The probability of an event. classical probability

12.10.2019

Classical probability and its properties

Probability is one of the basic concepts of probability theory. There are several definitions of this concept. Let us give a definition that is called classical.

Probability event is the ratio of the number of elementary outcomes that favor a given event to the number of all equally possible outcomes of experience in which this event can appear.

The probability of an event A is denoted by P(A)(here R- the first letter of the French word probability- probability).

According to the definition

where is the number of elementary test outcomes favoring the appearance of the event ;

The total number of possible elemental outcomes of the trial.

This definition of probability is called classic. It arose on initial stage development of the theory of probability.

The number is often referred to as the relative frequency of occurrence of the event. BUT in experience.

The greater the probability of an event, the more often it occurs, and vice versa, the lower the probability of an event, the less often it occurs. When the probability of an event is close to one or equal to one, then it occurs in almost all trials. Such an event is said to be almost certain, i.e., that one can certainly count on its offensive.

Conversely, when the probability is zero or very small, then the event occurs extremely rarely; such an event is said to be almost impossible.

Sometimes the probability is expressed as a percentage: R(A) 100% is the average percentage of the number of occurrences of the event A.

Example 2.13. When dialing a phone number, the subscriber forgot one digit and dialed it at random. Find the probability that the desired digit is dialed.

Decision.

Denote by BUT event - "the required number is dialed".

The subscriber could dial any of the 10 digits, so the total number of possible elementary outcomes is 10. These outcomes are incompatible, equally possible and form a complete group. Favors the event BUT only one outcome (the required number is only one).

The desired probability is equal to the ratio of the number of outcomes that favor the event to the number of all elementary outcomes:

The classical probability formula provides a very simple way to calculate probabilities that does not require experimentation. However, the simplicity of this formula is very deceptive. The fact is that when using it, as a rule, two very difficult questions arise:

1. How to choose a system of outcomes of experience so that they are equally likely, and is it possible to do this at all?

2. How to find numbers m and n?

If multiple subjects are involved in an experiment, it is not always easy to see equally likely outcomes.

The great French philosopher and mathematician d'Alembert entered the history of probability theory with his famous mistake, the essence of which was that he incorrectly determined the equiprobability of outcomes in an experiment with only two coins!

Example 2.14. ( d'Alembert error). Two identical coins are tossed. What is the probability that they fall on the same side?

d'Alembert's solution.

Experience has three equally possible outcomes:

1. Both coins will fall on the "eagle";

2. Both coins will fall on "tails";

3. One of the coins will land on heads, the other on tails.

Correct solution.

Experience has four equally possible outcomes:

1. The first coin will fall on the "eagle", the second also on the "eagle";

2. The first coin will fall on "tails", the second will also fall on "tails";

3. The first coin will land on heads, and the second on tails;

4. The first coin will land on tails, and the second on heads.

Of these, two outcomes will be favorable for our event, so the desired probability is equal to .

d'Alembert made one of the most common mistakes made when calculating probability: he combined two elementary outcomes into one, thereby making it unequal in probability to the remaining outcomes of the experiment.

Let's analyze the classical definition of probability using formulas and examples.

Random events are called incompatible if they cannot occur at the same time. For example, when we toss a coin, one thing will fall out - a "coat of arms" or a number "and they cannot appear at the same time, since it is logical that this is impossible. Events such as hit and miss after a shot is fired can be incompatible.

Random events of a finite set form full group pairwise incompatible events, if at each trial one appears, and only one of these events is the only possible one.

Consider the same coin tossing example:

First coin Second coin Events

1) "coat of arms" "coat of arms"

2) "coat of arms" "number"

3) "number" "coat of arms"

4) "number" "number"

Or abbreviated - "YY", - "MS", - "CH", - "CH".

The events are called equally possible, if the conditions of the study provide the same possibility of the appearance of each of them.

As you understand, when you toss a symmetrical coin, then it has the same possibilities, and there is a chance that both the “coat of arms” and the “number” will fall out. The same applies to throwing a symmetrical dice, since there is a possibility that faces with any number of 1, 2, 3, 4, 5, 6 may appear.

