Classical, statistical and geometric definitions of probability. Abstract: Statistical determination of probability

For practical activities it is necessary to be able to compare events according to the degree of possibility of their occurrence. Let's consider a classic case. There are 10 balls in the urn, 8 of them white, 2 black. Obviously, the event “a white ball will be drawn from the urn” and the event “a black ball will be drawn from the urn” have different degrees of possibility of their occurrence. Therefore, to compare events, a certain quantitative measure is needed.

A quantitative measure of the possibility of an event occurring is probability . The most widely used definitions of the probability of an event are classical and statistical.

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

Let the outcomes of some test form a complete group of events and are equally possible, i.e. uniquely possible, incompatible and equally possible. Such outcomes are called elementary outcomes, or cases. It is said that the test boils down to case scheme or " urn scheme", because Any probability problem for such a test can be replaced by an equivalent problem with urns and balls of different colors.

The outcome is called favorable event A, if the occurrence of this case entails the occurrence of the event A.

According to the classical definition probability of an 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 event A; m– number of cases favorable to the event A; n– total number of cases.

Example 1.1. When throwing a dice, there are six possible outcomes: 1, 2, 3, 4, 5, 6 points. What is the probability of getting an even number of points?

Solution. All n= 6 outcomes form a complete group of events and are equally possible, i.e. uniquely possible, incompatible and equally possible. Event A - “the appearance of an even number of points” - is favored by 3 outcomes (cases) - the loss of 2, 4 or 6 points. Using 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 stated earlier, the classical definition of probability is applicable only for those events that can arise as a result of tests that have symmetry of possible outcomes, i.e. reducible to a pattern of cases. However, there is a large class of events whose probabilities 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 a coat of arms” and “appearance of heads” cannot be considered equally possible. Therefore, the formula for determining the probability according to the classical scheme is not applicable in this case.

However, there is another approach to estimating the probability of events, based on how often a given event will occur in the trials performed. In this case, the statistical definition of probability is used.

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

,

(1.2)

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

Unlike mathematical probability P(A), considered in the classical definition, statistical probability P*(A) is a characteristic experienced, experimental. In other words, the statistical probability of an event A is the number around 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 they say about a shooter that he hits 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, when shooting is repeated many times under the same conditions, this percentage of hits will remain unchanged. The figure of 0.95, which serves as an indicator of the shooter's skill, is usually very stable, i.e. the percentage of hits in most shootings will be almost the same for a given shooter, only in rare cases deviating any significantly from its average value.

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

Let the flat figure g forms part flat figure G(Fig. 1.1). Fit G a dot is thrown at random. This means that all points in the region G“equal rights” with respect to whether a thrown random point hits it. Assuming that the probability of an event A– the thrown point hits the figure g– is proportional to the area of ​​this figure and does not depend on its location relative to G, neither from the form g, we'll find

Rice. 1.1 Fig 1.2

Example 1.2. Two students agreed to meet at a certain place between 10 and 11 o'clock in the afternoon. The first person to arrive waits for the second person for 15 minutes and then leaves. Find the probability that the meeting will take place if each student randomly chooses the time of his arrival between 10 and 11 o'clock.

Solution. Let us denote the moments of arrival of the first and second students at a certain place, respectively, by x And y. In a rectangular coordinate system Oxy Let's take 10 hours as the starting point, and 1 hour as the unit of measurement. By condition 0 ≤ x ≤ 1, 0 ≤ y≤ 1. These inequalities are satisfied by the coordinates of any point belonging to the square OKLM with a side equal to 1 (Fig. 1.2). Event A– meeting of two students – will happen if the difference between x and not y will exceed 1/4 hour (in absolute value), i.e. | y – x| ≤ 0,25.

The solution to this inequality is the strip x – 0,25 ≤ y ≤ x+ 0.25, which is inside the square G represents the shaded area g. According to formula (1.3)

Classic definition of probability.

Various definitions of probability.

Algebra of events.

In order to quantitatively compare events with each other according to the degree of their possibility, obviously, it is necessary to associate a certain number with each event, the more possible the event, the greater the number. We will call this number the probability of an event. Τᴀᴋᴎᴍ ᴏϬᴩᴀᴈᴏᴍ, probability of an event is a numerical measure of the degree of objective possibility of this event.

The first definition of probability should be considered classical, which arose from the analysis of gambling and was initially applied intuitively.

