In cases where the event of interest is the sum of other events, the addition formula is used to find its probability.

The addition formula has two main varieties - for joint and for non-joint events. You can justify these formulas using Venn diagrams (Fig. 21). Recall that in these diagrams the probabilities of events are numerically equal to the areas of the zones corresponding to these events.

For two incompatible events :

P(A+B) = P(A) + P(B).(8, a)

For N incompatible events , the probability of their sum is equal to the sum of the probabilities of these events:

= .(8b)

From the formula for adding incompatible events, there are two important consequences .

Consequence 1.For events that form a complete group, the sum of their probabilities is equal to one:

= 1.

This is explained as follows. For events that form a complete group, on the left side of expression (8b) is the probability that one of the events will occur And i , but since the complete group exhausts the entire list of possible events, one of such events will definitely occur. Thus, the left side contains the probability of an event that will definitely happen - a certain event. Its probability is equal to one.

Consequence 2.The sum of the probabilities of two opposite events is equal to one:

P(A) + P(Ā)= 1.

This consequence follows from the previous one, since opposite events always form a complete group.

Example 15

IN the probability of a working state of a technical device is 0.8. Find the probability of failure of this device for the same observation period.

R solution.

Important note. In reliability theory, it is customary to denote the probability of a working state by the letterR, and the probability of failure is a letter q. In what follows, we will use these notations. Both probabilities are functions of time. So, for long periods of time, the probability of an operable state of any object approaches zero. The probability of failure of any object is close to zero for small periods of time. In cases where the observation period is not specified in the tasks, it is assumed that it is the same for all objects under consideration.

Finding a device in the health and failure states are opposite events. Using Corollary 2, we obtain the probability of device failure:

q \u003d 1 - p \u003d 1 - 0.8 \u003d 0.2.

For two joint events probability addition formula looks like:

P (A + B) \u003d P (A) + P (B) - P (AB), (9)

which is illustrated by the Venn diagram (Fig. 22).

Indeed, to find the entire shaded area (it corresponds to the sum of events A + B), it is necessary to subtract the area of ​​the common zone from the sum of the areas of figures A and B (it corresponds to the product of events AB), since otherwise it will be counted twice.

For three joint events, the addition formula probabilities gets more complicated:

P (A + B + C) \u003d P (A) + P (B) + P (C) - P (AB) - P (AC) - P (BC) + P (ABC).(10)

In the Venn diagram (Fig. 23), the desired probability is numerically equal to the total area of ​​the zone formed by the events A, B and C (for simplicity, the unit square is not shown on it).

After subtracting the areas of zones AB, AC and CB from the sum of areas of zones A, B and C, it turned out that the area of ​​zone ABC was summed up three times and subtracted three times. Therefore, to account for this area, it must be added to the final expression.

With an increase in the number of terms, the addition formula becomes more and more cumbersome, but the principle of its construction remains the same: first, the probabilities of events taken singly are summed, then the probabilities of all pairwise combinations of events are subtracted, the probabilities of events taken by triples are added, the probabilities of combinations of events taken by fours and etc.

Finally, it should be emphasized : probability addition formula joint events with the number of terms from three or more is cumbersome and inconvenient to use, its use in solving problems is impractical.

Example 16

For the power supply scheme below (Fig. 24), determine the probability of failure of the system as a whole Q C by failure probabilities qi individual elements (generator, transformers and lines).

Failure states individual elements of the power supply system, as well as and health states are always pairwise joint events, since there are no fundamental obstacles to simultaneously repairing, for example, a line and a transformer. The failure of the system occurs when any of its elements fails: either the generator, or the 1st transformer, or the line, or the 2nd transformer, or the failure of any pair, any triple, or all four elements. Therefore, the desired event - system failure is the sum of failures of individual elements. To solve the problem, the formula for adding joint events can be used:

Q c \u003d q g + q t1 + q l + q t2 - q g q t1 - q g q l - q g q t2 - q t1 q l - q t1 q t2 - q l q t2 + q g q t1 q l + q g q l q t2 + q g q t1 q t2 + q t1 q t2 q l - q g q t1 q l q t2.

This solution once again convinces of the cumbersomeness of the addition formula for joint events. In the future, another more rational way of solving this problem will be considered.

