Theory of addition of probabilities. Theorem for adding probabilities of joint events

Directly counting cases favoring a given event may be difficult. Therefore, to determine the probability of an event, it can be advantageous to imagine this event as a combination of some other, simpler events. In this case, however, you need to know the rules that govern probabilities in combinations of events. It is to these rules that the theorems mentioned in the title of the paragraph refer.

The first of these relates to calculating the probability that at least one of several events will occur.

Addition theorem.

Let A and B be two incompatible events. Then the probability that at least one of these two events will occur is equal to the sum of their probabilities:

Proof. Let be a complete group of pairwise incompatible events. If then among these elementary events there are exactly events favorable to A and exactly events favorable to B. Since events A and B are incompatible, then no event can favor both of these events. An event (A or B), consisting in the occurrence of at least one of these two events, is obviously favored by both each of the events favoring A and each of the events

Favorable B. Therefore, the total number of events favorable to event (A or B) is equal to the sum which follows:

Q.E.D.

It is easy to see that the addition theorem formulated above for the case of two events can easily be transferred to the case of any finite number of them. Precisely if there are pairwise incompatible events, then

For the case of three events, for example, one can write

An important consequence of the addition theorem is the statement: if events are pairwise incompatible and uniquely possible, then

Indeed, the event either or or is by assumption certain and its probability, as indicated in § 1, is equal to one. In particular, if they mean two mutually opposite events, then

Let us illustrate the addition theorem with examples.

Example 1. When shooting at a target, the probability of making an excellent shot is 0.3, and the probability of making a “good” shot is 0.4. What is the probability of getting a score of at least “good” for a shot?

Solution. If event A means receiving an “excellent” rating, and event B means receiving a “good” rating, then

Example 2. In an urn containing white, red and black balls, there are white balls and I red balls. What is the probability of drawing a ball that is not black?

Solution. If event A consists of the appearance of a white ball, and event B consists of a red ball, then the appearance of the ball is not black

means the appearance of either a white or red ball. Since by definition of probability

then, by the addition theorem, the probability of a non-black ball appearing is equal;

This problem can be solved this way. Let event C consist in the appearance of a black ball. The number of black balls is equal so that P (C) The appearance of a non-black ball is the opposite event of C, therefore, based on the above corollary from the addition theorem, we have:

as before.

Example 3. In a cash-material lottery, for a series of 1000 tickets there are 120 cash and 80 material winnings. What is the probability of winning anything on one lottery ticket?

Solution. If we denote by A an event consisting of a monetary gain and by B a material gain, then from the definition of probability it follows

The event of interest to us is represented by (A or B), therefore it follows from the addition theorem

Thus, the probability of any winning is 0.2.

Before moving on to the next theorem, it is necessary to become familiar with a new important concept - the concept of conditional probability. For this purpose, we will start by considering the following example.

Suppose there are 400 light bulbs in a warehouse, manufactured in two different factories, with the first producing 75% of all light bulbs, and the second - 25%. Let us assume that among the light bulbs manufactured by the first plant, 83% satisfy the conditions of a certain standard, and for the products of the second plant this percentage is 63. Let us determine the probability that a light bulb randomly taken from the warehouse will satisfy the conditions of the standard.

Note that the total number of standard light bulbs available consists of the light bulbs manufactured by the first

factory, and 63 light bulbs manufactured by the second plant, that is, equal to 312. Since the choice of any light bulb should be considered equally possible, we have 312 favorable cases out of 400, so

where event B is that the light bulb we have chosen is standard.

During this calculation, no assumptions were made about the product of which plant the light bulb we selected belonged to. If we make any assumptions of this kind, then it is obvious that the probability we are interested in may change. So, for example, if it is known that the selected light bulb was manufactured at the first plant (event A), then the probability that it is standard will no longer be 0.78, but 0.83.

This kind of probability, that is, the probability of event B given that event A occurs, is called the conditional probability of event B given the occurrence of event A and is denoted

If in the previous example we denote by A the event that the selected light bulb is manufactured at the first plant, then we can write

Now we can formulate an important theorem related to calculating the probability of combining events.

