Events a and b are called dependent. Probability addition and multiplication theorems. Dependent and independent events. Problems on the theorems for adding probabilities of inconsistent and multiplying probabilities of independent events

IN Unified State Exam assignments in mathematics there are more complex tasks on probability (than we considered in part 1), where we have to apply the rule of addition, multiplication of probabilities, and distinguish between compatible and incompatible events.

So, the theory.

Joint and non-joint events

Events are called incompatible if the occurrence of one of them excludes the occurrence of others. That is, only one specific event or another can happen.

For example, when throwing a die, you can distinguish between events such as getting an even number of points and getting an odd number of points. These events are incompatible.

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

For example, when throwing a die, you can distinguish such events as rolling an odd number of points and rolling a number of points that are a multiple of three. When a three is rolled, both events occur.

Sum of events

The sum (or combination) of several events is an event consisting of the occurrence of at least one of these events.

Wherein sum of two incompatible events is the sum of the probabilities of these events:

For example, the probability of getting 5 or 6 points on a die with one throw will be , because both events (rolling 5, rolling 6) are inconsistent and the probability of one or the other event occurring is calculated as follows:

The probability sum of two joint events equal to the sum of the probabilities of these events without taking into account their joint occurrence:

For example, in a shopping center, two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Let's find the probability that by the end of the day the coffee will run out in at least one of the machines (that is, either one, or the other, or both at once).

The probability of the first event “coffee will run out in the first machine” as well as the probability of the second event “coffee will run out in the second machine” according to the condition is equal to 0.3. Events are collaborative.

The probability of the joint occurrence of the first two events according to the condition is 0.12.

This means that the probability that by the end of the day the coffee will run out in at least one of the machines is

Dependent and independent events

Two random events A and B are called independent if the occurrence of one of them does not change the probability of the occurrence of the other. Otherwise, events A and B are called dependent.

For example, when two dice are rolled simultaneously, one of them, say 1, and the other, 5, are independent events.

Product of probabilities

The product (or intersection) of several events is an event consisting of the joint occurrence of all these events.

If two occur independent events A and B with probabilities P(A) and P(B) respectively, then the probability of the occurrence of events A and B at the same time is equal to the product of the probabilities:

For example, we are interested in getting a six on a die twice in a row. Both events are independent and the probability of each of them occurring separately is . The probability that both of these events will occur will be calculated using the above formula: .

See a selection of tasks to practice the topic.

Classic definition of probability.

The probability of an event is a quantitative measure that is introduced to compare events according to the degree of possibility of their occurrence.

An event that can be represented as a collection (sum) of several elementary events is called composite.

An event that cannot be broken down into simpler ones is called elementary.

An event is called impossible if it never occurs under the conditions of a given experiment (test).

Certain and impossible events are not random.

Joint events– several events are called joint if, as a result of the experiment, the occurrence of one of them does not exclude the occurrence of others.

Incompatible events– several events are called incompatible in a given experiment if the occurrence of one of them excludes the occurrence of others. The two events are called opposite, if one of them occurs if and only if the other does not occur.

The probability of event A is P(A) – is called the number ratio m elementary events (outcomes) favorable to the occurrence of the event A, to the number n all elementary events under the conditions of a given probabilistic experiment.

The following properties of probability follow from the definition:

1. The probability of a random event is a positive number between 0 and 1:

(2)

2. The probability of a certain event is 1: (3)

3. If an event is impossible, then its probability is equal to

(4)

4. If events are incompatible, then

5. If events A and B are joint, then the probability of their sum 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)(6)

6. If and are opposite events, then (7)

7. Sum of event probabilities A 1, A 2, …, A n, forming a complete group, is equal to 1:

P(A 1) + P(A 2) + …+ P(A n) = 1.(8)

In economic studies, the values ​​and in the formula may be interpreted differently. At statistical definition The probability of an event is the number of observations of the experimental results in which the event occurred exactly once. In this case the relation is called relative frequency (frequency) of an event

Events A, B are called independent, if the probability of each of them does not depend on whether another event occurred or not. The probabilities of independent events are called unconditional.