The solution obtained above can be simplified taking into account the fact that the failure probabilities of individual elements of the power supply system for the period of one year usually used in reliability calculations are quite small (of the order of 10 -2). Therefore, all terms except for the first four can be discarded, which will practically not affect the numerical result. Then you can write:

Q with≈q g + q t1 + q l + q t2.

However, such simplifications must be treated with caution, carefully studying their consequences, since terms often discarded may turn out to be commensurate with the first.

Example 17

Determine the probability of a healthy state of the system R S, consisting of three elements reserving each other.

Solution. Elements reserving each other on the reliability analysis logic diagram are shown connected in parallel (Fig. 25):

A redundant system is operational when either the 1st, or the 2nd, or the 3rd element is operational, or any pair is operational, or all three elements together. Therefore, the operable state of the system is the sum of the operable states of individual elements. By the addition formula for joint events R c \u003d R 1 + R 2 + R 3 - R 1 R 2 - R 1 R 3 - R 2 R 3 + R 1 R 2 R 3. , Where R 1 , R 2 And R 3 are the probabilities of the operable state of elements 1, 2 and 3, respectively.

In this case, it is impossible to simplify the solution by discarding paired products, since such an approximation will give a significant error (these products are usually numerically close to the first three terms). As in Example 16, this problem has another more compact solution.

Example 18

For a double-circuit transmission line (Fig. 26), the probability of failure of each circuit is known: q 1 = q 2= 0.001. Determine the probabilities that the line will have one hundred percent throughput - P (R 100), fifty percent throughput - P (R 50), and the probability that the system will fail - Q.

The line has 100% capacity when both the 1st and 2nd circuits are operational:

P (100%) \u003d p 1 p 2 \u003d (1 - q 1) (1 - q 2) \u003d

= (1 – 0,001)(1 – 0,001) = 0,998001.

The line fails when both the 1st and 2nd circuits fail:

P(0%) \u003d q 1 q 2 \u003d 0.001 ∙ 0.001 \u003d 10 -6.

The line has fifty percent capacity when the 1st circuit is operational and the 2nd failed, or when the 2nd circuit is operational and the 1st failed:

P (50%) \u003d p 1 q 2 + p 2 q 1 \u003d 2 ∙ 0.999 ∙ 10 -3 \u003d 0.001998.

The last expression uses the addition formula for incompatible events, which they are.

The events considered in this problem form a complete group, so the sum of their probabilities is one.

The study of probability theory begins with solving problems for addition and multiplication of probabilities. It is worth mentioning right away that a student, when mastering this field of knowledge, may encounter a problem: if physical or chemical processes can be visualized and understood empirically, then the level of mathematical abstraction is very high, and understanding here comes only with experience.

However, the game is worth the candle, because the formulas - both those considered in this article and more complex ones - are used everywhere today and may well come in handy in work.

Origin

Oddly enough, the impetus for the development of this section of mathematics was ... gambling. Indeed, dice, coin toss, poker, roulette are typical examples that use addition and multiplication of probabilities. On the example of tasks in any textbook, this can be seen clearly. People were interested in learning how to increase their chances of winning, and I must say, some succeeded in this.

For example, already in the 21st century, one person, whose name we will not disclose, used this knowledge accumulated over the centuries to literally “cleanse” the casino, winning several tens of millions of dollars at roulette.

However, despite the increased interest in the subject, only by the 20th century a theoretical base was developed that made the "theorver" full-fledged. Today, in almost any science, one can find calculations using probabilistic methods.

Applicability

An important point when using the formulas for addition and multiplication of probabilities, conditional probability is the satisfiability of the central limit theorem. Otherwise, although it may not be realized by the student, all calculations, no matter how plausible they may seem, will be incorrect.

Yes, the highly motivated learner is tempted to use new knowledge at every opportunity. But in this case, one should slow down a little and strictly outline the scope of applicability.

Probability theory deals with random events, which in empirical terms are the results of experiments: we can roll a six-sided die, draw a card from a deck, predict the number of defective parts in a batch. However, in some questions it is categorically impossible to use formulas from this section of mathematics. We will discuss the features of considering the probabilities of an event, the theorems of addition and multiplication of events at the end of the article, but for now let's turn to examples.

Basic concepts

A random event is some process or result that may or may not appear as a result of an experiment. For example, we toss a sandwich - it can fall butter up or butter down. Either of the two outcomes will be random, and we do not know in advance which of them will take place.

