Rice. 7.2. Payment matrix taking into account the probabilities of event outcomes
p i – probability of the i-th outcome of events.
M j – mat. expectation of a criterion when choosing the j-th option of action alternatives, determined by the formula:
The two above approaches allow the implementation of four different solution selection algorithms.
1. Decision based on the maximum likelihood rule - maximizing the most probable values of the criterion (profit or income).
2. Decision based on the maximum likelihood rule - minimizing the most probable values of the criterion (possible losses or direct losses).
3. Decision based on the rule of maximizing the mathematical expectation (average value) of the criterion (profit or income).
4. Decision based on the rule of minimizing the mathematical expectation (average value) of the criterion (losses or damages).
The examples we have looked at so far in this chapter have involved a single solution. However, in practice, the result of one decision forces us to make the next one, and so on. This sequence cannot be expressed by a payoff matrix, so some other decision process must be used.
Scheme decision tree used when it is necessary to make several decisions under conditions of uncertainty, when each decision depends on the outcome of the previous one or the outcome of events.
When creating a decision tree, you need to draw a “trunk” and “branches” that reflect the structure of the problem.
· The “trees” are located from left to right. "Branches" represent possible alternative decisions that can be made and the possible outcomes that arise from those decisions.
· “Branches” come out of nodes. There are two types of nodes.
The square node represents the place where the decision is made.
The round node represents the place where different possible outcomes appear.
· The diagram uses two types of “branches”:
The first is dotted lines coming out of the squares of possible solutions; movement along them depends on the decisions made. All costs caused by the decision are indicated on the corresponding dotted “branch”.
The second is solid lines coming out of the circles of possible outcomes. Movement along them is determined by the outcome of events. The solid line indicates the probability of a given outcome.
decision making node.
branching node for possible outcomes of events.
branches, the movement along which depends on the decision being made.
branches, the movement along which depends on the outcome of events.
The search for a solution is divided into three stages.
Stage 1. A “tree” is being built (an example will be discussed in practical classes). When all the decisions and their outcomes are indicated on the “tree,” each of the options is calculated, and its monetary income is indicated at the end.
Stage 2. They are calculated and placed on the corresponding branches of the probability of each outcome.
Stage 3. At this stage, the monetary outcomes of each of the “nodes” are calculated and entered from right to left. Any expenses incurred are deducted from expected income.
After the “solution” squares have been completed, the “branch” leading to the highest possible expected income for a given decision is selected (an arrow is placed on this branch).
The other “branch” is crossed out, and the expected income is written above the solution square.
Thus, at the end of the third stage, a sequence of decisions leading to maximum income is formed.
In principle, the criterion can be maximization of mat. income expectations and minimizing swearing. expectations of losses.
a person contains a certain plan with which the soul came here, all the options for the development of events, including. We can go there and see the consequences of the important decisions we make. For example, about changing work and lifestyle. This can be done both in independent meditations and in joint leader-follower processes. Below is a description of how this was done in the session
Probability lines
I project three branches:
1) stay in Moscow at your existing job;
2) sell or rent out an apartment and go to Asia to visit friends to become a partner in their tourism business;
3) ideal option: I quit my job, participate in friends’ business on a project basis, and have 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 is a couple - their own partnerships, and they have their own business.
We build all three branches like roads, see if there are branches.
The Moscow line is a strong, thick gray rope, dull and reliable, you won’t come off, you won’t get lost. From the rope there are several thinner ropes, some brighter and more interesting, but none of them attract, call or glow. The feeling is that I still love Moscow, but this topic has become obsolete.
The thread with Asia and friends is very bright and visual, but short and thin, or something. It lacks the potential to confidently develop into the future. Insufficient resource.
The perfect third picture was divided into several geographical points on the map, each with its own specific flavor. The third branch, which contains my own story, is the most attractive, of course, for me. It is not as tangible now as the Moscow one and not as colorful as the second one, but it calls to itself. And it glows, filled from within. Like a thin living ray, it pulsates and shimmers.
Choosing your path
In this version of events, I am free to move around the world if I wish. My income is lower than in Moscow, but it is enough so that I don’t need anything and don’t deny myself anything, even in moderation. I come to my friends’ projects, they visit 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 provides a stable income.
