of a person contains a certain plan with which the soul came here, all variants of the development of events, including. You can go there and see the consequences of important decisions that we make. For example, about changing jobs and lifestyles. This can be done both in independent meditations and in joint master-slave processes. Below is a description of how this was done in a session
Probability Lines
I project three branches:
1) stay in Moscow at an existing job;
2) sell or rent an apartment and go to Asia with friends in order to become a partner in their tourism business;
3) ideal option: I leave work, participate in the business of friends on a project basis, while having my own house, but not in Moscow (either Asia too, but different, or Eastern Europe, or Latin America - a large bright villa in which you can receive guests and conduct retreats), there are a couple - their own partnerships, and they have their own business.
We build all three branches as roads, see if there are branches.
The Moscow branch is a strong thick gray rope, dull and reliable, you won't break away, you won't get lost. Several thinner ropes come from the rope, some of them are brighter and more interesting, but none of them attracts, calls or glows. The feeling is that I still love Moscow, but this topic has become obsolete.
The branch with Asia and friends is very bright and visual, but short and liquid, or something. It lacks the potential to confidently turn around in the future. Not enough resource.
The ideal third picture is divided into several geographical points on the map, each with its own specific flavor. The third branch, inside which there is my own story, is the most attractive, of course, for me. She is not as tangible now as Moscow and not as colorful as the second, But she calls to her. And it glows, filled from the inside. Like a thin living ray, it pulsates and shimmers.
Choosing your path
In this version of events, I freely move around the world at will. My income is lower than in Moscow, but it is enough to not need anything and not deny myself anything, albeit in moderation. I come to projects with friends, they stay with me. I write something and work with people, I do it for pleasure. There is also some kind of secular business project, which is also more or less successful, and gives a stable income.
At the same time, there is a close person with whom we will jointly realize this story, in a pair. In order for it to manifest, not only my intention is needed, and some payment will be required on both sides, of course, as for any choice. As soon as you choose something, you automatically refuse something .. It's always scary and unsafe, besides. Payment as a waiver of existing comfort or freedom. Payment as permission to enter into your life something completely new and unknown, albeit tempting. Pure free will and purity of intentions on both sides. And there too - how it will turn out .. In a different vein (not on a pure will), this topic simply will not take off.
This entire process is currently in development. This branch is in the maturing stage, and if everything goes well, then it can fully manifest itself in my reality. See if there are obstacles or stones on this ideal line for me. I see a fallen tree, right on the road. It is fear and self-doubt. From the series - it's too good for it all to turn out that way, it doesn't happen like that, these are all illusions and fairy tales invented by itself. I'm clearing the road.
The next important step is to make your own final decision - whether it is necessary to throw attention there at all, into this dream branch, since it will not be so easy to “rewind” later. I understand for myself that one way or another I have been energizing it for a long time and activating it internally. And this is not even due to stubbornness or the desire to have it my way.
Much more subtle things and signs that signal that this is fate, no matter how loud it may sound. This branch is gradually becoming more and more tangible. It condenses, slowly and surely. Although, of course, it is still extremely uncertain and can collapse at any moment, but there is a feeling that she herself is coming to me, this thread.
Since it has long been designed and predetermined, ordered, one might say. And I understand where this is leading. And how it develops. And that this is the correct development of events. Even though sometimes I'm afraid to believe it.
And still very much it would not be desirable to cement this branch. Make it rigid and unambiguous .. There is no need to build a rigid binding into it to a certain place or occupation, or to something else. I want it to have a lot of elements: air, water, fire, earth, so that it breathes, so that it is flexible and indestructible - mobile, transformable and reconfigurable. And so that everything that happens in it would be the result of co-creation, not autonomous actions. In any case, this is a paired story, it cannot be born as coercion, the maximum correctness is important here - in no case should you impose or pressure .. Everything is free will. And then - where will he call *
Strengthening the branch with attention
I stretch a ray from my Spark in the direction of this branch, to the point where it aspires, I connect with it with my attention. Thus, the spark begins to work towards the realization of this goal, anchors itself in it. I may not be aware of this, but the work will be carried out: the formation of events in space will take place in such a way that this goal is as close as possible to my reality, to its implementation.
The Spark Beam transforms into a gravitational beam and attracts objects and events from that branch of probabilities to me like a magnet. The goal is getting very close, you can say I'm in it now. Like a teleport, when you do not try to move to a new place with your whole body, but materialize the desired space around you: you tune in to the target and attract it to you. And the closer it is to you, the more your will extends to its implementation. And already Iskra is responsible for shaping those events that will entail the embodiment of this branch in reality, will allow it to play.
I paint my future with the light of my Spark. It’s so cool there, in this line of probabilities there is a very beautiful story where I want to invite everyone to visit .. A large bright room filled with life, sun and air .. I give it fuel, charge it with potential so that it has the opportunity to manifest itself in reality. When you are ready to make a final decision or you need to see some answers on the development of this branch, you can simply remember this state of attraction, soak up the emotional atmosphere and mood of this room, feel the emotion of creativity and partnership. The emotion of creation is always love.
Manifestation and consolidation of the result
To capture that picture that looks so attractive, but unsteady now, you need to let light through it, pour in emotion, charge it with positive. Enter the state of ananda - a joyful upsurge, a loving and beloved being, in love and filled with love, and redirect this internal fuel into an ideal scenario.
Clear the path and remove questions. Align with other branches of reality surrounding me and the players involved, so that all this is synchronized in place and in time. Coincided with intentions, will and freedom of choice. Saturate all this with your own light, warmth and love for the realization of your future. creativity in the way you like. Expose the desired result in such a way that the image is imprinted with light on a sensitive film - the canvas of future events, burns its imprint in it as a light projection. And hold for a little while so that the effect is as bright as possible.
Now you need to process the created dream imprint so that it passes into the layer of material reality. The next step is stabilization. It is necessary to add a little energy of darkness and cold to the picture so that it crystallizes and acquires a more solid outline, passes from the state of a magical mirage into denser layers, consolidates and manifests itself.
Working with a negative print .. The result is literally fixed on a sheet of reality, approximately the same as when we project an image from an analog photographic film onto analog photographic paper, and then pour the developer and fixer in turn so that we can see in detail what we captured with the help of light and intentions and enter there when appropriate and timely.
Because for communication with the world and creative realization The throat chakra answers, I send a ray from the throat chakra to the chosen branch. Behind him asked for a ray from the second chakra, followed by the third. Then the rest of the chakras were connected, it turned out such a ray shower, like from a seven-color flower. I wash and dry everything that has turned out, I fill it with movement, the material energy of the earth, vision, all the qualities of life force and magnetism, I attract the branch of probability into my reality even more, I connect it directly with each of the chakra centers, I prescribe it there in them ..
* a person forgets that the future is multivariate and often adheres to template models (these are usually determined by numerology, astrology, etc.). In fact, each of us is a flow, and the flow needs to flow, not to get hung up on the frames, to easily let go of the old and let in the new, to adapt. Therefore, if you do such practices, in no case "cement" your intention, as the world always offers even more cool options that we ourselves may not even be aware of, especially now.
Reality is multidimensional, opinions about it are multifaceted. Only one or a few faces are shown here. You should not take them as the ultimate truth, because, but for each level of consciousness and. We learn to separate what is ours from what is not ours, or to extract information autonomously)
THEMATIC SECTIONS:
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1. Ω = (11,12,13,14,15,16, 21, 22,..., 66),
2. Ω = (2,3,4,5,6, 7,8,9,10,11,12)
3. ● A = (16,61,34, 43, 25, 52);
● B = (11.12, 21.13,31.14, 41.15, 51.16, 61)
● C = (12, 21.36,63.45, 54.33.15, 51, 24.42.66).