When studying addition and multiplication of probabilities, we need two more concepts.

Joint events are such events, the occurrence of one of which does not exclude the occurrence of the other. Let's say two people shoot at a target at the same time. If one of them produces a successful one, it will not affect the ability of the second to hit the bull's-eye or miss.

Inconsistent events will be such events, the occurrence of which is simultaneously impossible. For example, by pulling out only one ball from the box, you cannot get both blue and red at once.

Designation

The concept of probability is denoted by the Latin capital letter P. Next, in parentheses, there are arguments denoting some events.

In the formulas of the addition theorem, conditional probability, multiplication theorem, you will see expressions in brackets, for example: A+B, AB or A|B. They will be calculated in various ways, and we will now turn to them.

Addition

Consider the cases in which the formulas for addition and multiplication of probabilities are used.

For incompatible events, the simplest addition formula is relevant: the probability of any of the random outcomes will be equal to the sum of the probabilities of each of these outcomes.

Suppose there is a box with 2 blue, 3 red and 5 yellow marbles. There are 10 items in total in the box. What is the percentage of the truth of the statement that we will draw a blue or red ball? It will be equal to 2/10 + 3/10, i.e. fifty percent.

In the case of incompatible events, the formula becomes more complicated, since an additional term is added. We will return to it in one paragraph, after considering one more formula.

Multiplication

Addition and multiplication of probabilities of independent events are used in different cases. If, according to the condition of the experiment, we are satisfied with either of the two possible outcomes, we will calculate the sum; if we want to get two certain outcomes one after the other, we will resort to using a different formula.

Returning to the example from the previous section, we want to draw the blue ball first and then the red one. The first number we know is 2/10. What happens next? There are 9 balls left, there are still the same number of red ones - three pieces. According to the calculations, you get 3/9 or 1/3. But what to do with two numbers now? The correct answer is to multiply to get 2/30.

Joint Events

Now we can again turn to the sum formula for joint events. Why are we digressing from the topic? To learn how probabilities are multiplied. Now we need this knowledge.

We already know what the first two terms will be (the same as in the addition formula considered earlier), but now we need to subtract the product of probabilities, which we just learned how to calculate. For clarity, we write the formula: P(A+B) = P(A) + P(B) - P(AB). It turns out that in one expression both addition and multiplication of probabilities are used.

Let's say we have to solve either of two problems in order to get credit. We can solve the first one with a probability of 0.3, and the second - 0.6. Solution: 0.3 + 0.6 - 0.18 = 0.72. Note that simply summing the numbers here will not be enough.

Conditional Probability

Finally, there is the concept of conditional probability, the arguments of which are indicated in brackets and separated by a vertical bar. The entry P(A|B) reads as follows: "probability of event A given event B".

Let's look at an example: a friend gives you some device, let it be a phone. It can be broken (20%) or good (80%). You are able to repair any device that falls into your hands with a probability of 0.4 or you are not able to do this (0.6). Finally, if the device is in working condition, you can get through to the right person with a probability of 0.7.

It's easy to see how conditional probability works in this case: you can't get through to the person if the phone is broken, and if it's good, you don't need to fix it. Thus, in order to get any results at the "second level", you need to know which event was executed at the first.

Calculations

Consider examples of solving problems for addition and multiplication of probabilities, using the data from the previous paragraph.

First, let's find the probability that you will repair the device given to you. To do this, firstly, it must be faulty, and secondly, you must cope with the repair. This is a typical multiplication problem: we get 0.2 * 0.4 = 0.08.

What is the probability that you will immediately get through to the right person? Easier than simple: 0.8 * 0.7 \u003d 0.56. In this case, you found that the phone is working and successfully made a call.

Finally, consider this scenario: you received a broken phone, fixed it, then dialed the number, and the person on the opposite end picked up the phone. Here, the multiplication of three components is already required: 0.2 * 0.4 * 0.7 \u003d 0.056.

But what if you have two non-working phones at once? How likely are you to fix at least one of them? on addition and multiplication of probabilities, since joint events are used. Solution: 0.4 + 0.4 - 0.4 * 0.4 = 0.8 - 0.16 = 0.64. Thus, if two broken devices fall into your hands, you will be able to fix it in 64% of cases.