At the same time, there is a close person with whom we are jointly implementing this story, as a couple. In order for it to manifest itself, not only my intention is needed, and on both my part a certain payment will be required, 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 renunciation of existing comfort or freedom. Payment is like allowing something completely new and unknown, albeit tempting, to enter your life. Pure free will and purity of intentions on both sides. And then - how will it turn out... In any other way (not based on pure expression of will), this topic simply will not take off.
This whole process is now in full swing. This thread is in its gestation stage, and if all goes well, it will be able to fully manifest itself in my reality. Let's see if there are any obstacles or stones on this ideal line for me. I see a fallen tree, right on the road. These are fears and self-distrust. From the series - it’s too good for it all to turn out that way, it doesn’t happen, these are all illusions and fairy tales invented for oneself. I'm clearing the way.
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 possible to “rewind” so easily later. I understand for myself that one way or another I have been imbuing it with energy for a long time and activating it internally. And this doesn’t even happen because of stubbornness or a 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 noticeable. It thickens, slowly and surely. Although, of course, everything is still extremely uncertain and could collapse at any moment, but there is a feeling that it is coming to me, this branch.
Since it had long been designed and predetermined, ordered, one might say. And I understand where this is leading. And how it turns out. And that this is the right development of events. Although sometimes I'm stupidly afraid to believe it..
And I really don’t want to cement this branch. Make it rigid and unambiguous.. There is no need to build into it a rigid link to a specific place or occupation, or to anything 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 happened 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, maximum correctness is important here - in no case impose or put pressure.. Everything is free will. And then - where will he call*
Strengthening a branch with attention
I extend a ray from my Spark in the direction of this branch, to the point where it tends, and connect with it with my attention. Thus, the Spark begins to work towards the realization of this goal and anchors itself in it. I may not realize this, but the work will be carried out: the formation of events in space will occur 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 probability to me, like a magnet. The goal is getting very close, you can say I’m there now. Like a teleport, when you don’t try to move to a new place with your whole body, but materialize the desired space around you: you tune in to the goal and attract it to you. And the closer it is to you, the more your will extends to its implementation. And Iskra is responsible for shaping the events that will lead to the embodiment of this branch in reality and 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 you 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 need to look at some answers on the development of this thread, you can simply remember this state of attraction, become emotionally saturated with the atmosphere and mood of this room, feel the emotion of creativity and partnership. The emotion of creation is always love..
Manifestation and consolidation of results
To capture that picture that looks so attractive, but is unsteady now, you need to pass light through it, pour in emotion, charge it with positivity. Enter the state of ananda - a joyful uplift, a loving and beloved being, in love and filled with love, and redirect this inner fuel into the ideal scenario for the development of events.
Clear the path and remove questions. To align with other branches of reality surrounding me and the players involved, so that all this is synchronized in place and time. Coincided with intentions, will and freedom of choice. Saturate all this with your own light, warmth and love in order to realize your creative potential in the future in the way that you like so much. Expose the desired result so that the image is imprinted with light onto the sensitive film - the outline of future events, and burns its imprint into it as a light projection. And wait a little so that the effect is as bright as possible.
Now you need to process the created dream imprint so that it moves into the layer of material reality. The next stage is stabilization. You need to add a little energy of darkness and cold to the picture so that it crystallizes and acquires more solid outlines, moves from the state of a magical mirage into denser layers, consolidates and manifests itself.
Working with a negative print.. The result is literally recorded on a sheet of reality, in much the same way as when we project an image from analog photographic film onto analog photographic paper, and then pour in 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 it is appropriate and timely.
Since the throat chakra is responsible for communication with the world and creative realization, I send a ray from the throat chakra there, to the chosen branch. A ray from the second chakra asked for it, then from the third. Then the rest of the chakras connected, and the result was a ray shower, like a seven-flowered flower. I wash and dry everything that has turned out, fill it with movement, the material energy of the earth, vision, all the qualities of vitality and magnetism, attract the branch of probability into my reality even more, connect it directly with each of the chakra centers, prescribe it there in them..
