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Classical definition of probability
The basic concept of probability theory is the concept of a random event. A random event is usually called an event, ĸᴏᴛᴏᴩᴏᴇ, under certain conditions, it may or may not occur. For example, hitting or missing an object when shooting at this object with a given weapon is a random event.
An event is usually called reliable if, as a result of the test, it necessarily occurs. It is customary to call an event impossible, ĸᴏᴛᴏᴩᴏᴇ cannot happen as a result of a test.
Random events are said to be inconsistent in a given trial if no two of them can appear together.
Random events form full group, in the event that any of them can appear at each test and no other event incompatible with them can appear.
Consider the complete group of equally possible incompatible random events. Such events will be called outcomes. An outcome is said to be favorable to the occurrence of event A if the occurrence of this event entails the occurrence of event A.
Geometric definition of probability
Let a random test be thought of as throwing a point at random into some geometric region G (on a line, plane, or space). Elementary outcomes - ϶ᴛᴏ separate points G, any event - a subset of this area, the space of elementary outcomes G. We can assume that all points G are ʼʼequalʼʼ and then the probability of a point falling into one of the ĸᴏᴛᴏᴩᴏᴇ subset is proportional to its measure (length, area , volume) and does not depend on its location and shape.
geometric probability event A is determined by the relation: , where m(G), m(A) are geometric measures (lengths, areas or volumes) of the entire space of elementary outcomes and event A.
Example. A circle of radius r () is randomly thrown onto a plane, delimited by parallel stripes of width 2d, the distance between the axial lines of which is 2D. Find the probability that the circle intersects some strip.
Solution. As an elementary outcome of this test, we will consider the distance x from the center of the circle to the center line of the strip closest to the circle. Then the entire space of elementary outcomes - ϶ᴛᴏ segment. The intersection of the circle with the strip will occur if its center falls into the strip, ᴛ.ᴇ. , or will be located from the edge of the strip at a distance less than the radius, ᴛ.ᴇ. .
For the desired probability, we obtain: .
5. The relative frequency of an event is the ratio of the number of trials in which the event occurred to the total number of practically performed trials. Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, the relative frequency A is given by:
(2)where m is the number of occurrences of the event, n is the total number of trials. Comparing the definition of probability and relative frequency, we conclude: the definition of probability does not require that tests be carried out in reality; the definition of the relative frequency assumes that the tests were actually carried out. In other words, the probability is calculated before the experience, and the relative frequency is calculated after the experience.
Example 2. Out of 80 randomly selected employees, 3 people have serious cardiac disorders. Relative frequency of people with heart disease
The relative frequency or a number close to it is taken as a static probability.
DEFINITION (statistical definition of probability). The number to which the stable relative frequency tends is commonly called the statistical probability of this event.
6. sum A+B two events A and B name an event consisting in the occurrence of event A, or event B, or both of these events. For example, if two shots were fired from the gun and A - hit on the first shot, B - hit on the second shot, then A + B - hit on the first shot, or on the second, or in both shots .
In particular, if two events A and B are incompatible, then A + B is an event consisting in the appearance of one of these events, no matter which one. The sum of several events called an event, ĸᴏᴛᴏᴩᴏᴇ consists in the occurrence of at least one of these events. For example, the event A + B + C consists in the appearance of one of the following events: A, B, C, A and B, A and C, B and C, A and B and C. Let the events A and B be incompatible, and the probabilities of these events are known. How to find the probability that either event A or event B will occur? The answer to this question is given by the addition theorem. Theorem. Probability of one of the two incompatible events, no matter what, is equal to the sum of the probabilities of these events:
P (A + B) = P (A) + P (B). Proof
Corollary. The probability of occurrence of one of several pairwise incompatible events, no matter which one, is equal to the sum of the probabilities of these events:
P (A 1 + A 2 + ... + A n) \u003d P (A 1) + P (A 2) + ... + P (A n).
Geometric definition of probability - concept and types. Classification and features of the category "Geometric definition of probability" 2017, 2018.
In practice, such trials are very often encountered, the number of possible outcomes of which is infinite. Sometimes in such cases it is possible to use the method of calculating the probability, in which the concept of the equiprobability of certain events still plays the main role .... .
In a certain square, a point is randomly selected, what is the probability that this point will be inside the region D., where SD is the area of \u200b\u200bthe region D, S is the area of the entire square. Under the classical, a certain zero probability had ... .
To overcome the disadvantage of the classical definition of probability, which is that it is not applicable to trials with an infinite number of outcomes, geometric probabilities are introduced - the probabilities of a point falling into an area. Let a flat figure g (segment or body)... .
Classical definition of probability LECTURE 1. PROBABILITY THEORIES. HISTORY OF ORIGIN. CLASSICAL DEFINITION OF PROBABILITY A.A. Khalafyan BIBLIOGRAPHICAL REFERENCES 1. Kolemaev V.A., Staroverov O.V., Turundaevsky V.B. Theory ... .[read more] .
This definition is used when an experience has an uncountable set of equally possible outcomes. In this case, the space of elementary events can be represented as a certain region G. Each point of this region corresponds to an elementary event. Hit... .
The geometric definition of probability is an extension of the concept of classical probability to the case of an uncountable set of elementary events. In the case when is an uncountable set, the probability is determined not on elementary events, but on their sets.... .