The classical method of determining probability is based on the concept of equally possible and incompatible events, which are the outcomes of a given experience and form a complete group of incompatible events.

Most simple example equally possible and incompatible events that form a complete group is the appearance of one or another ball from an urn containing several balls of the same size, weight and other tangible characteristics, differing only in color, thoroughly mixed before removal.

For this reason, a test whose outcomes form a complete group of incompatible and equally possible events is said to reduce to a pattern of urns, or case scheme, or fits into the classical scheme.

We will simply call the equally possible and incompatible events that make up the complete group cases or chances. Moreover, in each experiment, along with cases, more complex events can occur.

Example: When throwing a dice, along with the cases A i - the loss of i-points on the upper side, we can consider such events as B - the loss of an even number of points, C - the loss of a number of points that are a multiple of three...

In relation to each event that can occur during the experiment, cases are divided into favorable, in which this event occurs, and unfavorable, in which the event does not occur. In the previous example, event B is favored by cases A 2, A 4, A 6; event C – cases A 3, A 6.

Classical probability the occurrence of a certain event is usually called the ratio of the number of cases favorable to the occurrence of this event to the total number of equally possible, incompatible cases that make up the complete group in a given experiment:

Where P(A)– probability of occurrence of event A; m- the number of cases favorable to event A; n- total number of cases.

Examples:

1) (see example above) P(B)=, P(C)=.

2) The urn contains 9 red and 6 blue balls. Find the probability that one or two balls drawn at random will turn out to be red.

A- a red ball drawn at random:

m=9, n=9+6=15, P(A)=

B- two red balls drawn at random:

The following follows from the classical definition of probability: properties(show yourself):

1) The probability of an impossible event is 0;

2) The probability of a reliable event is 1;

3) The probability of any event lies between 0 and 1;

4) The probability of an event opposite to event A,

The classical definition of probability assumes that the number of outcomes of a trial is finite. In practice, very often there are tests, the number of possible cases of which is infinite. At the same time, weak side The classical definition is that very often it is impossible to present the result of a test in the form of a set of elementary events. It is even more difficult to indicate the reasons for considering the elementary outcomes of a test to be equally possible. Usually, the equipossibility of elementary test outcomes is concluded from considerations of symmetry. However, such tasks are very rare in practice. For these reasons, along with the classical definition of probability, other definitions of probability are used.

Statistical probability event A is usually called the relative frequency of occurrence of this event in the tests performed:

where is the probability of occurrence of event A;

– relative frequency of occurrence of event A;

The number of trials in which event A appeared;

Total number of trials.

Unlike classical probability, statistical probability is an experimental characteristic.

Example: To control the quality of products from the batch, 100 products were selected at random, among which 3 products turned out to be defective. Determine the probability of marriage.

.

The statistical method of determining probability is applicable only to those events that have the following properties:

· The events under consideration should be the outcomes of only those tests that can be reproduced an unlimited number of times under the same set of conditions.

· Events must have statistical stability (or stability of relative frequencies). This means that in various series tests, the relative frequency of the event changes slightly.

· The number of trials resulting in event A must be large enough.

It is easy to verify that the properties of probability arising from the classical definition are also preserved in the statistical definition of probability.

Statistical definition of probability. - concept and types. Classification and features of the category "Statistical determination of probability." 2017, 2018.

Consider a random experiment in which a die made of a heterogeneous material is tossed. Its center of gravity is not at the geometric center. In this case, we cannot consider the outcomes (losing a one, two, etc.) to be equally probable. It is known from physics that the bone will more often fall on the face that is closer to the center of gravity. How to determine the probability of getting, for example, three points? The only thing you can do is roll this die n times (where n is a fairly large number, say n=1000 or n=5000), count the number of three points rolled n 3 and consider the probability of the outcome of rolling three points to be n 3 /n – relative frequency of three points. In a similar way, you can determine the probabilities of other elementary outcomes - one, two, four, etc. In theory, this course of action can be justified by introducing a statistical definition of probability.

Probability P(wi) is defined as the limit on the relative frequency of occurrence of the outcome w i in the process of an unlimited increase in the number of random experiments n, that is

where m n (w i) is the number of random experiments (from total number n random experiments performed) in which the occurrence of an elementary outcome w i was recorded.

Since no proof is given here, we can only hope that the limit in the last formula exists, justifying the hope life experience and intuition.