* a person forgets that the future is multivariate and often refers 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 get hung up on boundaries, easily let go of the old and let in the new, adapt. Therefore, if you do such practices, do not “cement” your intention in any way, because the world always offers even better 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, and at each level of consciousness and. We learn to separate what is ours from what is not ours, or to obtain information autonomously)
THEMATIC SECTIONS:
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Night. The light of the full moon hanging in the starry sky through the stained glass windows illuminated the gloomy corridors of Zmiulan, from the walls of which the booming sound of running was reflected. -What a girl! - Fash muttered, catching his breath. - She was scared, you know... I just wasted my time! I hope I can still 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 didn’t bother to make windows), Dragotsiy ran into someone, hearing a familiar voice: “Who’s running around like crazy?!” "". The brunette summoned the hour arrow and lit the flame at its tip. Vasilisa fell into the light of an improvised lamp?! -You?! - these two exclaimed at the same time. Fash felt both surprise and relief: after all, they were on good terms with Ognevaya, and she wouldn’t give him up... well, he hoped so. The guy thought that the red-haired woman had experienced something similar. -What are you doing here? - Dragotsiy extended his hand to Vasilisa. She, having accepted the help, stood up and shook herself off: “I would like to ask you the same question.” “I asked first,” Fash crossed his arms over his chest. -Doesn't matter. Actually, it’s none of your business,” Vasilisa snapped. “Well, that means what I do is none of your business,” Dragotius calmly shrugged. The redhead pursed her lips and looked thoughtfully at the brunette: “I’ll tell you only after you.” “Well...I...” began Fash, trying to find words, but nothing came out. “Okay, I want to run away,” Dragotius blurted out. Vasilisa’s eyes widened: “Have you lost your mind?” Fash 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 redhead crossed her arms over her chest. Dragotsiy grimaced: “Listen, it’s better not to bother me.” Vasilisa looked thoughtfully at the brunette: “Okay, I won’t interfere... especially since I’m so kind today that I won’t even rat you out,” Ogneva giggled and, turning around, wanted to leave, but Fash stopped her with a call: “Vasilisa,” the girl turned around. and looked expectantly at the brunette, “Thank you,” Dragotius smiled and ran away. Ogneva smiled and headed towards her place... *** “It was a huge mistake, nephew,” Astragor towered over the lying half-naked Fesh. The students began to whisper quietly. - You tried to escape more than once and always received punishment... - Shuckle, who came specifically to carry out the reprisal, took out one of the rods and waved it a couple of times. A whipping sound was heard. “I hope you will understand that running is useless,” the great spirit of Ostala 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 passed along Fesh's back, leaving red, even bloody stripes. Blow after blow. The brunette stoically endured all the blows, only occasionally uttering a half-moan - half-roar. The students looked at this with some malice. Only Vasilisa and Zaharra looked excitedly at the brunette... *** Fash sat in the dungeon and thought. Previously, they simply put him in a dungeon, leaving him without food, but now, apparently, his uncle is tired of his nephew being punished so easily. The brunette shrugged his shoulder, grimacing painfully. He did not pay attention to the cold, dampness, immersed in his thoughts. He was brought out of his thoughts by the sound of footsteps along the corridor. Soon Vasilisa came out into the light of the torch. Flash immediately approached the bars: “What are you doing here?” “Here,” Ogneva stuck her hand between the bars and gave Dragotius a fairly decent piece of still warm bread with seeds. Fash took the food. -And what are these attacks of generosity? - he grinned. -Zakharra asked me to convey this. They didn’t let her through,” Ogneva shrugged. -That is, Zaharra was not allowed in, but you, the one who is not a relative of Astragor, were quietly allowed in? - the brunette grinned. “Well, it’s not me who decides,” Vasilisa shrugged her shoulders again, although Fesh noticed the excitement in her eyes. “Well, I’ll ask Zakharra about this later,” Dragotius said calmly, taking a bite of some bread. “Ask, but I have to go now,” Ogneva turned around and calmly walked to the corner and turned around it. Soon Fash heard the sounds of running and grinned. ,After all, this is her initiative. She probably ran to her sister to negotiate, just in case."
Candidate of Technical Sciences V. Chernobrov came to interesting conclusions during the study of the properties of time and the possibility of traveling into the past and future. So, in particular, he writes:
“The present is a transition, the transformation of a multivariate, easily changeable Future into a single-variant and unchangeable Past. It follows that flights to the Past (at a “negative” density-velocity t/to) and to the Future will occur differently.
To some extent, they can be compared to the movements of an ant along a tree: from any point on the tree (from the Present), only 1 path opens downward (to the Past) and many paths upward (to the Future).
However, among all the paths to the Future, there are undoubtedly the most probable options, the unlikely and almost incredible. The movement into the Future will be the more unstable and energy-intensive, the less probable this version of the Future turns out to be.