Multiplication theorem.

The probability of combining events A and B is equal to the product of the probability of one of the events and the conditional probability of the other, assuming that the first occurred:

In this case, the combination of events A and B means the occurrence of each of them, that is, the occurrence of both event A and event B.

Proof. Let us consider a complete group of equally possible pairwise incompatible events, each of which can be favorable or unfavorable for both event A and event B.

Let us divide all these events into four different groups as follows. The first group includes those events that favor both event A and event B; The second and third groups include those events that favor one of the two events of interest to us and do not favor the other, for example, the second group includes those that favor A but do not favor B, and the third group includes those that favor B but do not favor A; finally to

The fourth group includes those events that do not favor either A or B.

Since the numbering of events does not matter, we can assume that this division into four groups looks like this:

Group I:

Group II:

III group:

IV group:

Thus, among equally possible and pairwise incompatible events, there are events that favor both event A and event B, events that favor event A, but do not favor event A, events that favor B, but do not favor A, and, finally, events that do not favor neither A nor B.

Let us note, by the way, that any of the four groups we have considered (and even more than one) may not contain a single event. In this case, the corresponding number indicating the number of events in such a group will be equal to zero.

Our breakdown into groups allows you to immediately write

for the combination of events A and B is favored by the events of the first group and only by them. The total number of events favoring A is equal to the total number of events in the first and second groups, and those favoring B is equal to the total number of events in the first and third groups.

Let us now calculate the probability, that is, the probability of event B, provided that event A took place. Now the events included in the third and fourth groups disappear, since their occurrence would contradict the occurrence of event A, and the number of possible cases is no longer equal to . Of these, event B is favored only by events of the first group, so we get:

To prove the theorem, it is enough now to write the obvious identity:

and replace all three fractions with the probabilities calculated above. We arrive at the equality stated in the theorem:

It is clear that the identity we wrote above makes sense only if it is always true, unless A is an impossible event.

Since events A and B are equal, then, by swapping them, we get another form of the multiplication theorem:

However, this equality can be obtained in the same way as the previous one, if you notice that using the identity

Comparing the right-hand sides of the two expressions for the probability P(A and B), we obtain a useful equality:

Let us now consider examples illustrating the multiplication theorem.

Example 4. In the products of a certain enterprise, 96% of the products are considered suitable (event A). 75 products out of every hundred suitable ones turn out to belong to the first grade (event B). Determine the probability that a randomly selected product will be suitable and belong to the first grade.

Solution. The desired probability is the probability of combining events A and B. By condition we have: . Therefore the multiplication theorem gives

Example 5. The probability of hitting the target with a single shot (event A) is 0.2. What is the probability of hitting the target if 2% of the fuses fail (i.e., in 2% of cases the shot does not

Solution. Let the event B be that a shot will occur, and let B mean the opposite event. Then by condition and according to the corollary of the addition theorem. Further, according to the condition.

Hitting the target means the combination of events A and B (the shot will fire and hit), therefore, according to the multiplication theorem

An important special case of the multiplication theorem can be obtained by using the concept of independence of events.

Two events are called independent if the probability of one of them does not change as a result of whether the other occurs or does not occur.

Examples of independent events are the occurrence of a different number of points when throwing a dice again, or one or another side of coins when throwing a coin again, since it is obvious that the probability of getting a coat of arms on the second throw is equal regardless of whether the coat of arms came up or not on the first.

Similarly, the probability of drawing a white ball a second time from an urn containing white and black balls if the first ball drawn is previously returned does not depend on whether the ball was drawn the first time, white or black. Therefore, the results of the first and second removal are independent of each other. On the contrary, if the ball taken out first does not return to the urn, then the result of the second removal depends on the first, because the composition of the balls in the urn after the first removal changes depending on its outcome. Here we have an example of dependent events.

