We are often interested in the probability of several events happening at the same time, such as two heads on two coin tosses, or at least one six on two die rolls. Such situations are called situations with multiple possible outcomes.

Using Tree Diagrams

While it is fairly easy to understand that the probability of getting heads on one flip of a fair coin is ?, it is somewhat more difficult to intuitively determine the probability of four heads on four tosses of a fair coin. While the coin example may seem artificial, it is well suited to explain the combination of probabilities over multiple tries. Let's do the calculations. (Follow my reasoning, even if you are terribly afraid of mathematics. If you work through the examples, the calculations and mathematical reasoning will seem quite simple to you. No need to exclaim after looking at the next few numbers: “No, no way, I'll just skip it It's important to be able to think with numbers and about numbers.)

On the first roll, only one of two possible outcomes can occur; heads (O) or tails (P). What happens if a coin is tossed twice? There are four possible outcomes: heads both times (OO), heads the first time and tails the second time (OR), tails the first time and heads the second time (RO), and tails both times (RR). Since there are four possible outcomes and only one way of getting two heads, the probability of this event is 1/4 (again, we assume that the coin is "fair", i.e. heads and tails are equally likely). There is a general rule for calculating the probability of the joint occurrence of several events in any situation - the rule "and". If you want to find the probability of co-occurrence of the first and the second event (eagle at the first and on the second roll), we must multiply the probabilities of these events separately. Applying the “and” rule, we find that the probability of getting two tails when a coin is tossed twice is equal to? x? = 1/4 . Intuitively, it seems that the probability of the joint occurrence of two events should be less than the probability of each of them separately; so it turns out.

A simple way to calculate this probability is obtained by representing all possible events with tree diagram. Tree diagrams were used in Chapter 4 when we tested the validity of "if... then..." statements. In this chapter, we will assign probabilistic values ​​to the branches of the tree to determine the probabilities of various combinations of outcomes. In later chapters, I will return to tree diagrams when I look at ways to find creative solutions to problems.

The first time a coin is tossed, it will land either heads or tails. For a "fair" coin, heads and tails have the same probability of 0.5. Let's picture it like this:

When you toss a coin a second time, either the first heads will be followed by a second heads or tails, or the first tails will be followed by a second heads or tails. The probabilities of getting heads and tails on the second toss are still 0.5. The outcomes of the second roll are shown on the diagram as additional branches of the tree.

As you can see from the diagram, there are four possible outcomes. You can use this tree to find the probabilities of other events. What is the probability of getting one head on two tosses of a coin? Since there are two ways to get one tail (OR or RO), the answer is 2/4 or ?. If you want to find the probability of two or more different outcomes, add up the probabilities of all outcomes. This is called the "or" rule. In another way, this problem can be formulated as follows: “What is the probability of getting or first heads, then tails (1/4), or first tails, and then heads (1/4)?” The correct procedure for finding the answer is to add these values ​​together, resulting in ?. Intuitively, it seems that the probability of occurrence of one of several events must be greater than the probability of occurrence of each of them; so it turns out.

The rules "and" and "or" can be used only when the events of interest to us independent. Two events are independent if the occurrence of one of them does not affect the occurrence of the other. In this example, the result of the first coin toss does not affect the result of the second toss. In addition, for the "or" rule to apply, the events must be incompatible, that is, they cannot occur at the same time. In this example, the outcomes are incompatible because we cannot get heads and tails on the same toss.

Representing events as tree diagrams is useful in many situations. Let's expand our example. Suppose a man in a striped suit with a long, curled-up mustache and shifty little eyes stops you in the street and offers to play for money by tossing a coin. He always bets on the eagle. On the first toss, the coin lands heads up. The same thing happens on the second throw. On the third toss, heads come up again. When do you start to suspect that he has a "foul" coin? Most people have doubts on their third or fourth try. Calculate the probability of getting some heads on three and four fair coin tosses (the probability of getting heads is 0.5).

To calculate the probability of getting three heads in three attempts, you need to draw a tree with three rows of "nodes", with two "branches" coming out of each node.