In accordance with this “law of the tree crown,” returning to the Present is possible only if, while staying in the Past, the traveler does not interfere with what is happening around him and does not change the course of past History; otherwise, the time traveler will return to the parallel Present from the Past along another branch of History.
Penetration into the Future from the Present is complicated by the choice of branch of movement, but returning from any version of the Future to the Present is possible under any behavior scenario. If in front of you there are no mergers of different versions of History.”
Thus, even modern scientific research confirms the multidimensionality of time and the diversity of the future, as well as the possibility of moving to its various probabilities.
There is a hypothesis according to which the key moments in the fate of each person, the so-called “forks” of probabilities, give rise to different “branches” of reality depending on our actions.
All these “branches” exist in the Universe at the same time. But existence on only one such “branch” is available to a person, although sometimes cases of spontaneous transition from one “branch” of reality to another occur.
The existence of different probabilities of the future (“branches” of the Tree of Life, “grooves” of the Wheel of Time, etc.) is evidenced by the story that happened with Gustav and Johan Schroederman. It began in the spring of 1973, when the Schroederman family (husband, wife and son) moved from Berlin to a farm near Salzburg.
The youngest of the Schroedermans ran around the neighborhood all summer and one day discovered a rickety house in the forest, walking around it he almost fell into an overgrown well, but in time he grabbed onto a bush. Returning home, he experienced a strange dizziness and immediately went to bed. The next morning there was a knock on the door of the house, and when the boy opened it, he saw himself, wet and covered in dirt.
It turned out that the entire past of both boys completely coincides, the different probabilities of fate begin after an incident at the well, into which one of them fell into and the other survived.
It is possible that the severe stress and fear of the failed boy, thanks to an altered state of consciousness, moved him to another branch of reality, where he already existed, but not having fallen into the well.
It is characteristic that later the parents gave the boys new names and each of them lived their own destiny: one began exporting beer, the other became an architect.
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 OF POINTS IS 2 OR 3);
● E= (SUM OF POINTS IS 10).
Describe the event: WITH= (CIRCUIT CLOSED) for each case.
Solution. Let us introduce the following notation: event A- contact 1 is closed; event IN- contact 2 is closed; event WITH- 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 complete when both contacts are closed, so C = A B.
Task. 1.1.4 Two electrical diagrams have been compiled:
Event A - the circuit is closed, event A i - I-th contact is closed. For which of them is the relation valid?
A1 · (A2 + A3 · A4) · A5 = A?
Solution. For the first circuit, A = A1 · (A2 · A3 + A4 · A5), since a parallel connection corresponds to the sum of events, and a serial connection corresponds to the product of events. For the second scheme A = A1 (A2+A3 A4 A5). Therefore, this relationship is valid for the second scheme.
Task. 1.1.5 Simplify the expression (A + B)(B + C)(C+ A).
Solution. Let's use the properties of the addition and multiplication operations 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.
Task. 1.1.6Prove that 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 will show that these events are pairwise incompatible.
Now we will show that the sum of these events gives the space of elementary events.
Task. 1.1.7Using the Euler–Venn diagram, 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:
Task. 1.2.1In how many ways can 8 people be seated?
1. In one row?
2. At a round table?
Solution.
1. The required 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 at a round table the choice of the first person does not affect the alternation of elements, then anyone can be taken first, and the remaining ones will be ordered in relation to the chosen one. This action can be performed in 8!/8 = 5040 ways.
Task. 1.2.2The course covers 5 subjects. In how many ways can you create a schedule for Saturday if there are two different pairs on that day?
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Solution. The required number of ways is the number of placements
From 5 to 2, since you need to take into account the order of the pairs:
Task. 1.2.3How many examination committees consisting of 7 people can be composed of 15 teachers?
Solution. The required number of commissions (without taking into account the order) is the number of combinations of 15 to 7:
Task. 1.2.4 From a basket containing twenty numbered balls, 5 balls are selected for luck. Determine the number of elements of the space of elementary events of this experiment if:
Balls are selected sequentially one after the other, returning after each draw;
The balls are selected one by one without being returned;
Select 5 balls at once.
Solution.
The number of ways to remove the first ball from the basket is 20. Since the extracted ball returned to the basket, the number of ways to remove the second ball is also 20, etc. Then the number of ways to remove 5 balls in this case is 20 20 20 20 20 = 3200000.
