Theory of probability random events examples. Probability of an event

1.1. Some information from combinatorics

1.1.1. Accommodation

Consider the simplest concepts associated with the choice and location of a certain set of objects.
Counting the number of methods that these actions can be made are often produced when solving probabilistic tasks.
Definition. Accommodation out n. Elements in k. (k. ≤ N.) called any ordered subset of k.the elements of the set consisting of n. various elements.
Example.The following sequences of numbers are placed 2 elements of 3 sets of set (1; 2; 3): 12, 13, 23, 21, 31, 32.
Note that the placement is characterized by the procedure for the elements included in them and their composition. Placing 12 and 21 contain the same numbers, but the order of their location is different. Therefore, these accommodations are considered different.
The number of different accommodations from n. Elements in k. It is indicated and calculated by the formula:
,
Where n.! = 1∙2∙...∙(n. - 1)∙ N. (read " n. - factorial ").
The number of two-digit numbers that can be made from numbers 1, 2, 3, provided that no digit is repeated equal to :.

1.1.2. Rearranged

Definition. Permutations from n. elements are called such accommodations from n. Elements that differ only in the location of the elements.
The number of permutations is n. Elements P N. Calculated by the formula: P N.=n.!
Example.How many ways can there be a queue 5 people? The number of ways is equal to the number of permutations of 5 elements, i.e.
P. 5 =5!=1∙2∙3∙4∙5=120.
Definition. If among n. Elements k. the same, then the permutation of these n.elements are called rearrangement with repetitions.
Example.Let among 6 books 2 are the same. Any location of all books on the shelf - rearrangement with repetitions.
The number of different rearrangements with repetitions (from n. elements among which k.the same) is calculated by the formula :.
In our example, the number of ways that can be placed on the shelf, as well as :.

1.1.3. Combination

Definition . Combinations from n. Elements in k. are called such accommodations from n. Elements in k.which one of the other differ at least one element.
The number of different combinations from n. Elements in k. It is indicated and calculated by the formula :.
By definition 0! \u003d 1.
For combinations, the following properties are valid:
1.
2.
3.
4.
Example. There are 5 flowers of different colors. For a bouquet of 3 flower selected. The number of different bouquets of 3 flower out of 5 is :.

1.2. Random events

1.2.1. Events

The knowledge of reality in natural sciences occurs as a result of tests (experiment, observations, experience).
Test or experience is called the implementation of a certain set of conditions that can be reproduced arbitrarily a large number of times.
Random It is called an event that may occur or not happen as a result of a certain test (experience).
Thus, the event is considered as a test result.
Example. Throwing coins is a test. The appearance of an eagle when throwing - an event.
The events observed by us differ in the degree of possibility of their appearance and by the nature of their relationship.
The event is called reliable if it necessarily occurs as a result of this test.
Example. Obtaining a positive or negative assessment student on the exam is a reliable event if the exam proceeds according to the usual rules.
The event is called impossible If it cannot happen as a result of this test.
Example. Removing the white ball born, in which there are only colored (non-cheese) balls, there is an event impossible. Note that under other conditions, the experience of the appearance of a white ball is not excluded; Thus, this event is impossible only in the conditions of our experience.
Further, random events will be denoted by large Latin letters a, b, c ... a reliable event will be denoted by the letter Ω, the impossible - Ø.
Two or more events are called equal possible In this test, if there is reason to believe that none of these events are more possible or less possible than others.
Example.With one throwing a playing bone, the appearance of 1, 2, 3, 4, 5 and 6 points - all these events are equilibrium. Of course, the playing bone is made of homogeneous material and has the correct form.
Two events are called non-beds in this test, if one of them eliminates the appearance of another, and joint otherwise.
Example. There are standard and non-standard details in the box. We take one detail for good luck. The appearance of the standard part eliminates the appearance of a non-standard part. These events are incomplete.
Several events form full group of events In this test, if, as a result of this test, at least one of them will come.
Example.Events from the example form a complete group of equal and pairwise incomplete events.
Two incomplete events forming a complete group of events in this test are called opposite events.
If one of them is indicated through A., then another is customary to designate through (read "not A.»).
Example. Intelligence and misses at one point shot - events are opposite.

1.2.2. Classical probability definition

Probability of an event - Numerical measure of the possibility of its offensive.
Event BUT called favorable Event INif every time an event occurs BUToccurs the event IN.
Events BUT 1 , BUT 2 , ..., BUT N. Form scheme of cases , if they:
1) equilibrium;
2) in pairs are inconsistent;
3) form a complete group.
In the scheme of cases (and only in this scheme) there is a classic probability definition P.(A.) events BUT. Here, the case is called each of the events belonging to the allocated complete group of equivalence and pairwise incomplete events.
If a n. - the number of all cases in the scheme, and m. - the number of cases conducive to events BUTT. probability of an event BUT Determined by equality:

From the definition of the probability, the following properties flow:
1. The probability of a reliable event is equal to one.
Indeed, if the event is reliably, each case in the case scheme favors the event. In this case m. = n. And, therefore,

2. The probability of an impossible event is zero.
Indeed, if the event is impossible, no case from the case scheme does not favor an event. therefore m.\u003d 0 and therefore

The probability of a random event is a positive number concluded between zero and unit.
Indeed, only a part of the total incidence in the scheme of cases is conducive to a random event. Therefore 0.<m.<n., so, then 0<m./n.<1 и, следовательно, 0 < P (A) < 1.
So, the probability of any event satisfies inequalities
0 ≤ P (A) ≤ 1.
Currently, the probability properties are determined in the form of an axiom formulated by A.N. Kolmogorov
One of the main advantages of the classic probability determination is the ability to calculate the likelihood of an event directly, i.e. Without resorting to experiments that replace logical reasoning.

Tasks of direct probability calculation

Task 1.1.. What is the probability of the emergence of an even number of points (Event a) with one throwing a playing cube?
Decision. Consider events BUT I. - fell i. glasses i.\u003d 1, 2, ..., 6. Obviously, these events form the scheme of cases. Then the number of all cases n. \u003d 6. The experiment of an even number of points is favored by cases BUT 2 , BUT 4 , BUT 6, i.e. m.\u003d 3. then .
Task 1.2.. In the urn of 5 white and 10 black balls. The balls are thoroughly mixed and then rain 1 ball. What is the likelihood that the revealed ball will be white?
Decision. In total there are 15 cases that form the scheme of cases. And the expected event BUT - the appearance of a white bowl, 5 of them favors, so .
Task 1.3.. The child plays with six letters of the alphabet: a, a, e, k, r, T. Find the likelihood that he will be able to fold the challenge of the carriage (Event a).
Decision. The decision is complicated by the fact that among the letters there are the same - two letters "A". Therefore, the number of all possible cases in this test is equal to the number of permutations with repetitions of 6 letters:
.
These cases are equal, in pairs are inconsistent and form a complete group of events, i.e. Form the scheme of cases. Only one case favors the event BUT. therefore
.
Task 1.4.. Tanya and Vanya agreed to celebrate the New Year in the company of 10 people. They both really wanted to sit nearby. What is the likelihood of their desire, if there is a place among their friends to distribute by lots?
Decision. Denote by BUT Event "Execution of the desire of Tanya and Vanya". 10 people can pressent at the table 10! different ways. How many of these n. \u003d 10! equal ways are favorable for Tanya and Vanya? Tanya and Vanya sitting nearby, can take 20 different positions. At the same time, the eight of their friends can sit at the table 8! in different ways, so m. \u003d 20 ∙ 8!. Hence,
.
Task 1.5.. A group of 5 women and 20 men chooses three delegates. Considering that each of those present with the same probability can be selected, find the likelihood that they will choose two women and one man.
Decision. The total number of equilibrium test outcomes is equal to the number of ways that you can choose three delegates from 25 people, i.e. . We now calculate the number of favored cases, i.e. The number of cases in which the event that interests us is. The delegate can be chosen twenty ways. At the same time, the remaining two delegates should be women, and you can choose two women from five. Hence, . therefore
.
Task 1.6. Four balls randomly scatter on the four holes, each ball enters that or another well with the same probability and regardless of others (obstacles to getting into the same one and the same well of several balls). Find the chance that three balls will be in one of the wells, to the other one, and there will be no balls in the other left holes.
Decision. The total number of cases n.\u003d 4 4. The number of ways to choose one hole where there are three balls ,. The number of ways to choose a well where one ball will be ,. The number of ways to choose from four balls three to put them in the first hole ,. The total number of favorable cases. Event probability:
Task 1.7.In the drawer 10 of the same balls marked with numbers 1, 2, ..., 10. Six balls are extracted for good luck. Find the likelihood that among the extracted balls will be: a) Ball number 1; b) Balls №1 and №2.
Decision. a) the total number of possible elementary test outcomes is equal to the number of methods that can be removed six balls out of ten, i.e.
We will find the number of outcomes conducive to the event you are interested in: among the selected six balls there are ball number 1 and, therefore, the other five balls have other rooms. The number of such outcomes is obviously equal to the number of ways to select five balls from the remaining nine, i.e.
The desired probability is equal to the ratio of the number of outcomes, conducive to the event under consideration, to the total number of possible elementary outcomes:
b) the number of outcomes conducive to the event you are interested in (among selected balls there are balls No. 1 and No. 2, therefore four balls have other numbers), equal to the number of ways that you can remove four balls from the remaining eight, i.e. Saying probability

1.2.3. Statistical probability

The statistical definition of the probability is used in the case when the outcomes of experience are not equal.
Relative event frequency BUT Determined by equality:
,
Where m. - the number of tests in which the event BUT came n. - Total number of tests.
Ya. Bernoulli proved that with an unlimited increase in the number of experiments, the relative frequency of the event will almost be different from a certain constant number. It turned out that this constant number is the likelihood of an event. Therefore, naturally, the relative frequency of the event at a sufficiently large number of tests is called a statistical probability in contrast to the previously introduced probability.
Example 1.8.. How to approximately set the number of fish in the lake?
Let in the lake h. Fish. Throw the network and allow you to find in it n. Fish. Each of them is methim and release back. A few days later in the same weather and in the same place we throw the same network. Suppose that we find M fish in it, among which k. labeled. Let the event BUT - "Caught Fish Lady." Then to determine the relative frequency.
But if in the lake h. fish and we released it n. labeled, then.
As R * (BUT) » R(BUT), Then.