● D= (SUM POINTS IS 2 OR 3 );
● E= (TOTAL POINTS IS 10).
Describe the event: FROM= (CIRCUIT CLOSED) for each case.
Solution. Let's introduce the notation: event A- contact 1 is closed; event AT- contact 2 is closed; event FROM- the circuit is closed, the light is on.
1. For a parallel connection, the circuit is closed when at least one of the contacts is closed, so C = A + B;
2. For a series connection, the circuit is closed when both contacts are closed, so C \u003d A B.
A task. 1.1.4 Two electrical circuits have been drawn up:
Event A - the circuit is closed, event A i - I-th contact is closed. For which of them is the ratio
A1 (A2 + A3 A4) A5 = A?
Solution. For the first circuit, A = A1 · (A2 · A3 + A4 · A5), since the sum of events corresponds to the parallel connection, and the product of events to the serial connection. For the second scheme A = A1 (A2+A3 A4 A5). Therefore, this relation is valid for the second scheme.
A task. 1.1.5 Simplify the expression (A + B)(B + C)(C + A).
Solution. Let us use the properties of operations of addition and multiplication of events.
(A+ B)(B + C)(A + C) =
(AB+ AC + B B + BC)(A + C) =
= (AB+ AC + B + BC)(A + C) =
(AB + AC + B)(A + C) = (B + AC)(A + C) =
= BA + BC + ACA + ACC = B A + BC + AC.
A task. 1.1.6Prove that the events A, AB and A+B form a complete group.
Solution. When solving the problem, we will use the properties of operations on events. First, we show that these events are pairwise incompatible.
Let us now show that the sum of these events gives the space of elementary events.
A task. 1.1.7Using the Euler–Venn scheme, check the de Morgan rule:
A) Event AB is shaded.
B) Event A - vertical hatching; event B - horizontal hatching. Event
(A+B) - shaded area.
From a comparison of figures a) and c) it follows:
A task. 1.2.1In how many ways can 8 people be seated?
1. In one row?
2. Per round table?
Solution.
1. The desired number of ways is equal to the number of permutations out of 8, i.e.
P8 = 8! = 1 2 3 4 5 6 7 8 = 40320
2. Since the choice of the first person at the round table does not affect the alternation of elements, then anyone can be taken first, and the remaining ones will be ordered relative to the chosen one. This action can be done in 8!/8 = 5040 ways.
A task. 1.2.2The course covers 5 subjects. In how many ways can you make a schedule for Saturday if there are to be two different couples on that day?
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Solution. The desired number of ways is the number of placements
From 5 to 2, since you need to take into account the order of the pairs:
A task. 1.2.3How examination boards, consisting of 7 people, can be made up of 15 teachers?
Solution. The desired number of commissions (without regard to order) is the number of combinations of 15 to 7:
A task. 1.2.4 From a basket containing twenty numbered balls, 5 balls are chosen for good luck. Determine the number of elements in the space of elementary events of this experience if:
The balls are selected sequentially one after the other with a return after each extraction;
The balls are chosen one by one without returning;
5 balls are selected at once.
Solution.
The number of ways to extract the first ball from the basket is 20. Since the extracted ball is returned to the basket, the number of ways to extract the second ball is also 20, and so on. Then the number of ways to extract 5 balls in this case is 20 20 20 20 20 = 3200000.
The number of ways to extract the first ball from the basket is 20. Since the extracted ball did not return to the basket after the extraction, the number of ways to extract the second ball became 19, etc. Then the number of ways to extract 5 balls without replacement is 20 19 18 17 16 = A52 0
The number of ways to extract 5 balls from the basket at once is equal to the number of combinations of 20 by 5:
A task. 1.2.5 Two dice are thrown. Find the probability of event A that at least one 1 will be rolled.
Solution. Any number of points from 1 to 6 can fall on each die. Therefore, the space of elementary events contains 36 equally possible outcomes. Event A is favored by 11 outcomes: (1.1), (1.2), (2.1), (1.3), (3.1), (1.4), (4.1), (1 .5), (5.1), (1.6), (6.1), so
A task. 1.2.6 The letters y, i, i, k, c, f, n are written on red cards, the letters a, a, o, t, t, s, h are written on blue cards. After thorough mixing, which is more likely: from the first time from the letters to use the red cards to make the word "function" or the letters on the blue cards to make the word "frequency"?
Solution. Let event A be the word "function" randomly composed of 7 letters, event B - the word "frequency" randomly composed of 7 letters. Since two sets of 7 letters are ordered, the number of all outcomes for events A and B is n = 7!. Event A is favored by one outcome m = 1, since all the letters on the red cards are different. Event B is favored by m = 2! · 2! outcomes, since the letters "a" and "t" occur twice. Then P(A) = 1/7! , P(B) = 2! 2! /7! , P(B) > P(A).
A task. 1.2.7 At the exam, the student is offered 30 tickets; Each ticket has two questions. Of the 60 questions included in the tickets, the student knows only 40. Find the probability that the ticket taken by the student will consist of
1. from the issues known to him;
2. from questions unknown to him;
3. from one known and one unknown question.
Solution. Let A be the event that the student knows the answer to both questions; B - does not know the answer to both questions; C - he knows the answer to one question, he does not know the answer to another. The choice of two questions out of 60 can be done in n = C260 = 60 2 59 = 1770 ways.
1. There are m = C240 = 40 2 39 = 780 choices of questions known to the student. Then P(A) = M N = 17 78 70 0 = 0.44
2. The choice of two unknown questions from 20 can be done in m = C220 = 20 2 19 = 190 ways. In this case
P(B) = M N = 11 79 70 0 = 0.11
3. There are m = C14 0 C21 0 = 40 20 = 800 ways to choose a ticket with one known and one unknown question. Then P(C) = 18 70 70 0 = 0.45.
A task. 1.2.8Some information has been sent through three channels. Channels operate independently of each other. Find the probability that the information will reach the goal
1. Only on one channel;
2. At least one channel.
Solution. Let A be an event consisting in the fact that information reaches the goal through only one channel; B - at least one channel. Experience is the transmission of information through three channels. The outcome of the experience - the information has reached the goal. Denote Ai - information reaches the target through the i-th channel. The space of elementary events has the form:
Event B is favored by 7 outcomes: all outcomes except Then n = 8; mA = 3; mB = 7; P(A) = 3 8 ; P(B) = 7 8.
A task. 1.2.9A point randomly appears on a segment of unit length. Find the probability that the distance from the point to the ends of the segment is greater than 1/8.
Solution. According to the condition of the problem, the desired event is satisfied by all points that appear on the interval (a; b).
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Since its length is s = 1 - 1 8 + 1 8 = 3 4, and the length of the entire segment is S = 1, the required probability is P = s/S = 3/14 = 0.75.
A task. 1.2.10In a batch ofNproductsKproducts are defective. For control, m products are selected. Find the probability that from M Products L They turn out to be defective (event A).
Solution. The choice of m products from n can be done in ways, and the choice L defective out of k defective - in ways. After selection L defective products will remain (m - L) fit, located among (n - k) products. Then the number of outcomes favoring the event A is
And the desired probability 
A task. 1.3.1BAn urn contains 30 balls: 15 red, 10 blue and 5 white. Find the probability that a randomly drawn ball is colored.