The number of ways to remove the first ball from the basket is 20. Since the extracted ball did not return to the basket after removal, the number of ways to remove the second ball became 19, etc. Then the number of ways to remove 5 balls without returning is 20 19 18 17 16 = A52 0
The number of ways to extract 5 balls from the basket is immediately equal to the number of combinations of 20 by 5:
Task. 1.2.5 Two dice are thrown. Find the probability of event A that at least one one will appear.
Solution. Each dice can roll any number of points from 1 to 6. 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), therefore
Task. 1.2.6 The letters u, i, i, k, c, f, n are written on the red cards; the letters a, a, o, t, t, s, h are written on the blue cards. After thorough mixing, which is more likely: from the first time from the letters to Use the red cards to make up the word “function” or the letters on the blue cards to form the word “frequency”?
Solution. Let event A be the word “function” randomly composed of 7 letters, and event B be 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” appear twice. Then P(A) = 1/7! , P(B) = 2! 2! /7! , P(B) > P(A).
Task. 1.2.7 During the exam, the student is offered 30 tickets; Each ticket contains 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 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 - knows the answer to one question, 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 possibilities for choosing questions known to the student. Then P(A) = M N = 17 78 70 0 = 0.44
2. The choice of two unknown questions out of 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.
Task. 1.2.8Some information was sent through three channels. The channels operate independently of each other. Find the probability that the information will reach the goal
1. Only on one channel;
2. At least on one channel.
Solution. Let A be the event that information reaches the goal through only one channel; B - at least one channel. Experience is the transfer of information through three channels. The outcome of the experience is that the information has reached its goal. Let us denote Ai - information reaches the goal 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.
Task. 1.2.9A point appears randomly 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 conditions of the problem, the required event is satisfied by all points appearing 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, then the required probability is P = s/S = 3/14 = 0.75.
Task. 1.2.10In the party fromNproductsKproducts are defective. m products are selected for control. Find the probability that from M Products L They will 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 from k defective - in ways. After selection L defective products will remain (m - L) suitable, located among (n - k) products. Then the number of outcomes favorable to event A is equal to
And the desired probability 
Task. 1.3.1BThere are 30 balls in the urn: 15 red, 10 blue and 5 white. Find the probability that a ball drawn at random 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.
Task. 1.3.2Probability that there will be snow (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 simultaneous, so P(A + B) = P(A) + P(B) - P(AB) = 0.6 + 0.45 - 0.25 = 0.8
Task. 1.3.3BThe first box contains 2 white and 10 black balls, the second box contains 3 white and 9 black balls, and the third box contains 6 white and 6 black balls. A ball was taken from each box. Find the probability that all the drawn balls 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
The 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
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 are respectively equal 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) - fails I- th element. Events
Task. 1.3.5The circuit consists of independent blocks connected into a system with one input and one output.
Failure of various circuit elements within a time T are independent events with the following probabilitiesP 1 = 0.1; P 2 = 0.2; P 3 = 0.3; P 4 = 0.4. 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 BLOCK WORKS FAILLESSLY), then A = (A1 + A2)(A3 + A4). Events A1+A2, A3+A4 are independent, events A1 and A2, A3 and A4 are joint. Using the formulas for multiplying and adding probabilities
Task. 1.3.6A worker operates 3 machines. The probability that the machine will not require the worker’s attention within an hour is equal to 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 require attention.
Solution. Let Ai be the i-th machine that requires the attention of a worker; the i-th machine will not require the attention of a worker. Then
Space of elementary events:
1. Event A - the second machine will require attention: 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 states will require attention
kov:
Task. 1.3.7Bthe "Examiner" machine was introduced 50 Questions. The student is offered 5 Questions and a grade of “excellent” is given if all questions are answered correctly. Find the probability of getting “excellent” if the student prepared only 40 Questions.
Solution. A - (GRADE “EXCELLENT” RECEIVED), Ai - (ANSWERED TO THE i -th QUESTION). Then A = A1A2A3A4A5, we have:
Or, in another way - using the classical probability formula: 
AND
Task. 1.3.8The probability that the part needed by the assembler is inI, II, III, IVbox are respectively 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
Task. 1.4.1 A group of 10,000 people over 60 years of age was examined. It turned out that 4,000 people are regular smokers. 1,800 smokers showed serious changes in their lungs. Among non-smokers, 1,500 people had changes in their lungs. What is the probability that a randomly examined person with changes in the lungs is a smoker?