Let's say that now we throw the cube with a shift in the center of gravity, for example, towards the side with the number 1, then the opposite side, that is, the side with a different number, will most often fall out. Thus, in this model, the occurrence possibilities for each of the digits from 1 to 6 will be different.

Equally possible and uniquely possible random events are called cases.

There are random events that are cases, and there are random events that are not cases. Below are examples of these events.

Those cases, as a result of which a random event appears, are called favorable cases for this event.

If we denote by , which affect the event in all possible cases, and through - the probability of a random event , then we can write down the well-known classical definition of probability:

Definition

The probability of an event is the ratio of the number of cases favorable to this event to the total number of all possible cases, that is:

Probability Properties

The classical probability has been considered, and now we will analyze the main and important properties probabilities.

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

For example, if all the balls in the bucket are white, then the event , randomly select a white ball, is affected by the cases, .

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

Property 3. The probability of a random event is a positive number:

Hence, the probability of any event satisfies the inequality:

Now let's solve some examples on the classical definition of probability.

Examples of the classical definition of probability

Example 1

Task

There are 20 balls in a basket, 10 of them are white, 7 are red and 3 are black. One ball is chosen at random. A white ball (event ), a red ball (event ), and a black ball (event ) are selected. Find the probability of random events.

Decision

According to the condition of the problem, contribute to , and cases of possible, therefore, according to the formula (1):

is the probability of a white ball.

Similarly for red:

And for black: .

Answer

The probability of a random event , , .

Example 2

Task

There are 25 identical electric lamps in a box, 2 of them are defective. Find the probability that a randomly selected light bulb is not defective.

Decision

According to the condition of the problem, all lamps are the same and only one is selected. Total possibilities to choose . Among all 25 lamps, two are defective, which means that the remaining lamps are suitable. Therefore, according to formula (1), the probability of choosing a suitable electric lamp (event ) is equal to:

Answer

The probability that a randomly selected light bulb is not defective = .

Example 3

Task

Two coins are tossed at random. Find the probability of such events:

1) - on both coins the coat of arms fell out;

2) - on one of the coins a coat of arms fell out, and on the second - a number;

3) - numbers fell out on both coins;

4) - at least once the coat of arms fell out.

Decision

Here we are dealing with four events. Let us establish which cases contribute to each of them. The event is facilitated by one case, this is when the coat of arms fell out on both coins (abbreviated as “GG”).

To deal with the event, imagine that one coin is silver and the second is copper. When tossing coins, there may be cases:

1) on a silver coat of arms, on a copper coat of arms - a number (let's denote it as “MS”);

2) on a silver number, on a copper one - a coat of arms (- “ChG”).

Hence, the events are facilitated by the cases and .

The event is facilitated by one case: numbers fell out on both coins - “CH”.

Thus, the events or (YY, MG, TY, FF) form a complete group of events, all of these events are incompatible, since only one of them occurs as a result of the toss. In addition, for symmetric coins, all four events are equally likely, so they can be considered cases. There are four possible events.

An event is facilitated by only one event, so its probability is:

Two cases contribute to the event, so:

The probability of an event is the same as for:

Three cases contribute to the event: YY, YY, YY and therefore:

Since the events GY, MS, CH, CH are considered, which are equally probable and create a complete group of events, then the appearance of any of them is a reliable event (we denote it by the letter , which is facilitated by all 4 cases. Therefore, the probability:

Hence, the first property of probability is confirmed.

Answer

Probability of an event.

Example 4

Task

Two dice with the same and regular geometric shape are thrown. Find the probability of all possible sums on both sides that fall out.

Decision

To make it easier to solve the problem, imagine that one cube is white and the other is black. With each of the six sides of the white die and one of the six sides of the black die can also fall, so there will be all possible pairs.

Since the possibility of the appearance of faces on a separate die is the same (cubes of the correct geometric shape!), Then the probability of the appearance of each pair of faces will be the same, moreover, as a result of tossing, only one of the pairs falls out. Event values are incompatible, unique. These are cases, and there are 36 possible cases.