In practice, very often problems arise in which it is impossible or extremely difficult to find any other way to determine the probability of an event, other than a statistical determination.

Continuous probability space.

As mentioned earlier, the set of elementary outcomes can be more than countable (that is, uncountable). Thus, an experiment consisting of randomly throwing a point onto a segment has an innumerable number of outcomes. One can imagine that an experiment involving measuring temperature in given moment V given point also has an uncountable number of outcomes (indeed, temperature can take any value from a certain interval, although in reality we can measure it only with a certain accuracy, and the practical implementation of such an experiment will give a finite number of outcomes). In the case of an experiment with an uncountable set W of elementary outcomes, any subset of the set W cannot be considered an event. It should be noted that subsets of W that are not events are mathematical abstractions and do not occur in practical problems. Therefore, in our course this paragraph is optional.

To introduce the definition of a random event, consider a system (finite or countable) of subsets of the space of elementary outcomes W.

If two conditions are met:

1) from A’s membership in this system, it follows that A belongs to this system;

2) from belonging to this system it follows that A i A j belongs to this system

such a system of subsets is called algebra.

Let W be some space of elementary outcomes. Make sure that the two subset systems are:

1) W, Æ; 2) W, A, , Æ (here A is a subset of W) are algebras.

Let A 1 and A 2 belong to some algebra. Prove that A 1 \ A 2 belong to this algebra.

Let us call an s-algebra a system I of subsets of a set W that satisfies condition 1) and condition 2)¢:

2)¢ if the subsets A 1, A 2,¼, A n, ¼ belong to I, then their countable union (by analogy with summation, this countable union is briefly written by the formula) also belongs to I.

A subset A of the set of elementary outcomes W is an event if it belongs to some s-algebra.

It can be proven that if we choose any countable system of events belonging to some s-algebra and carry out with these events any operations accepted in set theory (union, intersection, difference and addition), then the result will be a set or event belonging to the same s-algebra algebra.

Let us formulate an axiom called the axiom of A.N. Kolmogorov.

Each event corresponds to a non-negative number P(A) that does not exceed one, called the probability of event A, and the function P(A) has the following properties:

2) if the events A 1 , A 2 ,..., A n , ¼ are inconsistent, then

If a space of elementary outcomes W, an algebra of events, and a function P defined on it that satisfies the conditions of the above axiom are given, then they say that a probability space is given.

This definition of a probability space can be extended to the case of a finite space of elementary outcomes W. Then the system of all subsets of the set W can be taken as an algebra.

Geometric probability

In one special case, we will give a rule for calculating the probability of an event for a random experiment with an uncountable set of outcomes.

If a one-to-one correspondence can be established between the set W of elementary outcomes of a random experiment and the set of points of some flat figure S (large sigma), and one-to-one correspondence can also be established between the set of elementary outcomes favorable to event A and the set of points of the flat figure s ( sigma small), which is part of the figure S, then

where s is the area of ​​the figure s, S is the area of ​​the figure S. Here, naturally, it is assumed that the figures S and s have areas. In particular, for example, the figure s can be a straight line segment with an area equal to zero.

Note that in this definition, instead of a flat figure S, we can consider the interval S, and instead of its part s, we can consider the interval s, which entirely belongs to the interval s, and the probability can be represented as the ratio of the lengths of the corresponding intervals.

Example. Two people have lunch in the dining room, which is open from 12 to 13 hours. Each of them comes at a random time and has lunch within 10 minutes. What is the probability of their meeting?

Let x be the time the first one arrives at the dining room, and y the time the second one arrives.

One can establish a one-to-one correspondence between all pairs of numbers (x;y) (or set of outcomes) and the set of points of a square with side equal to 1 on the coordinate plane, where the origin corresponds to the number 12 on the X axis and on the Y axis, as shown in Figure 6. Here, for example, point A corresponds to the outcome that the first one arrived at 12.30, and the second at 13.00. In this case, obviously, the meeting did not take place.

If the first one arrived no later than the second one (y ³ x), then the meeting will take place under the condition 0 £ y - x £ 1/6 (10 minutes is 1/6 hour).

If the second one arrived no later than the first one (x³y), then the meeting will occur under the condition 0 £ x – y £ 1/6..

A one-to-one correspondence can be established between the set of outcomes favorable to the meeting and the set of points in the region s shown in shaded form in Figure 7.