Solution. Let us introduce the hypotheses: H1 - the person examined is a constant smoker, H2 - is a non-smoker. Then, according to the conditions of the problem
P(H1)= ------- =0.4, P(H2)=--------- =0.6
Let us denote by A the event that the examined person has changes in the lungs. Then, according to the conditions of the problem
Using formula (1.15) we find
The desired probability that the examined person is a smoker, according to the Bayes formula, is equal to
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 plant contain 20% of televisions with hidden defects, the second - 10%, and the third - 5%. What is the probability of purchasing a working TV?
Solution. Let's consider the events: A - a working TV was purchased; hypotheses H1, H2, H3 - the TV went on sale from the first, second, third plant, respectively. According to the conditions of the problem
Using formula (1.15) we find
Task. 1.4.3There are three identical-looking boxes. The first has 20 white balls, the second has 10 white and 10 black balls, 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 event A - the white ball is taken out, hypotheses H1, H2, H3 - the ball is taken out from the first, second, third box, respectively. From the problem conditions we find
Then 
Using formula (1.15) we find
Using formula (1.16) we find
Task. 1.4.4A telegraph message consists of the dot and dash signals. The statistical properties of noise are such that they are distorted on average 2/5 Messages "dot" and 1/3 Messages "dash". It is known that among transmitted signals “dot” and “dash” occur in the ratio 5: 3. Determine the probability that the transmitted signal is received if:
A) the “dot” signal is received;
B)"dash" signal received.
Solution. Let event A mean a “dot” signal is received, and event B mean a “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
Probabilities of events A AND B We find using the total probability formula:
The required probabilities will be:
Task. 1.4.5Of the 10 radio channels, 6 channels are protected from interference. The probability that a secure channel over timeTwill not fail, is equal to 0.95, for an unprotected channel - 0.8. Find the probability that two randomly selected channels will not fail over timeT, and both channels are not protected from interference.
Solution. Let event A - both channels not fail during time t, event A1 - Protected channel selected A2 - An unprotected channel has been selected.
Let us write down 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 
Task. 1.5.1The communication channel transmits 6 Messages. Each message can be distorted by interference with a probability 0.2 Regardless of others. Find the probability that
1. 4 out of 6 messages are not distorted;
2. At least 3 out of 6 were transmitted distorted;
3. At least one message out of 6 is distorted;
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
ability to transmit 4 messages out of 6 without interference:
2. at least 3 out of 6 are transmitted distorted:
3. at least one message out of 6 is distorted:
4. at least one message out of 6 is distorted:
5. all messages are transmitted without distortion:
Task. 1.5.2The probability that a day will be clear in summer is 0.42; the probability of a cloudy day is 0.36 and partly cloudy is 0.22. How many days out of 59 can you expect clear and cloudy?
Solution. From the conditions of the problem it is clear that we need 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;
Therefore 20.16 ≤ M0 ≤ 21.60; → M0 = 21.
Thus, the most probable number of clear days Mo=25, cloudy days - M0 = 21. Then in summer you can expect Mo+ M0 =46 clear and cloudy days.
Task. 1.5.3There are 110 students attending 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 Poisson’s formula (1.22). Let's find the Poisson parameter. Because
N= 110, then λ = np = 110 1 /365 = 0.3.
Then, according to Poisson's formula 
Task. 1.5.4The probability that the part is not standard is equal to 0.1. How many parts 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 a constant probability p = 0.1 In absolute value no more than 0.01 ?
Solution.
Required number N Let's find it using formula (1.25). We have:
P = 1.1; q = 0.9; P= 0.96428. Let's substitute the data into the formula:
Where do we find it from?
According to the table of function values Φ( X) we find that
Task. 1.5.5The probability of failure of one capacitor during time T is 0.2. Determine the probability that during time T 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 Great, let's use the local theorem of Moivre - Laplace:
Let's calculate
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). Required probability
2. At least 20 capacitors;
The requirement that out of 100 capacitors at least 20 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.
Let's calculate x 1,x2:
Task. 1.5.6The probability of the occurrence of some event in one experiment is 0.6. What is the probability that this event will occur in the majority of 60 experiments?
Solution. Quantity M The occurrence of an event in a test series is between . "In most experiments" means that M Belongs to the interval By condition N= 60, P= 0.6, Q = 0.4, M1 = 30, m2 = 60. Let's calculate x1 and x2:
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Random variables and their distributions
Task. 2.1.1A table is given where the possible values of the random variable are indicated in the top line X , and at the bottom - their probabilities.
