Events A, B, C... are called dependent from each other if the probability of the occurrence of at least one of them changes depending on the occurrence or non-occurrence of other events. The events are called independent, if the probabilities of the appearance of each of them do not depend on the appearance or non-appearance of the others.

Conditional probability(PA (B) - conditional probability of event B relative to A) is the probability of event B, calculated under the assumption that event A has already occurred. example of conditional probability The conditional probability of event B, provided that event A has already occurred, by definition, is equal to PA (B) = P (AB) / P (A) (P (A)>0).

Multiplying the probabilities of dependent events: the probability of the joint occurrence of two events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the assumption that the first event has already occurred:
P (AB) = P (A) PA (B)

Example. The collector has 3 conical and 7 elliptical rollers. The picker took one roller, and then the second. Find the probability that the first of the taken rollers is conical, and the second is elliptical.

Solution: The probability that the first roller turns out to be conical (event A), P (A) = 3 / 10. The probability that the second roller turns out to be elliptical (event B), calculated under the assumption that the first roller is conical, i.e. conditional probability RA (B) = 7 / 9.
According to the multiplication formula, the desired probability is P (AB) = P (A) PA (B) = (3 / 10) * (7 / 9) = 7 / 30. Note that, keeping the notation, we can easily find: P (B) = 7 / 10, РB (A) = 3/9, Р (В) РB (А) = 7 / 30

Condition for the independence of events. Multiplying the probabilities of independent events. Examples.

Event B does not depend on event A if

P(B/A) = P(B) i.e. The probability of event B does not depend on whether event A occurs.

In this case, event A does not depend on event B, that is, the property of independence of events is mutual.

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

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

Example 1: A device operating for time t consists of three nodes, each of which, independently of the others, can fail (fail) during time t. The failure of at least one node leads to the failure of the device as a whole. During time t, the reliability (probability of failure-free operation) of the first node is p 1 = 0.8; second p 2 = 0.9 third p 3 = 0.7. Find the reliability of the device as a whole.

Solution. Denoting:

A – trouble-free operation of devices,

A 1 - trouble-free operation of the first node,

A 2 - trouble-free operation of the second node,

A 3 - trouble-free operation of the third node,

whence, by the multiplication theorem for independent events

P(A) = P(A 1)P(A 2)P(A 3) = p 1 p 2 p 3 = 0.504

Example 2. Find the probability of a number appearing together when two coins are tossed.

Solution. Probability of the appearance of the digit of the first coin (event A) P(A) = 1/2; the probability of the appearance of the digit of the second coin (event B) is P(B) = 1/2.

Events A and B are independent, so we will find the required probability

according to the formula:

P(AB) = P(A)P(B) = 1/2 *1/2 = 1/4

Compatibility and incompatibility of events. Adding the probabilities of two joint events. Examples.

The two events are called joint, if the appearance of one of them does not affect or exclude the appearance of the other. Joint events can occur simultaneously, such as, for example, the appearance of a number on one dice or

does not in any way affect the appearance of numbers on another die. Events are incompatible, if in one phenomenon or during one test they cannot be realized simultaneously and the appearance of one of them excludes the appearance of the other (hitting the target and missing are incompatible).

The probability of the occurrence of at least one of two joint events A or B 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).

Example. The probability of hitting the target for the first athlete is 0.85, and for the second - 0.8. Athletes independently of each other

fired one shot each. Find the probability that at least one athlete will hit the target?

Solution. Let us introduce the following notations: events A - “hit by the first athlete”, B - “hit by the second athlete”, C - “hit by at least one of the athletes”. Obviously, A + B = C, and events A and B are simultaneous. In accordance with the formula we get:

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

P(C) = P(A)+ P(B)-P(A)P(B),

since A and B are independent events. Substituting these values ​​P(A) = 0.85, P(B) = 0.8 into the formula for P(C), we find the desired probability

P(C) = (0.85 + 0.8) - 0.85·0.8 = 0.97

Definition 1. Event A is said to be dependent on event B if the probability of occurrence of event A depends on whether event B occurred or did not occur. The probability that event A occurred given that event B occurred will be denoted and called the conditional probability of event A subject to B.

Example 1. There are 3 white balls and 2 black balls in the urn. One ball is drawn from the urn (first draw) and then a second ball (second draw). Event B is the appearance of a white ball during the first draw. Event A is the appearance of a white ball during the second draw.

Obviously, the probability of event A, if event B occurs, will be

The probability of event A, provided that event B did not occur (a black ball appeared during the first draw), will be

We see that

Theorem 1. The probability of combining two events is equal to the product of the probability of one of them and the conditional probability of the second, calculated under the condition that the first event occurred, i.e.

Proof. We present the proof for events that reduce to the urn pattern (i.e., in the case when the classical definition of probability is applicable).

Let there be balls in the urn, white and black. Suppose that among the white balls there are balls marked “asterisk”, the rest are pure white (Fig. 408).

One ball is drawn from the urn. What is the probability of the event of taking out a white ball marked “star”?

Let B be an event consisting in the appearance of a (white ball), A - an event consisting in the appearance of a ball marked with an asterisk. Obviously,

The probability of a white ball with an asterisk appearing, given that a white ball appears, will be

The probability of a white ball with a star appearing is P (A and B). Obviously,

Substituting the left-hand sides of expressions (2), (3) and (4) into (5), we obtain

Equality (1) has been proven.