In this example, we are interested in the probability of getting three heads in a row, provided that the coin is fair. Look at the column labeled "outcome" and find the LLC outcome. Since this is the only outcome with three heads, multiply the probabilities along the 000 branch (circled in the diagram) and you get 0.5 x 0.5 x 0.5 = 0.125. A probability of 0.125 means that if the coin is "fair", then on average it will fall heads up three times in a row 12.5% ​​of the time. Since this probability is small, when three heads in a row fall out, most people begin to suspect that the coin is “with a secret”.

To calculate the probability of getting four heads in four attempts, add additional branches to the tree.

The probability of getting four heads is 0.5 x 0.5 x 0.5 x 0.5 = 0.0625, or 6.25%. As you already know, mathematically it is equal to 0.5 4 ; that is, multiplying a number by itself four times is the same as raising it to the fourth power. If you count on a calculator where there is an exponentiation operation, then you will get the same answer - 0.0625. Although such an outcome is possible and will someday happen, it is unlikely. In fact, he is so implausible and unusual that many would say that a person with shifty eyes is probably cheating. Undoubtedly, on the fifth head in a row, it would be reasonable to conclude that you are dealing with a scammer. For most scientific purposes, an event is considered "unusual" if it is expected to occur with a probability of less than 5%. (In the language of probability theory, this is written as p ‹ 0.05.)

Let's leave the artificial coin example and apply the same logic in a more useful context. I am sure that any student has ever come across multiple choice tests in which you need to choose the correct answers from the proposed options. In most of these tests, each question has five possible answers, of which only one is correct. Suppose the questions are so difficult that you can only randomly guess the correct answer. What is the probability of guessing correctly when answering the first question? If you have no idea which of the options is the correct answer, then you are equally likely to choose any of the five options, assuming that any of them may be correct. Since the sum of the probabilities of choosing all options should be equal to one, then the probability of choosing each of the options with the equiprobability of all options is 0.20. One of the options is correct and the rest are wrong, so the probability of choosing the correct option is 0.20. A tree diagram of this situation is shown below.

What is the probability of correctly guessing the answers to the first two questions of the test? We will have to add new branches to the tree, which will soon become very dense. To save space and simplify calculations, you can represent all incorrect options as a single branch, labeled "incorrect". The probability of making a mistake in answering one question is 0.8.

The probability of correctly guessing the answers to two questions is 0.2 x 0.2 = 0.04. That is, by chance it can happen only in 4% of attempts. Let's say we expand our example to three questions. I won't draw a tree, but you should already understand that the probability is 0.2 x 0.2 x 0.2 = 0.008. This is such an unusual event that it can happen by chance in less than 1% of attempts. What do you think of a person who managed to answer all three questions correctly? Most people (and educators are people too) will conclude that the student did not choose answers at random, but actually knew something. Of course, it is possible that he was just lucky, but this is extremely unlikely. Thus, we come to the conclusion that the result obtained cannot be explained by luck alone.

I would like to point out one curious aspect of such reasoning. Consider the deplorable situation that Sarah found herself in. She answered 15 test questions, where the answer to each question had to be chosen from five options. Sarah answered all 15 questions incorrectly. Can you determine the probability that this happened by chance? I won't draw a tree diagram to illustrate this situation, but it's easy to see that the probability of answering one question wrong is 0.8; so the probability of answering all 15 questions incorrectly is 0.8 15 . That number is 0.8 multiplied by itself 15 times, resulting in 0.0352. Since the probability of such an accident is 3.52%, maybe Sarah should tell the teacher that such an unusual result cannot be explained by chance? Sarah, of course, can make a similar argument, but would you believe her if you were a teacher? Suppose she claims to know the answers to all the questions. How else could she not choose the correct answer in 15 questions in a row? I don't know how many teachers would believe her claim that 15 incorrect answers prove she has knowledge, although in principle this line of reasoning is used to prove knowledge, since the probability of correctly guessing all the answers is about the same. (In this example, the probability of correctly answering all 15 questions at random is 0.2015, a number well below 0.0001.) If I were Sarah's teacher, I would give her high marks for her creativity and understanding of statistical principles. It is possible that Sarah really knew something about this topic, but there was a systematic error in this “something”. I would also point out to her that she may not have prepared for the test, and in addition, she was also unlucky, and she made 15 wrong guesses. In the end, sometimes it happens and very unusual events.

Before reading the next section, check that you understand how to use tree diagrams to calculate probabilities and account for all possible outcomes. I will return to such diagrams later in this chapter. Once you learn how to use them, you will be surprised how many situations they can be applied to.