Classical Definition of Probability PROBABILITY OF A RANDOM EVENT Set-Theoretic Interpretation of Operations on Events Let some experiment be carried out with a random outcome. Lots of &... .
The classical definition of probability has a limitation in its application. It is assumed that the set of elementary events Ω is finite or countable, i.e., Ω = ( ω 1 , ω 2 , … , ω n , …), and all ω i – equally possible elementary events. However, in practice there are tests for which the set of elementary outcomes is infinite. For example, when manufacturing a certain part on a machine, it is necessary to maintain a certain size. Here, the accuracy of manufacturing a part depends on the skill of the worker, the quality of the cutting tool, the perfection of the machine, etc. If a test is understood as the manufacture of a part, then as a result of such a test, an infinite number of outcomes are possible, in this case obtaining parts of the required size.
To overcome the shortcoming of the classical definition of probability, some concepts of geometry are sometimes used (if, of course, the circumstances of the test allow). In all such cases, the possibility of conducting (at least theoretically) any number of tests is assumed, and the concept equal opportunities also play a major role.
Let us consider a test with a space of events, the elementary outcomes of which are represented as points filling some area Ω (in the three-dimensional space R 3). Let the event BUT consists in hitting a randomly thrown point in the subdomain D domain Ω. event BUT favor elementary events in which the point falls into some subdomain D. Then under probability developments BUT we will understand the ratio of the volume of the subdomain D(highlighted area in Fig. 1.11) to the volume of the area Ω, R(BUT) = V(D) / V(Ω).
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Rice.1. 11 |
Here, by analogy with the concept of a favorable outcome, the area D will be called favorable to the appearance of the event BUT. The probability of an event is defined similarly BUT, when the set Ω is a certain area on a plane or a segment on a straight line. In these cases, the volumes of the regions are replaced by the areas of the figures or the lengths of the segments, respectively.
Thus, we come to a new definition - geometric probability for tests with an infinite uncountable set of elementary events, which is formulated as follows.
The geometric probability of an event A is the ratio of the measure of the subdomain that favors the occurrence of this event to the measure of the entire area, i.e.
p(A) =mesD / mesΩ,
where mes– measure of areas D and Ω , D Ì Ω.
The geometric probability of an event has all the properties inherent in the classical definition of probability. For example, the 4th property would be: R(BUT+ IN) = R(BUT) + R(IN).
Statistical definition of probability
Task 2. The shooter fires one shot at the target. Estimate the probability that he will hit the target.
Solution. In this experiment, two outcomes are possible: either the shooter hit the target (the event A), or he missed (event). Developments A and are incompatible and form a complete group. However, in the general case, it is not known whether they are equally possible or not. Therefore, in this case, the classical definition of the probability of a random event cannot be used. You can solve the problem using the statistical definition of the probability of a random event.
Definition 1.12. Relative event frequency A called the ratio of the number of trials in which the event A appeared, to the total number of tests actually performed.
Thus, the relative frequency of the event A can be calculated by the formula
where k– number of occurrences of the event A, l is the total number of trials.
Remark 1.2. The main difference in the relative frequency of the event A from its classical probability lies in the fact that the relative frequency is always found according to the results of the tests. To calculate the classical probability, it is not necessary to set up an experiment.
Long-term observations have shown that if a series of experiments are carried out under identical conditions, in each of which the number of tests is sufficiently large, then the relative frequency reveals stability property. This property consists in the fact that in different series of experiments the relative frequency W( A) changes little (the less, the more tests are performed), fluctuating around a certain constant number.
As statistical probability of an event take a relative frequency or a number close to it.
Let's return to problem 2 about calculating the probability of an event A(shooter will hit the target). To solve it, it is necessary to conduct several series of enough a large number shots at a target under the same conditions. This will allow you to calculate the relative frequency and estimate the probability of an event A.
The disadvantage of the statistical definition is the ambiguity of the statistical probability. For example, if W( A)»0.4, then as the probability of the event A you can take 0.4, and 0.39, and 0.41.
Remark 1.3. The statistical definition of probability overcomes the second shortcoming of the classical definition of probability.
Let there be figures on the plane G And g, and gÌ G(Fig. 1.1).