The required probability p is equal to the ratio of the area of ​​region s to the area of ​​the entire square. The area of ​​the square is equal to unity, and the area of ​​the region s can be defined as the difference between unity and the total area of ​​the two triangles shown in Figure 7. This implies:

Problems with solutions.

A coin with a radius of 1.5 cm is thrown onto a chessboard with a square 5 cm wide. Find the probability that the coin will not land on any cell boundary.

Task II.

A bridge spans the 100 m wide river. At some point, when there are two people on the bridge, the bridge collapses and both of them fall into the river. The first one knows how to swim and will be saved. The second one does not know how to swim, and will be saved only if it falls no further than 10 meters from the shore or no further than 10 meters from the first. What is the probability that the second person will be saved?

Task III.

Anti-tank mines are placed on a straight line 15 m apart. A tank 2 m wide drives perpendicular to this straight line. What is the probability that he will not be blown up by a mine?

Task VI.

In the interval (0; 2), two numbers are randomly selected. Find the probability that the square more less than the smaller number

Two points are thrown at random onto a segment. They break the segment into three parts. What is the probability that a triangle can be formed from the resulting segments?

Task VI.

Three points are thrown at random onto a segment, one after the other. What is the probability that the third point will fall between the first two?

Problem I. The position of a coin on a chessboard is completely determined by the position of its geometric center. The entire set of outcomes can be depicted as a square S with side 5. The entire set of favorable outcomes is then depicted as a square s lying inside the square S, as shown in Figure 1.

The desired probability is then equal to the ratio of the area of ​​the small square to the area of ​​the large square, that is, 4/25

Task II. Let us denote by x the distance from the left bank of the river to the point of fall of the first person, and by y the distance from the left bank to the point of fall of the second person. Obviously, both x and y belong to the interval (0;100). Thus, we can conclude that the entire set of outcomes can be mapped onto a square, the lower left corner of which lies at the origin of coordinates, and the upper right corner lies at the point with coordinates (100;100). Two lanes: 0 x, that is, the second fell closer to the right bank than the first, then in order for him to be saved, the condition y must be met<х+10. Если уx–10. From the above it follows that all outcomes favorable for the second person are displayed in the shaded area in Figure 2. The area of ​​this area is most easily calculated by subtracting the area of ​​two unshaded triangles from the area of ​​the entire square, which gives the result 10000–6400=3600. The required probability is 0.36.

Task III.

According to the conditions of the problem, the position of the tank in the gap between two adjacent mines is completely determined by the position of a straight line equidistant from the sides of the tank. This line is perpendicular to the line along which the mines are laid, and the tank is blown up by a mine if this line is located closer than 1 meter from the edge of the gap. Thus, the entire set of outcomes is mapped to an interval of length 15, and the set of favorable outcomes is mapped to an interval of length 13, as shown in Figure 3. The desired probability is 13/15.

Task IV.

Let's denote one of the numbers as x and the other as y. The entire set of possible outcomes is mapped into a square OBCD, two sides of which coincide with the coordinate axes and have a length equal to 2, as shown in Figure 4. Let us assume that y is a smaller number. Then the set of outcomes is mapped into triangle OCD with area equal to 2. The chosen numbers must satisfy two inequalities:

at<х, у>x 2

The set of numbers that satisfy these inequalities is displayed in the shaded area in Figure 4. The area of ​​this area is determined as the difference between the area of ​​the triangle OEG, equal to 1/2, and the area of ​​the curvilinear triangle OFEG. The area s of this curvilinear triangle is given by the formula

and is equal to 1/3. From this we find that the area of ​​the shaded figure OEF is 1/6. Thus, the desired probability is 1/12.

Let the length of the segment be l. If we take x and y to be the distances from the left end of the segment to the points mentioned in the problem statement, then the set of all outcomes can be mapped onto a square with side l, one of the sides of which lies on the x coordinate axis, and the other on the y coordinate axis . If we accept the condition y>x, then the set of outcomes will be mapped onto the triangle OBC shown in Figure 5. The area of ​​this triangle is l 2 /2. The resulting segments will have lengths: x, y–x and l-y. Now let's remember geometry. A triangle can be formed from three segments if and only if the length of each segment is less than the sum of the lengths of the other two segments. This condition in our case leads to a system of three inequalities

The first inequality is transformed to the form x l/2, and the third inequality takes the form y<х+l/2. Множество пар чисел х, у, являющееся решением системы неравенств отображается в заштрихованный треугольник на рисунке 5. Площадь этого треугольника в 4 раза меньше площади треугольника OВС. Отсюда следует, что ответ задачи составляет 1/4.