Events A, B are called dependent, if the probability of each of them depends on whether another event occurred or not. The probability of event B, calculated under the assumption that another event A has already occurred, is called conditional probability.

If two events A and B are independent, then the equalities are true:

P(B) = P(B/A), P(A) = P(A/B) or P(B/A) – P(B) = 0(9)

The probability of the product of two dependent events A, B is equal to the product of the probability of one of them by conditional probability another:

P(AB) = P(B) ∙ P(A/B) or P(AB) = P(A) ∙ P(B/A) (10)

Probability of event B given the occurrence of event A:

(11)

Probability of the product of two independent events A, B is equal to the product of their probabilities:

P(AB) = P(A) ∙ P(B)(12)

If several events are pairwise independent, then it does not follow that they are independent in the aggregate.

Events A 1, A 2, ..., A n (n>2) are called independent in the aggregate if the probability of each of them does not depend on whether any of the other events occurred or not.

The probability of the joint occurrence of several events that are independent in the aggregate is equal to the product of the probabilities of these events:

P(A 1 ∙A 2 ∙A 3 ∙…∙A n) = P(A 1)∙P(A 2)∙P(A 3)∙…∙P(A n). (13)

Probability addition and multiplication theorems.
Dependent and independent events

The title looks scary, but in reality everything is very simple. In this lesson we will get acquainted with the theorems of addition and multiplication of event probabilities, and also analyze typical problems that, along with problem on the classical determination of probability will definitely meet or, more likely, have already met on your way. To effectively study the materials in this article, you need to know and understand basic terms probability theory and be able to do the simplest arithmetic operations. As you can see, very little is required, and therefore a fat plus in the asset is almost guaranteed. But on the other hand, I again warn against a superficial attitude to practical examples - there are also plenty of subtleties. Good luck:

Theorem for adding probabilities of incompatible events: probability of occurrence of one of two incompatible events or (no matter what), is equal to the sum of the probabilities of these events:

A similar fact is true for a larger number of incompatible events, for example, for three incompatible events and:

The theorem is a dream =) However, such a dream is subject to proof, which can be found, for example, in textbook V.E. Gmurman.

Let's get acquainted with new, hitherto unknown concepts:

Dependent and independent events

Let's start with independent events. Events are independent , if the probability of occurrence any of them does not depend on the appearance/non-appearance of other events of the set under consideration (in all possible combinations). ...But why bother with general phrases:

Theorem for multiplying the probabilities of independent events: the probability of joint occurrence of independent events and is equal to the product of the probabilities of these events:

Let's return to the simplest example of the 1st lesson, in which two coins are tossed and the following events:

– heads will appear on the 1st coin;
– heads will appear on the 2nd coin.

Let's find the probability of the event (heads will appear on the 1st coin And an eagle will appear on the 2nd coin - remember how to read product of events!) . The probability of heads on one coin does not depend in any way on the result of throwing another coin, therefore, the events are independent.

Likewise:
– the probability that the 1st coin will land heads And on the 2nd tails;
– the probability that heads will appear on the 1st coin And on the 2nd tails;
– the probability that the 1st coin will show heads And on the 2nd eagle.

Notice that the events form full group and the sum of their probabilities is equal to one: .

The multiplication theorem obviously extends to a larger number of independent events, for example, if the events are independent, then the probability of their joint occurrence is equal to: . Let's practice with specific examples:

Problem 3

Each of the three boxes contains 10 parts. The first box contains 8 standard parts, the second – 7, the third – 9. One part is randomly removed from each box. Find the probability that all parts will be standard.

Solution: The probability of drawing a standard or non-standard part from any box does not depend on what parts are taken from other boxes, so the problem deals with independent events. Consider the following independent events:

– a standard part is removed from the 1st box;
– a standard part was removed from the 2nd box;
– a standard part is removed from the 3rd box.

According to the classical definition:
are the corresponding probabilities.

Event of interest to us (a standard part will be removed from the 1st box And from 2nd standard And from 3rd standard) is expressed by the product.

According to the theorem of multiplication of probabilities of independent events:

– the probability that one standard part will be removed from three boxes.

Answer: 0,504

After invigorating exercises with boxes, no less interesting urns await us:

Problem 4

Three urns contain 6 white and 4 black balls. One ball is drawn at random from each urn. Find the probability that: a) all three balls will be white; b) all three balls will be the same color.