Solution. Let event A - a red ball is drawn, event B - a blue ball is drawn. Then events (A + B) - a colored ball is drawn. We have P(A) = 1 3 5 0 = 1 2 , P(B) = 1 3 0 0 = 1 3. Since
Events A and B are incompatible, then P(A + B) = P(A) + P(B) = 1 2 + 1 3 = 5 6 = 0.83.
A task. 1.3.2The probability that it will snow (an event A ), is equal to 0.6, And the fact that it will rain (event B ), is equal to 0.45. Find the probability of bad weather if the probability of rain and snow (event AB ) is equal to 0.25.
Solution. Events A and B are joint, so P(A + B) = P(A) + P(B) - P(AB) = 0.6 + 0.45 - 0.25 = 0.8
A task. 1.3.3BThe first box contains 2 white and 10 black balls, the second - 3 white and 9 black balls, and the third - 6 white and 6 black balls. A ball was taken from each box. Find the probability that all the balls drawn are white.
Solution. Event A - a white ball is drawn from the first box, B - from the second box, C - from the third. Then P(A) = 12 2 = 1 6; P(B) = 13 2 = 1 4; P(C) = 16 2 = 1 2. Event ABC - all taken out
Balls are white. Events A, B, C are independent, therefore
P(ABC) = P(A) P(B) P(C) = 1 6 1 4 1 2 = 41 8 = 0.02
A task. 1.3.4Belectrical circuit connected in series 5 Elements that work independently of each other. The probability of failures of the first, second, third, fourth, fifth elements, respectively, are 0.1; 0.2; 0.3; 0.2; 0.1. Find the probability that there will be no current in the circuit (event A ).
Solution. Since the elements are connected in series, there will be no current in the circuit if at least one element fails. Event Ai(i =1...5) - will fail I-th element. Developments
A task. 1.3.5The circuit consists of independent blocks connected in a system with one input and one output.
Failure in time T of various circuit elements - independent events having the following probabilitiesP 1 = 0.1; P 2 = 0.2; P 3 = 0.3; P 4 = 0.4. The failure of any of the elements leads to an interruption of the signal in the branch of the circuit where this element is located. Find the reliability of the system.
Solution. If event A - (SYSTEM IS RELIABLE), Ai - (i - th UNIT WORKS FAULTY), then A = (A1 + A2)(A3 + A4). Events A1+A2, A3+A4 are independent, events A1 and A2, A3 and A4 are joint. According to the formulas for multiplication and addition of probabilities
A task. 1.3.6The worker serves 3 machines. The probability that within an hour the machine does not require the attention of a worker is 0.9 for the first machine, 0.8 for the second machine, and 0.7 for the third machine.
Find the probability that during some hour
1. The second machine will require attention;
2. Two machines will require attention;
3. At least two machines will need attention.
Solution. Let Ai - the i-th machine require the attention of the worker, - the i-th machine will not require the attention of the worker. Then
Space of elementary events:
1. Event A - will require the attention of the second machine: Then
Since the events are incompatible and independent. P(A) = 0.9 0.8 0.7 + 0.1 0.8 0.7 + 0.9 0.8 0.3 + 0.1 0.8 0.3 = 0.8
2. Event B - two machines will require attention:
3. Event C - at least two stuns will require attention
cov:
A task. 1.3.7Bmachine "Examiner" introduced 50 questions. The student is offered 5 Questions and an “excellent” mark is given if all questions are answered correctly. Find the probability of getting "excellent" if the student prepared only 40 questions.
Solution. A - (RECEIVED "EXCELLENT"), Ai - (ANSWERED TO i - th QUESTION). Then A = A1A2A3A4A5, we have:
Or, in another way - using the classical probability formula: 
And
A task. 1.3.8The probabilities that the part needed by the assembler is inI, II, III, IVbox, respectively, are equal 0.6; 0.7; 0.8; 0.9. Find the probability that the collector will have to check all 4 boxes (eventA).
Solution. Let Ai - (The part needed by the assembler is in the i-th box.) Then
Since the events are incompatible and independent, then
A task. 1.4.1 A group of 10,000 people over the age of 60 was examined. It turned out that 4000 people are permanent smokers. 1800 smokers showed serious changes in the lungs. Among non-smokers, 1500 people had changes in the lungs. What is the probability that a randomly examined person with lung changes is a smoker?
Solution. Let's introduce the hypotheses: H1 - the examined is a permanent smoker, H2 - is a non-smoker. Then by the condition of the problem
P(H1)= -------=0.4, P(H2)=---------=0.6
Denote by A the event that the examined person has changes in the lungs. Then by the condition of the problem
By formula (1.15) we find
The desired probability that the examined person is a smoker, according to the Bayes formula, is equal to
A task. 1.4.2Televisions from three factories go on sale: 30% from the first factory, 20% from the second, 50% from the third. The products of the first factory contain 20% of TVs with a hidden defect, the second - 10%, the third - 5%. What is the probability of getting a working TV?
Solution. Let's consider the following events: A - a serviceable TV was purchased; hypotheses H1, H2, H3 - the TV went on sale from the first, second, third factory, respectively. According to the task
By formula (1.15) we find
A task. 1.4.3There are three identical boxes. The first has 20 white balls, the second has 10 white and 10 black balls, and the third has 20 black balls. A white ball is drawn from a randomly selected box. Find the probability that this ball is from the second box.
Solution. Let the event A - a white ball is taken out, hypotheses H1, H2, H3 - the ball is taken out from the first, second, third boxes respectively. From the condition of the problem we find
Then 
By formula (1.15) we find
By formula (1.16) we find
A task. 1.4.4A telegraph message consists of the dot and dash signals. The statistical properties of interference are such that they are distorted on average 2/5 Dot messages and 1/3 Dash messages. It is known that among the transmitted signals "dot" and "dash" occur in the ratio 5: 3. Determine the probability that a transmitted signal is received if:
A) a "point" signal is received;
B)dash signal received.
Solution. Let the event A - the "dot" signal is received, and the event B - the "dash" signal is received.
Two hypotheses can be made: H1 - the "dot" signal is transmitted, H2 - the "dash" signal is transmitted. By condition P(H1) : P(H2) =5: 3. In addition, P(H1 ) + P(H2)= 1. Therefore P( H1 ) = 5/8, P(H2 ) = 3/8. It is known that
Event probabilities A And B We find by the formula of total probability:
The desired probabilities will be:
A task. 1.4.5Of the 10 radio channels, 6 channels are protected from interference. Probability that a secure channel over timeTwill not fail is 0.95, for an unprotected channel - 0.8. Find the probability that two randomly selected channels will not fail in timeT, and both channels are not protected from interference.
Solution. Let the event A - both channels will not fail during the time t, the event A1- Secure channel selected A2- An unsecured channel is selected.
Let's write the space of elementary events for the experiment - (two channels are selected):
Ω = (A1A1, A1A2, A2A1, A2A2)
Hypotheses:
H1 - both channels are protected from interference;
H2 - the first selected channel is protected, the second selected channel is not protected from interference;
H3 - the first selected channel is not protected, the second selected channel is protected from interference;
H4 - both selected channels are not protected from interference. Then
And 
A task. 1.5.1Transmitted over the communication channel 6 Messages. Each of the messages can be distorted by noise with a probability 0.2 Regardless of others. Find the probability that
1. 4 messages out of 6 are not distorted;
2. At least 3 out of 6 were transmitted distorted;
3. At least one message out of 6 is garbled;
4. No more than 2 out of 6 are not distorted;
5. All messages are transmitted without distortion.
Solution. Since the probability of distortion is 0.2, the probability of transmitting a message without interference is 0.8.
