then they say that the random variable ξ It has probability distribution density f (x).

Definition. Let be . For a continuous distribution function F theoretical α-quantile is called the solution to the equation.

This solution may not be the only one.

Quantile level ½ called theoretical median , level quantiles ¼ and ¾ -lower and upper quartiles respectively.

In actuarial applications, an important role is played by Chebyshev's inequality:

for any

The symbol of the expected value.

It reads like this: the probability that the modulus is greater than or equal to the mathematical expectation of modulus divided by.

Lifetime as a random variable

Uncertainty about the moment of death is a major risk factor in life insurance.

Nothing definite can be said about the moment of death of an individual. However, if we are dealing with a large homogeneous group of people and are not interested in the fate of individual people from this group, then we are within the framework of probability theory as a science of mass random phenomena that have the property of frequency stability.

Respectively, we can talk about life expectancy as a random variable T.

Survival function

In probability theory, they describe the stochastic nature of any random variable T distribution function F (x), which is defined as the probability that the random variable T less than number x:

.

In actuarial mathematics, it is pleasant to work not with a distribution function, but with an additional distribution function . In relation to long life, this is the likelihood that a person will live to age x years.

called survival function(survival function):

The survival function has the following properties:

In life tables, it is usually assumed that there is some age limit (limiting age) (as a rule, years) and, accordingly, at x>.

When describing mortality by analytical laws, it is usually considered that the lifetime is unlimited, however, the type and parameters of the laws are selected so that the probability of life above a certain age is negligible.

The survival function has a simple statistical meaning.

Let's say that we are observing a group of newborns (as a rule), whom we observe and can record the moments of their death.

Let us designate the number of living representatives of this group at the age through. Then:

.

Symbol E here and below is used to denote the mathematical expectation.

So, the survival function is equal to the average proportion of newborns surviving to age from a certain fixed group of newborns.

Actuarial mathematics often does not work with a survival function, but with the value just entered (by fixing the initial group size).

Survival function can be restored by density:

Life expectancy characteristics

From a practical point of view, the following characteristics are important:

1 . The average lifetime

,
2 . Dispersion lifetime

,
where
,

1.1. Some information from combinatorics

1.1.1. Accommodation

Consider the simplest concepts associated with the choice and location of a set of objects.
Counting the number of ways in which these actions can be performed is often done when solving probabilistic problems.
Definition... Accommodation from n elements by k (k ≤n) is called any ordered subset from k elements of the set consisting of n various elements.
Example. The following sequences of numbers are 2-element placements from 3 elements of the set (1; 2; 3): 12, 13, 23, 21, 31, 32.
Note that the placements differ in the order of their constituent elements and their composition. Placements 12 and 21 contain the same numbers, but their order is different. Therefore, these placements are considered to be different.
Number of different placements from n elements by k denoted and calculated by the formula:
,
where n! = 1∙2∙...∙(n - 1)∙n(reads " n- factorial ").
The number of two-digit numbers that can be formed from the digits 1, 2, 3, provided that no digit is repeated equals:.

1.1.2. Permutations

Definition... Permutations from n elements are called such placements from n elements that differ only in the arrangement of the elements.
Number of permutations from n elements P n calculated by the formula: P n=n!
Example. How many ways can 5 people queue up? 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 identical, then rearrange these n elements is called permutation with repetitions.
Example. Let there be 2 identical books among 6 books. Any arrangement of all books on the shelf - permutation with repetitions.
The number of different permutations with repetitions (from n elements, including k identical) is calculated by the formula:.
In our example, the number of ways in which you can arrange books on the shelf is:.

1.1.3. Combinations

Definition... Combinations of n elements by k such placements from n elements by k that differ from one another by at least one element.
Number of different combinations of n elements by k denoted and calculated by the formula:.
By definition, 0! = 1.
The following properties are valid for combinations:
1.
2.
3.
4.
Example. There are 5 flowers in different colors. 3 flowers are chosen for the bouquet. The number of different bouquets of 3 flowers out of 5 is equal to:.

1.2. Random events

1.2.1. Developments

Cognition of reality in natural sciences occurs as a result of tests (experiment, observation, experience).
Test or experience is the realization of a certain set of conditions, which can be reproduced as many times as desired.
Random is called an event that may or may not occur as a result of some test (experience).
Thus, the event is considered as a test result.
Example. Tossing a coin is a challenge. The appearance of an eagle when thrown is an event.
The events we observe differ in the degree of possibility of their occurrence and in the nature of their interconnection.
The event is called reliable if it will necessarily occur as a result of this test.
Example. Obtaining a positive or negative mark on the exam by a student is a reliable event if the exam proceeds according to the usual rules.
The event is called impossible if it cannot occur as a result of this test.
Example. Removing a white ball from an urn, which contains only colored (non-white) balls, is an impossible event. Note that under other experimental conditions the appearance of a white ball is not excluded; thus, this event is impossible only under the conditions of our experience.
In what follows, random events will be denoted by capital Latin letters A, B, C ... A reliable event will be denoted by the letter Ω, the impossible - Ø.
Two or more events are called equally possible in this test, if there is reason to believe that none of these events is more possible or less possible than others.
Example. With one throw of the dice, the appearance of 1, 2, 3, 4, 5 and 6 points - all these events are equally possible. It is, of course, assumed that the dice are made of a uniform material and have the correct shape.
Two events are called inconsistent in a given test, if the appearance of one of them precludes the appearance of the other, and joint otherwise.
Example. The box contains standard and non-standard parts. Let's take one detail for good luck. The appearance of a standard part eliminates the appearance of a non-standard part. These events are inconsistent.
Several events form complete group of events in this test, if as a result of this test at least one of them will necessarily occur.
Example. The events from the example form a complete group of equally possible and pairwise incompatible events.
Two incompatible events that form a complete group of events in a given trial are called opposite events.
If one of them is denoted by A, then the other is usually denoted by (read "not A»).
Example. Hitting and missing with one shot at a target are opposite events.