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Remark 1.4. In case when g And G- straight line segments, the probability of an event A is equal to the ratio of the lengths of these segments. If g And G are bodies in three-dimensional space, then the probability of an event A is found as the ratio of the volumes of these bodies. Therefore, in the general case where mes is the metric of the space under consideration. Remark 1.5. The geometric definition of probability applies to trials with an infinite number of outcomes. Example 1.13. Two persons agreed to meet at a certain place between 12 and 13 hours, and each person who came to the meeting waits for the other for 20 minutes, but no longer than until 13.00, after which he leaves. Find the probability of meeting these persons if each of them arrives at a random moment of time, not coordinated with the moment of arrival of the other. Solution. Let the event A- the meeting took place. Denote by x- the time of arrival of the first person to the meeting, y- arrival time of the second person. Then the set of all possible outcomes of the experience is the set of all pairs ( x, y), where x, yО . And the set of favorable outcomes is determined by the inequality |x – y| £20 (min). Both of these sets are infinite, so the classical definition for calculating probability cannot be applied. Let's use the geometric definition. On fig. 1.2 shows the sets of all possible outcomes (square OKMT) and favorable outcomes (hexagon OSLMNR). Using Definition 1.13, we get Sum and product of events. Theorems on the probability of the sum and product of events Definition 1.14.The sum of events A And B name the event consisting in the appearance of at least one of them. Designation: A + B. Definition 1.15.The product of events A And B call an event consisting in the simultaneous occurrence of these events in the same experience. Designation: AB. Example 1.14. From a deck of 36 cards, one card is drawn at random. Let us introduce the notation: A- the drawn card turned out to be a lady, B- they took out a card of spades. Find probabilities of events A + B And AB. Solution. Event A + B happens if the drawn card is of spades or a queen. This means that the event under consideration is favored by 13 outcomes (any of the 9 cards of spades, any of the 3 queens of another suit) out of 36 possible. Using the classical definition of the probability of a random event, we get Event AB occurs if the drawn card is of spades and a queen. Therefore, the event AB favors only one outcome of the experience (Queen of Spades) out of 36 possible. Taking into account Definition 1.11, we obtain Remark 1.6. The definitions of sum and product of events can be extended to any number of events. When calculating the probability of the sum and product of events, it is convenient to use the following statements. Theorem 1.1. The probability of the occurrence of one of two incompatible events, no matter which one, is equal to the sum of the probabilities of these events P( A+B)=P( A)+P( B). Corollary 1.1. The probability of occurrence of one of several pairwise incompatible events, no matter which one, is equal to the sum of the probabilities of these events P( A 1 +A 2 +…+A n)=P( A 1)+P( A 2)+…+P( A n). Corollary 1.2. The sum of the probabilities of pairwise incompatible events A 1 , A 2 ,…, A n, forming a complete group, is equal to one P( A 1)+P( A 2)+…+P( A n)=1. Corollary 1.3. Probability of the opposite event A random event was defined as an event that, as a result of experience, may or may not occur. If when calculating the probability of an event, no other restrictions (except for the conditions of the experiment) are imposed, then such a probability is called unconditional. If other additional conditions are imposed, then the probability of the event is called conditional. Definition 1.16.Conditional Probability P B(A) (or P( A|B)) is called the probability of an event A, calculated under the assumption that the event B already happened. Using the concept of conditional probability, we give a definition of the independence of events that differs from the one given earlier. Definition 1.17. Event A is independent of event B if the equality In practical questions, to determine the independence of these events, one rarely turns to checking the fulfillment of equalities (1.3) and (1.4) for them. Usually for this they use intuitive considerations based on experience. Definition 1.18. Several events are called pairwise independent if every two of them are independent. Definition 1.19. Several events are called collectively independent if they are pairwise independent and each event and all possible products of the others are independent. Theorem 1.2. The probability of the joint occurrence of two events is equal to the product of the probability of one of them by the conditional probability of the other, calculated on the assumption that the first event has already occurred. Depending on the choice of the sequence of events, Theorem 1.2 can be written as P( AB) = P( A)P A(B) P( AB) = P( B)P B(A). Corollary 1.4. The probability of the joint occurrence of several events is equal to the product of the probability of one of them by the conditional probabilities of all the others, and the probability of each subsequent event is calculated on the assumption that all previous events have already appeared In this case, the order in which the events are located can be chosen in any order. Example 1.15. An urn contains 6 white and 3 black balls. One ball is drawn at random from the urn until a black one appears. Find the probability that the fourth withdrawal will have to be carried out if the balls are not returned to the urn. Solution. In the experiment under consideration, it is necessary to carry out the fourth removal if the first three balls turn out to be white. Denote by A i an event that i-th drawing out a white ball will appear ( i= 1, 2, 3). The problem is to find the probability of an event A 1 A 2 A 3 . Since the drawn balls do not return back, the events A 1 , A 2 and A 3 are dependent (each previous affects the possibility of the next). To calculate the probability, we use Corollary 1.4 and the classical definition of the probability of a random event, namely Corollary 1.5. The probability of the joint occurrence of two independent events is equal to the product of their probabilities P( AB)=P( A)P( B). Corollary 1.6. The probability of the joint occurrence of several events that are independent in the aggregate is equal to the product of their probabilities P( A 1 A 2 …A n)=P( A 1)P( A 2)…P( A n). Example 1.16. Solve the problem from example 1.15, assuming that after each removal the balls are returned back to the urn. Solution. As before (Example 1.15), we need to find P( A 1 A 2 A 3). However, events A 1 , A 2 and A 3 are independent in the aggregate, since the composition of the urn is the same for each removal and, therefore, the result of a single test does not affect the others. Therefore, to calculate the probability, we use Corollary 1.6 and Definition 1.11 of the probability of a random event, namely P( A 1 A 2 A 3)=P( A 1)P( A 2)P( A 3)= = . Theorem 1.3. The probability of the occurrence of at least one of the two joint events is equal to the sum of the probabilities of these events without the probability of their joint occurrence