Using the notation adopted for conditional probabilities, we can write the condition for the independence of events A and B in the form

Using these equalities, we can reduce the multiplication theorem for independent events to the following form.

If events A and B are independent, then the probability of their combination is equal to the product of the probabilities of these events:

Indeed, it is enough to put in the initial expression of the multiplication theorem, which follows from the independence of events, and we will obtain the required equality.

Let us now consider several events: We will call them collectively independent if the probability of the occurrence of any of them does not depend on whether any other events under consideration occurred or not

In the case of events that are collectively independent, the multiplication theorem can be extended to any finite number of them, so it can be formulated as follows:

The probability of combining independent events in the aggregate is equal to the product of the probabilities of these events:

Example 6. A worker operates three automatic machines, each of which must be approached to correct a malfunction if the machine stops. The probability that the first machine will not stop within an hour is 0.9. The same probability for the second machine is 0.8 and for the third - 0.7. Determine the probability that within an hour the worker will not need to approach any of the machines he is servicing.

Example 7. Probability of shooting down a plane with a rifle shot What is the probability of destroying an enemy plane if 250 rifles are fired at the same time?

Solution. The probability that the plane will not be shot down with a single shot is equal to the addition theorem. Then we can calculate, using the multiplication theorem, the probability that the plane will not be shot down with 250 shots, as the probability of combining events. It is equal to After this, we can again use the addition theorem and find the probability that the plane will be shot down as the probability of the opposite event

From this it can be seen that, although the probability of shooting down a plane with a single rifle shot is negligible, nevertheless, when firing from 250 rifles, the probability of shooting down a plane is already very noticeable. It increases significantly if the number of rifles is increased. So, when shooting from 500 rifles, the probability of shooting down a plane, as is easy to calculate, is equal to when shooting from 1000 rifles - even.

The multiplication theorem proved above allows us to somewhat expand the addition theorem, extending it to the case of compatible events. It is clear that if events A and B are compatible, then the probability of the occurrence of at least one of them is not equal to the sum of their probabilities. For example, if event A means an even number

the number of points when throwing a die, and event B is the loss of a number of points that is a multiple of three, then the event (A or B) is favored by the loss of 2, 3, 4 and 6 points, that is

On the other hand, that is. So in this case

From this it is clear that in the case of compatible events the theorem of addition of probabilities must be changed. As we will now see, it can be formulated in such a way that it is valid for both compatible and incompatible events, so that the previously considered addition theorem turns out to be a special case of the new one.

Events that are not favorable to A.

All elementary events that favor an event (A or B) must favor either only A, or only B, or both A and B. Thus, the total number of such events is equal to

and the probability

Q.E.D.

Applying formula (9) to the above example of the number of points appearing when throwing a dice, we obtain:

which coincides with the result of direct calculation.

Obviously, formula (1) is a special case of (9). Indeed, if events A and B are incompatible, then the probability of combination

For example. Two fuses are connected in series to the electrical circuit. The probability of failure of the first fuse is 0.6, and the second is 0.2. Let us determine the probability of power failure as a result of failure of at least one of these fuses.

Solution. Since events A and B, consisting of the failure of the first and second of the fuses, are compatible, the required probability will be determined by formula (9):

Exercises

An experiment is being considered E. It is assumed that it can be carried out repeatedly. As a result of the experiment, various events may appear, making up a certain set F. Observable events are divided into three types: reliable, impossible, random.

Reliable an event that is sure to occur as a result of an experiment is called E. Denoted by Ω.

Impossible an event that is known to not occur as a result of an experiment is called E. Denoted by .

Random an event that may or may not occur as a result of an experiment is called E.

Additional (opposite) event A is an event, denoted by , that occurs if and only if the event does not occur A.

Sum (combination) events is an event that occurs if and only if at least one of these events occurs (Figure 3.1). Notation.

Figure 3.1

Product (intersection) events is an event that occurs if and only if all these events occur together (simultaneously) (Figure 3.2). Notation. It is obvious that events A and B incompatible , If .