1.2.2. Classical definition of probability

Event probability - a numerical measure of the possibility of its offensive.
Event A called favorable event V if whenever an event occurs A, the event also comes V.
Developments A 1 , A 2 , ..., An form case diagram , if they:
1) are equally possible;
2) pairwise incompatible;
3) form a complete group.
In the scheme of cases (and only in this scheme) there is a classical definition of the probability P(A) developments A... Here, a case is called each of the events belonging to the selected full group of equally possible and pairwise incompatible events.
If n Is the number of all cases in the scheme, and m- the number of cases favorable to the event A, then probability of event A is defined by the equality:

The following properties follow from the definition of probability:
1. The probability of a reliable event is equal to one.
Indeed, if the event is certain, then each event in the pattern of events favors the event. In this case m = n and therefore

2. The probability of an impossible event is zero.
Indeed, if an event is impossible, then none of the events in the scheme of events favors the event. That's why m= 0 and therefore

The probability of a random event is a positive number between zero and one.
Indeed, only a fraction of the total number of cases in the case scheme favors a random event. Therefore 0<m<n, and, therefore, 0<m/n<1 и, следовательно, 0 < P (A) < 1.
So, the probability of any event satisfies the inequalities
0 ≤ P (A) ≤ 1.
At present, the properties of probability are defined in the form of axioms formulated by A.N. Kolmogorov.
One of the main advantages of the classical definition of probability is the ability to calculate the probability of an event directly, i.e. without resorting to experiments, which are replaced by logical reasoning.

Problems of direct calculation of probabilities

Task 1.1... What is the probability of an even number of points (event A) occurring on one die roll?
Solution... Consider the events Ai- dropped i points, i= 1, 2, ..., 6. Obviously, these events form a case diagram. Then the number of all cases n= 6. An even number of points is favored by cases A 2 , A 4 , A 6, i.e. m= 3. Then .
Task 1.2... The urn contains 5 white and 10 black balls. The balls are mixed thoroughly and then 1 ball is taken out at random. What is the likelihood that a ball that is taken out turns out to be white?
Solution... There are 15 cases in total, which form a case diagram. Moreover, the expected event A- the appearance of a white ball, 5 of them favor, therefore .
Task 1.3... The child plays with six letters of the alphabet: A, A, E, K, P, T. Find the probability that he will be able to accidentally add the word CARETA (event A).
Solution... 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 equally possible, pairwise incompatible, and form a complete group of events, i.e. form a case diagram. Only one occasion favors an event A... That's why
.
Task 1.4... Tanya and Vanya agreed to celebrate the New Year in a company of 10 people. They both really wanted to sit next to each other. What is the probability of the fulfillment of their wishes, if it is customary to distribute places among their friends by lot?
Solution... Let us denote by A event "fulfillment of the wishes of Tanya and Vanya". 10 people can sit at 10 table! different ways. How many of these n= 10! are equally possible ways favorable for Tanya and Vanya? Tanya and Vanya, sitting next to each other, can take 20 different positions. At the same time, eight of their friends can sit at table 8! in different ways, therefore m= 20 ∙ 8 !. Hence,
.
Task 1.5... A group of 5 women and 20 men selects three delegates. Assuming that each of those present with the same probability can be chosen, find the probability that two women and one man will be chosen.
Solution... The total number of equally probable trial outcomes is equal to the number of ways in which three delegates can be selected out of 25 people, i.e. ... Let us now count the number of favorable cases, i.e. the number of cases in which the event of interest takes place. A male delegate can be selected in twenty ways. In this case, the other two delegates must be women, and you can choose two women out of five. Hence, . That's why
.
Task 1.6. Four balls are randomly scattered over four holes, each ball hits one hole or the other with the same probability and independently of the others (there are no obstacles to hitting the same hole for several balls). Find the probability that there will be three balls in one of the holes, one in the other, and no balls in the other two holes.
Solution. Total number of cases n= 4 4. The number of ways you can select one hole with three balls,. The number of ways in which you can select a hole where there will be one ball,. The number of ways you can choose from the four balls is three to put them in the first hole,. The total number of favorable cases. Event probability:
Task 1.7. There are 10 identical balls in the box, marked with numbers 1, 2,…, 10. Six balls have been drawn for luck. Find the probability that among the extracted balls there will be: a) ball # 1; b) balls # 1 and # 2.
Solution... a) The total number of possible elementary test outcomes is equal to the number of ways in which one can extract six balls out of ten, i.e.
Let's find the number of outcomes favorable to the event of interest to us: among the selected six balls there is ball # 1 and, therefore, the other five balls have different numbers. The number of such outcomes is obviously equal to the number of ways in which five balls can be selected from the remaining nine, i.e.
The desired probability is equal to the ratio of the number of outcomes favorable to the event under consideration to the total number of possible elementary outcomes:
b) The number of outcomes favorable to the event of interest to us (among the selected balls there are balls # 1 and # 2, therefore, four balls have different numbers) is equal to the number of ways in which four balls can be extracted from the remaining eight, i.e. Seeking probability