Based on the information received, guess how to deal with the “be” point ;-) An approximate example of a solution is designed in an academic style with a detailed description of all events.

Dependent Events. The event is called dependent , if its probability depends from one or more events that have already occurred. You don’t have to go far for examples - just go to the nearest store:

– tomorrow at 19.00 fresh bread will be on sale.

The likelihood of this event depends on many other events: whether fresh bread will be delivered tomorrow, whether it will be sold out before 7 pm or not, etc. Depending on various circumstances, this event can be either reliable or impossible. So the event is dependent.

Bread... and, as the Romans demanded, circuses:

– at the exam, the student will receive a simple ticket.

If you are not the very first, then the event will be dependent, since its probability will depend on what tickets have already been drawn by classmates.

How to determine the dependence/independence of events?

Sometimes this is directly stated in the problem statement, but most often you have to conduct an independent analysis. There is no unambiguous guideline here, and the fact of dependence or independence of events follows from natural logical reasoning.

In order not to lump everything into one pile, tasks for dependent events I will highlight the following lesson, but for now we will consider the most common set of theorems in practice:

Problems on addition theorems for incompatible probabilities
and multiplying the probabilities of independent events

This tandem, according to my subjective assessment, works in approximately 80% of tasks on the topic under consideration. Hit of hits and a real classic of probability theory:

Problem 5

Two shooters each fired one shot at the target. The probability of a hit for the first shooter is 0.8, for the second - 0.6. Find the probability that:

a) only one shooter will hit the target;
b) at least one of the shooters will hit the target.

Solution: One shooter's hit/miss rate is obviously independent of the other shooter's performance.

Let's consider the events:
– 1st shooter will hit the target;
– The 2nd shooter will hit the target.

By condition: .

Let's find the probabilities of opposite events - that the corresponding arrows will miss:

a) Consider the event: – only one shooter will hit the target. This event consists of two incompatible outcomes:

1st shooter will hit And 2nd one will miss
or
1st one will miss And The 2nd one will hit.

On the tongue event algebras this fact will be written by the following formula:

First, we use the theorem for adding the probabilities of incompatible events, then the theorem for multiplying the probabilities of independent events:

– the probability that there will be only one hit.

b) Consider the event: – at least one of the shooters hits the target.

First of all, LET’S THINK – what does the condition “AT LEAST ONE” mean? In this case, this means that either the 1st shooter will hit (the 2nd will miss) or 2nd (1st will miss) or both shooters at once - a total of 3 incompatible outcomes.

Method one: taking into account the ready probability of the previous point, it is convenient to represent the event as the sum of the following incompatible events:

someone will get there (an event consisting in turn of 2 incompatible outcomes) or
If both arrows hit, we denote this event with the letter .

Thus:

According to the theorem of multiplication of probabilities of independent events:
– probability that the 1st shooter will hit And The 2nd shooter will hit.

According to the theorem of addition of probabilities of incompatible events:
– the probability of at least one hit on the target.

Method two: Consider the opposite event: – both shooters will miss.

According to the theorem of multiplication of probabilities of independent events:

As a result:

Pay special attention to the second method - in general, it is more rational.

In addition, there is an alternative, third way of solving it, based on the theorem of addition of joint events, which was not mentioned above.

! If you are getting acquainted with the material for the first time, then in order to avoid confusion, it is better to skip the next paragraph.

Method three : the events are compatible, which means their sum expresses the event “at least one shooter will hit the target” (see. algebra of events). By the theorem for adding probabilities of joint events and the theorem of multiplication of probabilities of independent events:

Let's check: events and (0, 1 and 2 hits respectively) form a complete group, so the sum of their probabilities must equal one:
, which was what needed to be checked.

Answer:

With a thorough study of probability theory, you will come across dozens of problems with a militaristic content, and, characteristically, after this you will not want to shoot anyone - the problems are almost a gift. Why not simplify the template as well? Let's shorten the entry:

Solution: by condition: , – probability of hitting the corresponding shooters. Then the probabilities of their miss:

a) According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:
– the probability that only one shooter will hit the target.

b) According to the theorem of multiplication of probabilities of independent events:
– the probability that both shooters will miss.