1.2.4. Operations on events. Probability addition theorem

Sum, or association, several events are called an event consisting in the occurrence of at least one of these events (in the same test).
Sum BUT 1 + BUT 2 + … + BUT N. designated like this:
or .
Example. Two playing bones rush. Let the event BUT It consists in falling out 4 points on 1 bone, and an event IN - in falling out 5 points on another bone. Events BUT and IN jointly. Therefore event BUT +IN It consists in falling out 4 points on the first bone, or 5 points on the second bone, or 4 points on the first bone and 5 points on the second simultaneously.
Example. Event BUT - win 1 loan, event IN - Winning 2 loans. Then event A + B. - Winning at least one loan (perhaps on two immediately).
Work Or the intersection of several events is the event consisting in the joint appearance of all these events (in the same test).
Composition IN events BUT 1 , BUT 2 , …, BUT N. designated like this:
.
Example. Events BUT and IN Consist in successful passage of I and II tours, respectively, when entering the institute. Then event BUT× B. It consists in the successful passage of both tours.
The concepts of the amount and work of events have a visual geometric interpretation. Let the event BUT There is a point in the area BUT, and event IN - getting points to the area IN. Then event A + B. There is a point from entering the combination of these areas (Fig. 2.1), and the event BUTIN There is a point in the intersection of these areas (Fig. 2.2).

Fig. 2.1 Fig. 2.2.
Theorem. If events A I.(i. = 1, 2, …, n.) In pairs are inconsistent, the probability of the amount of events is equal to the sum of the probabilities of these events:
.
Let be BUT and Ā - opposite events, i.e. A + ā. \u003d Ω, where ω is a reliable event. From the addition theorem, it follows that
P (Ω) \u003d R(BUT) + R(Ā ) \u003d 1, so
R(Ā ) = 1 – R(BUT).
If events BUT 1 I. BUT 2 are together, the probability of the sum of two joint events is:
R(BUT 1 + BUT 2) = R(BUT 1) + R(BUT 2) - P ( BUT 1 × BUT 2).
Probability addition theorems allow you to move from directly calculating probabilities to determine the probability of occurrence of complex events.
Task 1.8.. The shooter produces one target shot. Probability knock out 10 points (event BUT), 9 points (event IN) and 8 points (event FROM) are equal, respectively, 0.11; 0.23; 0.17. Find the likelihood that with one shooter shooter will choose less than 8 points (Event D.).
Decision. Let us turn to the opposite event - with one shot shooter, there will take at least 8 points. Event comes if happens BUT or IN, or FROM. . Since events A, B., FROM in pairs are inconsistent, then, by the addition theorem,
From where.
Task 1.9.. From the team of a brigade, which consists of 6 men and 4 women, two people are chosen for the trade union conference. What is the likelihood that among the selected at least one woman (event BUT).
Decision. If an event occurs BUTThis will definitely happen one of the following incomplete events: IN - "Male and woman chosen"; FROM - "Two women chosen." Therefore, you can write: A \u003d B + C. Find the likelihood of events IN and FROM. Two people from 10 can be chosen in ways. Two women from 4 can be chosen in ways. Man and woman can choose 6 × 4 ways. Then. Since events IN and FROM inconsistent, then, by the addition theorem,
P (a) \u003d p (B + C) \u003d P (B) + P (with) = 8/15 + 2/15 = 2/3.
Task 1.10. On the rack in the library in random order, 15 textbooks are placed, and five of them are in the binding. The librarian takes the first textbook. Find the likelihood that at least one of the tanted textbooks will be in the binding (event BUT).
Decision. The first way. The requirement is at least one of the three binding textbooks - will be implemented if any of the following three inconsistent events occurs: IN - one binding tutorial FROM - Two binding textbooks, D. - Three binding textbooks.
Event you are interested in BUT You can imagine in the form of the amount of events: A \u003d b + c + d. By the addition theorem,
P (a) \u003d p (b) + p (c) + p (d). (2.1)
Find the likelihood of events B, C. and D. (See combinatorial schemes):

Presenting these probabilities into equality (2.1), finally get
P (A)= 45/91 + 20/91 + 2/91 = 67/91.
The second way. Event BUT (at least one of the three textbooks taken is a binding) and Ā (None of the tangled textbooks has binding) - opposite, therefore P (a) + p (ā) \u003d 1 (the sum of the probabilities of two opposite events is 1). From here P (A.) = 1 – P (ā). The likelihood of event appearance Ā (None of the tanted textbooks has binding)
Saying probability
P (A.) = 1 - P (ā) = 1 – 24/91 = 67/91.

1.2.5. Conditional probability. Probability multiplication theorem

Conditional probability P (B./BUT) It is called the likelihood of an event in calculated in the assumption that the event has already arrived.
Theorem. The probability of the joint appearance of two events is equal to the product of the probabilities of one of them on the conditional likelihood of the other, calculated in the assumption that the first event has already arrived:
P (A.∙C) \u003d p (a) ∙ P ( IN/BUT). (2.2)
Two events are called independent if the appearance of any of them does not change the likelihood of the other, i.e.
P (a) \u003d p (a / in) or P (B.) = P (B./BUT). (2.3)
If events BUT and IN Independent, then from formulas (2.2) and (2.3) follows
P (A.∙C) \u003d p (a)∙P (B.). (2.4)
Fair and reverse statement, i.e. If equality (2.4) is performed for two events, these events are independent. In fact, from formulas (2.4) and (2.2) flows
P (A.∙C) \u003d p (a)∙P (B.) = P (A.) × P (B./BUT), From P (A.) = P (B./BUT).
Formula (2.2) admits a generalization in case of a finite number of events BUT 1 , BUT 2 ,…,A N.:
P (A. 1 ∙BUT 2 ∙…∙A N.)=P (A. 1)∙P (A. 2 /BUT 1)∙P (A. 3 /BUT 1 BUT 2)∙…∙P (and n/BUT 1 BUT 2 …A N. -1).
Task 1.11. From the urn, in which 5 white and 10 black balls, take out two balls in a row. Find the chance that both white balls (event BUT).
Decision . Consider events: IN - the first revealed ball white; FROM - The second revealed ball white. Then A \u003d Sun..
Experience can be held in two ways:
1) WITH RETURN: The revealed ball after fixing the color returns to the urn. In this case, events IN and FROMindependent:
P (a) \u003d p (in)∙P (S.) \u003d 5/15 × 5/15 \u003d 1/9;
2) without return: the bowl is deposited to the side. In this case, events IN and FROM dependent:
P (a) \u003d p (in)∙P (S./IN).
For event IN conditions former, and for FROM the situation has changed. Occurred INTherefore 14 balls remained in the urn, among which are 4 whites.
So, .
Task 1.12.. Among the 50 light bulbs 3 non-standard. Find the likelihood that two non-standard light bulbs taken at the same time.
Decision . Consider events: BUT - the first lamp is non-standard, IN - the second lamp is non-standard, FROM - Both light bulbs. It's clear that C \u003d A.∙IN. Event BUT 3 cases are favorable from 50 possible, i.e. P (A.) \u003d 3/50. If the event BUT has already come, then the event IN Two cases favorably out of 49 possible, i.e. P (B./BUT) \u003d 2/49. Hence,
.
Task 1.13 . Two athletes independently shoot one target. The probability of hitting the target of the first athlete is 0.7, and the second is 0.8. What is the probability that the target will be amazed?
Decision . The target will be amazed if either the first arrows will fall into it, or the second, or both together, i.e. Event will occur A + B.Where event BUT lies in the first athlete in the target, and the event IN - Second. Then
P (A.+IN)=P (A.)+P (B.)–P (A.∙IN)=0, 7+0, 8–0, 7∙0,8=0,94.
Task 1.14.In the reading room there are six textbooks on the theory of probabilities, of which are three in the binding. The librarian of the muddy took two textbooks. Find the likelihood that two textbooks will be in the binding.
Decision. We introduce the designations of events : A. - The first tutorial has a binding, IN - The second textbook has a binding. The likelihood that the first textbook has a binding,
P (A.) = 3/6 = 1/2.
The likelihood that the second textbook has a binding, provided that the first tutorial was in the binding, i.e. Conditional probability of an event INSuch is: P (B./BUT) = 2/5.
The desired probability that both textbooks have a binding, on the multiplication theorem of the events of events is equal to
P (AB) = P (A.) ∙ P (B./BUT) \u003d 1/2 · ∙ 2/5 \u003d 0.2.
Task 1.15. In the workshop there are 7 men and 3 women. Three people were selected on the tablet numbers. Find the likelihood that all selected individuals will be men.
Decision. We introduce event notation: A. - the first is the man, IN - the second is selected a man, FROM - Third selected man. The likelihood that the man will be the first to be selected, P (A.) = 7/10.
The likelihood that the second is selected by a man, provided that the man was already the first were already selected, i.e. Conditional probability of an event IN Next : P (b / a) = 6/9 = 2/3.
The likelihood that the third will be selected by a man, provided that two men have already been selected, i.e. Conditional probability of an event FROM Such is: P (C./AU) = 5/8.
The desired probability that all three selected individuals will be men, P (ABC) \u003d P (a) P (B./BUT) P (C./AU) \u003d 7/10 · 2/3 · 5/8 \u003d 7/24.