Considerate use

As mentioned at the beginning of the article, the use of probability theory should be deliberate and conscious.

The larger the series of experiments, the closer the theoretically predicted value approaches the value obtained in practice. For example, we are tossing a coin. Theoretically, knowing about the existence of formulas for addition and multiplication of probabilities, we can predict how many times heads and tails will fall out if we conduct the experiment 10 times. We conducted an experiment, and by coincidence, the ratio of the sides that fell out was 3 to 7. But if you conduct a series of 100, 1000 or more attempts, it turns out that the distribution graph is getting closer and closer to the theoretical one: 44 to 56, 482 to 518, and so on.

Now imagine that this experiment is not carried out with a coin, but with the production of some new chemical substance, the probability of which we do not know. We would run 10 experiments and, without getting a successful result, we could generalize: "the substance cannot be obtained." But who knows, if we made the eleventh attempt, would we have reached the goal or not?

Thus, if you are going into the unknown, into an unexplored area, the theory of probability may not be applicable. Each subsequent attempt in this case may succeed, and generalizations like "X does not exist" or "X is impossible" will be premature.

Final word

So, we have considered two types of addition, multiplication and conditional probabilities. With further study of this area, it is necessary to learn to distinguish situations when each specific formula is used. In addition, you need to understand whether probabilistic methods are generally applicable in solving your problem.

If you practice, after a while you will begin to carry out all the required operations exclusively in your mind. For those who are fond of card games, this skill can be considered extremely valuable - you will significantly increase your chances of winning by just calculating the probability of a particular card or suit falling out. However, the acquired knowledge can easily be applied in other areas of activity.

The need for operations on probabilities comes when the probabilities of some events are known, and it is necessary to calculate the probabilities of other events that are associated with these events.

Probability addition is used when it is necessary to calculate the probability of a combination or a logical sum of random events.

Sum of events A And B designate A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B- an event that occurs if and only if an event occurs during the observation A or event B, or at the same time A And B.

If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.

The theorem of addition of probabilities. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:

For example, two shots were fired while hunting. Event A– hitting a duck from the first shot, event IN– hit from the second shot, event ( A+ IN) - hit from the first or second shot or from two shots. So if two events A And IN are incompatible events, then A+ IN- the occurrence of at least one of these events or two events.

Example 1 A box contains 30 balls of the same size: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball is taken without looking.

Solution. Let's assume that the event A– “the red ball is taken”, and the event IN- "The blue ball is taken." Then the event is “a colored (not white) ball is taken”. Find the probability of an event A:

and events IN:

Events A And IN- mutually incompatible, since if one ball is taken, then balls of different colors cannot be taken. Therefore, we use the addition of probabilities:

The theorem of addition of probabilities for several incompatible events. If the events make up the complete set of events, then the sum of their probabilities is equal to 1:

The sum of the probabilities of opposite events is also equal to 1:

Opposite events form a complete set of events, and the probability of a complete set of events is 1.

The probabilities of opposite events are usually denoted in small letters. p And q. In particular,

from which the following formulas for the probability of opposite events follow:

Example 2 The target in the dash is divided into 3 zones. The probability that a certain shooter will shoot at a target in the first zone is 0.15, in the second zone - 0.23, in the third zone - 0.17. Find the probability that the shooter hits the target and the probability that the shooter misses the target.

Solution: Find the probability that the shooter will hit the target:

Find the probability that the shooter misses the target:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

Addition of probabilities of mutually joint events

Two random events are said to be joint if the occurrence of one event does not preclude the occurrence of a second event in the same observation. For example, when throwing a dice, the event A is considered to be the occurrence of the number 4, and the event IN- dropping an even number. Since the number 4 is an even number, the two events are compatible. In practice, there are tasks for calculating the probabilities of the occurrence of one of the mutually joint events.

The theorem of addition of probabilities for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events is as follows:

Because the events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:

Event A occurs if one of two incompatible events occurs: or AB. However, the probability of occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:

Similarly:

Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:

When using formula (8), it should be taken into account that the events A And IN can be:

• mutually independent;
• mutually dependent.