1. Using the Bernoulli formula (1.17), we find the probability
transmission rate of 4 out of 6 messages without interference:
2. at least 3 out of 6 are transmitted distorted:
3. at least one message out of 6 is garbled:
4. at least one message out of 6 is garbled:
5. all messages are transmitted without distortion:
A task. 1.5.2The probability that the day will be clear in summer is 0.42; the probability of an overcast day is 0.36 and partly cloudy is 0.22. How many days out of 59 can be expected to be clear and overcast?
Solution. It can be seen from the condition of the problem that it is necessary to look for the most probable number of clear and cloudy days.
For clear days P= 0.42, N= 59. We compose inequalities (1.20):
59 0.42 + 0.42 - 1 < m0 < 59 0.42 + 0.42.
24.2 ≤ Mo≤ 25.2 → Mo= 25.
For cloudy days P= 0.36, N= 59 and
0.36 59 + 0.36 - 1 ≤ M0 ≤ 0.36 59 + 0.36;
Hence 20.16 ≤ M0 ≤ 21.60; → M0 = 21.
Thus, the most probable number of clear days Mo= 25, cloudy days - M0 = 21. Then in summer we can expect Mo+ M0 =46 clear and cloudy days.
A task. 1.5.3There are 110 students of the course at the lecture on probability theory. Find the probability that
1. k students (k = 0,1,2) of those present were born on the first of September;
2. at least one student of the course was born on the first of September.
P=1/365 is very small, so we use the Poisson formula (1.22). Let's find the Poisson parameter. Because
N= 110, then λ = np = 110 1 /365 = 0.3.
Then by the Poisson formula 
A task. 1.5.4The probability that a part is not standard is 0.1. How many details need to be selected so that with probability P = 0.964228 It could be argued that the relative frequency of occurrence of non-standard parts deviates from the constant probability p = 0.1 In absolute terms, no more than 0.01 ?
Solution.
Required number N We find by formula (1.25). We have:
P = 1.1; q = 0.9; P= 0.96428. Substitute the data in the formula:
Where do we find
According to the table of values of the function Φ( X) we find that
A task. 1.5.5The probability of failure in time T of one capacitor is 0.2. Determine the probability that in time T out of 100 capacitors will fail.
1. Exactly 10 capacitors;
2. At least 20 capacitors;
3. Less than 28 capacitors;
4. From 14 to 26 capacitors.
Solution. We have P = 100, P= 0.2, Q = 1 - P= 0.8.
1. Exactly 10 capacitors.
Because P Veliko, let's use the local de Moivre-Laplace theorem:
Compute
Since the function φ(x)- even, then φ (-2.5) = φ (2.50) = 0.0175 (we find from the table of function values φ(x). Desired probability
2. At least 20 capacitors;
The requirement that at least 20 out of 100 capacitors fail means that either 20, or 21, ..., or 100 will fail. Thus, T1 = 20, T 2=100. Then
According to the table of function values Φ(x) Let us find Φ(x1) = Φ(0) = 0, Φ(x2) = Φ(20) = 0.5. Required probability:
3. Less than 28 capacitors;
(here it was taken into account that the Laplace function Ф(x) is odd).
4. From 14 to 26 capacitors. By condition M1= 14, m2 = 26.
Calculate x 1,x2:
A task. 1.5.6The probability of occurrence of some event in one experiment is equal to 0.6. What is the probability that this event will occur in most of the 60 trials?
Solution. Quantity M The occurrence of an event in a series of tests is in the interval. "In most experiments" means that M Belongs to interval By condition N= 60, P= 0.6, Q = 0.4, M1 = 30, m2 = 60. Calculate x1 and x2:
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Random variables and their distributions
A task. 2.1.1Given a table where the top line indicates the possible values of a random variable X , and at the bottom - their probabilities.
Can this table be a distribution series X ?
Answer: Yes, since p1 + p2 + p3 + p4 + p5 = 1
A task. 2.1.2Released 500 Lottery tickets, and 40 Tickets will bring their owners a prize for 10000 Rub., 20 Tickets - by 50000 Rub., 10 Tickets - by 100000 Rub., 5 Tickets - by 200000 Rub., 1 Ticket - 500000 Rub., the rest - without a win. Find the winning distribution law for the owner of one ticket.
Solution.
Possible values of X: x5 = 10000, x4 = 50000, x3 = 100000, x2 = 200000, x1 = 500000, x6 = 0. The probabilities of these possible values are:
The desired distribution law:
A task. 2.1.3shooter, having 5 Cartridges, shoots until the first hit on the target. The probability of hitting each shot is 0.7. Construct the law of distribution of the number of cartridges used, find the distribution functionF(X) and plot its graph, find P(2< x < 5).
Solution.
Space of elementary events of experience
Ω = {1, 01, 001, 0001, 00001, 11111},
Where event (1) - hit the target, event (0) - did not hit the target. Elementary outcomes correspond to the following values of the random value of the number of cartridges used: 1, 2, 3, 4, 5. Since the result of each next shot does not depend on the previous one, the probabilities of possible values are:
P1 = P(x1= 1) = P(1)= 0.7; P2 = P(x2= 2) = P(01)= 0.3 0.7 = 0.21;
P3 = P(x3= 3) = P(001) = 0.32 0.7 = 0.063;
P4 = P(x4= 4) = P(0001) = 0.33 0.7 = 0.0189;
P5 = P(x5= 5) = P(00001 + 00000) = 0.34 0.7 + 0.35 = 0.0081.
The desired distribution law:
Find the distribution function F(X), Using formula (2.5)
X≤1, F(x)= P(X< x) = 0
1 < x ≤2, F(x)= P(X< x) = P1(X1 = 1) = 0.7
2 < x ≤ 3, F(x) = P1(X= 1) + P2(x = 2) = 0.91
3 < x ≤ 4, F(x) = P1 (x = 1) + P2(x = 2) + P3(x = 3) =
= 0.7 + 0.21 + 0.063 = 0.973
4 < x ≤ 5, F(x) = P1(x = 1) + P2(x = 2) + P3(x = 3) +
+ P4(x = 4) = 0.973 + 0.0189 = 0.9919
X >5, F(x) = 1
Find P(2< x < 5). Применим формулу (2.4): P(2 < X< 5) = F(5) - F(2) = 0.9919 - 0.91 = 0.0819
A task. 2.1.4DanaF(X) of some random variable:
Write down the distribution series for X.
Solution.
From properties F(X)
It follows that the possible values of the random variable X -
Function break points F(X),
And the corresponding probabilities are jumps of the function F(X).
Find the possible values of the random variable X=(0,1,2,3,4). 
A task. 2.1.5Set which function
Is a distribution function of some random variable.
If the answer is yes, find the probability that the corresponding random value takes values on[-3,2].
Solution. Let's plot the functions F1(x) and F2(x):
The function F2(x) is not a distribution function, since it is not non-decreasing. The function F1(x) is
The distribution function of some random variable, since it is non-decreasing and satisfies condition (2.3). Let's find the probability of hitting the interval:
A task. 2.1.6Given the probability density of a continuous random variable X :
Find:
1. Coefficient C ;
2. distribution function F(x) ;
3. The probability of a random variable falling into the interval(1, 3).
Solution. From the normalization condition (2.9) we find
Consequently,
By formula (2.10) we find:
In this way,
By formula (2.4) we find
A task. 2.1.7Random downtime of electronic equipment in some cases has a probability density
Where M = lge = 0.4343...
Find distribution function F(x) .
Solution. By formula (2.10) we find
Where 
A task. 2.2.1A distribution series of a discrete random variable is given X :
Find expected value, variance, standard deviation, M, D[-3X + 2].