1.2.3. Statistical probability

The statistical definition of probability is used when the outcomes of an experiment are not equally possible.
Relative frequency of the event A is defined by the equality:
,
where m- the number of trials in which the event A came n- the total number of tests performed.
J. Bernoulli proved that with an unlimited increase in the number of experiments, the relative frequency of occurrence of an event will practically differ little from a certain constant number. It turned out that this constant number is the probability of an event occurring. Therefore, naturally, the relative frequency of occurrence of an event with a sufficiently large number of tests is called the statistical probability, in contrast to the previously introduced probability.
Example 1.8... How to estimate the approximate number of fish in the lake?
Let in the lake NS fish. We throw the net and, let's say, we find in it n fish. We mark each of them and release them back. A few days later, in the same weather and in the same place, we cast the same net. Suppose that we find m fish in it, among which k labeled. Let the event A- “the caught fish is tagged”. Then by definition of the relative frequency.
But if in the lake NS fish and we released into it n labeled, then.
Because R * (A) » R(A), then .

1.2.4. Operations on events. Probability addition theorem

The sum, or unification, of several events is called an event consisting in the occurrence of at least one of these events (in the same test).
Sum A 1 + A 2 + … + An denoted as follows:
or .
Example... Two dice are thrown. Let the event A consists in the loss of 4 points on 1 die, and the event V- in the fall of 5 points on the other die. Developments A and V are joint. Therefore the event A +V consists in dropping 4 points on the first die, or 5 points on the second die, or 4 points on the first die and 5 points on the second at the same time.
Example. Event A- winnings for 1 loan, event V- winnings on the 2nd loan. Then the event A + B- winning at least one loan (possibly two at once).
By product or the intersection of several events is an event consisting in the joint appearance of all these events (in the same test).
Work V events A 1 , A 2 , …, An denoted as follows:
.
Example. Developments A and V consist in the successful completion of the I and II rounds, respectively, when entering the institute. Then the event A× B consists in successfully completing both rounds.
The concepts of sum and product of events have a clear geometric interpretation. Let the event A there is a hit of a point in the area A and the event V- hitting a point in the area V... Then the event A + B there is a hit of a point in the union of these areas (Fig. 2.1), and the event AV there is a hit of a point in the intersection of these areas (Fig. 2.2).

Rice. 2.1 Fig. 2.2
Theorem... If events A i(i = 1, 2, …, n) are pairwise inconsistent, then the probability of the sum of events is equal to the sum of the probabilities of these events:
.
Let be A and Ā - opposite events, i.e. A + Ā= Ω, where Ω is a valid event. From the addition theorem it follows that
P (Ω) = R(A) + R(Ā ) = 1, therefore
R(Ā ) = 1 – R(A).
If events A 1 and A 2 are consistent, then the probability of the sum of two joint events is:
R(A 1 + A 2) = R(A 1) + R(A 2) - P ( A 1 × A 2).
The theorems of addition of probabilities allow to go from direct calculation of probabilities to determination of probabilities of occurrence of complex events.
Task 1.8... The shooter fires one shot at the target. The probability of knocking out 10 points (event A), 9 points (event V) and 8 points (event WITH) are equal to 0.11, respectively; 0.23; 0.17. Find the probability that, with one shot, the shooter will score less than 8 points (event D).
Solution... Let's move on to the opposite event - with one shot, the shooter will knock at least 8 points. An event occurs if it happens A or V, or WITH, i.e. ... Since events A, B, WITH are pairwise inconsistent, then, by the addition theorem,
, where .
Task 1.9... From the team of the brigade, which consists of 6 men and 4 women, two people are selected to the trade union conference. What is the probability that among the selected at least one woman (event A).
Solution... If an event occurs A, then one of the following inconsistent events will surely occur: V- “a man and a woman were selected”; WITH- “two women were chosen”. Therefore, we can write: A = B + C... Find the probability of events V and WITH... Two people out of 10 can be selected in ways. Two women out of 4 can be chosen in ways. A man and a woman can be selected in 6 × 4 ways. Then . Since events V and WITH are inconsistent, then, by the addition theorem,
P (A) = P (B + C) = P (B) + P (C) = 8/15 + 2/15 = 2/3.
Task 1.10. Fifteen textbooks are randomly arranged on a shelf in the library, five of which are bound. The librarian takes three textbooks at random. Find the probability that at least one of the textbooks taken will be bound (event A).
Solution... The first way. The requirement - at least one of the three bound textbooks taken - will be met if any of the following three inconsistent events occurs: V- one bound textbook, WITH- two bound textbooks, D- three bound textbooks.
Event of interest to us A can be represented as a sum of events: A = B + C + D... By the addition theorem,
P (A) = P (B) + P (C) + P (D). (2.1)
Find the probability of events B, C and D(see combinatorial schemes):