Figure 3.2

Full group of events is a set of events whose sum is a certain event:

Event IN called a special case of an event A, if with the occurrence of an event IN the event appears A. They also say that the event IN entails an event A(Figure 3.3). Designation

Figure 3.3

Events A And IN are called equivalent , if they do or do not occur together during the experiment E. Designation Obviously, if...

A difficult event call an observed event expressed through other events observed in the same experiment using algebraic operations.

The probability of a particular complex event occurring is calculated using the formulas for adding and multiplying probabilities.

Probability addition theorem

Consequences:

1) if events A And IN are inconsistent, the addition theorem takes the form:

2) in the case of three terms, the addition theorem is written in the form

3) the sum of the probabilities of mutually opposite events is equal to 1:

The set of events ,, ..., is called full group of events , If

The sum of the probabilities of events forming a complete group is equal to 1:

Probability of event occurrence A provided that the event IN happened, they call it conditional probability and denote or.

A And IN – dependent events , If .

A And IN – independent events , If .

Probability multiplication theorem

Consequences:

1) for independent events A And IN

2) in the general case, for the product of three events, the probability multiplication theorem has the form:

Examples of problem solving

Example1 - Three elements are connected in series to the electrical circuit, operating independently of each other. The failure probabilities of the first, second and third elements are respectively equal to ,. Find the probability that there will be no current in the circuit.

Solution

First way.

Let us denote the events: - failure of the first, second and third elements occurred in the circuit, respectively.

Event A– there will be no current in the circuit (at least one of the elements will fail, since they are connected in series).

Event - there is current in the circuit (three elements are working), . The probability of opposite events is related by formula (3.4). An event is the product of three events that are pairwise independent. Using the theorem for multiplying the probabilities of independent events, we obtain

Then the probability of the desired event is .

Second way.

Taking into account the previously accepted notation, we write down the desired event A– at least one of the elements will fail:

Since the terms included in the sum are compatible, we should apply the probability addition theorem in general form for the case of three terms (3.3):

Answer: 0,388.

Problems to solve independently

1 The reading room has six textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that both textbooks will be bound.

2 There are threads mixed in the bag, 30% of which are white and the rest are red. Determine the probabilities that two threads drawn at random will be: the same color; different colors.

3 The device consists of three elements that work independently. The probabilities of failure-free operation for a certain period of time of the first, second and third elements, respectively, are 0.6; 0.7; 0.8. Find the probabilities that during this time only one element will work without failure; only two elements; all three elements; at least two elements.

4 Three dice are thrown. Find the probabilities of the following events:

a) five points will appear on each side drawn;

b) the same number of points will appear on all dropped sides;

c) one point will appear on two dropped sides, and another number of points will appear on the third side;

d) a different number of points will appear on all dropped faces.

5 The probability of a shooter hitting the target with one shot is 0.8. How many shots must the shooter fire so that with a probability of less than 0.4 it can be expected that there will be no misses?

6 From the numbers 1, 2, 3, 4, 5, one is first selected, and then the second digit is selected from the remaining four. It is assumed that all 20 possible outcomes are equally likely. Find the probability that an odd number will be chosen: for the first time; a second time; both times.

7 The probability that a pair of size 46 shoes will be sold again in the men's shoe section of the store is 0.01. How many pairs of shoes must be sold in a store so that with a probability of at least 0.9 one can expect that at least one pair of size 46 shoes will be sold?

8 There are 10 parts in the box, including two non-standard ones. Find the probability that out of six randomly selected parts there will be no more than one non-standard one.

9 The technical control department checks products for standardness. The probability that the product is non-standard is 0.1. Find the probability that:

a) out of three tested products, only two will turn out to be non-standard;

b) only the fourth product tested in order will turn out to be non-standard.

10 32 letters of the Russian alphabet are written on cut-out alphabet cards:

a) three cards are taken out at random one after another and placed on the table in the order of appearance. Find the probability that the word “world” will be obtained;

b) the three cards removed can be swapped in any way. What is the probability that the word “world” can be formed from them?