Now consider the possibility of the value of the sum on the faces. Obviously, the smallest sum is 1 + 1 = 2, and the largest is 6 + 6 = 12. The rest of the sum increases by one, starting from the second. Let's denote the events whose indices are equal to the sum of the points that fell on the faces of the dice. For each of these events, we write favorable cases using the notation , where is the sum, are the points on the upper face of the white die, and are the points on the face of the black die.

So for an event:

for – one case (1 + 1);

for – two cases (1 + 2; 2 + 1);

for – three cases (1 + 3; 2 + 2; 3 + 1);

for – four cases (1 + 4; 2 + 3; 3 + 2; 4 + 1);

for – five cases (1 + 5; 2 + 4; 3 + 3; 4 + 2; 5 + 1);

for – six cases (1 + 6; 2 + 5; 3 + 4; 4 + 3; 5 + 2; 6 + 1);

for – five cases (2 + 6; 3 + 5; 4 + 4; 5 + 3; 6 + 2);

for – four cases (3 + 6; 4 + 5; 5 + 4; 6 + 3);

for – three cases (4 + 6; 5 + 5; 6 + 4);

for – two cases (5 + 6; 6 + 5);

for – one case (6 + 6).

So the probabilities are:

Answer

Example 5

Task

Before the festival, three participants were offered to draw lots: each of the participants in turn approaches the bucket and randomly chooses one of three cards with numbers 1, 2 and 3, which means the serial number of the performance of this participant.

Find the probability of such events:

1) - the serial number in the queue coincides with the card number, that is, the serial number of the performance;

2) - no number in the queue matches the performance number;

3) - only one of the numbers in the queue matches the performance number;

4) – at least one of the numbers in the queue matches the performance number.

Decision

The possible results of choosing cards are permutations of three elements, the number of such permutations is equal to . Each permutation is an event. Let's denote these events as . We assign the corresponding permutation to each event in parentheses:

; ; ; ; ; .

The listed events are equally possible and uniform, that is, these are the cases. Denote as follows: (1h, 2h, 3h) - the corresponding numbers in the queue.

Let's start with the event. Favorable is only one case, therefore:

Favorable for the event are two cases and , therefore:

The event is facilitated by 3 cases: , therefore:

In addition to , the event also contributes to , that is:

Answer

The probability of an event is .

Probability of an event - updated: September 15, 2017 by: Scientific Articles.Ru

Fundamentals of Probability Theory

Plan:

1. Random events

2. Classical definition of probability

3. Calculation of event probabilities and combinatorics

4. Geometric probability

Theoretical information

Random events.

random phenomenon- a phenomenon, the outcome of which is unambiguously determined. This concept can be interpreted in a fairly broad sense. Namely: everything in nature is quite accidental, the appearance and birth of any individual is a random phenomenon, the choice of goods in a store is also a random phenomenon, getting a mark on an exam is a random phenomenon, illness and recovery are random phenomena, etc.

Examples of random phenomena:

~ Shooting is carried out from a gun set at a given angle to the horizon. Hitting it on the target is accidental, but hitting a projectile in a certain "fork" is a pattern. You can specify the distance closer than and beyond which the projectile will not fly. Get some "fork dispersion of shells"

~ The same body is weighed several times. Strictly speaking, different results will be obtained each time, albeit differing by a negligibly small amount, but different.

~ An aircraft flying along the same route has a certain flight corridor within which the aircraft can maneuver, but it will never have exactly the same route

~ An athlete will never be able to run the same distance with the same time. His results will also be within a certain numerical range.

Experience, experiment, observation are tests

Trial- observation or fulfillment of a certain set of conditions that are performed repeatedly, and regularly repeated in this and the same sequence, duration, while observing other identical parameters.

Let's consider performance by the sportsman of a shot on a target. In order for it to be produced, it is necessary to fulfill such conditions as the preparation of the athlete, loading the weapon, aiming, etc. "Hit" and "miss" are events as a result of a shot.

Event– qualitative test result.

An event may or may not occur Events are indicated by capital Latin letters. For example: D ="The shooter hit the target". S="White ball drawn". K="Random lottery ticket without winning.".

Tossing a coin is a test. The fall of her "coat of arms" is one event, the fall of her "number" is the second event.