Representing these probabilities in equality (2.1), we finally obtain
P (A)= 45/91 + 20/91 + 2/91 = 67/91.
Second way. Event A(at least one of the three textbooks taken is bound) and Ā (none of the textbooks taken are bound) are opposite, so P (A) + P (Ā) = 1 (the sum of the probabilities of two opposite events is 1). From here P (A) = 1 – P (Ā). Event probability Ā (none of the textbooks taken are bound)
Seeking probability
P (A) = 1 - P (Ā) = 1 – 24/91 = 67/91.

1.2.5. Conditional probability. Probability multiplication theorem

Conditional probability P (B/A) is the probability of event B, calculated under the assumption that event A has already occurred.
Theorem... The probability of the joint occurrence of two events is equal to the product of the probabilities of one of them by the conditional probability of the other, calculated under the assumption that the first event has already occurred:
P (A∙B) = P (A) ∙ Р ( V/A). (2.2)
Two events are called independent if the occurrence of any of them does not change the probability of the occurrence of the other, i.e.
P (A) = P (A / B) or P (B) = P (B/A). (2.3)
If events A and V are independent, then formulas (2.2) and (2.3) imply
P (A∙B) = P (A)∙P (B). (2.4)
The converse is also true, i.e. if equality (2.4) holds for two events, then these events are independent. Indeed, formulas (2.4) and (2.2) imply
P (A∙B) = P (A)∙P (B) = P (A) × P (B/A), where P (A) = P (B/A).
Formula (2.2) can be generalized to the case of a finite number of events A 1 , A 2 ,…,A n:
P (A 1 ∙A 2 ∙…∙A n)=P (A 1)∙P (A 2 /A 1)∙P (A 3 /A 1 A 2)∙…∙P (A n/A 1 A 2 …A n -1).
Task 1.11... From the urn, in which there are 5 white and 10 black balls, take out two balls in a row. Find the probability that both balls are white (event A).
Solution... Consider the events: V- the first ball taken out is white; WITH- the second ball removed is white. Then A = BC.
The experiment can be done in two ways:
1) with return: the removed ball, after fixing the color, is returned to the urn. In this case, the events V and WITH independent:
P (A) = P (B)∙P (C) = 5/15 × 5/15 = 1/9;
2) no return: the removed ball is laid to the side. In this case, the events V and WITH dependent:
P (A) = P (B)∙P (C/V).
For an event V the conditions are the same, and for WITH the situation has changed. Occurred V, therefore, there are 14 balls left in the urn, among which 4 are white.
So, .
Task 1.12... Among the 50 light bulbs, 3 are non-standard. Find the probability that two bulbs taken at the same time are non-standard.
Solution... Consider the events: A- the first light is non-standard, V- the second light is non-standard, WITH- both bulbs are non-standard. It's clear that C = A∙V... Event A 3 cases out of 50 are favorable, i.e. P (A) = 3/50. If the event A has already arrived, then the event V two cases out of 49 are favorable, i.e. P (B/A) = 2/49. Hence,
.
Target 1.13... Two athletes shoot independently of each other at one target. The probability of hitting the target of the first athlete is 0.7, and the second is 0.8. What is the likelihood that the target will be hit?
Solution... The target will be hit if it hits either the first shooter, or the second, or both together, i.e. an event will occur A + B where the event is A is the first athlete hitting the target, and the event V- the second. Then
P (A+V)=P (A)+P (B)–P (A∙V)=0, 7+0, 8–0, 7∙0,8=0,94.
Task 1.14. There are six textbooks on probability theory in the reading room, of which three are bound. The librarian took two textbooks at random. Find the probability that the two textbooks will be bound.
Solution... Let us introduce the notation of events : A- the first textbook taken is bound, V- the second textbook is bound. The probability that the first textbook is bound is
P (A) = 3/6 = 1/2.
The probability that the second textbook is bound, provided that the first textbook taken was bound, i.e. conditional probability of an event V, is this: P (B/A) = 2/5.
Seeking the probability that both textbooks are bound, by the multiplication theorem for the probabilities of events is equal to
P (AB) = P (A) ∙ P (B/A)= 1/2 · ∙ 2/5 = 0.2.
Task 1.15. The workshop employs 7 men and 3 women. Three people were randomly selected by personnel numbers. Find the probability that all selected individuals will be men.
Solution... Let us introduce the notation of events: A- the man was selected first, V- a man was selected second, WITH - the third is a man. The likelihood that a man will be selected first, P (A) = 7/10.
The probability that a man is selected second, provided that a man has already been selected first, i.e. conditional probability of an event V next : P (B / A) = 6/9 = 2/3.
The probability that a man will be selected third, provided that two men have already been selected, i.e. conditional probability of an event WITH is this: P (C/AB) = 5/8.
Seeking the likelihood that all three selected individuals will be men, P (ABC) = P (A) P (B/A) P (C/AB) = 7/10 2/3 5/8 = 7/24.