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 both 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 a dungeon, leaving him without food, but now, apparently, his uncle is tired of his nephew being 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 it, - Ogneva put her hand between the bars and gave Dragotsy a fairly decent piece of 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 "" ...

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 bob, 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 Charms, should come at midnight at White Castle . 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 ...

To build a probability tree, first of all, you need to draw the tree itself, then write down all the information known for this problem in the figure, and, finally, use the basic rules to calculate the missing numbers and complete the tree.

1. The probabilities are indicated at each of the endpoints and circled. At each level of the tree, the sum of these probabilities must equal 1 (or 100%). So, for example, in Fig. 6.5.1 The sum of probabilities at the first level is 0.20 + 0.80 = 1.00 and at the second level - 0.03 + 0.17 + 0.56 + 0.24 = 1.00. This rule helps to fill one empty circle in a column if the values ​​of all other probabilities of this level are known.

Rice. 6.5.1

2. Conditional probabilities are indicated next to each of the branches (except,
possibly first-level branches). For each of the groups of branches emerging from one point, the sum of these probabilities is also equal to 1 (or 100%).
For example, in fig. 6.5.1 for the first group of branches we get 0.15 + 0.85 =
1.00 and for the second group - 0.70 + 0.30 = 1.00. This rule allows
calculate one unknown value of conditional probability in a group of branches emanating from one point.

3. The circled probability at the beginning of the branch multiplied by the conditional
the probability next to this branch gives the probability written in a circle in
end of the branch. For example, in fig. 6.5.1 for the upper branch leading to the right
we have 0.20 x 0.15 = 0.03, for the next branch - 0.20 x 0.85 = 0.17; similar relationships hold for the other two branches. This rule can be used to compute a single unknown value
probabilities of three corresponding to some branch.

4. The value of the probability written in a circle is equal to the sum of the circled probabilities at the ends of all branches emerging from this circle
to the right. So, for example, for Fig. 6.5.1 exit the circle with a value of 0.20
two branches, at the ends of which are circled probabilities, the sum of which is equal to this value: 0.03 + 0.17 = 0.20. This rule allows you to find one unknown probability value in a group,
including this probability and all probabilities at the ends of the branches of the tree,
coming out of the corresponding circle.

Using these rules, it is possible, knowing all but one probability value for some branch or at some level, to find this unknown value.

37. What sample is called representative? How can a representative sample be taken?

Representativeness is the ability of the sample to represent the population under study. The more accurately the composition of the sample represents the population on the issues under study, the higher its representativeness.

A representative sample is one of the key concepts of data analysis. A representative sample is a sample from a population with a distribution F(x) representing the main features of the general population. For example, if there are 100,000 people in a city, half of which are men and half are women, then a sample of 1,000 people, of which 10 are men and 990 are women, will certainly not be representative. A public opinion poll built on its basis will, of course, contain a bias in estimates and lead to falsified results.

A necessary condition for the construction representative sample is the equal probability of inclusion in it of each element of the general population.

The sample (empirical) distribution function gives, with a large sample size, a fairly good idea of ​​the distribution function F(x) of the original general population.

The leading principle underlying such a procedure is the principle of randomization, randomness. A sample is said to be random (sometimes we'll say simple random or pure random) if two conditions are met. First, the sample must be designed in such a way that any person or object within the population has equal opportunity be selected for analysis. Second, the sample must be designed so that any combination of n items (where n is simply the number of items, or cases, in the sample) has an equal chance of being selected for analysis.

When examining populations that are too large to run a real lottery, simple random samples are often used. Writing down the names of several hundred thousand objects, putting them into a drum and selecting a few thousand is still not an easy job. In such cases, a different, but equally reliable method is used. Each object in the collection is assigned a number. The sequence of numbers in such tables is usually given by a computer program called a random number generator, which essentially puts a large number of numbers into a drum, draws them randomly, and prints them out in order. In other words, the same process that is characteristic of the lottery takes place, but the computer, using numbers rather than names, makes a universal choice. This choice can be used by simply assigning a number to each of our objects.

A table of random numbers like that one can be used in several different ways, and in each case three Decisions must be made. First, we need to decide how many digits We will use, and secondly, we need to develop decision rule for their use; thirdly, you need to choose the starting point and the method of passing through the table.