1.2.6. Formula of the full probability and Bayes formula

Let be B. 1 , B. 2 ,…, B N. - pairs of incomplete events (hypotheses) and BUT - An event that can happen only together with one of them.
Let, besides, we know P (B i) I. P (A./B I.) (i. = 1, 2, …, n.).
Under these conditions, Formulas are valid:
(2.5)
(2.6)
Formula (2.5) is called formula full probability . It calculates the likelihood of an event. BUT (full probability).
Formula (2.6) is called bayes formula . It allows you to restore the probabilities of the hypotheses if an event BUT occurred.
In the preparation of examples, it is convenient to assume that the hypothesis form a complete group.
Task 1.16.. In the basket apples with four trees of one variety. From the first - 15% of all apples, from the second - 35%, from the third - 20%, from the fourth - 30%. Ripe apples are 99%, 97%, 98%, 95%, respectively.
a) What is the likelihood that at random the apple will be ripe (event BUT).
b) Provided that at random, the apple was ripe, calculate the likelihood that it from the first tree.
Decision. a) We have 4 hypotheses:
B 1 - At ragged apple taken from the 1st tree;
B 2 - At ragged apple taken from the 2nd tree;
B 3 - At ragged apple taken from the 3rd tree;
B 4 - At ragged apple taken from the 4th tree.
Their probabilities under the condition: P (B. 1) = 0,15; P (B. 2) = 0,35; P (B. 3) = 0,2; P (B. 4) = 0,3.
Conditional probabilities event BUT:
P (A./B. 1) = 0,99; P (A./B. 2) = 0,97; P (A./B. 3) = 0,98; P (A./B. 4) = 0,95.
The likelihood that the border taken apple will be ripe, is in full probability formula:
P (A.)=P (B. 1)∙P (A./B. 1)+P (B. 2)∙P (A./B. 2)+P (B. 3)∙P (A./B. 3)+P (B. 4)∙P (A./B. 4)=0,969.
b) Bayes formula for our case has the form:
.
Task 1.17. In the urn, containing two balls, lowered a white ball, after which one ball was removed from it. Finding the likelihood that the removed ball will be white if all possible assumptions about the original composition of the balls (in color) are equal.
Decision. Denote by BUT Event - white ball extracted. The following assumptions (hypotheses) are possible on the initial composition of the balls: B 1. - No white balls, AT 2 - one white ball IN 3 - Two white balls.
Since all there are three hypotheses, and the probability of the hypotheses is 1 (as they form a complete group of events), then the probability of each of the hypotheses is 1/3, that is.
P (B. 1) = P (B. 2) \u003d P (B 3) = 1/3.
The conditional probability that the white ball will be extracted, provided that it was originally white balls in the urn, P (A./B. 1) \u003d 1/3. The conditional probability that a white ball will be extracted, provided that initially in the urn was one white ball, P (A./B. 2) \u003d 2/3. The conditional probability that the white ball will be extracted, provided that the original balls were initially in the urn P (A./B. 3)=3/ 3=1.
The desired probability that a white ball will be removed, we find the formula for a complete probability:
R(BUT)=P (B. 1)∙P (A./B. 1)+P (B. 2)∙P (A./B. 2)+P (B. 3)∙P (A./B. 3) \u003d 1/3 · 1/3 + 1/3 · 2/3 + 1/3 · 1 \u003d 2/3 .
Task 1.18.. Two automaton produce the same details that come to the general conveyor. The performance of the first machine twice the performance of the second. The first automatic machine produces an average of 60% of the details of excellent quality, and the second is 84%. Magnifying from the conveyor, the detail turned out to be excellent quality. Find the chance that this item is made by the first machine.
Decision. Denote by BUT Event - Detail of excellent quality. You can make two assumptions: B 1. - The part is made by the first automaton, and since the first machine produces twice the details than the second) P (A./B. 1) = 2/3; B. 2 - the part is made by the second automaton, and P (B. 2) = 1/3.
The conditional probability that the item will be excellent quality if it is made by the first machine, P (A./B. 1)=0,6.
The conditional probability that the item will be excellent quality if it is made by second automatics, P (A./B. 1)=0,84.
The likelihood that the border taken the detail will be excellent quality, according to the formula of a complete probability is equal to
P (A.)=P (B. 1) ∙P (A./B. 1)+P (B. 2) ∙P (A./B. 2) \u003d 2/3 · 0.6 + 1/3 · 0.84 \u003d 0.68.
The desired probability that the excellent item taken is made by the first automaton, the Bayes formula is equal to

Task 1.19. There are three parties of parts for 20 parts each. The number of standard parts in the first, second and third parties is equal to 20, 15, 10. From the selected batch, the detail that turned out to be standard is removed. Details are returned to the party and the second time from the same batch, the detail is retrieved, which also turns out to be standard. Find the probability that the details were extracted from the third party.
Decision. Denote by BUT Event - In each of the two tests (with the return), a standard part was retrieved. You can make three assumptions (hypotheses): B. 1 - details are extracted from the first batch, IN 2 - Details are extracted from the second game, IN 3 - Details are extracted from the third party.
Details were removed by a border from the batch taken, therefore the probabilities of the hypothesis are the same: P (B. 1) = P (B. 2) = P (B. 3) = 1/3.
Find a conditional probability P (A./B. 1), i.e. The likelihood that two standard parts will be consistently retrieved from the first batch. This event is reliably, because In the first batch, all the details are standard, so P (A./B. 1) = 1.
Find a conditional probability P (A./B. 2), i.e. The likelihood that from the second batch will be consistently extracted (with the return) two standard details: P (A./B. 2)= 15/20 ∙ 15/20 = 9/16.
Find a conditional probability P (A./B. 3), i.e. The likelihood that from the third party will be consistently extracted (with the return) two standard details: P (A./B. 3) \u003d 10/20 · 10/20 \u003d 1/4.
The desired probability that both extracted standard details are taken from the third party, by the Bayes formula is equal to

1.2.7. Repeated tests

If there are several tests, the probability of an event BUTin each test, it does not depend on the outcomes of other tests, then such tests are called independent on Event A. In different independent tests event BUTmay have either different probabilities or the same probability. Will further consider only such independent tests in which the event BUTit has one same probability.
Let it be produced pindependent tests in each of which is an event BUTmay appear either do not appear. Agree to believe that the probability of an event BUTin each test, the same, namely, is equal to r.Consequently, the likelihood of unactering an event BUTin each test is also constant and equal to 1- r. Such a probabilistic scheme is called bernoulli scheme. We set yourself the task to calculate the likelihood that ptests on the Bernoulli Scheme Event BUT Exact Rivne k. Once ( k. - the number of success) and, therefore, will not p- time. It is important to emphasize that it is not required to ever BUTrepeated exactly k. Once in a certain sequence. The desired probability is denoted P P (k). For example, symbol R 5 (3) means the likelihood that in five tests the event will appear exactly 3 times and, therefore, will not occur 2 times.
The task can be solved using the so-called bernoulli Formulas which has the form:
.
Task 1.20.The likelihood that electricity consumption in the continuation of one day will not exceed the established norm, is equal to r\u003d 0.75. Find the chance that in the next 6 days, electricity consumption for 4 days will not exceed the norm.
Decision. The probability of a normal consumption of electricity in continuation of each of the 6 days is constant and equal r\u003d 0.75. Consequently, the probability of electricity recalculation every day is also constant and equal to q \u003d1–r=1–0,75=0,25.
The desired probability by the Bernoulli formula is equal
.
Task 1.21. Two equivalent chess players play chess. What is more likely: to win two batches of four or three parties out of six (not taken into account)?
Decision. Play equivalent chess players, so the probability of winning r \u003d 1/2, therefore, the probability of losing q. Also equal to 1/2. Because In all parties, the probability of winning is constant and indifferent, in what sequence the party will be won, the formula of Bernoulli will be applied.
We find the likelihood that two parties from four will be won:

We find the likelihood that three parties from six will be won:

Because P. 4 (2) > P. 6 (3), it is likely to win two parties of four than three of six.
One-like to see that to use Bernoulli formula for large values n. It is quite difficult, since the formula requires actions on enormous numbers and therefore in the process of calculations accumulate errors; As a result, the final result can differ significantly from the true one.
To solve this problem, there are several limit theorems that are used for the case of a large number of tests.
1. Poisson theorem
When carrying out a large number of tests according to the Bernoulli scheme (when n. \u003d\u003e ∞) and with a small number of favorable outcomes k. (It is assumed that the probability of success p. Mala), Bernoulli formula is approaching the Formula of Poisson
.
Example 1.22. The probability of marriage when producing an enterprise unit of products is equal p.\u003d 0.001. What is the probability that the production of 5,000 units of products will be less than 4 defective (Event BUT Decision. Because n. Great, we use the local Laplace Theorem:

Calculate x.:
Function - even, therefore φ (-1.67) \u003d φ (1.67).
According to the table of Appendix, paragraph 1, we find φ (1.67) \u003d 0.0989.
Saying probability P. 2400 (1400) = 0,0989.
3. Integral Laplace Theorem
If probability r Event appearance A. In each test according to the Bernoulli scheme constant and different from zero and units, then with a large number of tests n. probability P P (k 1 K. 2) events A. In these tests from k. 1 BE k. 2 times approximately equal
P P.(k. 1 K. 2) \u003d φ ( x "") – Φ ( x "), where
- Laplace function,

A specific integral in the Laplace function is not calculated on the class of analytical functions, so it is used to calculate it. P.2, shown in the application.
Example 1.24.The likelihood of an event in each of the hundred independent tests is constant and equal p. \u003d 0.8. Find the likelihood that the event will appear: a) at least 75 times and no more than 90 times; b) at least 75 times; c) not more than 74 times.
Decision. We use the Laplace integral theorem:
P P.(k. 1 K. 2) \u003d φ ( x "") – Φ( x "), where f ( x.) - Laplace function,

a) under the condition n. = 100, p. = 0,8, q. = 0,2, k. 1 = 75, k. 2 \u003d 90. Calculate x "" and x " :


Considering that the Laplace function is odd, i.e. F (- x.) \u003d - f ( X.), we get
P. 100 (75; 90) \u003d F (2.5) - F (-1.25) \u003d Φ (2.5) + f (1.25).
Table. P. Applications will find:
F (2.5) \u003d 0.4938; F (1.25) \u003d 0.3944.
Saying probability
P. 100 (75; 90) = 0,4938 + 0,3944 = 0,8882.
b) the requirement that the event appears at least 75 times, indicates that the number of events may be 75, or 76, ..., or 100. So, in the case under consideration should be taken k. 1 = 75K. 2 \u003d 100. Then

.
Table. P. applications will find F (1.25) \u003d 0.3944; F (5) \u003d 0.5.
Saying probability
P. 100 (75;100) = (5) – (–1,25) = (5) + (1,25) = 0,5 + 0,3944 = 0,8944.
c) event - " BUT appeared at least 75 times "and" BUT no more than 74 times appeared "opposite, therefore the sum of the probabilities of these events is 1. Consequently, the desired probability
P. 100 (0;74) = 1 – P. 100 (75; 100) = 1 – 0,8944 = 0,1056.