Solution.
According to the formula (2.12) we find the mathematical expectation:
M[X] = x1p1 + x2p2 + x3p3 + x4p4 = 10 0.2 + 20 0.15 + 30 0.25 + 40 0.4 = 28.5
M = 2M[X] + M = 2M[X] + 5 = 2 28.5 + 5 = 62. Using formula (2.19), we find the dispersion:
A task. 2.2.2Find the mathematical expectation, variance and standard deviation of a continuous random variable X , whose distribution function
.
Solution. Find the probability density:
The mathematical expectation is found by the formula (2.13):
We find the dispersion by the formula (2.19):
Let us first find the mathematical expectation of the square of the random variable:
Standard deviation
A task. 2.2.3Xhas a number of distributions:
Find the mathematical expectation and variance of a random variableY = EX .
Solution. M[ Y] = M[ EX ] = e-- 1 0.2 + e0 0.3 + e1 0.4 + e2 0.1 =
0.2 0.3679 + 1 0.3 + 2.71828 0.4 + 7.389 0.1 = 2.2.
D[Y] = D = M[(eX)2 - M2[E X] =
[(e-1)2 0.2 + (e0)2 0.3 + (e1)2 0.4 + (e2)2 0.1] - (2.2)2 =
= (e--2 0.2 + 0.3 + e2 0.4 + e4 0.1) - 4.84 = 8.741 - 4.84 = 3.9.
A task. 2.2.4Discrete random variable X Can only take two values X1 And X2 , and X1< x2. Known Probability P1 = 0.2 Possible value X1 , expected value M[X] = 3.8 And dispersion D[X] = 0.16. Find the law of distribution of a random variable.
Solution. Since the random variable X takes only two values x1 and x2, then the probability p2 = P(X = x2) = 1 - p1 = 1 - 0.2 = 0.8.
By the condition of the problem, we have:
M[X] = x1p1 + x2p2 = 0.2x1 + 0.8x2 = 3.8;
D[X] = (x21p1 + x22p2) - M2[X] = (0.2x21 + 0.8x22) - (0.38)2 = 0.16.
Thus, we got the system of equations:
Condition x1
A task. 2.2.5The random variable X is subject to the distribution law, the density graph of which has the form:
Find the mathematical expectation, variance and standard deviation.
Solution.
Let us find the differential distribution function f(x). Outside the interval (0, 3) f(x) = 0. On the interval (0, 3) the density graph is a straight line with slope k = 2/9 passing through the origin. In this way, 
Expected value:
Find the variance and standard deviation:
A task. 2.2.6Find the mathematical expectation and variance of the sum of points on four dice in one roll.
Solution.
Let's denote A - the number of points on one die in one throw, B - the number of points on the second die, C - on the third die, D - on the fourth die. For random variables A, B, C, D, the distribution law 
one.
Then M[A] = M[B] = M[C] = M[D] = (1+2+3+4+5+6) / 6 = 3.5
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A task. 2.3.1The probability that a particle emitted from a radioactive source will be registered by a counter is equal to 0.0001. During the observation period, 30000 particles. Find the probability that the counter registered:
1. Exactly 3 particles;
2. Not a single particle;
3. At least 10 particles.
Solution.
By condition P= 30000, P= 0.0001. The events consisting in the fact that particles emitted from a radioactive source are registered are independent; number P Great, but the probability P Small, so we use the Poisson distribution: 
Let's find λ: λ
= n P =
30000 0.0001 = 3 = M[X]. Desired probabilities:
A task. 2.3.2There are 5% non-standard parts in the lot. 5 items were randomly selected. Write the distribution law of a discrete random variable X - the number of non-standard parts among the five selected; find the mathematical expectation and variance.
Solution. Discrete random variable X - the number of non-standard parts - has a binomial distribution and can take the following values: x1 = 0, x2 = 1, x3 = 2, x4 = 3, x5 = 4, x6 = 5. Probability of a non-standard part in a batch p = 5 /100 = 0.05. Let's find the probabilities of these possible values:
Let's write the desired distribution law:
Let's find numerical characteristics:
0 0.7737809 + 1 0.2036267 + 2 0.0214343+
3 0.0011281 + 4 0.0000297 + 5 0.0000003 = 0.2499999 ≈ 0.250
M[X] = Np= 5 0.05 = 0.25.
D[X] = M– M2 [X]= 02 0.7737809 + 12 0.2036267+
22 0.0214343 + 32 0.0011281 + 42 0.0000297 + 52 0.0000003- 0.0625 =
0.2999995 - 0.0625 = 0.2374995 ≈ 0.2375
Or D[ X] = np (1 - P) = 5 0.05 0.95 = 0.2375.
A task. 2.3.3The radar target detection time is distributed according to the exponential law
Where1/ λ = 10 Sec. - average target detection time. Find the probability that the target will be found within the time5 Before15 Sec. after the start of the search.
Solution. Probability of hitting a random variable X In interval (5, 15) Let us find by formula (2.8):
At 
We get
0.6065(1 - 0.3679) = 0.6065 0.6321 = 0.3834
A task. 2.3.4Random measurement errors are subject to the normal law with parameters a = 0, σ = 20 Mm. Write differential distribution functionF(X) and find the probability that the measurement made an error in the interval from 5 Before 10 Mm.
Solution. Let us substitute the values of the parameters a and σ into the differential distribution function (2.35):
Using formula (2.42), we find the probability of hitting a random variable X In the interval , i.e. A= 0, B= 0.1. Then the differential distribution function F(x) Will look like
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What is a probability?
Faced with this term for the first time, I would not understand what it is. So I'll try to explain in an understandable way.
Probability is the chance that the desired event will occur.
For example, you decided to visit a friend, remember the entrance and even the floor on which he lives. But I forgot the number and location of the apartment. And now you are standing on the stairwell, and in front of you are the doors to choose from.
What is the chance (probability) that if you ring the first doorbell, your friend will open it for you? Whole apartment, and a friend lives only behind one of them. With equal chance, we can choose any door.
But what is this chance?
Doors, the right door. Probability of guessing by ringing the first door: . That is, one time out of three you will guess for sure.
We want to know by calling once, how often will we guess the door? Let's look at all the options:
- you called to 1st Door
- you called to 2nd Door
- you called to 3rd Door
And now consider all the options where a friend can be:
a. Per 1st door
b. Per 2nd door
in. Per 3rd door
Let's compare all the options in the form of a table. A tick indicates the options when your choice matches the location of a friend, a cross - when it does not match.
How do you see everything Maybe options friend's location and your choice of which door to ring.
BUT favorable outcomes of all . That is, you will guess the times from by ringing the door once, i.e. .
This is the probability - the ratio of a favorable outcome (when your choice coincided with the location of a friend) to the number of possible events.
The definition is the formula. Probability is usually denoted p, so:
It is not very convenient to write such a formula, so let's take for - the number of favorable outcomes, and for - the total number of outcomes.
The probability can be written as a percentage, for this you need to multiply the resulting result by:
Probably, the word “outcomes” caught your eye. Since mathematicians call various actions (for us, such an action is a doorbell) experiments, it is customary to call the result of such experiments an outcome.
Well, the outcomes are favorable and unfavorable.
Let's go back to our example. Let's say we rang at one of the doors, but a stranger opened it for us. We didn't guess. What is the probability that if we ring one of the remaining doors, our friend will open it for us?
If you thought that, then this is a mistake. Let's figure it out.
We have two doors left. So we have possible steps:
1) Call to 1st Door
2) Call 2nd Door
A friend, with all this, is definitely behind one of them (after all, he was not behind the one we called):
a) a friend 1st door
b) a friend for 2nd door
Let's draw the table again:
As you can see, there are all options, of which - favorable. That is, the probability is equal.