Once this is done, we must develop a rule that links the numbers in the table to our object numbers. There are two possibilities here. The easiest way (although not necessarily the most correct) is to use only those numbers that fall into the number of numbers assigned to our objects. So if we have a population of 250 features (and thus use three-digit numbers) and decide to start at the top left corner of the table and move down the columns, we will include feature numbers 100, 084, and 128 in our sample, and let's skip the numbers 375 and 990, which do not correspond to our objects. This process will continue until the number of objects needed for our sample is determined.

A more time-consuming, but methodologically more correct procedure is based on the premise that in order to preserve the randomness characteristic of the table, every number of a given dimension (for example, every three-digit number) must be used. Following this logic, and again dealing with a collection of 250 objects, we must divide the region of three-digit numbers from 000 to 999 into 250 equal intervals. Since there are 1000 such numbers, we divide 1000 by 250 and find that each part contains four numbers. So table numbers from 000 to 003 will correspond to object 004 to 007 - object 2, and so on. Now, in order to determine which object number corresponds to the number in the table, you should divide the three-digit number from the table and round to the nearest whole number.

And finally, we must choose in the table the starting point and the method of passage. The starting point can be the top left corner (as in the previous example), the bottom right corner, the left edge of the second line, or anywhere else. This choice is completely arbitrary. However, when working with the table, we must act systematically. We could take the first three digits of each five-digit sequence, the middle three digits, the last three digits, or even the first, second, and fourth digits. (From the first five-digit sequence, these various procedures yield 100, 009, 097, and 109, respectively.) We could apply these procedures from right to left, getting 790, 900, 001, and 791. We could go along the rows , considering each next digit in turn and ignoring the division into fives (for the first row, the numbers 100, 973, 253, 376 and 520 will be obtained). We could only deal with every third group of digits (eg 10097, 99019, 04805, 99970). There are many different possibilities, and each next one is no worse than the previous one. However, once we have made a decision about one way or another, we must systematically follow it in order to respect the randomness of the elements in the table as much as possible.

38. What interval do we call confidence interval?

The confidence interval is the allowable deviation of the observed values ​​from the true values. The size of this assumption is determined by the researcher, taking into account the requirements for the accuracy of the information. If the margin of error increases, the sample size decreases even if the confidence level remains at 95%.

The confidence interval shows in what range the results of sample observations (surveys) will be located. If we conduct 100 identical surveys in identical samples from a single population (for example, 100 samples of 1000 people each in a city with a population of 5 million), then at a 95% confidence level, 95 out of 100 results will fall within the confidence interval (for example, from 28% to 32% with a true value of 30%).

For example, the true number of city residents who smoke is 30%. If we select 1000 people 100 times in a row and in these samples we ask the question "do you smoke?", in 95 of these 100 samples at a 2% confidence interval, the value will be from 28% to 32%.

39 What is called the level of confidence (confidence level)?

The confidence level reflects the amount of data needed by the evaluator in order to assert that the program being examined has the intended effect. AT social sciences 95% confidence level is traditionally used. However, for most community programs, 95% is overkill. A confidence level in the range of 80-90% is sufficient for an adequate assessment of the program. In this way, the size of the representative group can be reduced, thereby reducing the cost of the evaluation.

The statistical evaluation process tests the null hypothesis that the program did not have the intended effect. If the results obtained differ significantly from the initial assumptions about the correctness of the null hypothesis, then the latter is rejected.

40. Which of the two confidence intervals is larger: two-tailed 99% or two-tailed 95%? Explain.

The 2-sided 99% confidence interval is larger than the 95% because more values ​​fall into it. Doc-in:

Using z-scores, you can more accurately estimate the confidence interval and determine the overall shape of the confidence interval. The exact formulation of the confidence interval for the sample mean is as follows:

Thus, for a random sample of 25 observations satisfying a normal distribution, the confidence interval of the sample mean has the following form:

Thus, you can be 95% sure that the value lies within ±1.568 units of the sample mean. Using the same method, it can be determined that the 99% confidence interval lies within ±2.0608 units of the sample mean

value Thus, we have and from here , Similarly, we obtain the lower limit, which is equal to

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?

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).

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:

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

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