Probability formula for mutually independent events:

Probability formula for mutually dependent events:

If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is as follows:

Example 3 In auto racing, when driving in the first car, the probability of winning, when driving in the second car. Find:

• the probability that both cars will win;
• the probability that at least one car will win;

1) The probability that the first car will win does not depend on the result of the second car, so the events A(first car wins) and IN(second car wins) - independent events. Find the probability that both cars win:

2) Find the probability that one of the two cars will win:

More difficult tasks in which you need to apply both addition and multiplication of probabilities - on the page "Various tasks for addition and multiplication of probabilities" .

Solve the problem of addition of probabilities yourself, and then look at the solution

Example 4 Two coins are thrown. Event A- loss of coat of arms on the first coin. Event B- loss of coat of arms on the second coin. Find the probability of an event C = A + B .

Probability multiplication

Multiplication of probabilities is used when the probability of a logical product of events is to be calculated.

In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.

Probability multiplication theorem for independent events. The probability of the simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:

Example 5 The coin is tossed three times in a row. Find the probability that the coat of arms will fall out all three times.

Solution. The probability that the coat of arms will fall on the first toss of a coin, the second time, and the third time. Find the probability that the coat of arms will fall out all three times:

Solve problems for multiplying probabilities yourself, and then look at the solution

Example 6 There is a box with nine new tennis balls. Three balls are taken for the game, after the game they are put back. When choosing balls, they do not distinguish between played and unplayed balls. What is the probability that after three games there will be no unplayed balls in the box?

Example 7 32 letters of the Russian alphabet are written on cut alphabet cards. Five cards are drawn at random, one after the other, and placed on the table in the order in which they appear. Find the probability that the letters will form the word "end".

Example 8 From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards are of the same suit.

Example 9 The same problem as in example 8, but each card is returned to the deck after being drawn.

More complex tasks, in which you need to apply both addition and multiplication of probabilities, as well as calculate the product of several events - on the page "Various tasks for addition and multiplication of probabilities" .

The probability that at least one of the mutually independent events will occur can be calculated by subtracting the product of the probabilities of opposite events from 1, that is, by the formula:

Example 10 Cargoes are delivered by three modes of transport: river, rail and road transport. The probability that the cargo will be delivered by river transport is 0.82, by rail 0.87, by road 0.90. Find the probability that the goods will be delivered by at least one of the three modes of transport.

Theorems of addition and multiplication of probabilities.

The theorem of addition of probabilities of two events. The probability of the sum of two events is equal to the sum of the probabilities of these events without the probability of their joint occurrence:

P(A+B)=P(A)+P(B)-P(AB).

The theorem of addition of probabilities of two incompatible events. The probability of the sum of two incompatible events is equal to the sum of the probabilities of these:

P(A+B)=P(A)+P(B).

Example 2.16. The shooter shoots at a target divided into 3 areas. The probability of hitting the first area is 0.45, the second - 0.35. Find the probability that the shooter will hit either the first or the second area with one shot.

Solution.

Events A- "the shooter hit the first area" and IN- “the shooter hit the second area” - are inconsistent (hitting in one area excludes getting into another), so the addition theorem is applicable.

The desired probability is equal to:

P(A+B)=P(A)+P(B)= 0,45+ 0,35 = 0,8.

Addition theorem P incompatible events. The probability of the sum of n incompatible events is equal to the sum of the probabilities of these:

P (A 1 + A 2 + ... + A p) \u003d P (A 1) + P (A 2) + ... + P (A p).

The sum of the probabilities of opposite events is equal to one:

Event Probability IN assuming an event has occurred A, is called the conditional probability of the event IN and is marked like this: P(B/A), or R A (B).

. The probability of the product of two events is equal to the product of the probability of one of them by the conditional probability of the other, provided that the first event occurred:

P(AB)=P(A)P A(B).

Event IN does not depend on the event A, If

P A (B) \u003d P (B),

those. event probability IN does not depend on whether the event occurred A.

The theorem of multiplication of probabilities of two independent events.The probability of the product of two independent events is equal to the product of their probabilities:

P(AB)=P(A)P(B).

Example 2.17. The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 = 0,7; p 2= 0.8. Find the probability of hitting with one volley (from both guns) by at least one of the guns.

Solution.

The probability of hitting the target by each of the guns does not depend on the result of firing from the other gun, so the events A- "First gun hit" and IN– “second gun hit” are independent.