11 A fighter attacks a bomber and fires two independent bursts at it. The probability of shooting down a bomber with the first burst is 0.2, and the second - 0.3. If the bomber is not shot down, it fires at the fighter from its rear guns and shoots it down with a probability of 0.25. Find the probability that a bomber or fighter is shot down as a result of an air battle.

Homework

1 Total probability formula. Bayes' formula.

2 Solve problems

Task1 . A worker operates three machines that operate independently of each other. The probability that the first machine will not require the worker’s attention within an hour is 0.9, the second is 0.8, and the third is 0.85. Find the probability that within an hour at least one machine will require the attention of a worker.

Task2 . The computer center, which must continuously process incoming information, has two computing devices. It is known that each of them has a probability of failure over some time equal to 0.2. You need to determine the probability:

a) the fact that one of the devices will fail, and the second will be operational;

b) trouble-free operation of each device.

Task3 . Four hunters agreed to shoot at game in a certain sequence: the next hunter fires a shot only if the previous one misses. The probability of a hit for the first hunter is 0.6, for the second – 0.7, for the third – 0.8. Find the probability that shots will be fired:

d) four.

Task4 . The part goes through four processing operations. The probability of receiving a defect during the first operation is 0.01, during the second - 0.02, during the third - 0.03, and during the fourth - 0.04. Find the probability of receiving a part without defects after four operations, assuming that the events of receiving defects in individual operations are independent.

Let events A And IN- inconsistent, and the probabilities of these events are known. Question: how to find the probability that one of these incompatible events will occur? The answer to this question is given by the addition theorem.

Theorem.The probability of one of two incompatible events occurring is equal to the sum of the probabilities of these events:

p(A + IN) = p(A) + p(IN) (1.6)

Proof. Indeed, let n– the total number of all equally possible and incompatible (i.e. elementary) outcomes. Let the event A favors m 1 outcomes, and the event IN – m 2 outcomes. Then, according to the classical definition, the probabilities of these events are equal: p(A) = m 1 / n, p(B) = m 2 / n .

Since events A And IN incompatible, then none of the outcomes favorable to the event A, not conducive to the event IN(see diagram below).

Therefore the event A+IN will be favorable m 1 + m 2 outcomes. Therefore, for the probability p(A + B) we get:

Corollary 1. The sum of the probabilities of events forming a complete group is equal to one:

p(A) + p(IN) + p(WITH) + … + p(D) = 1.

Indeed, let events A,IN,WITH, … , D form a complete group. Because of this, they are incompatible and the only possible ones. Therefore the event A + B + C + …+D, consisting in the occurrence (as a result of testing) of at least one of these events, is reliable, i.e. A+B+C+…+D = And p(A+B+C+ …+D) = 1.

Due to the incompatibility of events A,IN,WITH, … , D the formula is correct:

p(A+B+C+ …+D) = p(A) + p(IN) + p(WITH) + … + p(D) = 1.

Example. There are 30 balls in an urn, of which 10 are red, 5 are blue and 15 are white. Find the probability of drawing a red or blue ball, provided that only one ball is drawn from the urn.

Solution. Let the event A 1 – drawing the red ball, and the event A 2 – extraction of the blue ball. These events are incompatible, and p(A 1) = 10 / 30 = 1 / 3; p(A 2) = 5 / 30 = 1 /6. By the addition theorem we get:

p(A 1 + A 2) = p(A 1) + p(A 2) = 1 / 3 + 1 / 6 = 1 / 2.

Note 1. We emphasize that, according to the meaning of the problem, it is necessary, first of all, to establish the nature of the events under consideration - whether they are incompatible. If the above theorem is applied to joint events, the result will be incorrect.

At When assessing the probability of the occurrence of any random event, it is very important to have a good understanding of whether the probability () of the occurrence of the event we are interested in depends on how other events develop.

In the case of the classical scheme, when all outcomes are equally probable, we can already estimate the probability values ​​of the individual 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 we are interested in happening?