Then: – the probability that at least one of the shooters will hit the target.

Answer:

In practice, you can use any design option. Of course, much more often they take the short route, but we must not forget the 1st method - although it is longer, it is more meaningful - it is clearer, what, why and why adds and multiplies. In some cases, a hybrid style is appropriate, when it is convenient to use capital letters to indicate only some events.

Similar tasks for independent decision:

Problem 6

To signal a fire, two independently operating sensors are installed. The probabilities that the sensor will operate in the event of a fire are 0.5 and 0.7, respectively, for the first and second sensors. Find the probability that in a fire:

a) both sensors will fail;
b) both sensors will work.
c) Using the theorem for adding the probabilities of events forming a complete group, find the probability that in a fire only one sensor will work. Check the result by directly calculating this probability (using addition and multiplication theorems).

Here, the independence of the operation of the devices is directly stated in the condition, which, by the way, is an important clarification. The sample solution is designed in an academic style.

What if in a similar problem the same probabilities are given, for example, 0.9 and 0.9? You need to decide exactly the same! (which, in fact, has already been demonstrated in the example with two coins)

Problem 7

The probability of hitting the target by the first shooter with one shot is 0.8. The probability that the target is not hit after the first and second shooters fire one shot each is 0.08. What is the probability of the second shooter hitting the target with one shot?

And this is a small puzzle, which is designed in a short way. The condition can be reformulated more succinctly, but I will not redo the original - in practice, I have to delve into more ornate fabrications.

Meet him - he is the one who has planned an enormous amount of details for you =):

Problem 8

A worker operates three machines. The probability that during a shift the first machine will require adjustment is 0.3, the second - 0.75, the third - 0.4. Find the probability that during the shift:

a) all machines will require adjustment;
b) only one machine will require adjustment;
c) at least one machine will require adjustment.

Solution: since the condition does not say anything about a single technological process, then the operation of each machine should be considered independent of the operation of other machines.

By analogy with Problem No. 5, here you can enter into consideration the events that the corresponding machines will require adjustments during the shift, write down the probabilities, find the probabilities of opposite events, etc. But with three objects, I don’t really want to format the task like this anymore – it will turn out long and tedious. Therefore, it is noticeably more profitable to use the “fast” style here:

According to the condition: – the probability that during the shift the corresponding machines will require tuning. Then the probabilities that they will not require attention are:

One of the readers found a cool typo here, I won’t even correct it =)

a) According to the theorem of multiplication of probabilities of independent events:
– the probability that during the shift all three machines will require adjustments.

b) The event “During the shift, only one machine will require adjustment” consists of three incompatible outcomes:

1) 1st machine will require attention And 2nd machine won't require And 3rd machine won't require
or:
2) 1st machine won't require attention And 2nd machine will require And 3rd machine won't require
or:
3) 1st machine won't require attention And 2nd machine won't require And 3rd machine will require.

According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:

– the probability that during a shift only one machine will require adjustment.

I think by now you should understand where the expression comes from

c) Let’s calculate the probability that the machines will not require adjustment, and then the probability of the opposite event:
– that at least one machine will require adjustment.

Answer:

Point “ve” can also be solved through the sum , where is the probability that during a shift only two machines will require adjustment. This event, in turn, includes 3 incompatible outcomes, which are described by analogy with the “be” point. Try to find the probability yourself to check the whole problem using equality.

Problem 9

A salvo was fired from three guns at the target. The probability of a hit with one shot from only the first gun is 0.7, from the second – 0.6, from the third – 0.8. Find the probability that: 1) at least one projectile will hit the target; 2) only two shells will hit the target; 3) the target will be hit at least twice.

The solution and answer are at the end of the lesson.

And again about coincidences: if, according to the condition, two or even all values ​​of the initial probabilities coincide (for example, 0.7, 0.7 and 0.7), then exactly the same solution algorithm should be followed.

To conclude the article, let’s look at another common puzzle:

Problem 10

The shooter hits the target with the same probability with each shot. What is this probability if the probability of at least one hit with three shots is 0.973.