Why not?
The situation we have considered is example of dependent events. The first event is the first doorbell, the second event is the second doorbell.
And they are called dependent because they affect the following actions. After all, if a friend opened the door after the first ring, what would be the probability that he was behind one of the other two? Correctly, .
But if there are dependent events, then there must be independent? True, there are.
A textbook example is tossing a coin.
- We toss a coin. What is the probability that, for example, heads will come up? That's right - because the options for everything (either heads or tails, we will neglect the probability of a coin to stand on edge), but only suits us.
- But the tails fell out. Okay, let's do it again. What is the probability of coming up heads now? Nothing has changed, everything is the same. How many options? Two. How much are we satisfied with? One.
And let tails fall out at least a thousand times in a row. The probability of falling heads at once will be the same. There are always options, but favorable ones.
Distinguishing dependent events from independent events is easy:
- If the experiment is carried out once (once a coin is tossed, the doorbell rings once, etc.), then the events are always independent.
- If the experiment is carried out several times (a coin is tossed once, the doorbell is rung several times), then the first event is always independent. And then, if the number of favorable or the number of all outcomes changes, then the events are dependent, and if not, they are independent.
Let's practice a little to determine the probability.
Example 1
The coin is tossed twice. What is the probability of getting heads up twice in a row?
Solution:
Consider all possible options:
- eagle eagle
- tails eagle
- tails-eagle
- Tails-tails
As you can see, all options. Of these, we are satisfied only. That is the probability:
If the condition asks simply to find the probability, then the answer must be given as a decimal fraction. If it were indicated that the answer must be given as a percentage, then we would multiply by.
Answer:
Example 2
In a box of chocolates, all candies are packed in the same wrapper. However, from sweets - with nuts, cognac, cherries, caramel and nougat.
What is the probability of taking one candy and getting a candy with nuts. Give your answer in percentage.
Solution:
How many possible outcomes are there? .
That is, taking one candy, it will be one of those in the box.
And how many favorable outcomes?
Because the box contains only chocolates with nuts.
Answer:
Example 3
In a box of balls. of which are white and black.
- What is the probability of drawing a white ball?
- We added more black balls to the box. What is the probability of drawing a white ball now?
Solution:
a) There are only balls in the box. of which are white.
The probability is:
b) Now there are balls in the box. And there are just as many whites left.
Answer:
Full Probability
| The probability of all possible events is (). |
For example, in a box of red and green balls. What is the probability of drawing a red ball? Green ball? Red or green ball?
Probability of drawing a red ball
Green ball:
Red or green ball:
As you can see, the sum of all possible events is equal to (). Understanding this point will help you solve many problems.
Example 4
There are felt-tip pens in the box: green, red, blue, yellow, black.
What is the probability of drawing NOT a red marker?
Solution:
Let's count the number favorable outcomes.
NOT a red marker, that means green, blue, yellow, or black.
Probability of all events. And the probability of events that we consider unfavorable (when we pull out a red felt-tip pen) is .
Thus, the probability of drawing NOT a red felt-tip pen is -.
Answer:
| The probability that an event will not occur is minus the probability that the event will occur. |
Rule for multiplying the probabilities of independent events
You already know what independent events are.
And if you need to find the probability that two (or more) independent events will occur in a row?
Let's say we want to know what is the probability that by tossing a coin once, we will see an eagle twice?
We have already considered - .
What if we toss a coin? What is the probability of seeing an eagle twice in a row?
Total possible options:
- Eagle-eagle-eagle
- Eagle-head-tails
- Head-tails-eagle
- Head-tails-tails
- tails-eagle-eagle
- Tails-heads-tails
- Tails-tails-heads
- Tails-tails-tails
I don't know about you, but I made this list wrong once. Wow! And only option (the first) suits us.
For 5 rolls, you can make a list of possible outcomes yourself. But mathematicians are not as industrious as you.
Therefore, they first noticed, and then proved, that the probability of a certain sequence of independent events decreases each time by the probability of one event.
In other words,
Consider the example of the same, ill-fated, coin.
Probability of coming up heads in a trial? . Now we are tossing a coin.
What is the probability of getting tails in a row?
This rule does not only work if we are asked to find the probability that the same event will occur several times in a row.
If we wanted to find the TAILS-EAGLE-TAILS sequence on consecutive flips, we would do the same.
The probability of getting tails - , heads - .
The probability of getting the sequence TAILS-EAGLE-TAILS-TAILS:
You can check it yourself by making a table.
The rule for adding the probabilities of incompatible events.
So stop! New definition.
Let's figure it out. Let's take our worn out coin and flip it once.
Possible options:
- Eagle-eagle-eagle
- Eagle-head-tails
- Head-tails-eagle
- Head-tails-tails
- tails-eagle-eagle
- Tails-heads-tails
- Tails-tails-heads
- Tails-tails-tails
So here are incompatible events, this is a certain, given sequence of events. are incompatible events.
If we want to determine what is the probability of two (or more) incompatible events, then we add the probabilities of these events.
You need to understand that the loss of an eagle or tails is two independent events.
If we want to determine what is the probability of a sequence falling out) (or any other), then we use the rule of multiplying probabilities.
What is the probability of getting heads on the first toss and tails on the second and third?
But if we want to know what is the probability of getting one of several sequences, for example, when heads come up exactly once, i.e. options and, then we must add the probabilities of these sequences.
Total options suits us.
We can get the same thing by adding up the probabilities of occurrence of each sequence:
Thus, we add probabilities when we want to determine the probability of some, incompatible, sequences of events.
There is a great rule to help you not get confused when to multiply and when to add:
Let's go back to the example where we tossed a coin times and want to know the probability of seeing heads once.
What is going to happen?
Should drop:
(heads AND tails AND tails) OR (tails AND heads AND tails) OR (tails AND tails AND heads).
And so it turns out:
Let's look at a few examples.
Example 5
There are pencils in the box. red, green, orange and yellow and black. What is the probability of drawing red or green pencils?
Solution:
What is going to happen? We have to pull out (red OR green).
Now it’s clear, we add up the probabilities of these events:
Answer:
Example 6
A die is thrown twice, what is the probability that a total of 8 will come up?
Solution.
How can we get points?
(and) or (and) or (and) or (and) or (and).
The probability of falling out of one (any) face is .
We calculate the probability:
Answer:
Workout.
I think now it has become clear to you when you need to how to count the probabilities, when to add them, and when to multiply them. Is not it? Let's get some exercise.
Tasks:
Let's take a deck of cards in which the cards are spades, hearts, 13 clubs and 13 tambourines. From to Ace of each suit.
- What is the probability of drawing clubs in a row (we put the first card drawn back into the deck and shuffle)?
- What is the probability of drawing a black card (spades or clubs)?
- What is the probability of drawing a picture (jack, queen, king or ace)?
- What is the probability of drawing two pictures in a row (we remove the first card drawn from the deck)?
- What is the probability, taking two cards, to collect a combination - (Jack, Queen or King) and Ace The sequence in which the cards will be drawn does not matter.