Can this table be a distribution row X ?
Answer: Yes, since p1 + p2 + p3 + p4 + p5 = 1
Task. 2.1.2Released 500 Lottery tickets, and 40 Tickets will bring their owners winnings 10000 Rub., 20 Tickets - per 50000 Rub., 10 Tickets - per 100000 Rub., 5 Tickets - per 200000 Rub., 1 Ticket - 500000 Rub., the rest - no winnings. Find the law of distribution of winnings for the owner of one ticket.
Solution.
Possible values for X: x5 = 10000, x4 = 50000, x3 = 100000, x2 = 200000, x1 = 500000, x6 = 0. The probabilities of these possible values are:
The required distribution law:
Task. 2.1.3Shooter having 5 Cartridges, fires until the first hit on the target. The probability of hitting with each shot is 0.7. Construct a distribution law for the number of cartridges used, find the distribution functionF(X) and build 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. The following values of the random variable of the number of cartridges used correspond to elementary outcomes: 1, 2, 3, 4, 5. Since the result of each next shot does not depend on the previous one, the probabilities of the 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 required distribution law:
Let's 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
Let's find P(2< x < 5). Применим формулу (2.4): P(2 < X< 5) = F(5) - F(2) = 0.9919 - 0.91 = 0.0819
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 function jumps F(X).
We find possible values of the random variable X=(0,1,2,3,4). 
Task. 2.1.5Set which function
Is the distribution function of some random variable.
If the answer is yes, find the probability that the corresponding random variable 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 falling into the interval:
Task. 2.1.6Given the probability density of a continuous random variable X :
Find:
1. Coefficient C ;
2. Distribution function F(x) ;
3. Probability of a random variable falling into the interval(1, 3).
Solution. From the normalization condition (2.9) we find
Hence,
Using formula (2.10) we find:
Thus,
Using formula (2.4) we find
Task. 2.1.7The random downtime of electronic equipment in some cases has a probability density
Where M = lge = 0.4343...
Find the distribution function F(x) .
Solution. Using formula (2.10) we find
Where 
Task. 2.2.1Given a distribution series of a discrete random variable X :
Find the mathematical expectation, variance, standard deviation, M, D[-3X + 2].
Solution.
Using 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 variance:
Task. 2.2.2Find the expectation, variance and standard deviation of a continuous random variable X , whose distribution function
.
Solution. Let's find the probability density:
We find the mathematical expectation using formula (2.13):
We find the variance using formula (2.19):
Let us first find the mathematical expectation of the square of the random variable:
Standard deviation
Task. 2.2.3Xhas a distribution series:
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.
Task. 2.2.4Discrete random variable X Can only take two values X1 AND X2 , and X1< x2. Known probability P1 = 0.2 Possible meaning X1 , expected value M[X] = 3.8 And variance 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.
According to the problem conditions 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 obtained a system of equations:
Condition x1
Task. 2.2.5The random variable X is subject to a distribution law, the density graph of which has the form:
Find the expected value, variance and standard deviation.
Solution.
Let's 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. Thus, 
Expected value:
Let's find the variance and standard deviation:
Task. 2.2.6Find the mathematical expectation and variance of the sum of points that appear on four dice in one throw.
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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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, it flew out from the source 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 that consist of particles emitted from a radioactive source being detected 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]. Searched probabilities:
Task. 2.3.2The batch contains 5% non-standard parts. 5 parts were selected at random. Write the law of distribution of a discrete random variable X - 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. The probability of a non-standard part in a batch is p = 5 /100 = 0.05. Let's find the probabilities of these possible values:
Let us write the required distribution law:
Let's find the 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] = n p (1 - P) = 5 0.05 0.95 = 0.2375.
Task. 2.3.3The time of detection of a target by a radar is distributed according to the exponential law
Where1/ λ = 10 Sec. - average target detection time. Find the probability that the target will be detected in time from5 Before15 Sec. after starting the search.
Solution. Probability of hitting a random variable X In the interval (5, 15) Let us find using formula (2.8):
At 
We get
0.6065(1 - 0.3679) = 0.6065 0.6321 = 0.3834
Task. 2.3.4Random measurement errors are subject to the normal law with parameters a = 0, σ = 20 Mm. Write the differential distribution functionF(X) and find the probability that there was an error in the measurement in the range from 5 Before 10 Mm.
Solution. Let us substitute the values of 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) It will look like
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