Remark 1.7. When using formula (1.5), one must keep in mind that the events A And B can be either dependent or independent. Example 1.17. Two shooters fired one shot each at the target. It is known that the probability of hitting the target for one of the shooters is 0.6, and for the other - 0.7. Find the probability that a) both shooters hit the target (event D); b) only one of the shooters will hit the target (event E); c) at least one of the shooters will hit the target (the event F). Solution. Let us introduce the notation: A- the first shooter hit the target, B The second shooter hit the target. By condition P( A) = 0.6 and P( B) = 0.7. We will answer the questions. a) Event D will happen if an event occurs AB. Because the events A And B are independent, then, taking into account Corollary 1.5, we obtain P( D) = P( AB) = P( A)P( B) = 0.6×0.7 = 0.42. b) Event E happens if one of the events occurs A or B. These events are incompatible, and the events A() And B() are independent, therefore, by Theorem 1.1, Corollaries 1.3 and 1.5, we have P( E) = P( A+ B) = P( A) + P( B) = P( A)P() + P()P( B) = 0.6×0.3 + 0.4×0.7 = 0.46. c) Event F will occur if at least one of the events occurs A or B. These events are shared. Therefore, by Theorem 1.3, we have P( F) = P( A+B) = P( A) + P( B) - P( AB) = 0,6 + 0,7 - 0,42 = 0,88. Note that the probability of an event F could have been calculated differently. Namely P( F) = P( A+ B + AB) = P( A) + P( B) + P( AB) = 0,88 P( F) = 1 - P() = 1 - P()P() = 1 - 0.4×0.3 = 0.88. Total Probability Formula. Bayes formulas Let the event A can occur if one of the incompatible events occurs B 1 , B 2 ,…, B n, forming a complete group. Since it is not known in advance which of these events will occur, they are called hypotheses. Estimate the probability of an event occurring A before the experiment, you can use the following statement. Theorem 1.4. Event Probability A, which can occur only if one of the incompatible events occurs B 1 , B 2 ,…, B n, forming a complete group, is equal to
Formula (1.6) is called total probability formulas. Example 1.18. To pass the exam, students had to prepare 30 questions. Out of 25 students, 10 prepared all questions, 8 - 25 questions, 5 - 20 questions and 2 - 15 questions. Find the probability that a randomly selected student will answer the given question. Solution. Let us introduce the following notation: A- an event consisting in the fact that a randomly called student answered the question posed, B 1 - randomly called student knows the answers to all questions, B 2 - randomly called student knows the answers to 25 questions, B 3 - randomly called student knows the answers to 20 questions and B 4 - randomly called student knows the answers to 15 questions. Note that the events B 1 ,B 2 ,B 3 and B 4 are incompatible, form a complete group, and the event A can occur if one of these events occurs. Therefore, to calculate the probability of an event A we can use the total probability formula (1.6): According to the condition of the problem, the probabilities of the hypotheses are known P( B 1) = , P( B 2) = , P( B 3) = , P( B 4) = and conditional probabilities (probabilities for students of each of the four groups to answer the question) 1, = , = , = . In this way, P( A) = ×1 + × + × + × = . Assume that a test has been performed, as a result of which an event has occurred A, and which of the events B i (i =1, 2,…, n) occurred is not known to the researcher. To estimate the probabilities of hypotheses after the result of the test becomes known, you can use Bayes formulas
Here P( A) is calculated by the total probability formula (1.6). Example 1.19. In a certain factory, machine I produces 40% of all output, and machine II produces 60%. On average, 9 out of 1,000 units produced by machine I are defective, and machine II has 4 out of 500 defective units. What is the probability that it was produced by machine II? Solution. Let us introduce the notation: A- an event consisting in the fact that a unit of production, selected at random from a daily production, turned out to be a defect, B i- a unit of production, chosen at random, is made by a machine i(i= I, II). Developments B 1 and B 2 are incompatible and form a complete group, and the event A can only occur as a result of the occurrence of one of these events. It is known that the event A happened (a randomly selected unit of production turned out to be a defect). Which of the events B 1 or B 2 at the same time, it is unknown, because it is not known on which of the two machines the selected item was made. Estimating the likelihood of a hypothesis B 2 can be carried out using the Bayes formula (1.7): where the probability of a random selection of a defective product is calculated by the total probability formula (1.6): Considering that, according to the condition of the problem P( B 1) = 0.40, P( B 2) = 0,60, = 0,009, = 0,008, Sequence of independent trials In scientific and practical activities, it is constantly necessary to carry out repeated tests under similar conditions. As a rule, the results of previous tests do not affect the subsequent ones. The simplest type of such tests is very important, when in each of the tests some event A may appear with the same probability, and this probability remains the same, regardless of the results of previous or subsequent tests. This type of test was first explored by Jacob Bernoulli and is therefore called Bernoulli schemes. Bernoulli scheme. Let it be produced n independent tests under similar conditions (or the same experiment is carried out n times), in each of which the event A may or may not appear. In this case, the probability of occurrence of an event A in each trial is the same and equal p. Therefore, the probability of non-occurrence of the event A in each individual test is also constant and equal to q= 1 - p. The probability that under these conditions an event A will come true exactly k times (and, therefore, will not be realized n – k times) can be found by Bernoulli formula
In this case, the order of occurrence of the event A in the specified n tests can be arbitrary. Example 1.20. The probability that a customer will require size 41 shoes is 0.2. Find the probability that out of the first 5 buyers shoes of this size will be needed: a) one; b) at least one; c) at least three; d) more than one and less than four. Solution. In this example, the same experiment (choosing shoes) is carried out 5 times, and the probability of the event is A- shoes of the 41st size are chosen - it is constant and equal to 0.2. In addition, the result of each individual test does not affect other experiments, because. buyers choose shoes independently of each other. Therefore, we have a sequence of tests carried out according to the Bernoulli scheme, in which n = 5, p = 0,2, q= 0.8. To answer the questions posed, it is necessary to calculate the probabilities P 5 ( k). We use formula (1.8). a) P 5 (1) = = 0.4096; b) P 5 ( k³ 1) = 1 - P 5 ( k < 1) = 1 - P 5 (0) = 1- = 0,67232; c) P 5 ( k³ 3) \u003d P 5 (3) + P 5 (4) + P 5 (5) \u003d + + \u003d \u003d 0.5792; d) P 5 (1< k < 4) = P 5 (2) + P 5 (3) = + = 0,256. The use of the Bernoulli formula (1.32) for large values of n and m causes great difficulties, since this involves cumbersome calculations. Thus, at n = 200, m = 116, p = 0.72, the Bernoulli formula takes the form P 200 (116) = (0.72) 