Answers:
- In a deck of cards of each value, it means:
- The events are dependent, since after the first card drawn, the number of cards in the deck has decreased (as well as the number of “pictures”). Total jacks, queens, kings and aces in the deck initially, which means the probability of drawing the “picture” with the first card:
Since we are removing the first card from the deck, it means that there is already a card left in the deck, of which there are pictures. Probability of drawing a picture with the second card:
Since we are interested in the situation when we get from the deck: “picture” AND “picture”, then we need to multiply the probabilities:
Answer:
- After the first card is drawn, the number of cards in the deck will decrease. Thus, we have two options:
1) With the first card we take out Ace, the second - jack, queen or king
2) With the first card we take out a jack, queen or king, the second - an ace. (ace and (jack or queen or king)) or ((jack or queen or king) and ace). Don't forget about reducing the number of cards in the deck!
If you were able to solve all the problems yourself, then you are a great fellow! Now tasks on the theory of probability in the exam you will click like nuts!
PROBABILITY THEORY. AVERAGE LEVEL
Consider an example. Let's say we throw a die. What kind of bone is this, do you know? This is the name of a cube with numbers on the faces. How many faces, so many numbers: from to how many? Before.
So we roll a die and want it to come up with an or. And we fall out.
In probability theory they say what happened favorable event(not to be confused with good).
If it fell out, the event would also be auspicious. In total, only two favorable events can occur.
How many bad ones? Since all possible events, then the unfavorable of them are events (this is if it falls out or).
Definition:
Probability is the ratio of the number of favorable events to the number of all possible events.. That is, the probability shows what proportion of all possible events are favorable.
They denote the probability with a Latin letter (apparently, from the English word probability - probability).
It is customary to measure the probability as a percentage (see topics and). To do this, the probability value must be multiplied by. In the dice example, probability.
And in percentage: .
Examples (decide for yourself):
- What is the probability that the toss of a coin will land on heads? And what is the probability of a tails?
- What is the probability that an even number will come up when a dice is thrown? And with what - odd?
- In a drawer of plain, blue and red pencils. We randomly draw one pencil. What is the probability of pulling out a simple one?
Solutions:
- How many options are there? Heads and tails - only two. And how many of them are favorable? Only one is an eagle. So the probability
Same with tails: .
- Total options: (how many sides a cube has, so many different options). Favorable ones: (these are all even numbers :).
Probability. With odd, of course, the same thing. - Total: . Favorable: . Probability: .
Full Probability
All pencils in the drawer are green. What is the probability of drawing a red pencil? There are no chances: probability (after all, favorable events -).
Such an event is called impossible.
What is the probability of drawing a green pencil? There are exactly as many favorable events as there are total events (all events are favorable). So the probability is or.
Such an event is called certain.
If there are green and red pencils in the box, what is the probability of drawing a green or a red one? Yet again. Note the following thing: the probability of drawing green is equal, and red is .
In sum, these probabilities are exactly equal. That is, the sum of the probabilities of all possible events is equal to or.
Example:
In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of not drawing green?
Solution:
Remember that all probabilities add up. And the probability of drawing green is equal. This means that the probability of not drawing green is equal.
Remember this trick: The probability that an event will not occur is minus the probability that the event will occur.
Independent events and the multiplication rule
You flip a coin twice and you want it to come up heads both times. What is the probability of this?
Let's go through all the possible options and determine how many there are:
Eagle-Eagle, Tails-Eagle, Eagle-Tails, Tails-Tails. What else?
The whole variant. Of these, only one suits us: Eagle-Eagle. So, the probability is equal.
Good. Now let's flip a coin. Count yourself. Happened? (answer).
You may have noticed that with the addition of each next throw, the probability decreases by a factor. The general rule is called multiplication rule:
The probabilities of independent events change.
What are independent events? Everything is logical: these are those that do not depend on each other. For example, when we toss a coin several times, each time a new toss is made, the result of which does not depend on all previous tosses. With the same success, we can throw two different coins at the same time.
More examples:
- A die is thrown twice. What is the probability that it will come up both times?
- A coin is tossed times. What is the probability of getting heads first and then tails twice?
- The player rolls two dice. What is the probability that the sum of the numbers on them will be equal?
Answers:
- The events are independent, which means that the multiplication rule works: .
- The probability of an eagle is equal. Tails probability too. We multiply:
- 12 can only be obtained if two -ki fall out: .
Incompatible events and the addition rule
Incompatible events are events that complement each other to full probability. As the name implies, they cannot happen at the same time. For example, if we toss a coin, either heads or tails can fall out.
Example.
In a box of pencils, among them are blue, red, green, simple, yellow, and the rest are orange. What is the probability of drawing green or red?
Solution .
The probability of drawing a green pencil is equal. Red - .
Auspicious events of all: green + red. So the probability of drawing green or red is equal.
The same probability can be represented in the following form: .
This is the addition rule: the probabilities of incompatible events add up.
Mixed tasks
Example.
The coin is tossed twice. What is the probability that the result of the rolls will be different?
Solution .
This means that if heads come up first, tails should be second, and vice versa. It turns out that there are two pairs of independent events here, and these pairs are incompatible with each other. How not to get confused about where to multiply and where to add.
There is a simple rule for such situations. Try to describe what should happen by connecting the events with the unions "AND" or "OR". For example, in this case:
Must roll (heads and tails) or (tails and heads).
Where there is a union "and", there will be multiplication, and where "or" is addition:
Try it yourself:
- What is the probability that two coin tosses come up with the same side both times?
- A die is thrown twice. What is the probability that the sum will drop points?
Solutions:
- (Heads up and heads up) or (tails up and tails up): .
- What are the options? and. Then:
Rolled (and) or (and) or (and): .
Another example:
We toss a coin once. What is the probability that heads will come up at least once?
Solution:
Oh, how I don’t want to sort through the options ... Head-tails-tails, Eagle-heads-tails, ... But you don’t have to! Let's talk about full probability. Remembered? What is the probability that the eagle will never drop? It's simple: tails fly all the time, that means.
PROBABILITY THEORY. BRIEFLY ABOUT THE MAIN
Probability is the ratio of the number of favorable events to the number of all possible events.
Independent events
Two events are independent if the occurrence of one does not change the probability of the other occurring.
Full Probability
The probability of all possible events is ().
The probability that an event will not occur is minus the probability that the event will occur.
Rule for multiplying the probabilities of independent events
The probability of a certain sequence of independent events is equal to the product of the probabilities of each of the events
Incompatible events
Incompatible events are those events that cannot possibly occur simultaneously as a result of an experiment. A number of incompatible events form a complete group of events.
The probabilities of incompatible events add up.
Having described what should happen, using the unions "AND" or "OR", instead of "AND" we put the sign of multiplication, and instead of "OR" - addition.