1.2.6. Total probability formula and Bayes formula

Let be B 1 , B 2 ,…, B n- pairwise incompatible events (hypotheses) and A- an event that can occur only in conjunction with one of them.
Let, in addition, we know P (B i) and P (A/B i) (i = 1, 2, …, n).
Under these conditions, the following formulas are valid:
(2.5)
(2.6)
Formula (2.5) is called total probability formula ... It calculates the probability of an event A(full probability).
Formula (2.6) is called Bayes' formula ... It allows you to recalculate the probabilities of hypotheses if the event A occurred.
When compiling examples, it is convenient to assume that the hypotheses form a complete group.
Target 1.16... The basket contains apples from four trees of the same sort. From the first - 15% of all apples, from the second - 35%, from the third - 20%, from the fourth - 30%. Ripe apples account for 99%, 97%, 98%, 95%, respectively.
a) What is the probability that an apple taken at random will be ripe (event A).
b) Provided that the apple taken at random turned out to be ripe, calculate the probability that it is from the first tree.
Solution... a) We have 4 hypotheses:
B 1 - an apple taken at random is removed from the 1st tree;
B 2 - an apple taken at random is removed from the 2nd tree;
B 3 - an apple taken at random is removed from the 3rd tree;
B 4 - the apple taken at random is taken from the 4th tree.
Their probabilities by condition: P (B 1) = 0,15; P (B 2) = 0,35; P (B 3) = 0,2; P (B 4) = 0,3.
Conditional event probabilities A:
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 probability that an apple taken at random turns out to be ripe is found by the formula for the total probability:
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 looks like:
.
Task 1.17. A white ball is dropped into an urn containing two balls, after which one ball is taken at random from it. Find the probability that the removed ball will turn out to be white if all possible assumptions about the initial composition of the balls (by color) are equally possible.
Solution... Let us denote by A event - the white ball is removed. The following assumptions (hypotheses) about the initial composition of the balls are possible: B 1- there are no white balls, IN 2- one white ball, AT 3- two white balls.
Since there are three hypotheses in total, and the sum of the probabilities of the hypotheses is 1 (since they form a complete group of events), the probability of each of the hypotheses is 1/3, i.e.
P (B 1) = P (B 2)= P (B 3) = 1/3.
The conditional probability that a white ball will be drawn, provided that there were no white balls initially in the urn, P (A/B 1) = 1/3. The conditional probability that a white ball will be drawn, given that there was originally one white ball in the urn, P (A/B 2) = 2/3. The conditional probability that a white ball will be drawn, provided that there were two white balls in the urn initially P (A/B 3)=3/ 3=1.
We find the desired probability that the white ball will be drawn using the formula for the total probability:
R(A)=P (B 1)∙P (A/B 1)+P (B 2)∙P (A/B 2)+P (B 3)∙P (A/B 3) = 1/3 1/3 + 1/3 2/3 + 1/3 1 = 2/3 .
Target 1.18... Two machines produce identical parts that go to a common conveyor. The productivity of the first machine is twice that of the second. The first automatic machine produces on average 60% of the parts of excellent quality, and the second - 84%. The part taken at random from the assembly line turned out to be of excellent quality. Find the probability that this part was produced by the first machine.
Solution... Let us denote by A the event is an excellent quality item. Two assumptions can be made: B 1- the part is produced by the first machine, moreover (since the first machine produces twice as many parts as the second) P (A/B 1) = 2/3; B 2 - the part is produced by the second machine, and P (B 2) = 1/3.
The conditional probability that the part will be of excellent quality if it is produced by the first automatic machine, P (A/B 1)=0,6.
The conditional probability that the part will be of excellent quality if it is produced by the second automatic machine, P (A/B 1)=0,84.
The probability that a randomly taken part turns out to be of excellent quality, according to the formula for the total probability, is
P (A)=P (B 1) ∙P (A/B 1)+P (B 2) ∙P (A/B 2) = 2/3 0.6 + 1/3 0.84 = 0.68.
Seeking the probability that the taken excellent part is produced by the first automaton, according to the Bayesian formula is equal to

Target 1.19... There are three lots of parts with 20 parts each. The number of standard parts in the first, second and third lots is, respectively, 20, 15, 10. A part that turned out to be standard was randomly extracted from the selected lot. The parts are returned to the batch and, for the second time, a part is randomly removed from the same batch, which also turns out to be standard. Find the probability that the parts were recovered from the third batch.
Solution... Let us denote by A event - in each of the two tests (with return) a standard part was removed. Three assumptions (hypotheses) can be made: B 1 - parts are removed from the first batch, V 2 - parts are removed from the second batch, V 3 - parts are removed from the third batch.
The details were taken at random from the batch taken, so the probabilities of the hypotheses are the same: P (B 1) = P (B 2) = P (B 3) = 1/3.
Find the conditional probability P (A/B 1), i.e. the probability that two standard parts will be removed successively from the first batch. This event is reliable because in the first batch, all parts are standard, therefore P (A/B 1) = 1.
Find the conditional probability P (A/B 2), i.e. the probability that two standard parts will be sequentially removed (with return) from the second batch: P (A/B 2)= 15/20 ∙ 15/20 = 9/16.
Find the conditional probability P (A/B 3), i.e. the probability that two standard parts will be sequentially removed (with return) from the third batch: P (A/B 3) = 10/20 10/20 = 1/4.
Looking for the probability that both extracted standard parts are from a third batch, using Bayes's formula is

