Another scheme for describing experiments with ambiguously predicted outcomes, which makes it quite easy to introduce a quantitative characteristic of the feasibility of an event, is the scheme of geometric probabilities, which, like the scheme of cases considered above, exploits the idea of the equal possibility of the outcomes of the experiment. Just as it was done in the scheme of cases, the quantitative characteristic of the feasibility of an event - its probability - is defined as a value normalized in some way, proportional to the stock of outcomes that favor the implementation of the event. Let the set of outcomes of the experiment under study be described as a set of P points of some "geometric continuum" - each outcome corresponds to a certain point and each point corresponds to a certain outcome. The “geometric continuum” Q can be a segment on a straight line, an arc of a rectifiable curve on a plane or in space, a squaring set on a plane (triangle, rectangle, circle, ellipse, etc.) or a part of a squaring surface, some volume in space ( a polyhedron - a prism, a pyramid, a ball, an ellipsoid, etc.) An event is any squaring subset of a set. (length, area, volume) we can measure. Assuming the equiprobability of outcomes, let's call the probability of the event A a number proportional to the measure of the subset A of the set P: Geometric probabilities in this case will be between zero - the probability of an impossible event, and one - the probability of a reliable event4*. The normalization condition allows you to find the constant k - the coefficient of proportionality that sets the probability. It turns out to be equal to Thus, in the scheme of geometric probabilities, the probability of any event is defined as the ratio of the measure of the subset A, describing the event, to the measure of the set il, describing the experiment as a whole: contained within another cannot be greater than the latter. As in the scheme of cases, events in the scheme of geometric probabilities can be combined, combined and built on their basis of opposite ones - in this case, generally speaking, events different from the original events will be obtained. The next property is very important. 3. If the events are incompatible, then, in particular, the principle of complementarity is valid: This property, usually called the rule of addition of probabilities, obviously follows from the additivity of the measure5*. In conclusion, we note that the probability of any outcome in the scheme of geometric probabilities is always equal to zero, as well as the probability of any event described by a “skinny” set of points, i.e. set, the measure of which (respectively - length, area, volume) is equal to zero. Let us consider several examples illustrating the calculation of probabilities in the scheme of geometric probabilities. Example 1. The experiment consists in random selection of a point from the segment [a, 6|. Find the probability that the selected point lies in the left half of the considered segment. 4 By definition, the probability of choosing a point from any set on a segment is greater than zero, and their product is negative.
Answer: 0;25.
4.6. During combat training, the n-th bomber squadron received the task of attacking the “enemy” oil depot. On the territory of the oil depot, which has the shape of a rectangle with sides of 30 and 50 m, there are four round oil tanks with a diameter of 10 m each. Find the probability of a direct hit of the oil tanks by a bomb that hit the territory of the oil depot, if the bomb hits any point of this base with equal probability.
Answer: π/15.
4.7. Two real numbers x and y are chosen at random so that the sum of their squares is less than 100. What is the probability that the sum of the squares of these numbers is greater than 64?
Answer: 0;36.
4.8. The two friends agreed to meet between 13:00 and 14:00. The first person to arrive waits for the second person for 20 minutes and then leaves. Determine the probability of meeting friends if the moments of their arrival in the specified time interval are equally probable.
Answer: 5/9.
4.9. Two steamboats must come to the same pier. The time of arrival of both ships is equally possible during the given day. Determine the probability that one of the steamers will have to wait for the berth to be released if the first steamer stays for one hour and the second for two hours.
Answer: ≈ 0;121.
4.10. Two positive numbers x and y are taken at random, each of which does not exceed two. Find the probability that the product x y is at most one and the quotient y/x is at most two.
Answer: ≈ 0;38.
4.11. In the region G bounded by the ellipsoid 
, a point is fixed at random. What is the probability that the coordinates (x; y; z) of this point will satisfy the inequality x 2 + y 2 + z 2 ≤4?
Answer: 1/3.
4.12. A point is thrown into a rectangle with vertices R(-2;0), L(-2;9), M (4;9), N (4;0). Find the probability that its coordinates will satisfy the inequalities 0 ≤ y ≤ 2x – x 2 +8.
Answer: 2/3.
4.13. The region G is bounded by the circle x 2 + y 2 = 25, and the region g is bounded by this circle and the parabola 16x - 3y 2 > 0. Find the probability of falling into the region g.
Answer: ≈ 0;346.
4.14. Two positive numbers x and y are taken at random, each of which does not exceed one. Find the probability that the sum x + y does not exceed 1 and the product x · y is not less than 0.09.
Answer: ≈ 0;198.
Statistical definition of probability
Task 2. The shooter fires one shot at the target. Estimate the probability that he will hit the target.
Decision. In this experiment, two outcomes are possible: either the shooter hit the target (the event A), or he missed (event). Events 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 a sufficiently large number of shots at the target in 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 a 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. Decision. 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. Decision. 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 the other 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. Decision. 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. Decision. 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). Decision. 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. Decision. 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, = , = , = . Thus, 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 items. What is the probability that it was produced by machine II? Decision. 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). Events 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 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 indicated 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. Decision. In this example, the same experience (choosing shoes) is performed 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 the event A in each trial decreases indefinitely (p), but in such a way that their product pr is a constant value (pr = a = 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 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. Decision. 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 1,000 bottles of 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. Decision. 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 of random variables with very different distribution laws leads to the normal distribution of this sum. 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 the distribution of the individual components, the distribution of the resulting consumption will be close to normal. |