Solution: let us denote by – the probability of hitting the target with each shot.
and through - the probability of a miss with each shot.

And let’s write down the events:
– with 3 shots the shooter will hit the target at least once;
– the shooter will miss 3 times.

By condition, then the probability of the opposite event:

On the other hand, according to the theorem of multiplication of probabilities of independent events:

Thus:

- the probability of a miss with each shot.

As a result:
– the probability of a hit with each shot.

Answer: 0,7

Simple and elegant.

In the problem considered, additional questions can be asked about the probability of only one hit, only two hits, and the probability of three hits on the target. The solution scheme will be exactly the same as in the two previous examples:

However, the fundamental substantive difference is that here there are repeated independent tests, which are performed sequentially, independently of each other and with the same probability of outcomes.

It is unlikely that many people think about whether it is possible to calculate events that are more or less random. To put it simply in simple words, is it really possible to know which side of the cube will come up next time? It was this question that two great scientists asked themselves, who laid the foundation for such a science as the theory of probability, in which the probability of an event is studied quite extensively.

Origin

If you try to define such a concept as probability theory, you will get the following: this is one of the branches of mathematics that studies the constancy of random events. Of course, this concept does not really reveal the whole essence, so it is necessary to consider it in more detail.

I would like to start with the creators of the theory. As mentioned above, there were two of them, and they were one of the first to try to calculate the outcome of this or that event using formulas and mathematical calculations. In general, the beginnings of this science appeared in the Middle Ages. At that time, various thinkers and scientists tried to analyze gambling, such as roulette, dice, and so on, thereby establishing the pattern and percentage of a particular number falling out. The foundation was laid in the seventeenth century by the above-mentioned scientists.

At first, their works could not be considered great achievements in this field, because all they did were simply empirical facts, and experiments were carried out visually, without using formulas. Over time, it was possible to achieve great results, which appeared as a result of observing the throwing of dice. It was this tool that helped to derive the first intelligible formulas.

Like-minded people

It is impossible not to mention such a person as Christiaan Huygens in the process of studying a topic called “probability theory” (the probability of an event is covered precisely in this science). This person is very interesting. He, like the scientists presented above, tried in the form mathematical formulas derive a pattern of random events. It is noteworthy that he did not do this together with Pascal and Fermat, that is, all his works did not intersect with these minds. Huygens deduced

An interesting fact is that his work came out long before the results of the discoverers’ work, or rather, twenty years earlier. Among the identified concepts, the most famous are:

• the concept of probability as the value of chance;
• mathematical expectation for discrete cases;
• theorems of multiplication and addition of probabilities.

It is also impossible not to remember who also made a significant contribution to the study of the problem. Conducting his own tests, independent of anyone, he was able to present a proof of the law of large numbers. In turn, the scientists Poisson and Laplace, who worked at the beginning of the nineteenth century, were able to prove the original theorems. It was from this moment that probability theory began to be used to analyze errors in observations. Russian scientists, or rather Markov, Chebyshev and Dyapunov, could not ignore this science. Based on the work done by great geniuses, they established this subject as a branch of mathematics. These figures worked already at the end of the nineteenth century, and thanks to their contribution, the following phenomena were proven:

• law of large numbers;
• Markov chain theory;
• central limit theorem.

So, with the history of the birth of science and with the main people who influenced it, everything is more or less clear. Now the time has come to clarify all the facts.

Basic Concepts

Before touching on laws and theorems, it is worth studying the basic concepts of probability theory. The event plays a leading role in it. This topic quite voluminous, but without it you won’t be able to figure out everything else.

An event in probability theory is any set of outcomes of an experiment. There are quite a few concepts of this phenomenon. Thus, the scientist Lotman, working in this area, said that in this case we are talking about what “happened, although it might not have happened.”

Random events (the theory of probability pays special attention to them) is a concept that implies absolutely any phenomenon that has the opportunity to occur. Or, conversely, this scenario may not happen if many conditions are met. It is also worth knowing that they capture the entire scope of the phenomena that occurred precisely random events. The theory of probability indicates that all conditions can be repeated constantly. It is their conduct that is called “experience” or “test”.