1.2.7. Repeated tests

If several tests are carried out, and the probability of an event A in each trial does not depend on the outcomes of other trials, then such trials are called independent regarding the event A. In various independent trials, the event A can have either different probabilities or the same probability. We will further consider only such independent tests in which the event A has the same probability.
Let it be produced NS independent tests, in each of which an event A may or may not appear. Let us agree to assume that the probability of an event A in each test is the same, namely equal to R. Therefore, the probability of non-occurrence of the event A in each test is also constant and equal to 1– R. Such a probabilistic scheme is called Bernoulli scheme... Let us set ourselves the task of calculating the probability that for NS Bernoulli test event A will come true exactly k once ( k Is the number of successes) and, therefore, will not be realized NS- once. It is important to emphasize that it is not required that the event A repeated exactly k times in a specific sequence. The required probability is denoted by P n (k). For example, the symbol R 5 (3) means the probability that in five tests the event will appear exactly 3 times and, therefore, will not occur 2 times.
The problem can be solved using the so-called Bernoulli formulas, which looks like:
.
Task 1.20. The probability that the consumption of electricity during one day will not exceed the established norm is equal to R= 0.75. Find the probability that in the next 6 days the power consumption for 4 days will not exceed the norm.
Solution. The probability of normal electricity consumption during each of 6 days is constant and equal to R= 0.75. Consequently, the probability of excessive consumption of electricity per day is also constant and equal to q = 1–R=1–0,75=0,25.
The desired probability by the Bernoulli formula is
.
Target 1.21... Two equivalent chess players play chess. Which is more likely: to win two games out of four or three games out of six (draws are not taken into account)?
Solution... Equivalent chess players play, so the probability of winning R= 1/2, therefore, the probability of losing q is also 1/2. Because in all games the probability of winning is constant and it does not matter in what sequence the games will be won, then the Bernoulli formula is applicable.
Let's find the probability that two games out of four will be won:

Let us find the probability that three games out of six will be won:

Because P 4 (2) > P 6 (3), it is more likely to win two games out of four than three out of six.
However, one can see that using Bernoulli's formula for large values n it is quite difficult, since the formula requires performing actions on huge numbers and therefore in the process of calculations errors accumulate; as a result, the final result may 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's theorem
When carrying out a large number of tests according to the Bernoulli scheme (with n=> ∞) and with a small number of favorable outcomes k(in this case, it is assumed that the probability of success p is small), Bernoulli's formula approaches Poisson's formula
.
Example 1.22. The probability of a marriage when a unit of production is produced by an enterprise is equal to p= 0.001. What is the probability that with the release of 5000 units of products there will be less than 4 defective ones (event A Solution... Because n is large, we use the local Laplace theorem:

Let's calculate x:
Function - even, therefore φ (–1.67) = φ (1.67).
According to the table in Appendix A.1, we find φ (1.67) = 0.0989.
Seeking probability P 2400 (1400) = 0,0989.
3. Laplace's integral theorem
If the probability R occurrence of an event A in each test according to the Bernoulli scheme is constant and different from zero and one, then with a large number of tests n, probability P n (k 1 , k 2) the occurrence of the event A in these tests from k 1 to k 2 times is approximately equal to
R p(k 1 , k 2) = Φ ( x "") – Φ ( x "), where
- Laplace function,

The definite integral in the Laplace function is not calculated in the class of analytic functions, therefore, to calculate it, Table 1 is used. A.2, given in the appendix.
Example 1.24. The probability of occurrence of an event in each of one hundred independent tests is constant and equal to p= 0.8. Find the probability that the event will occur: a) at least 75 times and not more than 90 times; b) at least 75 times; c) no more than 74 times.
Solution... We will use Laplace's integral theorem:
R p(k 1 , k 2) = Φ ( x "") – Φ( x "), where Ф ( x) Is the Laplace function,

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


Considering that the Laplace function is odd, i.e. Ф (- x) = - Ф ( x), we get
P 100 (75; 90) = Ф (2.5) - Ф (–1.25) = Ф (2.5) + Ф (1.25).
According to the table. A.2. we will find applications:
F (2.5) = 0.4938; Ф (1.25) = 0.3944.
Seeking probability
P 100 (75; 90) = 0,4938 + 0,3944 = 0,8882.
b) The requirement that the event appears at least 75 times means that the number of occurrences of the event can be equal to 75, or 76, ..., or 100. Thus, in this case, one should accept k 1 = 75, k 2 = 100. Then

.
According to the table. A.2. applications, we find Ф (1.25) = 0.3944; Ф (5) = 0.5.
Seeking probability
P 100 (75;100) = (5) – (–1,25) = (5) + (1,25) = 0,5 + 0,3944 = 0,8944.
c) Event - " A appeared at least 75 times "and" A appeared no more than 74 times "are opposite, so the sum of the probabilities of these events is 1. Therefore, the desired probability
P 100 (0;74) = 1 – P 100 (75; 100) = 1 – 0,8944 = 0,1056.