Let's take the length of the segment to be l. Let the distance from the left end of the segment to the first point be x, to the second point – y, and to the third point – z. Then the entire set of outcomes is mapped into a cube, three edges of which lie on the x, y and z axes of the rectangular coordinate system, and with an edge of length l. Let's assume that y>x. Then the set of outcomes will be mapped into the direct prism ABCA 1 B 1 C 1 shown in Figure 6. The condition z>x means that all outcomes will be mapped to the region lying above the plane AD 1 C 1 B shown in Figure 7. This plane is now all valid outcomes will be mapped into a pyramid with a square AA 1 B 1 B at the base and a height B 1 C 1 . All outcomes satisfying the condition z

Problems for independent solution.

1. Two ships must approach the same pier. The arrival times of both ships are independent and equally possible during a given day. Determine the probability that one of the steamships will have to wait for the berth to clear if the first steamer's stay time is one hour, and the second one is two hours. Answer: 139/1152.

2. An automatic traffic light is installed at the intersection, in which the light is green for one minute and red for half a minute, then green again for one minute and red for half a minute, etc. At a random moment in time, a car approaches the intersection. What is the probability that he will cross the intersection without stopping? Answer: 2/3

3. A coin of radius 1.5 cm is thrown onto an infinite chessboard with a square 5 cm wide. Find the probability that a coin will be located in no more than two squares of the chessboard. Answer: 16/25.

4. A triangle is randomly fit into the circle. What is the probability that it is acute? Answer: 1/4.

5. A triangle is randomly fit into a circle. What is the probability that it is rectangular? Answer: 0.

6. A rod of length a is randomly broken into three parts. Find the probability that the length of each part is greater than a/4. Answer: 1/16.

The classical definition of probability assumes that all elementary outcomes equally possible. The equality of the outcomes of an experiment is concluded due to considerations of symmetry (as in the case of a coin or a dice). Problems in which symmetry considerations can be used are rare in practice. In many cases it is difficult to provide reasons for believing that all elementary outcomes are equally possible. In this regard, it became necessary to introduce another definition of probability, called statistical. To give this definition, the concept of relative frequency of an event is first introduced.

Relative frequency of the event, or frequency, is the ratio of the number of experiments in which this event occurred to the number of all experiments performed. Let us denote the frequency of the event by , then by definition

(1.4.1)
where is the number of experiments in which the event occurred and is the number of all experiments performed.

The event frequency has the following properties.

Observations made it possible to establish that the relative frequency has the properties of statistical stability: in various series of polynomial tests (in each of which this event may or may not appear), it takes values ​​quite close to some constant. This constant, which is an objective numerical characteristic of a phenomenon, is considered the probability of a given event.

Probability event is the number around which the values ​​of the frequency of a given event are grouped in various series of a large number of tests.

This definition of probability is called statistical.

In the case of a statistical definition, probability has the following properties:
1) the probability of a reliable event is equal to one;
2) the probability of an impossible event is zero;
3) the probability of a random event lies between zero and one;
4) the probability of the sum of two incompatible events is equal to the sum of the probabilities of these events.

Example 1. Out of 500 parts taken at random, 8 were defective. Find the frequency of defective parts.

Solution. Since in this case = 8, = 500, then in accordance with formula (1.4.1) we find

Example 2. The dice is tossed 60 times, while six appeared 10 times. What is the frequency of occurrence sixes?

Solution. From the conditions of the problem it follows that = 60, = 10, therefore

Example 3. Among 1000 newborns, there were 515 boys. What is the birth rate of boys?
Solution. Since in this case, , then .

Example 4. As a result of 20 shots at the target, 15 hits were obtained. What is the hit rate?

Solution. Since = 20, = 15, then

Example 5. When shooting at a target, hit rate = 0.75. Find the number of hits with 40 shots.

Solution. From formula (1.4.1) it follows that . Since = 0.75, = 40, then . Thus, 30 hits were received.

Example 6. www.. Of the sown seeds, 970 germinated. How many seeds were sown?

Solution. From formula (1.4.1) it follows that . Since , , then . So, 1000 seeds were sown.

Example 7. On a segment of the natural series from 1 to 20, find the frequency of prime numbers.