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Your number is the twelfth, - said the fir, writing down something in a little book. Flash thanked the man and flew off to his cabin. , Now the main thing is not to adjust. I hope the fairy will not let you down when we perform…"" - with these thoughts, the brunette landed on a branch next to the gazebo, where two people were already waiting for him. “Finally, you have come,” one of those waiting, Nick, waved to him with a smile. The gray-eyed girl with a dark square, who is the second person, only nodded as a sign of greeting, going straight to the point: - And under what number do we perform? she asked, placing cups of aromatic coffee on the table. - Twelve, - sitting down at the table, the guy answered. - We need to rehearse: we need to know how the three of us sound. -We don't have to perform very well, Dragotius, - the girl immediately cooled him, - this is a cover. After the performance, you will simply receive the key from our mistress, as promised, - at these words, Flash grimaced as if he had eaten a lemon, - and Nick will be initiated. “I don’t want to lose face in front of the whole court,” Dragotius answered. “Fash, Diana,” Nick begged, looking at the two in turn, “please stop. I think we really should rehearse. - The mood is not song, - Fash muttered and, without even eating, went to his room. *a few days ago* - So, - Konstantin said with a joyful smile, having gathered Fash and Nick in the workshop, - I have two news. First, I arranged with the White Queen for your initiation, Nick. - How did you do it? Flash looked at Lazarev in surprise. “I’ll tell you later,” Nick’s father smiled. - son, could you leave us? The blonde left the room, closing the door behind him. Konstantin grew serious, shifting his gaze to the brunette: -Fesh, Astarius asked me to tell you that the White Queen promised him the Silver Key. You must go to Charodol, participate in Enchantments and take the Silver Key from the Queen, - Dragotius was amazed that Astarius entrusted him to carry this key, even though he heard about it the second time. The teacher had already warned him, explaining that the brunette had escaped from Astrogor... *** Their performance made a splash in the kingdom of fairies: six-winged creatures lifted the clock arrows, applauded and shouted enthusiastically. Flash's fears were unfounded, which he was glad about. Soon he received a letter on the watchlist, which said that he, as the winner of the Enchantment, should come to the White Castle at midnight. The brunette approached the gazebo, where Nick and Diana were already sitting, who were also glad that the performance was successful. “Well,” he turned to Fraser in a playful manner, “would you escort us to the White Castle, madam maid of honor?” - Nick snorted into the cup, and Diana just smiled. Why didn't you say you were a lady-in-waiting? - Fash sat down at the table - I felt like a fool when they approached me and said that my performance with Mrs. Diana Fraser, Her Majesty's maid of honor, made a splash! - neither Nick nor Diana could not help laughing... *midnight* -Fashiar Dragotsiy, - the White Queen, who got up from the throne, decorated on the back with golden twigs with emerald leaves, waved her hand to one of the girls, - for the victory in the enchantments and promises to Astarius, I will give you the Silver Key. I think you know it's a huge responsibility. Protect him, keep him like the apple of an eye. "I promise," Flash nodded, looking confidently at the Fairy Queen. The door opened and the girl brought in the Silver Key resting on a cushion of red silk. The fairy approached him and stopped in a bow, holding out a pillow with a key. Flash carefully took the key and bowed to the Queen: - I humbly thank you for the honor done to me. The Fairy Ruler nodded and waved her hand, allowing Fash to go to the rest house. Nick was taken away at the beginning in order for him to undergo initiation. *** -…and they gave me some kind of time potion. Well, I drank it. As a result, the third hour degree, - Nick smiled happily, telling his friend about what happened to him in the White Castle. Diana sat with them and calmly drank coffee, eating a bun. - By the way, I also have some news. Putting the cup aside, Diana smiled, putting a small iron key on the table. For a second, Flash and Nick looked in surprise at the key, then at the girl, but the next moment Dragotsy jumped up from his seat and rushed to hug Diana, smiling joyfully. -I knew! he exclaimed. the blushing fairy barely escaped from the guy's arms: -Firstly, let go, you'll strangle me! Second, how did you know? - -Guess, of course, it was not difficult, - said a pleased Fash. - The court fairy, the best student, and even desperate ... I guessed that you were also a housekeeper, as soon as I saw you. - Yes, - drawled Nick, who had recovered from surprise, - meeting in the forest with you was a little unexpected. - What was so unexpected? Diana looked at her friend with interest. “For example, the fact that you suddenly jumped out of the darkness at us,” Flash put in. - Yes, - the younger-now-already-watchmaker Lazarev nodded, - Of course, we knew that we would meet you in the forest, but it was not worth jumping out of the darkness so unexpectedly at us. “But it’s good that we immediately went to Charodol,” Dragotius chuckled. The guys nodded in agreement and continued breakfast ...
Night. The light of the full moon, hanging in the starry sky, through the stained-glass windows on the windows illuminated the gloomy corridors of Zmiulan, from the walls of which the echoing sound of running was reflected. - Well, what a girl! Flash muttered breathlessly. - She was frightened, you know ... Only time wasted in vain! I hope I still manage to escape... this time... Rushing towards the Stone Hall, he prayed that no one would get in his way. But everything happened exactly the opposite. In the darkness of the corridors (where they did not bother to make windows) Dragotsy collided with someone, hearing a familiar voice: “Who is running around here like crazy?! "". The brunet summoned an hour arrow and lit a light on its tip. In the light of an impromptu lamp hit ... Vasilisa ?! -You?! the two exclaimed at the same time. Flash was simultaneously surprised and relieved: after all, they are on good terms with Ogneva, and she will not betray him ... well, he hoped so. The guy thought that the redhead experienced something similar. -What are you doing here? Dragotsy held out his hand to Vasilisa. Having accepted help, she got up and brushed herself off: - I would like to ask you the same question. "I was the first to ask," Flash crossed his arms. -Doesn't matter. In general, it's none of your business, - Vasilisa snapped. “Well, that means that what I do is none of your business,” Dragotius shrugged calmly. The redhead pursed her lips and thoughtfully looked at the brunette: - I'll tell you only after you. "Well…I…" Flash began, trying to find the words, but nothing came out. “Okay, I want to run away,” Dragotius blurted out. Vasilisa's eyes widened: -Are you crazy? Flash rolled his eyes and looked irritably at Ogneva: -No, but I don't want to stay here. - If you are caught, you will be punished. Remember what happened last time, - the red-haired woman crossed her arms over her chest. Dragotius grimaced: -Listen, it's better not to bother me. Vasilisa thoughtfully looked at the brunette: - Well, I won’t interfere ... all the more, I’m so kind today that I won’t even betray you, - Ogneva giggled and, turning around, wanted to leave, but Flash stopped her with a hail: - Vasilisa, - the girl turned around and looked expectantly at the brunette, - thank you, - Dragotius smiled and ran away. Ogneva smiled and went towards her... *** -It was a huge mistake, nephew, - Astragor towered over the lying half-naked Fesh. The students began to whisper softly. - You tried to escape more than once and always got punished ... - Shackle, who came specifically to carry out the massacre, took out one of the rods and waved a couple of times. A crackling sound was heard. - I hope you will understand that it is useless to run, - the great spirit of Osla turned his back to the offender, his face - to the rest of the students: - I think this will serve as an example for you too. The rod, cutting through the air, immediately went over the back of Flash, leaving red, even bloody streaks. Blow after blow. The brunette stoically endured all the blows, only occasionally emitting a half-groan - half-roar. The disciples looked at it with a kind of malice. Only Vasilisa and Zakharra looked excitedly at the brunette... *** Flash sat in the dungeon and thought. Previously, they simply put him in the dungeon, leaving him without food, but now, apparently, his uncle is tired that his nephew is punished so lightly. The brunet shrugged, grimacing in pain. He did not pay attention to the cold, dampness, immersed in his thoughts. He was snapped out of his thoughts by the sound of footsteps echoing down the corridor. Soon Vasilisa came out under the light of the torch. Flash immediately went to the bars: -What are you doing here? - Hold, - Ogneva put her hand between the bars and gave Dragotsy a rather decent piece of still warm bread with seeds. Flash took the food. - And what are these bouts of generosity? he chuckled. - This Zakharra asked me to pass. They didn't let her through, - Ogneva shrugged her shoulders. - That is, Zakharra was not allowed in, but you, the one who is not a relative of Astragor, were quietly let in? The brunette chuckled. “Well, I don’t decide,” Vasilisa shrugged her shoulders again, however, Flash noticed excitement in her eyes. “Well, I’ll ask Zaharra about it later,” Dragotius said calmly, biting off some bread. “Ask me, but I have to go already,” Ogneva turned around and calmly walked to the corner and turned behind him. Soon, Flash heard the sound of running and chuckled. However, this is her initiative. Probably, she ran to her sister to negotiate just in case "" ...