116 (0.28) 84 . It is almost impossible to calculate the result. The calculation of P n (m) also causes difficulties for small values of p (q). There is a need to find approximate formulas for calculating P n (m), providing the necessary accuracy. Such formulas give us limit theorems; they contain the so-called asymptotic formulas, which, for large test values, give an arbitrarily small relative error. Consider three limit theorems containing asymptotic formulas for calculating the binomial probability P n (m) as n. Theorem 1.5. If the number of trials increases indefinitely (n) and the probability p of the occurrence of event A in each trial decreases indefinitely (p), but in such a way that their product pr is constant value(pr \u003d a \u003d const), then the probability P n (m) satisfies the limit equality Expression (1.9) is called the asymptotic Poisson formula. From the limit equality (1.9) for large n and small p follows the approximate Poisson formula Formula (1.10) is used when the probability p = const of success is extremely small, i.e. success itself (the appearance of event A) is a rare event (for example, winning a car with a lottery ticket), but the number of trials n is large, the average number of successes pr = a slightly. The approximate formula (1.10) is usually used when n 50, and pr 10. Poisson's formula finds application in queuing theory. A stream of events is a sequence of events occurring at random times (for example, a stream of visitors in a hairdresser, a stream of calls at a telephone exchange, a stream of element failures, a stream of subscribers served, etc.). The flow of events, which has the properties of stationarity, ordinariness and absence of consequences, is called the simplest (Poisson) flow. The property of stationarity means that the probability of occurrence of k events in a time segment of length depends only on its length (i.e., does not depend on its origin). Consequently, the average number of events that appear per unit of time, the so-called flow intensity, is a constant value: ( t) = . The property of ordinary means that the event does not appear in groups, but one by one. In other words, the probability of occurrence of more than one event for a small period of time t is negligibly small compared to the probability of occurrence of only one event (for example, the flow of boats approaching the pier is ordinary). The property of the absence of a consequence means that the probability of occurrence k of events at any time interval of length does not depend on how many events appeared on any other segment that does not intersect with it (they say: the "future" of the flow does not depend on the "past", for example, the flow of people, included in the supermarket). It can be proved that the probability of the occurrence of m events of the simplest flow in a time of duration t is determined by the Poisson formula. Use the Bernoulli formula for large values n difficult enough, because in this case, one has to perform operations on huge numbers. You can simplify calculations using factorial tables or using technical means (calculator, computer). But in this case, errors accumulate in the calculation process. Therefore, the final result may differ significantly from the true one. There is a need to apply approximate (asymptotic) formulas. Remark 1.8. Function g(x) are called asymptotic approximation of the function f(x), if. Theorem 1.6. (Local Moivre-Laplace theorem) If the probability p occurrence of an event A in each trial is constant and different from 0 and 1, and the number of independent trials is large enough, then the probability that the event A will appear in n tests carried out according to the Bernoulli scheme, exactly k times, approximately equal (the more accurate, the more n) The graph of the function has the form shown in Fig. 1.3. It should be taken into account that: a) the function φ(x) is even, i.e. φ(-x) = φ(x); For function j(x) tables of values are compiled for x³ 0. For x< 0 пользуются теми же таблицами, т.к. функция j(x) is even. Theorem 1.7. (Moivre-Laplace integral theorem) If the probability p event A in each trial is constant and different from 0 and 1, then the probability P n(k 1 , k 2) that the event A will appear in n tests carried out according to the Bernoulli scheme, from k 1 to k 2 times, approximately equal Here z 1 and z 2 are defined in (1.14). Example 1.21. Seed germination is estimated with a probability of 0.85. Find the probability that out of 500 sown seeds there will sprout: a) 425 seeds; b) from 425 to 450 seeds. Solution. Here, as in the previous example, there is a sequence of independent tests carried out according to the Bernoulli scheme (experiment - planting one seed, event A- seed sprouted n = 500, p = 0,85, q= 0.15. Since the number of trials is large ( n> 100), we use the asymptotic formulas (1.10) and (1.13) to calculate the required probabilities. b) »F(3.13)–F(0)»0.49. If the number of trials n, carried out according to the Bernoulli scheme, is large, and the probability p occurrence of an event A in each of them is small ( p£ 0,1), then Laplace's asymptotic formula is unsuitable. In this case, use asymptotic Poisson formula
where l = np. Example 1.22. The store received 1000 bottles mineral water. The probability that a bottle will be broken in transit is 0.003. Find the probability that the store will receive broken bottles: a) exactly 2; b) less than 2; c) at least one. Solution. In this problem, there is a sequence of independent tests carried out according to the Bernoulli scheme (experiment - checking one bottle for integrity, event A- the bottle is broken n = 1000, p = 0,003, q= 0.997. Because the number of trials is large ( n> 100), and the probability p small ( p < 0,1) воспользуемся при вычислении требуемых вероятностей формулой Пуассона (1.14), учитывая, что l=3. a) = 4.5 e-3 » 0.224; b) P 1000 ( k < 2) = P 1000 (0) + P 1000 (1) » + = 4e-3 » 0.199; c) P 1000 ( k³ 1) = 1 - P 1000 ( k < 1) = 1 - P 1000 (0) » 1 - = 1 - e-3 » 0.95. The local and integral Moivre–Laplace theorems are corollaries of a more general central limit theorem. Many continuous random variables have normal distribution. This circumstance is largely determined by the fact that the summation of a large number random variables with very different distribution laws leads to normal distribution this amount. Theorem . If a random variable is the sum of a very large number of mutually independent random variables, the influence of each of which is negligible on the entire sum, then it has a distribution close to normal . The central limit theorem is of great practical importance. Suppose some economic indicator is determined, for example, electricity consumption in the city for the year. The value of total consumption is the sum of energy consumption by individual consumers, which has random values with different distributions. The theorem states that in this case, whatever distribution the individual components have, the distribution of the resulting consumption will be close to normal. |
IV. PROBABILITY AND MATHEMATICAL
STATISTICS
Reference material and principles of problem solving
The classical definition of probability
By experience or experiment we will understand any realization of a complex of certain conditions, as a result of which the phenomenon of interest to us will occur.