Any test involves the occurrence of several events. Some of them may be needed at a given time by the researcher, while others may not be needed.

The event is called random, if under the implementation of a certain set of conditions S it can either happen or not happen. In what follows, instead of saying "the set of conditions S is fulfilled," we will say briefly: "the test was carried out." Thus, the event will be considered as the result of the test.

~ The shooter shoots at a target divided into four areas. The shot is a test. Hitting a certain area of the target is an event.

~ There are colored balls in the urn. One ball is drawn at random from the urn. Removing a ball from an urn is a test. The appearance of a ball of a certain color is an event.

Types of random events

1. Events are said to be incompatible if the occurrence of one of them excludes the occurrence of other events in the same trial.

~ A part was taken at random from a box with parts. The appearance of a standard part excludes the appearance of a non-standard part. Events € a standard part appeared" and with a non-standard part appeared" - incompatible.

~ A coin is thrown. The appearance of the "coat of arms" excludes the appearance of the inscription. The events "a coat of arms appeared" and "an inscription appeared" are incompatible.

Several events form full group, if at least one of them appears as a result of the test. In other words, the occurrence of at least one of the events of the complete group is a certain event.

In particular, if the events that form a complete group are pairwise incompatible, then one and only one of these events will appear as a result of the test. This special case is of greatest interest to us, since it is used below.

~ Two tickets of the money and clothing lottery were purchased. One and only one of the following events must occur:

1. "the winnings fell on the first ticket and did not fall on the second",

2. "the winnings did not fall on the first ticket and fell on the second",

3. "the winnings fell on both tickets",

4. "both tickets did not win."

These events form a complete group of pairwise incompatible events,

~ The shooter fired a shot at the target. One of the following two events is sure to occur: hit, miss. These two disjoint events also form a complete group.

2. Events are called equally possible if there is reason to believe that neither is more possible than the other.

~ The appearance of a "coat of arms" and the appearance of an inscription when a coin is tossed are equally possible events. Indeed, it is assumed that the coin is made of a homogeneous material, has a regular cylindrical shape, and the presence of a coinage does not affect the loss of one or another side of the coin.

~ The appearance of one or another number of points on a thrown dice is an equally probable event. Indeed, it is assumed that dice is made of a homogeneous material, has the shape of a regular polyhedron, and the presence of glasses does not affect the loss of any face.

3. The event is called authentic, if it cannot happen

4. The event is called not reliable if it can't happen.

5. The event is called opposite to some event if it consists of the non-occurrence of the given event. Opposite events are not compatible, but one of them must necessarily occur. Opposite events are commonly referred to as negations, i.e. a dash is written above the letter. The events are opposite: A and Ā; U and Ū, etc. .

The classical definition of probability

Probability is one of the basic concepts of probability theory.

There are several definitions of this concept. Let us give a definition that is called classical. Next, we indicate weak sides of this definition and we give other definitions that allow us to overcome the shortcomings of the classical definition.

Consider the situation: A box contains 6 identical balls, 2 being red, 3 being blue and 1 being white. Obviously, the possibility of drawing a colored (i.e., red or blue) ball at random from an urn is greater than the possibility of drawing a white ball. This possibility can be characterized by a number, which is called the probability of an event (the appearance of a colored ball).

Probability- a number characterizing the degree of possibility of occurrence of the event.

In the situation under consideration, we denote:

Event A = "Pulling out a colored ball".

Each of the possible outcomes of the test (the test consists in extracting a ball from the urn) is called elementary (possible) outcome and event. Elementary outcomes can be denoted by letters with indexes below, for example: k 1 , k 2 .

In our example, there are 6 balls, so there are 6 possible outcomes: a white ball appeared; a red ball appeared; a blue ball appeared, and so on. It is easy to see that these outcomes form a complete group of pairwise incompatible events (only one ball will necessarily appear) and they are equally probable (the ball is taken out at random, the balls are the same and thoroughly mixed).