If I roll a die several times and want a six to come up, and I keep getting unlucky, does that mean I should increase my bet because, according to probability theory, I'm about to get lucky? Alas, the theory of probability does not state anything like this. No dice, no cards, no coins can't remember what they showed us last time. It doesn’t matter to them at all whether it’s the first time or the tenth time I’m testing my luck today. Every time I repeat the roll, I know only one thing: and this time the probability of getting a six is ​​again one sixth. Of course, this does not mean that the number I need will never come up. This 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 in any way affect the probability of another event. For example, the probabilities of hitting a target with the first of two weapons do not depend on whether the target was hit by the other weapon, so the events “the first weapon hit the target” and “the second weapon 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 the simultaneous occurrence of both event A and event B (denoted AB) can be calculated using the following theorem.

Probability multiplication theorem for independent events

Example.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 a hit with one salvo by both guns simultaneously.

Solution: 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)=p 1 *p 2 =0.56.

What happens to our estimates if the initial events are not independent? Let's change the previous example a little.

Example.Two shooters shoot at targets at a competition, and if one of them shoots accurately, the opponent begins to get nervous and his results worsen. 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 options for the development of events, to essentially create two scenarios, two different tasks. In the first case, if the opponent missed, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent has taken his chance decently, the probability of hitting the target for the second athlete decreases.

To separate possible scenarios (often called hypotheses) for the development of events, we will often use a “probability tree” diagram. This diagram is similar in meaning to the decision tree that you have probably already dealt with. Each branch represents a separate scenario for the development of events, only now it has its own meaning of the so-called conditional probabilities (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 the probabilities come from? real situations ? After all, probability theory doesn’t work with just coins and dice? Usually these estimates are taken from statistics, and when statistical information is not available, we conduct our own research. And we often have to start not with collecting data, but with the question of what information we actually need.

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

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

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

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

4) of them, only 10% can muster the courage to try a new product,

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

Solution: According to the law of multiplication of probabilities, we determine the probability of the event we are interested in A = (a city resident buys this new balm from us) = 0.00045.

Let's multiply this probability value by the number of city residents. As a result, we have only 45 potential customers, and considering that one bottle of this product lasts for several months, the trade is not very lively.

And yet there is some benefit from our assessments.

Firstly, we can compare forecasts of different business ideas; they will have different “forks” in 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. Just her exact the 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 suit us particularly, on those factors that we are able to influence.

Let's look at another quantitative example of consumer behavior research.

Example. On average, 10,000 people visit the food market per day. The probability that a market visitor enters the dairy products pavilion is 1/2. It is known that this pavilion sells an average of 500 kg of various products per day.

Can we say that the average purchase in the 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, to answer the question about the average weight of a purchase, we must find an answer to the question, what is the probability that a person entering the pavilion will buy something there. If we do not have such data at our disposal, but we need it, we will have to obtain it ourselves by observing the visitors to the pavilion for some time. Let’s say our observations showed that only a fifth of pavilion visitors buy something.

Once we have obtained these estimates, the task becomes simple. Out of 10,000 people who come to the market, 5,000 will go to the dairy products pavilion; there will be only 1,000 purchases. The average purchase weight is 500 grams. It is interesting to note that in order to build a complete picture of what is happening, the logic of conditional “branching” must be defined at each stage of our reasoning as clearly as if we were working with a “specific” situation, and not with probabilities.

Self-test tasks

1. Let there be an electrical circuit consisting of n elements connected in series, each of which operates independently of the others.

The probability p of failure of each element is known. Determine the probability of proper operation of the entire section of the circuit (event A).

2. The student knows 20 out of 25 exam questions. Find the probability that the student knows the three questions given to him by the examiner.

3. Production consists of four successive stages, at each of which equipment operates, for which the probabilities of failure over 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 stoppages due to equipment failure in a month.