Theory of probability is a rather extensive independent section of mathematics. In the school year, the theory of probability is considered very superficially, however, there are tasks for this topic. However, it is not so difficult to solve the tasks of the school course (at least what concerns arithmetic operations) - here you do not need to consider derivatives, take integrals and solve complex trigonometric transformations - the main thing is to be able to handle simple numbers and fractions.

Probability Theory - Basic Terms

The main terms of the theory of probability are testing, an outcome and a random event. The test in the theory of probability is called an experiment - to throw a coin, pull the card, draw the draw - all this tests. The result of the test, as you already guessed, is called an outcome.

What is the event random? In the theory of probability it is assumed that the test is carried out by many times a lot of outcomes. Random event is called a lot of test outcomes. For example, if you throw a coin, two random events may occur - eagle or rush falls.

Do not confuse the outcome and random event. The outcome is one result of one test. Random event is a variety of possible outcomes. There is, by the way, and such a term as an impossible event. For example, the Event "Dropped Number 8" on the standard game cube is impossible.

How to find a probability?

We all understand what is probability, and quite often use this word in your vocabulary. In addition, we can even make some conclusions regarding the likelihood of a particular event, for example, if behind the snow window, we can be likely to say that now is not summer. However, how to express this assumption numerically?

In order to introduce the formula for finding a probability, we introduce another concept - a favorable outcome, i.e. the outcome that is favorable for a particular event. The definition is rather ambiguous, of course, however, by the condition of the problem, it is always clear which of the outcomes is favorable.

For example: in class 25 people, three of them Kati. The teacher appoints Olya duty, and she needs a partner. What is the likelihood that the partner will Katya become?

In this example, a favorable outcome - Katya's partner. A little later we will solve this task. But first we introduce with the help of an additional definition formula for finding a probability.

• P \u003d a / n, where p is the probability, A is the number of favorable outcomes, N is the total number of outcomes.

All school challenges are spinning around one of this formula, and the main difficulty usually consists in finding outcomes. Sometimes they are simple to find, sometimes - not very.

How to solve tasks for the likelihood?

Task 1.

So, now let's decide the above task.

The number of favorable outcomes (the teacher chooses Katya) is equal to three, because Cat in class three, and total outcomes - 24 (25-1, because Olya is already chosen). Then the probability is equal: p \u003d 3/2 \u003d 1/8 \u003d 0.125. Thus, the likelihood that Katya will turn out to be 12.5%. Is it easy? Let's wonder something more comprehensive.

Task 2.

The coin was thrown twice, what is the likelihood of a combination: one eagle and one rush?

So, we consider total outcomes. How can the coins fall out - Eagle / Eagle, Rushka / Rushka, Eagle / Rush, Rushka / Eagle? So, the total number of outcomes - 4. How many favorable outcomes? Two - eagle / rush and rush / eagle. Thus, the probability of a combination of an eagle / rush is equal to:

• P \u003d 2/4 \u003d 0.5 or 50 percent.

And now consider such a task. Masha in the pocket of 6 coins: two - a denomination of 5 rubles and four - a denomination of 10 rubles. Masha shifted 3 coins to another pocket. What is the likelihood that 5-ruble coins will be in different pockets?

For simplicity, we denote the coins with numbers - 1.2 - five-membered coins, 3,4,5,6 - ten-meter coins. So how can the coins lie in your pocket? There are 20 combinations in total:

• 123, 124, 125, 126, 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256, 345, 346, 356, 456.

At first glance, it may seem that some combinations are disappeared, for example, 231, however, in our case, combinations 123, 231 and 321 are equivalent.

Now we consider how much favorable outcomes we have. For them we take those combinations in which there are either a number 1, or a number 2: 134, 135, 136, 145, 146, 156, 234, 235, 236, 245, 246, 256. They are 12. Thus, the probability is equal to:

• P \u003d 12/20 \u003d 0.6 or 60%.

Tasks on the theory of probability presented here are quite simple, but do not think that the theory of probability is a simple section of mathematics. If you decide to continue education in the university (with the exception of humanitarian specialties), you will definitely have a couple of higher mathematics on which you will be familiar with more complex terms of this theory, and the tasks there will be much more difficult.

Brief theory

For a quantitative comparison of events in the degree of possibility of their appearance, a numerical measure is introduced, which is called the probability of an event. Probability of a random event The number that is an expression of the measure of an objective possibility of an event appearance is called.

The values \u200b\u200bthat determine how significant objective grounds are to count on the events are characterized by the probability of an event. It is necessary to emphasize that the probability is an objective value that is existing independently of the learning and due to the entire set of conditions that contribute to the emergence of an event.

The explanations that we gave the concept of probability are not a mathematical definition, since they do not determine this concept quantitatively. There are several definitions of the probability of random event that are widely used in solving specific tasks (classical, geometrical definition of probability, statistical, etc.).

Classic definition of the probability of an event Supports this concept to a more elementary concept of equilibrium events, which is no longer defined and is assumed to be intuitive. For example, if the playing bone is a homogeneous cube, then the fallout of any of the edges of this cube will be equal to events.

Let a reliable event disintegrate on equilibrium cases, the amount of which gives an event. That is, cases of which disintegrate are called conducive to the event, since the appearance of one of them provides an offensive.

The probability of events will be denoted by the symbol.

The probability of an event is equal to the ratio of the number of cases conducive to it, from the total number of the only possible, equal and inconsistencies to the number, i.e.

This is a classic probability definition. Thus, to find the probability of an event, it is necessary, having considered various outcomes of the test, to find the set of the only possible, equal and inconsistent cases, to calculate their total number N, the number of M cases conducive to this event, and then calculate the calculation according to the above formula.

The probability of an event equal to the ratio of the number of favorable events of the experience of experience to the total number of outcomes of experience is called classic probability Random event.

The determination flows the following probability properties:

Property 1. The probability of a reliable event is equal to one.

Property 2. The probability of an impossible event is zero.

Property 3. The probability of a random event is a positive number concluded between zero and unit.

Property 4. The probability of occurrence of events forming a complete group is equal to one.

Property 5 The likelihood of an opposite event is defined in the same way as the probability of an occurrence of event A.

The number of cases conducive to the appearance of the opposite event. From here the probability of the opposite event is equal to the difference between unit and the likelihood of event A:

An important advantage of the classical definition of the probability of an event is that with its help, the probability of an event can be determined without resorting to the experiment, and on the basis of logical reasoning.

When performing a complex of conditions, a reliable event will definitely happen, and the impossible will not necessarily occur. Among the events that, when creating a complex of conditions, can occur, and may not happen, on the appearance of some can count on a large base, to the appearance of others with a smaller base. If, for example, in the urn of white balls more than black, then hope for the appearance of a white bowl when removing from the urn a lot more reasons than on the appearance of a black bowl.

On the next page is considered.

An example of solving the problem

Example 1.

There are 8 white, 4 black and 7 red balls in the box. Route retrieved 3 balls. Find the probabilities of the following events: - at least 1 red ball is extracted - there are at least 2 balls of one color, - there is at least 1 red and 1 white ball.

The solution of the problem

The total number of test outcomes will find as a number of combinations of 19 (8 + 4 + 7) elements of 3:

Find the likelihood of an event - extracted at least 1 red ball (1.2 or 3 red balls)

The desired probability:

Let the event - There are at least 2 bowls of one color (2 or 3 white balls, 2 or 3 black balls and 2 or 3 red balls)

The number of outcomes conducive to events:

The desired probability:

Let the event - There is at least one red and 1 white ball

(1 red, 1 white, 1 black or 1 red, 2 white or 2 red, 1 white)

The number of outcomes conducive to events:

The desired probability:

Answer:P (a) \u003d 0.773; p (c) \u003d 0.7688; P (D) \u003d 0.6068

Example 2.

Two playing bones were thrown. Find the likelihood that the amount of points is not less than 5.

Decision

Let the event - the amount of points at least 5

We use the classic probability definition:

Total number of possible test outcomes

The number of tests conducive to the event you are interested in

On the fallen face of the first playing cube, one point may appear, two points ..., six points. Similarly, six outcomes are possible when throwing a second cube. Each of the outcomes of throwing the first dice can be combined with each of the outcomes of the second. Thus, the total number of possible elementary test outcomes is equal to the number of placements with repetitions (selection with placing 2 elements from a set of volume 6):

Find the likelihood of the opposite event - the amount of points is less than 5

Favorite the event will be the following combinations of glowing points:

The price strongly affects the urgency of the solution (from day to several hours). Online assistance on the exam / standings is carried out by appointment.

The application can be left directly in the chat, having previously throwing the condition of tasks and informing the decision you need. Answer time - a few minutes.