Elementary outcomes, in which the event of interest to us occurs, we will call favorable outcomes this event. In our example, the event is favored BUT(the appearance of a colored ball) the following 5 outcomes:

Thus the event BUT observed if one occurs in the test, no matter which, of the elementary outcomes that favor BUT. This is the appearance of any colored ball, of which there are 5 pieces in the box

In the considered example of elementary outcomes 6; of which 5 favor the event BUT. Hence, P(A)= 5/6. This number gives that quantification of the degree of possibility of the appearance of a colored ball.

Probability definition:

Probability of event A is the ratio of the number of outcomes favorable to this event to the total number of all equally possible incompatible elementary outcomes that form a complete group.

P(A)=m/n or P(A)=m: n, where:

m is the number of elementary outcomes that favor BUT;

P- the number of all possible elementary outcomes of the test.

It is assumed here that the elementary outcomes are incompatible, equally probable and form a complete group.

The following properties follow from the definition of probability:

1. The probability of a certain event is equal to one.

Indeed, if the event is reliable, then each elementary outcome of the test favors the event. In this case m = n hence p=1

2. The probability of an impossible event is zero.

Indeed, if the event is impossible, then none of the elementary outcomes of the trial favors the event. In this case m=0, hence p=0.

3.The probability of a random event is a positive number between zero and one. 0t< n.

In subsequent topics, theorems will be given that allow, from the known probabilities of some events, to find the probabilities of other events.

Measurement. There are 6 girls and 4 boys in the group of students. What is the probability that a randomly selected student will be a girl? will it be a young man?

p dev = 6 / 10 = 0.6 p jun = 4 / 10 = 0.4

The concept of "probability" in modern rigorous courses of probability theory is built on a set-theoretic basis. Let's take a look at some of this approach.

Suppose that as a result of the test one and only one of the following events occurs: w i(i=1, 2, .... n). Events w i, is called elementary events (elementary outcomes). O it follows that the elementary events are pairwise incompatible. The set of all elementary events that can appear in a trial is called elementary event spaceΩ (Greek letter omega capital), and the elementary events themselves - points in this space..

Event BUT identified with a subset (of the space Ω) whose elements are elementary outcomes favoring BUT; event AT is a subset Ω whose elements are outcomes that favor AT, etc. Thus, the set of all events that can occur in the test is the set of all subsets of Ω. Ω itself occurs for any outcome of the test, therefore Ω is a certain event; an empty subset of the space Ω is an -impossible event (it does not occur for any outcome of the test).

Elementary events are distinguished from among all events by topics, "each of them contains only one element Ω

To every elementary outcome w i match a positive number p i is the probability of this outcome, and the sum of all p i equal to 1 or with the sign of the sum, this fact will be written as an expression:

By definition, the probability P(A) events BUT is equal to the sum of the probabilities of elementary outcomes favoring BUT. Therefore, the probability of a certain event is equal to one, impossible - to zero, arbitrary - is between zero and one.

Let us consider an important particular case, when all outcomes are equally probable. The number of outcomes is equal to n, the sum of the probabilities of all outcomes is equal to one; hence the probability of each outcome is 1/n. Let the event BUT favors m outcomes.

Event Probability BUT is equal to the sum of the probabilities of outcomes favoring BUT:

P(A)=1/n + 1/n+…+1/n = n 1/n=1

The classical definition of probability is obtained.

There is still axiomatic approach to the concept of "probability". In the system of axioms proposed. Kolmogorov A.N., undefined concepts are elementary event and probability. The construction of a logically complete probability theory is based on the axiomatic definition of a random event and its probability.

Here are the axioms that define the probability:

1. Every event BUT assigned a non-negative real number P(A). This number is called the probability of the event. BUT.

2. The probability of a certain event is equal to one:

3. The probability of occurrence of at least one of the pairwise incompatible events is equal to the sum of the probabilities of these events.

Based on these axioms, the properties of probabilities for the relationship between them are derived as theorems.

Brief theory

For a quantitative comparison of events according to the degree of possibility of their occurrence, a numerical measure is introduced, which is called the probability of an event. The probability of a random event a number is called, which is an expression of a measure of the objective possibility of the occurrence of an event.

The values that determine how significant are the objective grounds for counting on the occurrence of an event are characterized by the probability of the event. It must be emphasized that probability is an objective quantity that exists independently of the cognizer and is conditioned by the totality of conditions that contribute to the occurrence of an event.