Example 1 Experience σ: target shooting. Event A - hitting the target. Event B is a miss.
Example 2 Experience σ: selection of products from a finished batch. Event A - the product is defective. Event B is a standard product.
An elementary event (or elementary outcome) is any elementary, that is, indivisible within the framework of a given experiment, the outcome of an experience. The set of all elementary outcomes will be called elementary event space and denote Ω. That is, the set of outcomes of experiments forms the space of elementary events if:
As a result of experience, one of the outcomes necessarily occurs;
The appearance of one of the outcomes of experience excludes the appearance of the others;
Within the framework of this experience, it is impossible to divide an elementary outcome into smaller components.
Write it down like this:
Ω =(w 1, w 2, …w n ,…)=(w k , k=1…n, …).
Example 3 Experience: coin toss 1 time.
Here Ω=(w g, w c ), where w g is the loss of the coat of arms, w c is the loss of the number.
Experience: A coin is tossed 2 times. In this case, the space of elementary events Ω=(w g g, w g c, w c g, w c c ).
The experiment consists in determining the number of calls received by the telephone exchange during the time T. Here Ω=(0,1,2.…n,… ).
Any set of elementary outcomes or an arbitrary subset A Ω is called random event.
Let Ω be the space of elementary events, S be some subset of random events satisfying the following conditions:
The set S is closed under the operations of addition, multiplication and negation.
Certain E and impossible events belong to S.
Sometimes more is required: for any infinite sequence of events 
A subset of S that satisfies these conditions is called σ - algebra .
Let a function be given that for each random event from S matches a number from the interval ; R:S → , and the following axioms hold:
,
P(E)=1, P(Ø)=0,
For any sequence А 1 ,…А n … pairwise incompatible events А i нS,
"i, j, i≠ј,
.
A function P that satisfies these axioms is called probability, and the value Р(A) is called the probability of the event BUT.
Definition. Three of objects (Ω, S, P), where Ω is the space of elementary events, S– σ-algebra, Р – probability, is called probability space.
The classical definition of probability serves as a good mathematical model of those random phenomena for which the outcomes of an experiment are a finite number n and all outcomes are equally possible. In the classical definition of probability, one assumes:
;
The probability of an event equal to
In other words, the probability of an event is equal to the ratio of the number of elementary events included in to the total number of elementary events in .
The following formulation of the classical definition of probability is also generally accepted: the probability of an event is the ratio of the number of outcomes of experience that favor the appearance of an event to the total number of equally possible outcomes of experience.
That is, the probability of an event is defined as .
Example 4 What is the probability that the coat of arms will appear at least once when the coin is tossed twice?
Solution. The space of equally probable elementary events of this experiment consists of the following events: The event = (when a coin is tossed twice, the coat of arms will appear at least once) consists of incompatible elementary events 
. Consequently, .
In this way, 
.
Example 5 What is the probability that a randomly named two-digit number will be divisible by eleven without a remainder?
Solution. Since there are 90 of all two-digit numbers, the number of equally possible outcomes of this experience is . Of these numbers, 11, 22, 33, 44, 55, 66, 77, 88, 99 are divisible by 11 without a remainder. Therefore, the number of outcomes that favor the event (a two-digit number will be divisible by eleven without a remainder). The desired probability will be equal to 
.
Example 6 What is the probability that a randomly chosen year will have 5 Sundays in September?
Solution. September of any year has 30 days. The number of Sundays in September depends on what day of the week September 1st is. September 1st can be any day of the week. Since there are 7 days in a week, the number of all possible outcomes is . If September starts on Monday, Tuesday, Wednesday, Thursday or Friday, then there will be 4 Sundays. If September starts on Saturday or Sunday, then there will be 5 Sundays. Among 7 equally possible outcomes, 2 will be favorable to the event (in September, a randomly chosen year will have 5 Sundays), Consequently, . Desired probability 
.