Initially, being just a meeting of information and empirical observations of the game in the bone, the theory of probability has become a solid science. The first who gave her mathematical framework was the farm and Pascal.

From thinking about eternal to probability theory

Two personalities who are obliged by many fundamental formulas, Blaise Pascal and Thomas Bayes, are known as deeply believers, the latter was the Presbyterian priest. Apparently, the desire of these two scientists to prove the fallacy of views on some kind of fortune, giving good luck to their pets, gave impetus to research in this area. After all, in fact, any gambling with its winnings and losses is just a symphony of mathematical principles.

Thanks to the Azart Cavaller, who was equally a player and a person who is not indifferent to science, Pascal was forced to find a way to calculate the probability. Deverage was interested in such a question: "How many times should you throw two bones in pairs so that the likelihood of getting 12 points exceeded 50%?". The second question is extremely interested in Cavallar: "How to share a bet between participants of an unfinished game?" Of course, Pascal successfully responded to both questions, which became the involuntary impetus for the development of probability theory. Interestingly, the person's person remained known in the art, and not in the literature.

Previously, no mathematician has yet made attempts to calculate the probabilities of events, since it was believed that this is just a gady decision. Blaise Pascal gave the first definition of the likelihood of an event and showed that this is a specific figure that can be justified with mathematical means. Probability Theory has become the basis for statistics and is widely used in modern science.

What is accidents

If we consider the test that you can repeat the infinite number of times, then you can define a random event. This is one of the probable outcomes of experience.

Experience is the implementation of concrete actions in constant conditions.

To work with the results of experience, events are usually denoted by letters a, b, c, d, e ...

Probability of a random event

So that you can begin the mathematical part of the probability, you need to define all its components.

The likelihood of an event is pronounced in the numerical form of the measure of the appearance of a certain event (A or B) as a result of experience. The probability of as p (a) or p (b) is indicated.

In probability theory, distinguish:

• reliable The event is guaranteed as a result of the experiment P (ω) \u003d 1;
• impossible The event can never occur p (Ø) \u003d 0;
• random The event lies between reliable and impossible, that is, the probability of its appearance is possible, but not guaranteed (the probability of a random event is always within 0≤p (a) ≤ 1).

Relationships between events

Consider both the same and the sum of the events of A + B, when the event is counted in the implementation of at least one of the components, A or B, or both - A and V.

In relation to each other, events can be:

• Equilibrium.
• Compatible.
• Incompatible.
• Opposite (mutually exclusive).
• Dependent.

If two events can occur with an equal probability, then they equilibrium.

If the appearance of an event and does not reduce the likelihood of the appearance of an event b, then they compatible.

If events A and B never happen simultaneously in the same experience, they are called incompatible. Throwing coins is a good example: the appearance of the rush is automatically the fault of the eagle.

The probability for the amount of such incompatible events consists of the probability of each of the events:

P (a + c) \u003d p (a) + p (c)

If the onset of one event makes it impossible to occur other, they are called opposite. Then one of them is designated as a, and the other - ā (read as "not a"). The appearance of an event A means that ā did not happen. These two events form a complete group with the sum of probability equal to 1.

Dependent events have a mutual influence, reducing or increasing each other's probability.

Relationships between events. Examples

Examples are much easier to understand the principles of probability theory and events combinations.

The experience that will be conducted is to pull out the balls from the box, and the result of each experience is an elemental outcome.

Event is one of the possible outcomes of the experience - a red ball, a blue ball, a ball with a number six, etc.

Test number 1. 6 balls are involved, three of which are painted into blue, odd numbers are applied on them, and three others are red with even numbers.

Test number 2. 6 balls of blue with numbers from one to six are involved.

Based on this example, you can call combinations:

• Reliable event. In №2 Event "Get a blue ball" is reliable, since the probability of its appearance is equal to 1, since all the balls blue and miss can not be. Whereas the event "get a ball with a number 1" is random.
• Impossible event. In №1 With blue and red balls The event "get a purple ball" is impossible, since the probability of its appearance is 0.
• Equal events. In №1 Events "get a ball with a number 2" and "get a ball with a number 3" equilibrium, and the events "get a ball with an even number" and "get a ball with a number 2" have a different probability.
• Compatible events. Two times in a row to get a six in the process of throwing a playing bone - these are compatible events.
• Incompatible events. In the same isp. №1 Events "get a red ball" and "get a ball with an odd number" cannot be combined in the same experience.
• Opposite events. The most striking example of this is to throw coins when the eagle pulling is tantamount to the non-captivity of the river, and the sum of their probabilities is always 1 (full group).
• Dependent events. So, in the isp. №1 You can set the goal to remove the red balloon twice in a row. His extraction or unknown for the first time affects the probability of extracting a second time.

It can be seen that the first event significantly affects the probability of the second (40% and 60%).

Event probability formula

The transition from gadetting reflections to exact data is due to the translating theme into the mathematical plane. That is, judgments about a random event like "high probability" or "minimum probability" can be transferred to specific numerical data. Such material is permissible to evaluate, compare and introduce into more complex calculations.

From the point of view of calculation, the definition of the probability of an event is the ratio of the number of elementary positive outcomes to the amount of all possible outcomes of the experience of a relatively specific event. It is indicated by the probability of p (a), where R means the word "probabilite", which is translated from the French as "probability."

So, the probability formula event:

Where M is the number of favorable outcomes for the event A, N - the sum of all outcomes possible for this experience. In this case, the probability of events always lies between 0 and 1:

0 ≤ p (a) ≤ 1.

Calculation of the probability of an event. Example

Take the spell. №1 with balls, which is previously described: 3 blue balls with numbers 1/3/5 and 3 red with 2/4/6 numbers.

Based on this test, several different tasks can be viewed:

• A - Loss of the Red Bowl. Red Balls 3, and Total options 6. This is the simplest example in which the probability of the event is p (a) \u003d 3/6 \u003d 0.5.
• B - the loss of an even number. In total even numbers 3 (2,4,6), and the total number of possible numeric variants is 6. The probability of this event is p (b) \u003d 3/6 \u003d 0.5.
• C is a loss of a number greater than 2. Total options 4 (3,4,5,6) from the total amount of possible outcomes 6. The probability of an event with equal to P (C) \u003d 4/6 \u003d 0.67.

As can be seen from the calculations, the event C has a greater probability, since the number of probable positive outcomes is higher than in A and V.

Invalid events

Such events cannot simultaneously appear in the same experience. Like in №1 It is impossible to simultaneously reach the blue and red ball. That is, you can get either blue or red ball. In the same way in the playing bone, an even and odd number can be at the same time.

The probability of two events is considered as the probability of their sum or work. The amount of such events A + B is considered to be such an event that consists in the emergence of an event A or B, and the work of them aw is in the appearance of both. For example, the appearance of two sixes immediately on the edges of two cubes in one throw.

The sum of several events is an event involving the emergence of at least one of them. The work of several events is the joint appearance of them all.

In the theory of probability, as a rule, the use of the Union "and" denotes the amount, the union "or" - multiplication. Formulas with examples will help to understand the logic of addition and multiplication in the theory of probability.

The probability of incomplete events

If the probability of inconsistent events is considered, the probability of the amount of events is equal to the addition of their probability:

P (a + c) \u003d p (a) + p (c)

For example: I calculate the likelihood that in the PC. No. 1 With blue and red balls, the number of 1 and 4. Calculate not in one action, but the sum of the probabilities of the elementary components. So, in this experience only 6 balls or 6 of all possible outcomes. The numbers that satisfy the condition - 2 and 3. The likelihood of figure 2 is 1/6, the probability of the figure 3 is also 1/6. The probability that the digit will fall out between 1 and 4 are:

The probability of incompatible events of the complete group is equal to 1.

So, if in the experiment with a cube, lay the probabilities of the fallout of all numbers, then as a result we obtain a unit.

It is also true for opposite events, for example, experience with a coin, where one side is an event A, and the other is the opposite event ā, as is known,

P (a) + p (ā) \u003d 1

The probability of the work of non-prominent events

Multiplication of probabilities apply when they consider the emergence of two or more incomplete events in one observation. The probability that events A and B will appear simultaneously, equal to the product of their probabilities, or:

P (A * B) \u003d P (a) * P (B)

For example, the likelihood that in the isp. №1 As a result of two attempts, a blue ball will appear twice, equal to

That is, the probability of an occurrence of an event, when, as a result of two attempts with the removal of the balls, only blue balls will be extracted, equal to 25%. It is very easy to do practical experiments of this task and see if it really is.

Joint events

Events are considered jointly when the appearance of one of them may coincide with the emergence of another. Despite the fact that they are joint, the likelihood of independent events is considered. For example, throwing two playing bones can give a result when the number 6 falls on both of them. Although the events coincided and appeared simultaneously, they are independent of each other - only one six, the second bone does not have an influence on it.

The probability of joint events is considered as the probability of their sum.

The probability of the sum of joint events. Example

The probability of the amount of events A and B, which in relation to each other joints, equals the sum of the probability of the event with a deduction of the probability of their work (that is, their joint implementation):

P joint. (A + c) \u003d p (a) + P (B) - P (AV)

Suppose that the likelihood of getting into the target with one shot is 0.4. Then the event A - hitting the target in the first attempt, in - in the second. These events are joint, since it is possible that the target can be hit and from the first and from the second shot. But events are not dependent. What is the probability of an occurrence of a target defeat from two shots (at least one)? According to the formula:

0,4+0,4-0,4*0,4=0,64

The answer to the question is as follows: "The probability of getting into a goal from two shots is 64%."

This event probability formula can also be applicable to incomplete events, where the probability of the appearance of the event p (AV) \u003d 0. This means that the probability of incomplete events can be considered a special case of the proposed formula.