Event Probability AB- "both guns hit":

Desired probability

P (A + B) \u003d P (A) + P (B) - P (AB)= 0,7 + 0,8 – 0,56 = 0,94.

Probability multiplication theorem P events.The probability of a product of n events is equal to the product of one of them by the conditional probabilities of all the others, calculated under the assumption that all previous events have occurred:

Example 2.18. An urn contains 5 white, 4 black and 3 blue balls. Each test consists in the fact that one ball is drawn at random without returning it back. Find the probability that a white ball will appear on the first trial (event A), a black ball on the second trial (event B), and a blue ball on the third trial (event C).

Solution.

Probability of a white ball appearing in the first trial:

The probability of a black ball appearing in the second trial, calculated assuming that a white ball appeared in the first trial, i.e. the conditional probability:

The probability of a blue ball appearing in the third trial, calculated assuming that a white ball appeared in the first trial and a black one in the second, i.e. the conditional probability:

The desired probability is equal to:

Probability multiplication theorem P independent events.The probability of a product of n independent events is equal to the product of their probabilities:

P (A 1 A 2 ... A p) \u003d P (A 1) P (A 2) ... P (A p).

The probability that at least one of the events will occur. The probability of occurrence of at least one of the events A 1 , A 2 , ..., A p, independent in the aggregate, is equal to the difference between unity and the product of the probabilities of opposite events:

.

Example 2.19. The probabilities of hitting the target when firing from three guns are as follows: p 1 = 0,8; p 2 = 0,7;p 3= 0.9. Find the probability of at least one hit (event A) with one salvo from all guns.

Solution.

The probability of hitting the target by each of the guns does not depend on the results of firing from other guns, so the events under consideration A 1(hit by the first gun), A 2(hit by the second gun) and A 3(hit of the third gun) are independent in the aggregate.

Probabilities of events opposite to events A 1, A 2 And A 3(i.e. miss probabilities), respectively, are equal to:

, , .

The desired probability is equal to:

If independent events A 1, A 2, ..., A p have the same probability R, then the probability of occurrence of at least one of these events is expressed by the formula:

Р(А)= 1 – q n ,

Where q=1-p

2.7. Total Probability Formula. Bayes formula.

Let the event A can occur if one of the incompatible events occurs N 1, N 2, ..., N p, forming a complete group of events. Since it is not known in advance which of these events will occur, they are called hypotheses.

Probability of an event occurring A calculated by total probability formula:

P (A) \u003d P (N 1) P (A / N 1) + P (N 2) P (A / N 2) + ... + P (N p) P (A / N p).

Let us assume that an experiment has been carried out, as a result of which the event A happened. Conditional event probabilities N 1, N 2, ..., N p regarding the event A determined Bayes formulas:

,

Example 2.20. In a group of 20 students who came to the exam, 6 are excellent, 8 are good, 4 are satisfactory and 2 are poorly prepared. There are 30 questions in the exam papers. A well-prepared student can answer all 30 questions, a well-prepared student can answer 24, a satisfactory student can answer 15, and a poor student can answer 7.

A randomly selected student answered three random questions. Find the probability that this student is prepared: a) excellent; b) bad.

Solution.

Hypotheses - "the student is well prepared";

– “the student is well prepared”;

– “the student is prepared satisfactorily”;

- "the student is poorly prepared."

Before experience:

; ; ; ;

7. What is called a complete group of events?

8. What events are called equally likely? Give examples of such events.

9. What is called an elementary outcome?

10. What outcomes do I call favorable to this event?

11. What operations can be performed on events? Give them definitions. How are they designated? Give examples.

12. What is called probability?

13. What is the probability of a certain event?

14. What is the probability of an impossible event?

15. What are the limits of probability?

16. How is the geometric probability on the plane determined?

17. How is probability defined in space?

18. How is the probability on a straight line determined?

19. What is the probability of the sum of two events?

20. What is the probability of the sum of two incompatible events?

21. What is the probability of the sum of n incompatible events?

22. What is the conditional probability? Give an example.

23. Formulate the probabilities multiplication theorem.

24. How to find the probability of occurrence of at least one of the events?

25. What events are called hypotheses?

26. When are the total probability formula and Bayes formulas applied?

Addition and multiplication of probabilities. This article will focus on solving problems in probability theory. Earlier, we have already analyzed some of the simplest tasks, to solve them it is enough to know and understand the formula (I advise you to repeat it).