Example 7 There are five segments of length 3, 5, 6, 9 and 11 cm. Determine the probability that from three randomly taken segments (of these five) it is possible to construct a triangle.
Solution. There are equally possible outcomes of this experience: , , , , , , , , , .
In order for three segments to be able to build a triangle, it is necessary that the larger segment be less than the sum of the other two segments. This condition is satisfied by the following outcomes , , , , . The number of such outcomes. Consequently,
.
In cases where a direct enumeration of all possible outcomes becomes cumbersome, it is advisable to use combinatorics.
Elements of combinatorics
Let a set consisting of elements be given. There are two fundamental various ways selection of elements from the set : selection of elements without return and selection of elements with return.
The first way of choosing elements leads to the notions of permutations, placements, and combinations without repetition, or simply permutations, placements, and combinations; the second - to the concepts of permutations, placements and combinations with repetitions.
permutation of elements is any ordered set of these elements. Each permutation contains elements. Permutations differ only in the order in which the elements are arranged. The number of different permutations of elements is calculated by the formula
.
Accommodation of elements by is any ordered set of different elements selected from the total set of elements. Placements differ from each other either by the order of the elements, or at least one element.
The number of placements is calculated by the formula 
.
combination of elements by is any unordered set of different elements selected from the total set of elements. Combinations differ from each other by at least one element.
The number of combinations is calculated by the formula
.
Combination properties:
Example 8 Let there be a set 
from three elements. Then all arrangements of two elements from three are as follows: All permutations of the set have the form: and All combinations of two elements from the set are as follows: 
Placements and combinations with repetitions differ from placements and combinations without repetitions only in that repeating elements can be present in these connections.
The number of placements from elements by with repetitions is calculated by the formula .
The number of combinations of elements with repetitions is calculated by the formula 
.
Since all elements of the set participate in such a type of connections as permutations with repetitions, then the repetition of elements must be embedded in the elements of the set. So, if it contains elements of the first type, elements of the second type, ..., elements of the th type, then the number of permutations with repetitions is calculated by the formula 
.
When solving combinatorial problems, the following two rules can be useful:
Sum rule: if an object can be chosen in ways, and an object can be chosen in ways, then the choice "either or" can be done in ways.
Product rule: if an object can be chosen in ways, and after each of these choices, object , in turn, can be chosen in ways, then the choice of "and" in in that order can be done in ways.
Example 9 Let there be groups of elements, and the -th group consists of - elements. We choose one element from each group, then the total number of ways in which such a choice can be made according to the product rule
. (1)
If 
, then we can assume that the selection is made from the same group, and the element after the selection is returned to the group again. Then .
Example 10 The teacher offers each of the three students to think of any number from 1 to 10. Assuming that each student's choice of any number from the given ones is equally possible, find the probability that one of the three conceived numbers will coincide.
Solution. First, let's calculate the total number of outcomes. The first student chooses one of 10 numbers, the second and third do the same, 
According to formula (1), the total number of ways will be equal to Let's calculate the number of favorable outcomes. To do this, first we find the total number of combinations of the intended numbers in which there are no matches. The first student can choose any of the 10 numbers, the second student can choose any of the 9 numbers, and the third student can choose any of the remaining 8 numbers. Therefore, the total number of combinations of conceived numbers, in which there are no matches, according to the formula (1) is equal. Other cases (1000 - 720 = 280) are characterized by the presence of at least one match. Therefore, the desired probability is equal to 
Example 11. All letters of the Russian alphabet are transmitted randomly over the communication line. Find the probability that a sequence of letters will appear on the tape that begins with the word "world".
Solution. The Russian alphabet contains 33 letters. Since all letters are transmitted over the communication line, the number of equally possible outcomes of the experience 
. Of these outcomes, favorable for the appearance of the event (a sequence of letters that begins with the word “world” will appear) will be all outcomes in which the word “world” will be in the first three positions (one outcome corresponds to such a choice), and the remaining positions will be filled in any way ( the number of such options). According to the product rule, the number of favorable outcomes 
.
Consequently,
Example 12. From an urn containing 3 balls, one ball is taken out at random three times and returned each time. Find the probability that all the balls are in the hand.
Solution. By the condition of the problem, the balls are returned to the urn, therefore, we have a scheme for choosing elements with a return.
The number of all possible outcomes of a given experience is the number of placements of three elements of three with repetitions, that is
.
Favorable event A=( ) there will be those outcomes in which the elements (balls) will not be repeated. The number of such outcomes is the number of placements of three elements of three, or the number of permutations of three elements, that is 
. Since all outcomes of the experiment are equally likely, then
.
Example 13 Technical control checks out of a batch of 500 parts 20 parts taken at random. The batch contains 15 non-standard parts. What is the probability that there will be exactly two non-standard parts among the tested parts?
Solution. Since, according to the condition of the problem, 20 parts out of 500 are extracted at random, it is natural to consider all possible options for extracting 20 parts out of 500 equally possible and to find the required probability, use the classical scheme (the classical definition of probability).
The order of standard and non-standard parts in the retrieved 20 does not matter. What matters is the number of standard and non-standard parts. Therefore, the number of all possible ways in which this can be done is , that is, .
The event = (among the checked parts there will be exactly two non-standard ones) (therefore, the remaining 18 must be standard), will correspond to the (product rule) of outcomes, that is 
. In this way,
.