A reliable event is a phenomenon that is one hundred percent likely to happen in a given test. Accordingly, an impossible event is one that will not happen.

The combination of a pair of actions (conditionally, case A and case B) is a phenomenon that occurs simultaneously. They are designated as AB.

The sum of pairs of events A and B is C, in other words, if at least one of them happens (A or B), then C will be obtained. The formula for the described phenomenon is written as follows: C = A + B.

Incongruent events in probability theory imply that two cases are mutually exclusive. Under no circumstances can they happen at the same time. Joint events in probability theory are their antipode. What is meant here is that if A happened, then it does not prevent B in any way.

Opposite events (probability theory considers them in great detail) are easy to understand. The best way to understand them is by comparison. They are almost the same as incompatible events in probability theory. But their difference lies in the fact that one of many phenomena must happen in any case.

Equally probable events are those actions whose repetition is equal. To make it clearer, you can imagine tossing a coin: the loss of one of its sides is equally likely to fall out of the other.

It is easier to consider an auspicious event with an example. Let's say there is an episode B and an episode A. The first is the roll of the dice with an odd number appearing, and the second is the appearance of the number five on the die. Then it turns out that A favors B.

Independent events in probability theory are projected only onto two or more cases and imply the independence of any action from another. For example, A is the loss of heads when tossing a coin, and B is the drawing of a jack from the deck. They are independent events in probability theory. At this point it became clearer.

Dependent events in probability theory are also permissible only for a set of them. They imply the dependence of one on the other, that is, phenomenon B can only occur if A has already happened or, conversely, has not happened, when this is the main condition for B.

Exodus random experiment, consisting of one component, is elementary events. The theory of probability explains that this is a phenomenon that happened only once.

Basic formulas

So, the concepts of “event” and “probability theory” were discussed above; a definition of the basic terms of this science was also given. Now it’s time to get acquainted directly with the important formulas. These expressions mathematically confirm all the main concepts in such a complex subject as probability theory. The probability of an event plays a huge role here too.

It’s better to start with the basic ones. And before you start with them, it’s worth considering what they are.

Combinatorics is primarily a branch of mathematics; it deals with the study of a huge number of integers, as well as various permutations of both the numbers themselves and their elements, various data, etc., leading to the appearance of a number of combinations. In addition to probability theory, this branch is important for statistics, computer science and cryptography.

So, now we can move on to presenting the formulas themselves and their definition.

The first of them will be the expression for the number of permutations, it looks like this:

P_n = n ⋅ (n - 1) ⋅ (n - 2)…3 ⋅ 2 ⋅ 1 = n!

The equation is applied only if the elements differ only in the order of their arrangement.

Now the placement formula will be considered, it looks like this:

A_n^m = n ⋅ (n - 1) ⋅ (n-2) ⋅ ... ⋅ (n - m + 1) = n! : (n - m)!

This expression is applicable not only to the order of placement of the element, but also to its composition.

The third equation from combinatorics, and it is also the last, is called the formula for the number of combinations:

C_n^m = n ! : ((n - m))! :m!

A combination refers to selections that are not ordered; accordingly, this rule applies to them.

It was easy to understand the combinatorics formulas; now you can move on to the classical definition of probabilities. This expression looks like this:

In this formula, m is the number of conditions favorable to event A, and n is the number of absolutely all equally possible and elementary outcomes.

There are a large number of expressions; the article will not cover all of them, but the most important ones will be touched upon, such as, for example, the probability of the sum of events:

P(A + B) = P(A) + P(B) - this theorem is for adding only incompatible events;

P(A + B) = P(A) + P(B) - P(AB) - and this one is for adding only compatible ones.

Probability of events occurring:

P(A ⋅ B) = P(A) ⋅ P(B) - this theorem is for independent events;

(P(A ⋅ B) = P(A) ⋅ P(B∣A); P(A ⋅ B) = P(A) ⋅ P(A∣B)) - and this one is for the dependent.

The list of events will be completed by the formula of events. Probability theory tells us about Bayes' theorem, which looks like this:

P(H_m∣A) = (P(H_m)P(A∣H_m)) : (∑_(k=1)^n P(H_k)P(A∣H_k)),m = 1,..., n

In this formula, H 1, H 2, ..., H n is a complete group of hypotheses.