There are tasks that are a little more complicated, for their solution you need to know and understand: the rule of addition of probabilities, the rule of multiplication of probabilities, the concepts of dependent and independent events, opposite events, joint and incompatible events. Do not be afraid of definitions, everything is simple)).In this article, we will consider just such tasks.

Some important and simple theory:

incompatible if the occurrence of one of them excludes the occurrence of the others. That is, only one particular event can occur, or another.

A classic example: when throwing a dice (dice), only one can fall out, or only two, or only three, etc. Each of these events is incompatible with the others, and the occurrence of one of them excludes the occurrence of the other (in one test). The same with the coin - the loss of "eagle" eliminates the possibility of loss of "tails".

This also applies to more complex combinations. For example, two lighting lamps are lit. Each of them may or may not burn out for some time. There are options:

1. The first burns out and the second burns out
2. The first burns out and the second does not burn out
3. The first does not burn out and the second burns out
4. The first does not burn out and the second burns out.

All these 4 variants of events are incompatible - they simply cannot happen together and none of them with any other ...

Definition: Events are called joint if the occurrence of one of them does not exclude the occurrence of the other.

Example: a queen will be taken from a deck of cards and a spade card will be taken from a deck of cards. Two events are considered. These events are not mutually exclusive - you can draw the Queen of Spades and thus both events will occur.

On the sum of probabilities

The sum of two events A and B is called the event A + B, which consists in the fact that either the event A or the event B or both will occur at the same time.

If occur incompatible events A and B, then the probability of the sum of these events is equal to the sum of the probabilities of the events:

Dice example:

We throw a dice. What is the probability of getting a number less than four?

Numbers less than four are 1,2,3. We know that the probability of getting a 1 is 1/6, a 2 is 1/6, and a 3 is 1/6. These are incompatible events. We can apply the addition rule. The probability of getting a number less than four is:

Indeed, if we proceed from the concept of classical probability: then the number of possible outcomes is 6 (the number of all faces of the cube), the number of favorable outcomes is 3 (one, two or three). The desired probability is 3 to 6 or 3/6 = 0.5.

* The probability of the sum of two joint events is equal to the sum of the probabilities of these events without taking into account their joint occurrence: P (A + B) \u003d P (A) + P (B) -P (AB)

On multiplication of probabilities

Let two incompatible events A and B occur, their probabilities are respectively P(A) and P(B). The product of two events A and B is called such an event A B, which consists in the fact that these events will occur together, that is, both event A and event B will occur. The probability of such an event is equal to the product of the probabilities of events A and B.Calculated according to the formula:

As you have already noticed, the logical connective "AND" means multiplication.

An example with the same dice:Throw a die twice. What is the probability of rolling two sixes?

The probability of rolling a six for the first time is 1/6. The second time is also equal to 1/6. The probability of getting a six both the first time and the second time is equal to the product of the probabilities:

In simple terms: when an event occurs in one test, AND then another (others) occurs, then the probability that they will occur together is equal to the product of the probabilities of these events.

We solved problems with dice, but we used only logical reasoning, we did not use the product formula. In the problems considered below, one cannot do without formulas, or rather, it will be easier and faster to get the result with them.

It is worth mentioning one more nuance. When reasoning in solving problems, the concept of SIMULTANEOUSITY of events is used. Events occur SIMULTANEOUSLY - this does not mean that they occur in one second (at one moment in time). This means that they occur in a certain period of time (with one test).

For example:

Two lamps burn out within a year (it can be said - simultaneously within a year)

Two automata break down within a month (it can be said - simultaneously within a month)

The dice is thrown three times (points fall out at the same time, which means in one test)

Biathlete makes five shots. Events (shots) occur during one test.

Events A and B are independent if the probability of either of them does not depend on the occurrence or non-occurrence of the other event.

Consider the tasks:

Two factories produce the same glass for car headlights. The first factory produces 35% of these glasses, the second - 65%. The first factory produces 4% of defective glasses, and the second - 2%. Find the probability that a glass accidentally bought in a store will be defective.

The first factory produces 0.35 products (glasses). The probability of buying defective glass from the first factory is 0.04.

The second factory produces 0.65 glasses. The probability of buying defective glass from the second factory is 0.02.