Basic Concepts
Events are called incompatible if the occurrence of one of them excludes the occurrence of other events in the same trial. Otherwise they are called joint.
A complete group is a set of events, the combination of which is a reliable event.
The only two possible events that form a complete group are called opposite.
Events are called dependent if the probability of the occurrence of one of them depends on the occurrence or non-occurrence of other events.
Events are called independent if the probability of one of them does not depend on the occurrence or non-occurrence of others.
Theorem for adding probabilities of incompatible events
P(A+B)=P(A)+P(B),
where A, B are incompatible events.

Theorem for adding probabilities of joint events
P(A+B)=P(A)+P(B)-P(AB), where A and B are joint events.

Theorem for multiplying the probabilities of independent events
,
where A and B are independent events.
Theorem for multiplying probabilities of dependent events
P(AB)=P(A)P A (B),
where P A (B) is the probability of the occurrence of event B, provided that event A has occurred; A and B are dependent events.

Task 1.
The shooter fires two shots at the target. The probability of hitting each shot is 0.8. Compose a complete group of events and find their probabilities. Solution.
Test - Two shots are fired at a target.
Event A- missed both times.
Event IN- hit once.
Event WITH- hit both times.
.

Control: P(A) +P(B) +P(C) = 1.
Task 2.
According to meteorologists' forecast, P(rain)=0.4; P(wind)=0.7; R(rain and wind)=0.2. What is the probability that it will rain or wind? Solution. By the theorem of addition of probabilities and due to the compatibility of the proposed events, we have:
P(rain or wind or both)=P(rain) +P(wind) –P(rain and wind)=0.4+0.7-0.2=0.9.
Task 3.
At the departure station there are 8 orders for goods to be sent: five for domestic shipments and three for export. What is the probability that two orders chosen at random will be for domestic consumption? Solution. Event A– the first order taken at random is within the country. Event IN– the second is also intended for domestic consumption. We need to find the probability. Then, by the theorem on the multiplication of probabilities of dependent events, we have

Task 4.
From a batch of products, the merchandiser randomly selects the highest grade products. The probability that the selected item will be of the highest quality is 0.8; first grade – 0.7; second grade – 0.5. Find the probability that out of three randomly selected products there will be:
a) only two premium grades;
b) everyone is different. Solution. Let the event be a product of the highest quality; event - first-class product; the event is a second-class product.
According to the conditions of the problem; ; Events are independent.
a) Event A– only two top-grade products will look like this then

b) Event IN– all three products are different - let’s put it this way: , Then .
Task 5.
The probabilities of hitting the target when firing from three guns are as follows: p1= 0,8; p2=0,7; p3=0.9. Find the probability of at least one hit (event A) with one salvo from all guns. Solution. The probability of each gun hitting the target does not depend on the results of firing from other guns, therefore the events under consideration (hit by the first gun), (hit by the second gun) and (hit by the third gun) are independent in the aggregate.
The probabilities of events opposite to events (i.e., the probability of misses) are respectively equal to:

Required probability
Task 6.
The printing house has 4 printing machines. For each machine, the probability that it is currently running is 0.9. Find the probability that at least one machine is currently working (event A). Solution. The events “the machine is working” and “the machine is not working” (at the moment) are opposite, therefore the sum of their probabilities is equal to one:
Hence the probability that the machine is currently not working is equal to
The required probability. Problem 7. In the reading room there are 6 textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that both textbooks will be bound.

Solution. Consider the following events:
A1 - the first bound textbook taken;
A2 is the second bound textbook taken.
An event consisting in the fact that both taken textbooks are bound. Events A1 and A2 are dependent, since the probability of the occurrence of event A2 depends on the occurrence of event A1. To solve this problem, we use the theorem for multiplying the probabilities of dependent events: .
The probability of the occurrence of event A1 p(A1) in accordance with the classical definition of probability:
P(A1)=m/n=3/6=0.5.
The probability of the occurrence of event A2 is determined by the conditional probability of the occurrence of event A2 subject to the occurrence of event A1, i.e. (A2)==0.4.
Then the desired probability of the event occurring:
P(A)=0.5*0.4=0.2.