For practical activity, it is necessary to be able to compare events according to the degree of their possibility of occurrence. Consider the classic case. The urn contains 10 balls, 8 of them are white, 2 are black. Obviously, the event “a white ball will be removed from the urn” and the event “a black ball will be removed from the urn” have different degrees of possibility of their occurrence. Therefore, a certain quantitative measure is needed to compare events.

A quantitative measure of the possibility of an event occurring is probability ... The most widespread are two definitions of the probability of an event: classical and statistical.

Classic definition probability is associated with the concept of a favorable outcome. Let's dwell on this in more detail.

Let the outcomes of some test form a complete group of events and are equally possible, i.e. are the only possible, incompatible and equally possible. Such outcomes are called elementary outcomes, or cases... At the same time, they say that the test is reduced to case diagram or " urn scheme", Because any probabilistic problem for such a test can be replaced by an equivalent problem with urns and balls of different colors.

Exodus is called favorable event A if the occurrence of this event entails the occurrence of an event A.

According to the classical definition probability of event A is equal to the ratio of the number of outcomes favorable to this event to the total number of outcomes, i.e.

where P (A)- probability of an event A; m- the number of cases favorable to the event A; n- the total number of cases.

Example 1.1. When throwing a dice, six outcomes are possible - 1, 2, 3, 4, 5, 6 points are dropped. What is the probability that an even number of points will appear?

Solution. Everything n= 6 outcomes form a complete group of events and are equally possible, i.e. are the only possible, incompatible and equally possible. Event A - “the appearance of an even number of points” - 3 outcomes (cases) are favored - 2, 4 or 6 points are dropped. According to the classical formula for the probability of an event, we obtain

P (A) = = . ◄

Based on the classical definition of the probability of an event, we note its properties:

1. The probability of any event lies between zero and one, i.e.

0 ≤ R(A) ≤ 1.

2. The probability of a reliable event is equal to one.

3. The probability of an impossible event is zero.

As mentioned earlier, the classical definition of probability is applicable only to those events that can appear as a result of trials with the symmetry of possible outcomes, i.e. reduced to the scheme of cases. However, there is a large class of events, the probabilities of which cannot be calculated using the classical definition.

For example, if we assume that the coin is flattened, then it is obvious that the events “appearance of the coat of arms” and “appearance of tails” cannot be considered equally possible. Therefore, the formula for determining the probability according to the classical scheme is inapplicable in this case.

However, there is a different approach to assessing the likelihood of events based on how often the event will occur in the tests performed. In this case, a statistical definition of probability is used.

Statistical probabilityevent A is called the relative frequency (frequency) of the occurrence of this event in n tests performed, i.e.

,

(1.2)

where P * (A)- statistical probability of an event A; w (A)- relative frequency of the event A; m- the number of trials in which the event occurred A; n- the total number of tests.

Unlike mathematical probability P (A) considered in the classical definition, the statistical probability P * (A) is a characteristic experienced, experimental... In other words, the statistical probability of the event A is the number relative to which the relative frequency is stabilized (set) w (A) with an unlimited increase in the number of tests carried out under the same set of conditions.

For example, when a shooter is said to hit the target with a probability of 0.95, this means that out of hundreds of shots fired by him under certain conditions (the same target at the same distance, the same rifle, etc. .), on average there are about 95 successful ones. Naturally, not every hundred will have 95 successful shots, sometimes there will be fewer, sometimes more, but on average, with multiple repetitions of shooting under the same conditions, this percentage of hits will remain unchanged. The number 0.95, which is an indicator of the skill of the shooter, is usually very stable, i.e. the percentage of hits in most shooting will be almost the same for a given shooter, only in rare cases deviating somewhat significantly from its average value.

Another drawback of the classical definition of probability ( 1.1 ), limiting its use, is that it assumes a finite number of possible trial outcomes. In some cases, this disadvantage can be overcome by using a geometric definition of probability, i.e. finding the probability of a point hitting a certain area (segment, part of a plane, etc.).

Let a flat figure g forms part of a flat figure G(fig. 1.1). On the figure G a point is thrown at random. This means that all points of the region G"Equal" in relation to hitting it with a thrown random point. Assuming that the probability of an event A- hitting the thrown point on the figure g- proportional to the area of ​​this figure and does not depend on its location relative to G nor from the form g, find

When assessing the probability of the occurrence of any random event, it is very important to have a good preliminary understanding of whether the probability (probability of an event) of the occurrence of an event of interest to us depends on how other events develop. In the case of the classical scheme, when all outcomes are equally probable, we can already estimate the values ​​of the probability of the particular event of interest to us independently. We can do this even if the event is a complex collection of several elementary outcomes. What if several random events occur simultaneously or sequentially? How does this affect the likelihood of the event of interest to us? If I roll the dice several times and want a six to come up, but I’m not lucky all the time, does this mean that I need to increase the bet, because, according to the theory of probability, I’m about to get lucky? Alas, the theory of probability does not state anything of the kind. Neither dice, nor cards, nor coins can remember what they showed us last time. It doesn't matter to them at all whether for the first time or for the tenth time today I try my fate. Every time I repeat the throw, I only know one thing: and this time the probability of getting a "six" is again equal to one-sixth. Of course, this does not mean that the number I need will never fall out. It only means that my loss after the first throw and after any other throw are independent events. Events A and B are called independent if the implementation of one of them does not affect the probability of another event in any way. For example, the probabilities of hitting a target with the first of two guns do not depend on whether the target was hit by another gun, so the events "the first gun hit the target" and "the second gun hit the target" are independent. If two events A and B are independent, and the probability of each of them is known, then the probability of simultaneous occurrence of both events A and events B (denoted by AB) can be calculated using the following theorem.