Examples

If you carefully study any section of mathematics, it is not complete without exercises and sample solutions. So is the theory of probability: events and examples here are an integral component that confirms scientific calculations.

Formula for the number of permutations

Let's say in deck of cards there are thirty cards, starting with a value of one. Next question. How many ways are there to stack the deck so that cards with value one and two are not next to each other?

The task has been set, now let's move on to solving it. First you need to determine the number of permutations of thirty elements, for this we take the formula presented above, it turns out P_30 = 30!.

Based on this rule, we find out how many options there are to fold the deck in different ways, but we need to subtract from them those in which the first and second cards are next to each other. To do this, let's start with the option when the first is above the second. It turns out that the first card can take up twenty-nine places - from the first to the twenty-ninth, and the second card from the second to the thirtieth, making a total of twenty-nine places for a pair of cards. In turn, the rest can accept twenty-eight places, and in any order. That is, to rearrange twenty-eight cards, there are twenty-eight options P_28 = 28!

As a result, it turns out that if we consider the solution when the first card is above the second, there will be 29 ⋅ 28 extra possibilities! = 29!

Using the same method, you need to calculate the number of redundant options for the case when the first card is under the second. It also turns out to be 29 ⋅ 28! = 29!

It follows from this that there are 2 ⋅ 29 extra options!, while the necessary ways to assemble a deck are 30! - 2 ⋅ 29!. All that remains is to count.

30! = 29! ⋅ 30; 30!- 2 ⋅ 29! = 29! ⋅ (30 - 2) = 29! ⋅ 28

Now you need to multiply all the numbers from one to twenty-nine, and then finally multiply everything by 28. The answer is 2.4757335 ⋅〖10〗^32

Example solution. Formula for placement number

In this problem, you need to find out how many ways there are to put fifteen volumes on one shelf, but provided that there are thirty volumes in total.

The solution to this problem is a little simpler than the previous one. Using the already known formula, it is necessary to calculate the total number of arrangements of thirty volumes of fifteen.

A_30^15 = 30 ⋅ 29 ⋅ 28⋅... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ 16 = 202 843 204 931 727 360 000

The answer, accordingly, will be equal to 202,843,204,931,727,360,000.

Now let's take a slightly more difficult task. You need to find out how many ways there are to arrange thirty books on two bookshelves, provided that only fifteen volumes can be placed on one shelf.

Before starting the solution, I would like to clarify that some problems can be solved in several ways, and this one has two methods, but both use the same formula.

In this problem, you can take the answer from the previous one, because there we calculated how many times you can fill a shelf with fifteen books in different ways. It turned out A_30^15 = 30 ⋅ 29 ⋅ 28 ⋅ ... ⋅ (30 - 15 + 1) = 30 ⋅ 29 ⋅ 28 ⋅ ...⋅ 16.

We will calculate the second shelf using the permutation formula, because fifteen books can be placed in it, while only fifteen remain. We use the formula P_15 = 15!.

It turns out that the total will be A_30^15 ⋅ P_15 ways, but, in addition to this, the product of all numbers from thirty to sixteen will need to be multiplied by the product of numbers from one to fifteen, in the end you will get the product of all numbers from one to thirty, that is, the answer equals 30!

But this problem can be solved in another way - easier. To do this, you can imagine that there is one shelf for thirty books. All of them are placed on this plane, but since the condition requires that there be two shelves, we saw one long one in half, so we get two of fifteen. From this it turns out that there can be P_30 = 30 options for arrangement!.

Example solution. Formula for combination number

Now we will consider a version of the third problem from combinatorics. It is necessary to find out how many ways there are to arrange fifteen books, provided that you need to choose from thirty absolutely identical ones.

To solve, of course, the formula for the number of combinations will be applied. From the condition it becomes clear that the order of the identical fifteen books is not important. Therefore, initially you need to find out the total number of combinations of thirty books of fifteen.

C_30^15 = 30 ! : ((30-15)) ! : 15 ! = 155 117 520

That's all. Using this formula, we were able to solve this problem in the shortest possible time; the answer, accordingly, is 155,117,520.