The explanations that we have given to the concept of probability are not a mathematical definition, since they do not define this concept quantitatively. There are several definitions of the probability of a random event, which are widely used in solving specific problems (classical, axiomatic, statistical, etc.).

The classical definition of the probability of an event reduces this concept to a more elementary concept of equally probable events, which is no longer subject to definition and is assumed to be intuitively clear. For example, if a dice is a homogeneous cube, then the fallout of any of the faces of this cube will be equally probable events.

Let a certain event be divided into equally probable cases, the sum of which gives the event. That is, the cases from , into which it breaks up, are called favorable for the event, since the appearance of one of them ensures the offensive.

The probability of an event will be denoted by the symbol .

The probability of an event is equal to the ratio of the number of cases favorable to it, out of the total number of unique, equally possible and incompatible cases, to the number, i.e.

This is the classical definition of probability. Thus, to find the probability of an event, it is necessary, after considering the various outcomes of the test, to find a set of the only possible, equally possible and incompatible cases, calculate their total number n, the number of cases m that favor this event, and then perform the calculation according to the above formula.

The probability of an event equal to the ratio of the number of outcomes of experience favorable to the event to the total number of outcomes of experience is called classical probability random event.

The following properties of probability follow from the definition:

Property 1. The probability of a certain 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 between zero and one.

Property 4. The probability of the occurrence of events that form a complete group is equal to one.

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

The number of occurrences that favor the occurrence of the opposite event. Hence, the probability of the opposite event occurring is equal to the difference between unity and the probability of the event A occurring:

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 experience, but on the basis of logical reasoning.

When a set of conditions is met, a certain event will definitely happen, and the impossible will definitely not happen. Among the events that, when a complex of conditions is created, may or may not occur, the appearance of some can be counted on with more reason, on the appearance of others with less reason. If, for example, there are more white balls in the urn than black ones, then there are more reasons to hope for the appearance of a white ball when taken out of the urn at random than for the appearance of a black ball.

Problem solution example

Example 1

A box contains 8 white, 4 black and 7 red balls. 3 balls are drawn at random. Find the probabilities of the following events: - at least 1 red ball is drawn, - there are at least 2 balls of the same color, - there are at least 1 red and 1 white ball.

The solution of the problem

We find the total number of test outcomes as the number of combinations of 19 (8 + 4 + 7) elements of 3 each:

Find the probability of an event– drawn at least 1 red ball (1,2 or 3 red balls)

Required probability:

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

Number of outcomes favoring the event:

Required probability:

Let the event– there is at least one red and one white ball

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

Number of outcomes favoring the event:

Required probability:

Answer: P(A)=0.773;P(C)=0.7688; P(D)=0.6068

Example 2

Two dice are thrown. Find the probability that the sum of the points is at least 5.

Decision

Let the event be the sum of points not less than 5

Let's use the classical definition of probability:

Total number of possible trial outcomes

The number of trials that favor the event of interest to us

On the dropped face of the first dice, one point, two points ..., six points can appear. similarly, six outcomes are possible on the second die roll. Each of the outcomes of the first die can be combined with each of the outcomes of the second. Thus, the total number of possible elementary outcomes of the test is equal to the number of placements with repetitions (selection with placements of 2 elements from a set of volume 6):

Find the probability of the opposite event - the sum of points is less than 5

The following combinations of dropped points will favor the event:

№

1st bone

2nd bone

The geometric definition of probability is presented and the solution of the well-known meeting problem is given.

3) P (Æ )=0.

We will say what is given probability space, if the space of elementary outcomes9 is given and the correspondence

w i ® P(w i ) =Pi .

The question arises: how to determine the probability P (w i ) of individual elementary outcomes from the specific conditions of the problem being solved?

The classic definition of probability.

The probabilities P (w i ) can be calculated using an a priori approach, which consists in analyzing the specific conditions of a given experiment (before the experiment itself).