Multiplication theorem for probabilities for independent events

P (AB) = P (A) * P (B) the probability of the simultaneous occurrence of two independent events is equal to the product of the probabilities of these events.

Example 1... The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 = 0.7; p 2 = 0.8. Find the probability of hitting with one volley with both guns simultaneously.

as we have already seen events A (hit by the first gun) and B (hit by the second gun) are independent, i.e. P (AB) = P (A) * P (B) = p1 * p2 = 0.56. What happens to our estimates if the initiating events are not independent? Let's modify the previous example a bit.

Example 2. Two shooters in competitions shoot at targets, and if one of them shoots accurately, the opponent becomes nervous and his results deteriorate. How to turn this everyday situation into a mathematical problem and outline ways to solve it? It is intuitively clear that it is necessary to somehow separate the two scenarios for the development of events, to draw up essentially two scenarios, two different tasks. In the first case, if the opponent misses, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent has decently realized his chance, the probability of hitting the target for the second athlete is reduced. To separate possible scenarios (often called hypotheses) of events, we will often use a "probability tree" scheme. This scheme is similar in meaning to the decision tree that you have probably already dealt with. Each branch represents a separate scenario of the development of events, only now it has its own value of the so-called conditional probability (q 1, q 2, q 1 -1, q 2 -1).

This scheme is very convenient for analyzing sequential random events. It remains to clarify one more important question: where do the initial values ​​of probabilities in real situations come from? After all, the theory of probability works not with the same coins and dice? Usually these estimates are taken from statistics, and when statistics are not available, we conduct our own research. And we often have to start it not with collecting data, but with the question of what information we generally need.

Example 3. Let's say we need to estimate the market size in a city with a population of one hundred thousand inhabitants for a new product that is not an essential item, for example, a balm for the care of colored hair. Consider a "probability tree" scheme. In this case, the value of the probability on each "branch" we need to approximately estimate. So, our estimates of the market size:

1) 50% of all residents of the city are women,

2) of all women, only 30% dye their hair often,

3) only 10% of them use balms for colored hair,

4) of these, only 10% can get up the courage to try a new product,

5) 70% of them usually buy everything not from us, but from our competitors.

According to the law of multiplication of probabilities, we determine the probability of the event of interest to us A = (a resident of the city buys this new balm from us) = 0.00045. Let's multiply this value of the probability by the number of inhabitants of the city. As a result, we have only 45 potential customers, and if we consider that one bubble of this money is enough for several months, the trade is not very lively. Still, there is some benefit from our assessments. First, we can compare the forecasts of different business ideas, they will have different “forks” on the diagrams, and, of course, the probability values ​​will also be different. Secondly, as we have already said, a random variable is not called random because it does not depend on anything at all. It's just that its exact meaning is not known in advance. We know that the average number of buyers can be increased (for example, by advertising a new product). So it makes sense to focus our efforts on those “forks” where the probability distribution does not particularly suit us, on those factors that we are able to influence. Consider another quantitative example of shopping behavior research.

Example 3. On average, 10,000 people visit the food market per day. The probability that a market visitor enters a dairy pavilion is 1/2. It is known that in this pavilion, on average, 500 kg of various products are sold per day. Can we say that the average purchase in a pavilion weighs only 100 g?

Discussion.

Of course not. It is clear that not everyone who entered the pavilion ended up buying something there.

As shown in the diagram, in order to answer the question about the average purchase weight, we must find the answer to the question, what is the probability that a person who enters the pavilion will buy something there. If we do not have such data at our disposal, and we need them, we will have to obtain them ourselves, after observing the visitors of the pavilion for some time. Let's say our observations have shown that only one-fifth of the visitors to the pavilion buy something. As soon as we have received these estimates, the task becomes already simple. Out of 10,000 people who came to the market, 5,000 will enter the pavilion of dairy products, and there will be only 1,000 purchases. The average purchase weight is 500 grams. It is interesting to note that to build a complete picture of what is happening, the logic of conditional "branches" must be defined at each stage of our reasoning as clearly as if we were working with a "specific" situation, and not with probabilities.

Tasks for self-testing.

1. Let there be an electrical circuit consisting of n series-connected elements, each of which works independently of the others. The probability p of failure of each element is known. Determine the probability of correct operation of the entire section of the chain (event A).

2. The student knows 20 of the 25 exam questions. Find the probability that the student knows the three questions suggested by the examiner.

3. Production consists of four successive stages, at each of which the equipment operates, for which the probabilities of failure within the next month are equal to p 1, p 2, p 3 and p 4, respectively. Find the probability that there will be no production interruption due to equipment failure in a month.