Example solution. Classic definition of probability

Using the formula above, you can find the answer to a simple problem. But this will help to clearly see and track the progress of actions.

The problem states that there are ten absolutely identical balls in the urn. Of these, four are yellow and six are blue. One ball is taken from the urn. You need to find out the probability of getting blue.

To solve the problem, it is necessary to designate getting the blue ball as event A. This experiment can have ten outcomes, which, in turn, are elementary and equally possible. At the same time, out of ten, six are favorable to event A. We solve using the formula:

P(A) = 6: 10 = 0.6

Applying this formula, we learned that the probability of getting the blue ball is 0.6.

Example solution. Probability of the sum of events

An option will now be presented that is solved using the sum-of-events probability formula. So, the condition is given that there are two boxes, the first contains one gray and five white balls, and the second contains eight gray and four white balls. As a result, they took one of them from the first and second boxes. You need to find out what is the chance that the balls you get will be gray and white.

To solve this problem, it is necessary to identify events.

• So, A - took a gray ball from the first box: P(A) = 1/6.
• A’ - took a white ball also from the first box: P(A") = 5/6.
• B - a gray ball was removed from the second box: P(B) = 2/3.
• B’ - took a gray ball from the second box: P(B") = 1/3.

According to the conditions of the problem, it is necessary for one of the phenomena to happen: AB’ or A’B. Using the formula, we get: P(AB") = 1/18, P(A"B) = 10/18.

Now the formula for multiplying the probability has been used. Next, to find out the answer, you need to apply the equation of their addition:

P = P(AB" + A"B) = P(AB") + P(A"B) = 11/18.

This is how you can solve similar problems using the formula.

Bottom line

The article presented information on the topic "Probability Theory", in which the probability of an event plays a vital role. Of course, not everything was taken into account, but, based on the presented text, you can theoretically familiarize yourself with this section of mathematics. The science in question can be useful not only in professional work, but also in Everyday life. With its help, you can calculate any possibility of any event.

The text also touched upon significant dates in the history of the formation of the theory of probability as a science, and the names of the people whose work was invested in it. This is how human curiosity led to the fact that people learned to calculate even random events. Once upon a time they were simply interested in this, but today everyone already knows about it. And no one will say what awaits us in the future, what other brilliant discoveries related to the theory under consideration will be made. But one thing is for sure - research does not stand still!

General statement of the problem: the probabilities of some events are known, and you need to calculate the probabilities of other events that are associated with these events. In these problems, there is a need for operations with probabilities such as addition and multiplication of probabilities.

For example, while hunting, two shots are fired. Event A- hitting a duck with the first shot, event B- hit from the second shot. Then the sum of events A And B- hit with the first or second shot or with two shots.

Problems of a different type. Several events are given, for example, a coin is tossed three times. You need to find the probability that either the coat of arms will appear all three times, or that the coat of arms will appear at least once. This is a probability multiplication problem.

Addition of probabilities of incompatible events

Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.

Sum of events A And B denote 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 the event occurred during observation A or event B, or simultaneously 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.

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

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

Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.

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

and events IN:

Events A And IN– mutually incompatible, since if one ball is taken, then it is impossible to take balls of different colors. Therefore, we use the addition of probabilities:

The theorem for adding probabilities for several incompatible events. If events constitute a 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.

Probabilities of opposite events are usually indicated 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 shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the 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 will hit the target and the probability that the shooter will miss the target.

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

Let's find the probability that the shooter will miss the target:

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

Addition of probabilities of mutually simultaneous events

Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing dice event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since 4 is an even number, the two events are compatible. In practice, there are problems of calculating the probabilities of the occurrence of one of the mutually simultaneous events.

Probability addition theorem 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 has the following form:

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

Likewise:

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 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:

Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive 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(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:

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

More complex problems, in which you need to use both addition and multiplication of probabilities, can be found on the page "Various problems involving addition and multiplication of probabilities".

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

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

Multiplying Probabilities

Probability multiplication is used when the probability of a logical product of events must 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. Probability of 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 appear all three times.

Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:

Solve probability multiplication problems on your own and then look at the solution

Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games Are there any unplayed balls left in the box?

Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. 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 will be of different suits.

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

More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".

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