Example 14 A three-digit number is composed as follows: three dice are thrown: white, blue and red; the number of points rolled on the white die is the number of hundreds, the number of points rolled on the blue die is the number of tens, and the number of points rolled on the red die is the number of units of the three-digit number. What is the probability that the resulting number will be greater than 456?
Solution. The number of all numbers that can be obtained in this way, in accordance with the product rule, will be equal to 
.
Let's calculate the number of outcomes of the experience favorable for the occurrence of event A. Numbers greater than 456 will be obtained if the number of hundreds is greater than 4, that is, 5 or 6, or the number of hundreds is 4, and the number of tens is greater than 5, that is, 6. Let the number of hundreds will be 5. There will be such experiments since the number of tens and units can arbitrarily change from 1 to 6. The same reasoning is valid if the number of hundreds is 6. Experiments in which the first two digits of 45 will be 6. Using the rules of product and sum , find the number of such numbers 
. Since all outcomes of the experiment are equally likely, the desired probability 
.
Example 15 Three radio stations are allowed to operate on six different frequencies. Determine the probability that at least two radio stations will operate on the same frequencies if the choice of frequencies is made at random.
Solution. The number of all equally possible outcomes of the experiment is the number of placements of six elements (frequencies) three each with repetitions, that is 
. Favorable event A= (at least two radio stations will operate on the same frequencies) there will be outcomes in which elements (frequencies) will be repeated. The number of such outcomes is the sum of the outcomes in which two radios operate on the same frequency – and three radios operate on the same frequency – . The number of outcomes in which two of the three radios can operate on one of the six frequencies is . The number of different frequencies is 6. The third radio station can operate on one of the five “unoccupied” frequencies. According to the product rule 
. Obviously, the number of outcomes (three radio stations will operate on the same frequency) is 6.
In this way, .
Consequently, 
.
Geometric definition of probability
The geometric definition generalizes the classical definition of probability to the case where the elementary event space is a subset of the space.
Is the concept of a random event. A random event is an event that, under certain conditions, may or may not occur. For example, hitting or missing an object when firing at this object with a given weapon is a random event.
An event is called certain if, as a result of the test, it necessarily occurs. An impossible event is an event that, as a result of a test, cannot occur.
Random events are said to be inconsistent in a given trial if no two of them can appear together.
Random events form a complete group if, on each trial, any one of them can appear and no other event cannot appear that is incompatible with them.
Consider the complete group of equally possible incompatible random events. Such events will be called outcomes. An outcome is called favorable to the occurrence of event A if the occurrence of this event entails the occurrence of event A.
Geometric definition of probability
Let a random test be thought of as throwing a point at random into some geometric region G (on a line, plane, or space). Elementary outcomes are individual points G, any event is a subset of this area, the space of elementary outcomes G. We can assume that all points G are “equal” and then the probability of a point falling into a certain subset is proportional to its measure (length, area, volume) and independent of its location and shape.
geometric probability event A is determined by the relation:
,
where m(G), m(A) are geometric measures (lengths, areas or volumes) of the entire space of elementary outcomes and event A.
Example. A circle of radius r () is randomly thrown onto a plane delimited by parallel strips of width 2d, the distance between the axial lines of which is equal to 2D. Find the probability that the circle intersects some strip.
Solution. As an elementary outcome of this test, we will consider the distance x from the center of the circle to the center line of the strip closest to the circle. Then the entire space of elementary outcomes is a segment 
. The intersection of the circle with the strip will occur if its center falls into the strip, i.e. , or will be located from the edge of the strip at a distance less than the radius, i.e. .
For the desired probability, we obtain: .
The relative frequency of an event is the ratio of the number of trials in which the event occurred to the total number of practically performed trials. Thus, the relative frequency A is determined by the formula:
(2) where m is the number of occurrences of the event, n is the total number of trials. Comparing the definition of probability and relative frequency, we conclude: the definition of probability does not require that tests be carried out in reality; the determination of the relative frequency assumes that the tests were actually carried out. In other words, the probability is calculated before the experience, and the relative frequency - after the experience.
Example 2. Out of 80 randomly selected employees, 3 people have serious cardiac disorders. Relative frequency of people with heart disease
The relative frequency or a number close to it is taken as a static probability.
DEFINITION (by the statistical definition of probability). The number towards which the steady relative frequency tends is called the statistical probability of that event. 
sum A+B two events A and B name an event consisting in the occurrence of event A, or event B, or both of these events. For example, if two shots are fired from a gun and A is a hit on the first shot, B is a hit on the second shot, then A + B is a hit on the first shot, or the second, or both shots.
In particular, if two events A and B are incompatible, then A + B is an event consisting in the appearance of one of these events, no matter which one. The sum of several events name an event that consists in the occurrence of at least one of these events. For example, the event A + B + C consists in the occurrence of one of the following events: A, B, C, A and B, A and C, B and C, A and B and C. Let the events A and B be incompatible, and the probabilities these events are known. How to find the probability that either event A or event B will occur? The answer to this question is given by the addition theorem.
Theorem. The probability of occurrence of one of two incompatible events, no matter which one, is equal to the sum of the probabilities of these events:
P (A + B) = P (A) + P (B). Proof
Corollary. The probability of occurrence of one of several pairwise incompatible events, no matter which one, is equal to the sum of the probabilities of these events:
P (A 1 + A 2 + ... + A n) \u003d P (A 1) + P (A 2) + ... + P (A n).