Solution. On the indicated segment of the natural series of numbers there are the following prime numbers: 2, 3, 5, 7, 11, 13, 17, 19; there are 8 of them in total. Since = 20, = 8, then the desired frequency

.

Example 8. Three series of multiple tosses of a symmetrical coin were carried out, the number of appearances of the coat of arms was calculated: 1) = 4040, = 2048, 2) = 12000, = 6019; 3) = 24000, = 12012. Find the frequency of the appearance of the coat of arms in each series of tests.

Solution. In accordance with formula (1.4.1) we find:

Comment. These examples indicate that over repeated trials, the frequency of an event differs little from its probability. The probability of a coat of arms appearing when tossing a coin is p = 1/2 = 0.5, since in this case n = 2, m = 1.

Example 9. Among the 300 parts produced on an automatic machine, there were 15 that did not meet the standard. Find the frequency of occurrence of non-standard parts.

Solution. In this case n = 300, m = 15, so

Example 10. The inspector, checking the quality of 400 products, found that 20 of them belonged to the second grade, and the rest - to the first. Find the frequency of products of the first grade, the frequency of products of the second grade.

Solution. First of all, let’s find the number of products of the first grade: 400 - 20 = 380. Since n = 400, = 380, then the frequency of products of the first grade

Similarly, we find the frequency of products of the second grade:

Tasks

1. The technical control department discovered 10 non-standard products in a batch of 1000 products. Find the frequency of manufacturing defective products.
2. To determine the quality of seeds, 100 seeds were selected and sown in laboratory conditions. 95 seeds sprouted normally. What is the frequency of normal seed germination?
3. Find the frequency of occurrence of prime numbers in the following segments of the natural series: a) from 21 to 40; b) from 41 to 50; c) from 51 to 70.
4. Find the frequency of occurrence of the digit in 100 tosses of a symmetrical coin. (Conduct the experiment yourself).
5. Find the frequency of a six in 90 tosses of a die.
6. By surveying all students in your course, determine the frequency of birthdays that fall in each month of the year.
7. Find the frequency of five-letter words in any newspaper text.

Answers

1. 0.01. 2. 0.95; 0.05. 3. a) 0.2; b) 0.3; c) 0.2.

Questions

1. What is event frequency?
2. What is the frequency of a reliable event?
3. What is the frequency of an impossible event?
4. What are the limits of the frequency of a random event?
5. What is the frequency of the sum of two incompatible events?
6. What definition of probability is called statistical?
7. What properties does statistical probability have?

Tags. Look .

The randomness of the occurrence of events is associated with the impossibility of predicting in advance the outcome of a particular test. However, if we consider, for example, a test: repeated coin toss, ω 1, ω 2, ..., ω n, then it turns out that in approximately half of the outcomes ( n / 2) a certain pattern is discovered that corresponds to the concept of probability.

Under probability events A is understood as a certain numerical characteristic of the possibility of an event occurring A. Let us denote this numerical characteristic R(A). There are several approaches to determining probability. The main ones are statistical, classical and geometric.

Let it be produced n tests and at the same time some event A it has arrived n A times. Number n A is called absolute frequency(or simply the frequency) of the event A, and the relation is called relative frequency of occurrence of event A. Relative frequency of any event characterized by the following properties:

The basis for applying the methods of probability theory to the study of real processes is the objective existence of random events that have the property of frequency stability. Multiple trials of the event being studied A show that at large n relative frequency ( A) remains approximately constant.

The statistical definition of probability is that the probability of event A is taken to be a constant value p(A), around which the values ​​of relative frequencies fluctuate (A) with an unlimited increase in the number of testsn.

Note 1. Note that the limits of change in the probability of a random event from zero to one were chosen by B. Pascal for the convenience of its calculation and application. In correspondence with P. Fermat, Pascal indicated that any interval could be chosen as the indicated interval, for example, from zero to one hundred and other intervals. In the problems below in this manual, probabilities are sometimes expressed as percentages, i.e. from zero to one hundred. In this case, the percentages given in the problems must be converted into shares, i.e. divide by 100.

Example 1. 10 series of coin tosses were carried out, each with 1000 tosses. Magnitude ( A) in each of the series is equal to 0.501; 0.485; 0.509; 0.536; 0.485; 0.488; 0.500; 0.497; 0.494; 0.484. These frequencies are grouped around R(A) = 0,5.

This example confirms that the relative frequency ( A) is approximately equal R(A), i.e.