A situation is possible when the space of elementary outcomes consists of a finite number N of elementary outcomes, and a random experiment is such that the probabilities of each of these N elementary outcomes appear to be equal. Examples of such random experiments are: tossing a symmetrical coin, throwing a regular dice, randomly removing a playing card from a shuffled deck. By virtue of the introduced axiom, the probability of each elementary

outcomes in this case are equal to N . It follows from this that if event A contains N A elementary outcomes, then in accordance with the definition (*)

P(A) = A

In this class of situations, the probability of an event is defined as the ratio of the number of favorable outcomes to the total number of all possible outcomes.

Example. From a set containing 10 identical-looking electric lamps, among which 4 are defective, 5 lamps are randomly selected. What is the probability that among the selected lamps there will be 2 defective ones?

First of all, we note that the choice of any five lamps has the same probability. In total, there are C 10 5 ways to make such a five, that is, a random experiment in this case has C 10 5 equiprobable outcomes.

How many of these outcomes satisfy the condition "there are two defective lamps in the five", that is, how many outcomes belong to the event of interest to us?

Each five we are interested in can be composed as follows: choose two defective lamps, which can be done in a number of ways equal to C 4 2 . Each pair of defective lamps can occur as many times as there are ways to complement it with three non-defective lamps, that is, 6 3 times. It turns out that the number of fives containing two

Statistical definition of probability.

Consider a random experiment in which a dice made of a non-homogeneous material is tossed. Its center of gravity is not in the geometric center. In this case, we cannot consider the outcomes (rolling one, two, etc.) equally probable. It is known from physics that the bone will fall more often 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 toss that die n times (where n is a big enough number, say n=1000 or n=5000), count the number of rolls of three n 3 and calculate the probability of a three roll outcome as n 3 /n - the relative frequency of getting three points. Similarly, you can determine the probabilities of the remaining elementary outcomes - ones, twos, fours, etc. Theoretically, this course of action can be justified by introducing statistical definition of probability.

The probability P(M i ) is defined as the limit of the relative frequency of occurrence of the outcome M i in the process of an unlimited increase in the number of random experiments n , that is

P i = P (M i ) = lim m n (M i ) , n ®¥n

where m n (M i ) is the number of random experiments (out of the total number n of random experiments performed) in which the occurrence of an elementary outcome M i is registered.

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

geometric probability

In one special case, let us define 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 a one-to-one correspondence can also be established between the set of elementary outcomes that favor the event A and the set of points of a flat figure I (small sigma) , which is part of the figure S , then

P(A) = S ,

where s is the area of figure s, S is the area of figure S.

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 for 10 minutes. What is the probability of their meeting?

Let x be the arrival time of the first in the canteen, аy be the arrival time of the second

£12 x £13; £12y £13.

You can establish a one-to-one correspondence between all pairs of numbers (x ;y ) (or a 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 the figure 6. Here, for example, point A corresponds to the outcome, which consists in the fact that the first came 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 occur 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 (x ³ y ), then

the meeting will occur under the condition 0 £ x - y £ 1/6..

Between many favorable outcomes

meeting, and the set of points of the region s depicted in

Figure 7 in shaded form, you can install

one-to-one correspondence.

The desired probability p is equal to the ratio of the area

area s to the area of the whole square.. Area of the square

equals unity, and the area of the region s can be defined as

the difference between a unit and the total area of two

triangles shown in Figure 7. It follows from this:

p=1 -

Continuous probability space.

As mentioned earlier, the set of elementary outcomes can be more than countable (that is, uncountable). In this case, any subset of the set W cannot be considered an event.

To introduce the definition of a random event, consider a system (finite or countable) of subsets A 1 , A 2 ,... A n of the space of elementary outcomes W .

If three conditions are met: 1) W belongs to this system;

2) membership of A in this system implies membership of A in this system;

3) membership of A i and A j in this system implies membership of A i U A j in this system

such a system of subsets is called an algebra.

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

1) W ,Æ ; 2) W , A , 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 and A 1 ∩ A 2 belong to this algebra.

A subset A of an uncountable set of elementary outcomes 9 is an event if it belongs to some algebra.

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

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

1) P (9)=1

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

P (A 1 U A 2 U ... U A n) \u003d P (A 1) + P (A 2) + ... + P (A n)

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

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

Probability addition formulas.