Probability geometry for clarity

Interestingly, the probability of the amount of joint events can be represented as two regions A and B, which intersect together. As can be seen from the picture, the area of \u200b\u200btheir association is equal to the total area per minute of their intersection areas. This geometric explanation makes more understandable illogical at first glance formula. Note that geometric solutions are not uncommon in the theory of probability.

Determination of the probability of the sum of the set (more than two) joint events is quite cumbersome. To calculate it, you need to use the formulas that are provided for these cases.

Dependent events

The dependent events are called if the offensive of one (a) of them affects the likelihood of another (B). Moreover, the influence of both the events A and its faults is taken into account. Although events are called dependent on the definition, but only one of them (B) is dependent. The usual probability was designated as P (B) or the likelihood of independent events. In the case of dependent, a new concept is introduced - the conditional probability P A (B), which is the probability of the dependent event in provided that the event A (hypothesis) occurred from which it depends.

But after all, an event is also by chance, so it also has a chance that you need and can be taken into account in the calculated calculations. Next, the example will be shown how to work with dependent events and hypothesis.

An example of calculating the probability of dependent events

A good example for calculating dependent events can be a standard deck of cards.

On the example of a deck in 36 cards, consider dependent events. It is necessary to determine the likelihood that the second card extracted from the deck will be a tambourine, if the first extracted:

1. Bubnovy.
2. Another suit.

It is obvious that the probability of the second event is depends on the first A. So, if the first option is true that the deck has become 1 card (35) and 1 tambourine (8) less, the probability of an event in:

P A (B) \u003d 8/35 \u003d 0.23

If the second option is fair, the deck has become 35 cards, and the total number of tambourine (9) is still preserved, then the likelihood of the next event in:

P A (B) \u003d 9/35 \u003d 0.26.

It can be seen that if the event is agreed in the fact that the first card is a tambourine, then the probability of an event in decreases, and vice versa.

Multiplying dependent events

Guided by the previous chapter, we accept the first event (a) as a fact, but if we say in essence, it has a random character. The probability of this event, namely the extraction of the tambourine from the deck of cards, is equal to:

P (a) \u003d 9/36 \u003d 1/4

Since the theory does not exist in itself, but is designed to serve for practical purposes, it is right to note that the likelihood of a product of dependent events is most often needed.

According to the theorem on the product of the probabilities of dependent events, the probability of the appearance of jointly dependent events A and B is equal to the likelihood of one event A, multiplied by the conditional probability of an event in (dependent A):

P (AB) \u003d P (a) * P A (B)

Then in the example with a deck, the probability of extracting two cards with a mahi of the tambourine is:

9/36 * 8/35 \u003d 0.0571, or 5.7%

And the probability of extracting is not a tambourine first, and then the tambourines are equal to:

27/36 * 9/35 \u003d 0.19, or 19%

It can be seen that the probability of the appearance of an event in more, provided that the first extraction card is extracted from the tambourine. This result is quite logical and understandable.

Full probability of event

When the problem with conditional probabilities becomes multifaceted, it is impossible to calculate the usual methods. When the hypotheses are more than two, namely a1, a2, ..., and n, .. Cooling a complete group of events provided:

• P (A i)\u003e 0, i \u003d 1,2, ...
• A I ∩ A J \u003d Ø, I ≠ J.
• Σ k a k \u003d ω.

So, the formula for the full probability for the event in the complete group of random events A1, A2, ..., and n is:

A look into the future

The probability of a random event is extremely necessary in many areas of science: econometric, statistics, physics, etc. As some processes cannot be determined, as they themselves have probabilistic nature, special methods of work are needed. The theory of the probability of an event can be used in any technological sphere as a way to determine the possibility of an error or malfunction.

It can be said that, learning the likelihood, we are doing theoretical step into the future in some way, looking at it through the prism of the formula.

ChapterI.. Random events. PROBABILITY

1.1. Pattern and randomness, random variability in the exact sciences, biology and medicine

Probability Theory is a mathematics area that studies patterns in random phenomena. A random phenomenon is a phenomenon that, with repeated reproduction of the same experiment, can flow every time somewhat differently.

Obviously, there is not a single phenomenon in nature in nature, in one way or another elements of chance, but in various situations we take into account them in different ways. Thus, in a number of practical problems, they can be neglected and considered instead of a real phenomenon its simplified scheme - "model", assuming that in these conditions of experience, the phenomenon proceeds quite in a certain way. This highlights the most important, decisive factors characterizing the phenomenon. It is such a scheme of studying phenomena most often applied in physics, technician, mechanics; This is how the main pattern is detected. , Prophurated with this phenomenon and gives the ability to predict the result of the experience of the specified source conditions. And the effect of random, secondary, factors on the result of experience is taken into account by random errors of measurements (we will consider the method of calculation further).

However, the described classic scheme of the so-called accurate sciences is poorly adapted to solve many tasks in which numerous, closely intertwined random factors play a noticeable (often determining) role. Here, the random nature of the phenomenon, which is no longer neglected. This phenomenon must be studied from the point of view of the patterns inherent in both a random phenomenon. In physics, examples of such phenomena are Brownian movement, a radioactive decay, a number of quantum-mechanical processes, etc.

The subject to study biologists and physicians - a living organism, the origin, the development and existence of which is determined by very many and diverse, often by random external and internal factors. That is why phenomena and events of the living world are in many ways also random by nature.

Elements of uncertainty, complexity, multi-range inherent in random phenomena determine the need to create special mathematical methods to study these phenomena. Development of such methods, the establishment of specific patterns peculiar to random phenomena, - the challenges of probability theory. It is characteristic that these patterns are performed only with the mass of random phenomena. Moreover, the individual features of individual cases are mutually repaid, and the average result for the mass of random phenomena is no longer random, but quite natural . To a large extent, this circumstance was the reason for the widespread dissemination of probabilistic research methods in biology and medicine.

Consider the basic concepts of probability theory.

1.2. Probability of a random event

Each science developing a general theory of any circle of phenomena is based on a number of basic concepts. For example, in geometry - this is the concepts of a point, straight line; In mechanics - the concepts of force, mass, velocity, etc. The main concepts exist in the theory of probabilities, one of them is a random event.

Random event is any phenomenon (fact), which, as a result of experience (test), may occur or not happen.

Random events are denoted by letters A, B, with ... etc. Let's give a few examples of random events:

BUT- Eagle (coat of arms) when throwing a standard coin;

IN - the birth of a girl in this family;

FROM - Birth of a child with a predetermined body weight;

D. - emergence of the epidemic disease in the region at a certain period of time, etc.

The main quantitative characteristic of the random event is its probability. Let be BUT - some random event. The probability of a random event A is a mathematical value that determines the possibility of its appearance.It is denoted R(BUT).

Consider two basic methods for determining this value.

Classical definition of the probability of a random eventit is usually based on the results of the analysis of speculative experiments (tests), the essence of which is determined by the condition of the task. In this case, the probability of a random event R (a)equal to:

where m. - the number of cases that conducive events BUT; n. - The total number of equilibrium cases.

Example 1. The laboratory rat is placed in a labyrinth, in which only one of four possible paths leads to encouraging in the form of food. Determine the likelihood of choosing a rat of such a path.

Decision: under the condition of the problem of four equilibrium cases ( n.\u003d 4) Event BUT(Rat finds food)
only one favors, that is, m. \u003d 1 then R(BUT) = R (The rat finds food) \u003d \u003d 0.25 \u003d 25%.

Example 2. In the urn of 20 black and 80 white balls. One ball is removed from it. Determine the likelihood that this ball will be black.

Decision: The number of all balls in the urn is the total number of equilibrium cases n., i.e. n. = 20 + 80 = 100, of which the event BUT (Black ball extraction) is possible only at 20, i.e. m. \u003d 20. Then R(BUT) = R(h. sh.) \u003d \u003d 0.2 \u003d 20%.

We list the probability properties following its classical definition - formula (1):

1. The probability of a random event is the value of dimensionless.

2. The probability of a random event is always positive and less than one, i.e. 0< P. (A.) < 1.

3. The likelihood of a reliable event, i.e. the events that will necessarily occur as a result of experience ( m. = n.) is equal to one.

4. Probability of the Impossible Event ( m. \u003d 0) equal to zero.

5. The likelihood of any event is not negative and not exceeding one:
0 £ P. (A.) £ 1.

Statistical definition of the probability of a random eventit is used when it is impossible to useclassical definition (1). This often takes place in biology and medicine. In this case, the probability R(BUT) Determine by generalizing the results of actually conducted test series (experiments).

We introduce the concept of the relative frequency of the occurrence of a random event. Let a series of consisting of N. experiments (number N. can be selected in advance); Event you are interested in BUT happened in M. of them ( M. < N.). Attitude of the number of experiences M.in which this event occurred, to the total number of experiments N. Call the relative frequency of random event BUT In this series of experiments - R* (BUT)

R*(BUT) = .

It is experimentally established that if a series of tests (experiments) are carried out in the same conditions and in each of them the number N. large enough, then relative frequency detects the property of stability : From the series to the series she changes little , Approaching with an increase in the number of experiments to a certain constant value . It is accepted for the statistical probability of a random event. BUT:

R(BUT) \u003d LIM, with N. , (2)

So, statistical probability R(BUT) Random Event BUT they call the limit to which the relative frequency of the appearance of this event seeks with an unlimited increase in the number of tests (when N. → ∞).

Approximately the statistical probability of a random event is equal to the relative frequency of the appearance of this event with a large number of tests:

R(BUT) ≈ P *(BUT) \u003d (at large N.) (3)

For example, in the experiments on the cast of coins, the relative frequency of the emblem of the coat of arms at 12,000 times was equal to 0.5016, and at 24,000 throws - 0,5005. In accordance with formula (1):

P.(coat of arms) \u003d \u003d 0.5 \u003d 50%

Example . In a medical examination, 500 people have 5 of them discovered a tumor in the lungs (about l.). Determine the relative frequency and probability of this disease.