The probability that the glass was bought at the first factory AND at the same time it will be defective is 0.35∙0.04 = 0.0140.

The probability that the glass was bought at the second factory AND at the same time it will be defective is 0.65∙0.02 = 0.0130.

Buying defective glass in a store implies that it (defective glass) was purchased EITHER from the first factory OR from the second. These are incompatible events, that is, we add the resulting probabilities:

0,0140 + 0,0130 = 0,027

Answer: 0.027

If grandmaster A. plays white, then he wins grandmaster B. with a probability of 0.62. If A. plays black, then A. beats B. with a probability of 0.2. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.

The chances of winning the first and second games are independent of each other. It is said that a grandmaster must win both times, that is, win the first time AND at the same time win the second time. In the case when independent events must occur together, the probabilities of these events are multiplied, that is, the multiplication rule is used.

The probability of producing these events will be equal to 0.62∙0.2 = 0.124.

Answer: 0.124

In the geometry exam, the student gets one question from the list of exam questions. The probability that this is an inscribed circle question is 0.3. The probability that this is a Parallelogram question is 0.25. There are no questions related to these two topics at the same time. Find the probability that the student will get a question on one of these two topics on the exam.

That is, it is necessary to find the probability that the student will get a question EITHER on the topic “Inscribed circle”, OR on the topic “Parallelogram”. In this case, the probabilities are summed up, since these events are incompatible and any of these events can occur: 0.3 + 0.25 = 0.55.

*Disjoint events are events that cannot happen at the same time.

Answer: 0.55

The biathlete shoots five times at the targets. The probability of hitting the target with one shot is 0.9. Find the probability that the biathlete hit the targets the first four times and missed the last one. Round the result to the nearest hundredth.

Since the biathlete hits the target with a probability of 0.9, he misses with a probability of 1 - 0.9 = 0.1

*A miss and a hit are events that cannot occur simultaneously with one shot, the sum of the probabilities of these events is 1.

We are talking about the commission of several (independent) events. If an event occurs and at the same time another (subsequent) occurs at the same time (test), then the probabilities of these events are multiplied.

The probability of producing independent events is equal to the product of their probabilities.

Thus, the probability of the event "hit, hit, hit, hit, missed" is equal to 0.9∙0.9∙0.9∙0.9∙0.1 = 0.06561.

Rounding up to hundredths, we get 0.07

Answer: 0.07

The store has two payment machines. Each of them can be faulty with a probability of 0.07, regardless of the other automaton. Find the probability that at least one automaton is serviceable.

Find the probability that both automata are faulty.

These events are independent, so the probability will be equal to the product of the probabilities of these events: 0.07∙0.07 = 0.0049.

This means that the probability that both automata are working or one of them will be equal to 1 - 0.0049 = 0.9951.

* Both are serviceable and some one is completely - meets the condition "at least one".

One can present the probabilities of all (independent) events to test:

1. “faulty-faulty” 0.07∙0.07 = 0.0049

2. “Good-Faulty” 0.93∙0.07 = 0.0651

3. "Faulty-Faulty" 0.07∙0.93 = 0.0651

4. “healthy-healthy” 0.93∙0.93 = 0.8649

To determine the probability that at least one automaton is in good condition, it is necessary to add the probabilities of independent events 2,3 and 4: a certain event An event is called an event that is certain to occur as a result of an experience. The event is called impossible if it never happens as a result of experience.

For example, if one ball is randomly drawn from a box containing only red and green balls, then the appearance of a white ball among the drawn balls is an impossible event. The appearance of the red and the appearance of the green balls form a complete group of events.

Definition: The events are called equally possible , if there is no reason to believe that one of them will appear as a result of the experiment with a greater probability.

In the above example, the appearance of red and green balls are equally likely events if the box contains the same number of red and green balls. If there are more red balls in the box than green ones, then the appearance of a green ball is less probable than the appearance of a red one.

In we will consider more problems where the sum and product of the probabilities of events are used, do not miss it!

That's all. I wish you success!

Sincerely, Alexander Krutitskikh.

Maria Ivanovna scolds Vasya:
Petrov, why weren't you at school yesterday?!
My mom washed my pants yesterday.
- So what?
- And I was walking past the house and saw that yours were hanging. I thought you weren't coming.

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