From point 2 of the above axiom it follows that if A 1 and A2 are incompatible events, then

P (A 1 U A 2) \u003d P (A 1) + P (A 2)

If A 1 and A 2 are joint events, then A 1 U A 2 =(A 1 \A 2 )U A 2 , and it is obvious that A 1 \A 2 and A 2 are incompatible events. This implies:

P (A 1 U A 2 ) =P (A1 \A 2 ) +P (A2 )

Further, it is obvious: A 1 = (A1 \A 2 )U (A 1 ∩ A 2 ), and A1 \A 2 and A 1 ∩ A 2 are incompatible events, whence follows: P (A 1 ) =P (A1 \A 2 ) +P (A 1 ∩ A 2 ) Find an expression for P (A1 \A 2 ) from this formula and substitute it into the right side of the formula (*). As a result, we obtain the formula for adding probabilities:

P (A 1 U A 2 ) =P (A 1 ) +P (A 2 ) –P (A 1 ∩ A 2 )

From the last formula, it is easy to obtain a formula for adding probabilities for incompatible events by setting A 1 ∩ A 2 =Æ .

Example. Find the probability of drawing an ace or a suit of hearts by randomly selecting one card from a deck of 32 cards.

P (ACE) \u003d 4/32 \u003d 1/8; P (HEART SUIT) \u003d 8/32 \u003d 1/4;

P (ACE OF HEARTS) = 1/32;

P ((ACE) U (HEART SUIT)) \u003d 1/8 + 1/4 - 1/32 \u003d 11/32

The same result could be achieved using the classical definition of probability by counting the number of favorable outcomes.

Conditional probabilities.

Let's consider the problem. Before the exam, the student learned from 30 tickets tickets with numbers from 1 to 5 and from 26 to 30. It is known that the student pulled out a ticket with a number not exceeding 20 during the exam. What is the probability that the student pulled out the learned ticket?

Let's define the space of elementary outcomes: W =(1,2,3,...,28,29,30). Let event A be that the student pulled out a learned ticket: A = (1,...,5,25,...,30,), and event B is that the student pulled out a ticket from the first twenty: B = ( 1,2,3,...,20)

The event A ∩ B consists of five outcomes: (1,2,3,4,5), and its probability is 5/30. This number can be represented as the product of 5/20 and 20/30. The number 20/30 is the probability of event B . The number 5/20 can be considered as the probability of event A, provided that event B happened (let's denote it as P (A / B)). Thus, the solution to the problem is determined by the formula

P (A ∩ B) \u003d P (A / B) P (B)

This formula is called the probability multiplication formula, and the probability P (A / B) is the conditional probability of the event A.

Example .. From an urn containing 7 white and 3 black balls, two balls are randomly drawn one after the other (without replacement). What is the probability that the first ball is white and the second black?

Let X be the event that the first draw is a white ball, and Y be the event that the second draw is a black ball. Then X ∩ Y is the event that the first ball is white and the second is black. P (Y /X ) =3/9 =1/3 is the conditional probability that the second ball will draw a black ball, if the white ball was drawn first. Considering that P (X ) = 7/10, according to the probability multiplication formula we get: P (X ∩ Y ) = 7/30

Event A is called independent of event B (in other words: events A and B are called independent) if P (A / B) = P (A ). For the definition of independent events, we can take the consequence of the last formula and the multiplication formula

P (A ∩ B) \u003d P (A) P (B)

Prove for yourself that if A and B are independent events, then A and B are also independent events.

Example. Consider a problem similar to the previous one, but with one additional condition: after drawing the first ball, remember its color and return the ball to the urn, after which we mix all the balls. In this case, the result of the second extraction does not depend in any way on which ball - black or white - appeared during the first extraction. The probability of a white ball appearing first (event A) is 7/10. The probability of event B - the appearance of the second black ball - is 3/10. Now the multiplication formula gives: P (A ∩ B) = 21/100.

Extracting balls in the manner described in this example is called fetch with return or return sampling.

It should be noted that if in the last two examples we set the initial numbers of white and black balls equal to 7000 and 3000, respectively, then the results of calculating the same probabilities will differ negligibly small for the return and irrevocable samples.