Decision: under the condition of the task M. = 5, N. \u003d 500, relative frequency R* (o. l.) \u003d M./N. \u003d 5/500 \u003d 0.01; insofar as N. It is large enough, it is possible to believe with good accuracy that the probability of the presence of a tumor in the lungs is equal to the relative frequency of this event:

R(about l.) \u003d R* (o. l.) \u003d 0.01 \u003d 1%.

The previously listed properties of the probability of a random event are preserved with the statistical determination of this value.

1.3. Types of random events. The main theorems of probability theory

All random events can be divided into:

¾ incomplete;

¾ independent;

¾ dependent.

For each type of events, their characteristics and the theorems of probability theory are characteristic.

1.3.1. Incomplete random events. Probability addition theorem

Random events (A, B, C,D. ...) are called incomplete , if the appearance of one of them eliminates the emergence of other events in the same test.

Example1 . The coin was added. With its fall, the appearance of "coat of arms" excludes the appearance of the "Dish" (the inscriptions determining the price of the coin). The events "fell out the coat of arms" and "dropped the rush" incomplete.

Example 2. . Obtaining a student on a single Evaluation exam "2", or "3", or "4", or "5" - events inconsistencies, since one of these estimates excludes another on the same exam.

For incomplete random events are performed probability Addition Theorem: Probability of Appearance one, but still what, of several incomplete events A1, A2, A3 ... Ak. equal to the sum of their probability:

P (a1i a2 ... or ak.) \u003d P (A1) + P (A2) + ... + p (ak.). (4)

Example 3. There are 50 balls in the urn: 20 white, 20 black and 10 red. Find the probability of white (event BUT) or red ball (event IN) When the bowl of at random is getting out of the urn.

Solution: R.(A or B.) \u003d R.(BUT) + R.(IN);

R(BUT) = 20/50 = 0,4;

R(IN) = 10/50 = 0,2;

R(BUT or IN) \u003d R.(b. sh. or to. sh.) = 0,4 + 0,2 = 0,6 = 60%.

Example 4. . In class 40 children. Of them aged 7 to 7.5 years old, 8 boys ( BUT) and 10 girls ( IN). Find the probability of presence in the class of children of this age.

Solution: R.(BUT) \u003d 8/40 \u003d 0.2; R(IN) = 10/40 = 0,25.

P (a or c) \u003d 0.2 + 0.25 \u003d 0.45 \u003d 45%

Next important concept - full Event Group: Several inconsistent events form a complete group of events, if only one of the events of this group may appear as a result of each test and no other.

Example 5. . The arrows made a shot of a target. One of the following events will take place: getting into the "top ten", in the "nine", in the "eight", .., in the "unit" or misses. These 11 incomplete events form a complete group.

Example 6. . On the exam in the university, the student can get one of the following four estimates: 2, 3, 4 or 5. These four inconsideration events also form a complete group.

If incomplete events A1, A2 ... Ak. Form a complete group, then the sum of the probabilities of these events is always equal to one:

R(A1.) + R.(A2.)+ ... R.(BUTk.) = 1, (5)

This statement is often used in solving many applied tasks.

If two events are single and inconsistent, they are called opposite and denote BUT and . Such events make up a complete group, so the sum of their probabilities is always equal to one:

R(BUT) + R.() = 1. (6)

Example 7. Let R(BUT) - the likelihood of death with some disease; It is known and is equal to 2%. Then the likelihood of a prosperous outcome in this disease is 98% ( R() = 1 – R(BUT) \u003d 0.98), since R(BUT) + R() = 1.

1.3.2. Independent random events. Probability multiplication theorem

Random events are called independent if the appearance of one of them does not affect the likelihood of other events.

Example 1. . If there are two or more urns with colored balls, then the extraction of any ball from one branch will not affect the likelihood of extracting other balls from the remaining urns.

For independent events is valid Probability Multiplication Theorem: Probability of Joint(simultaneous) The appearance of several independent random events is equal to the product of their probability:

P (a1i a2 and a3 ... and ak.) \u003d P (A1) ∙ P (A2) ∙ ... ∙ P (Ak.). (7)

Joint (simultaneous) appearance of events means that events occur and A1, and A2,and A3.... I. BUTk. .

Example 2. . There are two urns. One is 2 black and 8 white balls, in the other - 6 black and 4 white. Let the event BUT - Chief of the White Bowl of the first urn, IN - From the second. What is the probability of choosing at the same time from these urns on the white ball, i.e. what is equal R (BUT and IN)?

Decision: Probability to get a white ball from the first urn
R(BUT) \u003d \u003d 0.8 from the second - R(IN) \u003d \u003d 0.4. The probability of simultaneously get over the white ball from both urns -
R(BUT and IN) = R(BUT)· R(IN) = 0,8∙ 0,4 = 0,32 = 32%.

Example 3. The diet with a reduced content of iodine causes an increase in the thyroid gland in 60% of animals of a large population. For the experiment, 4 enlarged glands are needed. Find the likelihood that 4 randomly selected animals will have increased thyroid gland.

Decision: Random Event BUT - Choosing an animal raw with an increased thyroid gland. By the condition of the task, the likelihood of this event R(BUT) \u003d 0.6 \u003d 60%. Then the likelihood of the joint appearance of four independent events is the choice of at random 4 animals with an increased thyroid gland - will be equal to:

R(BUT1 I. BUT2 I. BUT3 I. BUT4) = 0,6 ∙ 0,6 ∙0,6 ∙ 0,6=(0,6)4 ≈ 0,13 = 13%.

1.3.3. Dependent events. Probability multiplication theorem for dependent events

Random events A and B are called dependent, if the appearance of one of them, for example, and changes the likelihood of the emergence of another event - V.Therefore, two probabilities are used for dependent events: unconditional and conditional probability .

If a BUT and IN dependent events, then the likelihood of an event IN first (i.e. before the event BUT) Called unconditional probability This event is indicated R(IN). The likelihood of an event IN provided that the event BUT already happened, called conditional probability events IN And denotes R(IN/BUT) or R.(IN).

The same meaning has unconditional - R(BUT) and conditional - R(A / B.) probability for event BUT.

Probability multiplication theorem for two dependent events: the probability of the simultaneous occurrence of two dependent events A and B is equal to the undisputed probability of the first event on the conditional probability of the second:

R(A and B.) \u003d R.(BUT) ∙ R.(V / A.) , (8)

BUT, or

R(A and B.) \u003d R.(IN) ∙ R.(A / c), (9)

if the first event comes IN.

Example 1. In the urn of 3 black balls and 7 white. Find the likelihood that from this urn one by one (and the first ball is not returned to the urn) 2 white balls will be taken out.

Decision: Probability to get the first white ball (event BUT) Equal to 7/10. After it is removed, 9 balls remain in the urn, of which 6 are white. Then the likelihood of the appearance of the second white ball (event IN) Equal R(IN/BUT{!LANG-4d80f131c7d4d61b4f74da467ab77437!}

R(BUT and IN) = R(BUT)∙R(IN/BUT) = = 0,47 = 47%.

{!LANG-21052485544f08f44823760f3b569446!}

R(BUTand {!LANG-d8ad5985031a23f23242ed3936785046!}and {!LANG-0835fa868f22dd9b3cdab5d40f496659!}) \u003d R.(BUT){!LANG-4b08d458f2c0fdf6f558332b8b6859c0!}(V / A.){!LANG-4b08d458f2c0fdf6f558332b8b6859c0!}({!LANG-daf8f7482bcb106387eaca9cacdeee05!}). (10)

{!LANG-f9f4371fa70ece15f36c7d4d476f5fcc!}

{!LANG-647fb10f412d71571e55c78b3879d393!} BUT{!LANG-d6ec57d8ccc66516ad4134c913a126a2!} IN).

{!LANG-561829c38c901dfade93231e2c630d4f!} FROM{!LANG-d3951ac29bf1b13e795c69c1ca754397!} D.{!LANG-348c8dca6999e606b448c546de54fcf2!} {!LANG-4a60b344e75cf5d278fd2000ae9e9060!}).

Decision. {!LANG-7825252080980117d2c35ca494feaabf!}

R(BUT and IN) = R(BUT) ∙ R(IN/BUT) = = 0,1 = 10%.

{!LANG-3ba2ca7f9606ba96d1daa6360a0ff85f!}

R(FROM and D. and {!LANG-4a60b344e75cf5d278fd2000ae9e9060!}) = R(FROM) ∙ R(D./{!LANG-b39bfc0e26a30024c76e4dcb8a1eae87!}) ∙ R({!LANG-4a60b344e75cf5d278fd2000ae9e9060!}/{!LANG-3a9334fdb39fd04400e4163ea4e3ef96!}) = = 5%.

