problem solved.

  • 06:19

    Another problem, to the solution of which the same key is drawing.
    There are 26 students in the class, among them 2 twins - Andrey and Sergey.
    The class is randomly divided into two groups of 13 people each.
    The question is, what is the probability that Andrey and Sergey will be in the same group?

    It's pretty difficult task if you don't see quick decision.
    Suppose we have 13 seats for one group - these are the chairs on which the students of this group sit.
    All others will automatically fall into the second group. And let Andrei sit on one of these chairs.
    In order for Sergey to be in the same group, he must take any of the remaining 12 chairs.
    And they claim to be in the same group with Andrei, 25 people, that is, the desired probability:

  • 07:17

    Next task. The clock with a circular dial is broken.
    What is the probability that the minute hand stops
    between the twelfth and third hour marks?

    Here we are talking about a fundamentally different probability. The number of positions in which the minute hand can freeze is infinite.
    Therefore, our classical approach is to take the ratio of two natural numbers- won't work here.
    The concept of geometric probability comes to the rescue. The ratio of areas is taken as the probability. The probability that the arrow will freeze in the marked sector is the ratio of the area of ​​this sector to the area of ​​the entire circle.
    It can be seen that the area of ​​our sector (3 hour divisions) is equal to one fourth of the area of ​​the circle
    (12 hour divisions).
    This means that the desired probability is equal to:

    Problem solved.

    This concludes our conversation about tips and secrets for solving problems in the theory of probability of the Unified State Exam.
    Good luck!