{!LANG-4666ebe634fb6c7b8f6f5a1ebe08661e!}

{!LANG-0af4e6db192fc450f2e894a586e68a79!} BUT and IN{!LANG-12abf01dcf09b5af16094bc608584a41!} R(BUTand {!LANG-d8ad5985031a23f23242ed3936785046!}) \u003d R.(BUT) ∙ R.(V / A.) \u003d R.(IN) × R(A / B.). {!LANG-8c26619e773cd66f9a18f1272eb7e80b!}

R(V / A.) = (11)

{!LANG-785892621595bc7b7b0de83402cb67ae!}

{!LANG-21713e17ada9e754c81ecfc20a9e7d99!} n.{!LANG-87a098820251b596c97ffbc7f9208264!} {!LANG-9cfb64221a1076c6f02e36a33ed25a3e!}n.{!LANG-deacd4c28761768826e9ed421e940e97!} R({!LANG-726913fa091173cc2c6fbd7d18118926!}){!LANG-3778ef2d038cfc11147fbca572d0b161!}({!LANG-6e44a1362960245cbf09e03533164d46!}){!LANG-0b5cc613972ccad7a80a293b13b47e08!}({!LANG-0407f93d642c4349aaf560037338096f!}n.{!LANG-e0c979442e017ff968f731f9a37cccb8!}

{!LANG-1053213e3c5f215303fcc9942082e6c6!} BUT{!LANG-2cb18e454ca6a9b474cc1a4b36e959ea!} {!LANG-9cfb64221a1076c6f02e36a33ed25a3e!}n.{!LANG-4d44344d86941354165a7967b8964009!} BUT{!LANG-2b996806fe870fdc54d75324d337d3cd!} {!LANG-0407f93d642c4349aaf560037338096f!}i.{!LANG-4a9a006740322bdf9ce6a3997a2e6bdb!} R({!LANG-7e239500a51a46d65ec9411e522f3bf2!}){!LANG-3778ef2d038cfc11147fbca572d0b161!}({!LANG-46819926aa4e6f80cbe1eea7f7b31fd4!}){!LANG-0b5cc613972ccad7a80a293b13b47e08!}({!LANG-74967277e30aa34b291c8e1f9ab71b16!}n.{!LANG-a1ca96c68fcef9ad1e8bca8ca0c13007!} R({!LANG-74967277e30aa34b291c8e1f9ab71b16!}i.{!LANG-383be15ce58d3635eb6bddbc4aab0324!} ≠ 1.

{!LANG-a63d162637c57227f1bca874682dac2c!} {!LANG-0407f93d642c4349aaf560037338096f!}i.{!LANG-2cfc429a1ca7bfae2071af651a4668a4!} BUT{!LANG-7275dbffa3152facd5d01df00f25a8dd!} BUT{!LANG-587db789c5f1042a66ea5719b1cd16d3!} :

{!LANG-93ec9706b7e37f8cd262e50360b92809!} .

{!LANG-ea27de14dbc1d38f1bc985f35ab9cce4!} {!LANG-0407f93d642c4349aaf560037338096f!}1,…, {!LANG-0407f93d642c4349aaf560037338096f!}n.{!LANG-20bf2accd0a126447a1485de76963a92!} BUT{!LANG-80dd75a8ed143aca68c5be30cb6f9ed0!} R({!LANG-74967277e30aa34b291c8e1f9ab71b16!}i.{!LANG-f85c47e064840e3049cc7516f9bc6b99!} {!LANG-0407f93d642c4349aaf560037338096f!}i. (i. = 1,2,3,…n.{!LANG-275db9aee2046cee678a63abee90f04a!} R({!LANG-0407f93d642c4349aaf560037338096f!}i.{!LANG-78713c262ed41d7cfbf4ec1d92d26e6b!}{!LANG-39d7bd2f6c082f20e5816c63e4a80892!} BUT{!LANG-b2d8a3e84f08d2b1ecf9d7b56e4dba8a!}

{!LANG-83b0fa7d655b14b4802749934d65d6f1!} {!LANG-0407f93d642c4349aaf560037338096f!}1, {!LANG-0407f93d642c4349aaf560037338096f!}2, {!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-a0edd64692ccb6254f8a8757424195ac!} R({!LANG-0407f93d642c4349aaf560037338096f!}1) = 0,5; R({!LANG-0407f93d642c4349aaf560037338096f!}2) = 0,17; R({!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-d42373bc1f608b1771fc62dd15275018!} BUT{!LANG-40a78a588f67c4073fbac74dc152dfce!}

R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}1) = 0,1; R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}2) = 0,2; R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}3) = 0,9.

{!LANG-776f86bd4769b3d763f0c32aad7a8345!} BUT{!LANG-79e5e14ad7f219e75df0675f05cf3045!} R({!LANG-0407f93d642c4349aaf560037338096f!}1/BUT) = 0,13; R({!LANG-0407f93d642c4349aaf560037338096f!}2/BUT) = 0,09;
R({!LANG-0407f93d642c4349aaf560037338096f!}3/BUT{!LANG-7e2fb904047b3a980d94ea050ea35a38!}

{!LANG-4679efde207401c6841129624b9df72a!}

{!LANG-9555b98fe1baf60f42656cc578fdff4b!}

Decision{!LANG-11bf366b28abf00ba9121372adfc8182!} {!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-dc89846e5d67687a4f0795e5e5a27c23!} R({!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-af3e39c4bd6c2da9800ff222e823ad7d!} {!LANG-6e44a1362960245cbf09e03533164d46!}{!LANG-dd641eb63fe5abdc37bd7f35b4e47675!} R({!LANG-0407f93d642c4349aaf560037338096f!}2) = 1 – 0,975 = 0,025 = 2,5 %.

{!LANG-e5cf82f92097240048a70848434ec5ab!} BUT{!LANG-d76baaa3818e5b1f4e9bcd11afba0a52!} R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-b0881904f31ddd51a28c94f2e43e276f!} R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-21e37bdd3a40159da23b995ca8ac42cb!} R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-b088121f09ad44e47302e804a505953f!}
R(BUT/{!LANG-0407f93d642c4349aaf560037338096f!}{!LANG-523b4cf83bd306b227e88fa9a07a8d4d!}

{!LANG-74778c68bc9f07d4740c263ec35ec577!}

{!LANG-9dc38505159dd350b9252d1d91b9ade5!}

{!LANG-b46809a5bc04f720910d148e7c660b49!}

{!LANG-72573d63df58070b727d59e526c9f170!}

{!LANG-8dcfe245ab0d94573494641563ded740!}

{!LANG-98adbb74557d1207984538029ffa53cb!} {!LANG-5706d4630d8994aff6d379c92f4da816!}{!LANG-ef32ec0d001fd9c9e299116aebf8f5ca!}

{!LANG-3999aa7422de49bf1df2e4d6db6d55e3!} BUT{!LANG-41e66e7350690ea5aad09633cdd643e8!} {!LANG-62c44a534fdf1713cd44c13fd3b8cf70!} R() » 1.

Chapter{!LANG-561884418c8cd7ed1a49a30caf520595!}{!LANG-10f388dcf981e8da3bbd7429d686cf9c!}

{!LANG-07a646e0f2ff68d80c36595cc7d3a09a!}

{!LANG-a97588678c4d267ac5c2c8d85ff554c3!}

{!LANG-45aaaf0c620f548b58b46b868693bfbe!} {!LANG-52895d319fd1a2a84bcff3d974536e7c!}{!LANG-8e5ca12af91ab9f8b2216c1ebac64394!}

{!LANG-51c054bb50b617cc8f48a3f21c1da644!} .

{!LANG-ef32a438835267ae996cc137174a04d7!}

{!LANG-bfc4deec643b99b9f0c37f2618a9c2c3!}

{!LANG-1a07b33bddbd98d7edeaafa606aeeae0!}

{!LANG-023e78c29f4bb00a8897afeb61e49cdf!}

{!LANG-a62e3eb4adee46f44788fe4540c73b14!}

{!LANG-fae208a1ba7a79e884d3b59b20c07216!}

{!LANG-9c9a59b8a16923cfb389b6c85dd36aab!}*. {!LANG-0dea9fd02d31a58ef84834b26a255a6c!} r{!LANG-c9f139f2f6ef28f8c6e9e03b67100b2e!}

{!LANG-1e105a497485d1ccc8648befec5f62a3!}

{!LANG-02f977720e21b967f46e4abc06f678a5!}

{!LANG-c2f2320a80346febc619dc187eac2300!}

{!LANG-dd9203a25e2ed7551c99786cf570e2a3!}

{!LANG-da9a8b29499ed867aba6a3ef9a89ad32!}

{!LANG-adc68dc03dcbb949bd71a21b43585fd0!} {!LANG-1a50f56c1cb183beffb60597d8779183!}{!LANG-661bdb6c0f989a804f78ecfad8c4d389!} h.i.{!LANG-ae27d6f927571413fd214d68dc1d9ea8!} ri. *. {!LANG-75cf43fdcb76692d5c52ceb7c8799832!}

{!LANG-8e4d88cb3ececdbf076f9f22ff6a57b9!}{!LANG-5839bef9419186acdf20af4b334cfb7f!} {!LANG-1a50f56c1cb183beffb60597d8779183!}{!LANG-df6d1e1eb3f47383ce58f082243b210e!} R({!LANG-1a50f56c1cb183beffb60597d8779183!}):

{!LANG-1a50f56c1cb183beffb60597d8779183!}

…..

…..

P.({!LANG-253bcac7dd806bb7cf57dc19f71f2fa0!})

…..

…..

{!LANG-53dee19d4ab3e6567ae26778f47d7fad!} ri.{!LANG-921f9a9c99316562a8d0fdf059116e86!}

ri. = p.1 + p.2 + ... + {!LANG-709b8a73ecd44d1c54f064821d8802a7!} = 1. (13)

{!LANG-7744182404e831d476b855e182ea111b!}{!LANG-cfe130db053b5c778e49698b5dcacea0!} h.i., , {!LANG-20e5e53bb8d1da6ca9c9d6e37384f397!} ri.

{!LANG-b7710f76cc70fdfface8e1fe2a7a7a2c!}{!LANG-dec02c5b602d1e2cd8d1a39f8b6072d4!} {!LANG-ce81a2dfede1ad48123cd1fc73f8fc40!}{!LANG-4ec91cfc3c025eea69b307984b22be27!} n.{!LANG-b2dae1b3e1c8ef3a5aa6e117c2258517!} R(n.) = n. {!LANG-20de7467143d7421122daecb28b0d727!}-1 × p.{!LANG-619f042ce33a01a2d041fea9173cc356!} q.{!LANG-f6d23804981ef6c4f13d5d814def25fd!}{!LANG-9663759be92cf2926d9d36c502b5e7e8!}

{!LANG-ec1009156082349d71cf458102599670!}

{!LANG-ce8e36c79eda071e87563d2cc5e551e4!}

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