If the events under consideration do not fit into the classical scheme, then formula (1) serves to determine the conditional probability. Namely, the conditional probability of event A given the occurrence of event B is determined using

Remark 1. Apply the last formula to the expression:

In equalities (1) and (6), the left sides are equal, since this is the same probability; therefore, the right sides are also equal. Therefore we can write the equality

Example 2. For the case of example 1 given at the beginning of this section, we have According to formula (1) we obtain Probability P(A and B) is easily calculated and directly.

Example 3. The probability of producing a suitable product using this machine is 0.9. The probability of a 1st grade product appearing among suitable products is 0.8. Determine the probability of producing a 1st grade product using this machine.

Solution. Event B is the production of a suitable product using this machine, event A is the appearance of a 1st grade product. Here, substituting into formula (1), we obtain the desired probability

Theorem 2. If event A can only occur if one of the events that form a complete group of incompatible events occurs, then the probability of event A is calculated by the formula

Formula (8) is called the total probability formula. Proof. Event A can occur when any of the combined events occur

Therefore, by the theorem on the addition of probabilities we obtain

Replacing the terms on the right side according to formula (1), we obtain equality (8).

Example 4. Three consecutive shots are fired at the target. Probability of hitting with the first shot with the second with the third With one hit, the probability of hitting the target with two hits, with three hits Determine the probability of hitting the target with three shots (event A).

Solution. Let's consider the complete group of incompatible events:

There was one hit;

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 the basic terms probability theory and be able to perform simple 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. IN bon voyage:

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 extracting 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.

According to the 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 of addition of 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: , is the 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: is 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 solution:

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 a 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 came 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:

The “ve” point 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

Three guns fired a salvo 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.

The dependence of events is understood in probabilistic sense, not functional. This means that the occurrence of one of the dependent events cannot be used to unambiguously judge the occurrence of another. Probabilistic dependence means that the occurrence of one of the dependent events only changes the probability of the occurrence of the other. If the probability does not change, then the events are considered independent.

Definition: Let be an arbitrary probability space, and be some random events. They say that event A does not depend on the event IN , if its conditional probability coincides with the unconditional probability:

If , then they say that the event A depends on the event IN.

The concept of independence is symmetrical, that is, if an event A does not depend on the event IN, then the event IN does not depend on the event A. Indeed, let it be. Then . Therefore they simply say that events A And IN independent.

The following symmetric definition of the independence of events follows from the rule of multiplication of probabilities.

Definition: Events A And IN, defined on the same probability space are called independent, If

If , then events A And IN are called dependent.

Note that this definition is also valid in the case when or .

Properties of independent events.

1. If events A And IN are independent, then the following pairs of events are also independent: .

▲ Let us prove, for example, the independence of events. Let's imagine an event A in the form: . Since the events are incompatible, then , and due to the independence of the events A And IN we get that . This is what independence means. ■

2. If the event A does not depend on events B 1 And B 2, which are inconsistent () , that event A does not depend on the amount.

▲ Indeed, using the axiom of additivity of probability and independence of the event A from events B 1 And B 2, we have:

The relationship between the concepts of independence and incompatibility.

Let A And IN- any events that have a non-zero probability: , so . If the events A And IN are inconsistent (), then equality can never take place. Thus, incompatible events are dependent.

When more than two events are considered simultaneously, their pairwise independence does not sufficiently characterize the relationship between the events of the entire group. In this case, the concept of independence in the aggregate is introduced.

Definition: Events defined on the same probability space are called collectively independent, if for any 2 £ m £ n and any combination of indices the equality is true:

At m = 2 From independence in the aggregate follows pairwise independence of events. The reverse is not true.

Example. (Bernstein S.N.)

A random experiment involves tossing a regular tetrahedron (tetrahedron). A face that has fallen down is observed. The faces of the tetrahedron are colored as follows: 1st face - white, 2nd face - black,
The 3rd side is red, the 4th side contains all colors.

Let's consider the events:

A= (white dropout); B= (black dropout);

C= (Red drop).

Therefore, events A, IN And WITH are pairwise independent.

However, .

Therefore events A, IN And WITH are not collectively independent.

In practice, as a rule, the independence of events is not established by checking it by definition, but on the contrary: events are considered independent from any external considerations or taking into account the circumstances random experiment, and use independence to find the probabilities of events occurring.

Theorem (multiplication of probabilities for independent events).

If events defined on the same probability space are independent in the aggregate, then the probability of their product is equal to the product of the probabilities:

▲ The proof of the theorem follows from the definition of the independence of events in the aggregate or from the general theorem of multiplication of probabilities, taking into account the fact that in this case

Example 1 (typical example on finding conditional probabilities, the concept of independence, the theorem of addition of probabilities).

Electrical diagram consists of three independently working elements. The failure probabilities of each element are respectively equal.

1) Find the probability of failure of the circuit.

2) It is known that the circuit has failed.

What is the probability that it refused:

a) 1st element; b) 3rd element?

Solution. Consider the events = (Refused k th element), and event A= (The circuit has failed). Then the event A is presented as:

1) Since the events are not incompatible, the axiom of additivity of probability P3) is not applicable and to find the probability one should use the general theorem of addition of probabilities, according to which

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 the conditional probability of the other:

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:

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)

End of work -

This topic belongs to the section:

Lecture notes: basic concepts of probability theory and statistics used in econometrics

Kazan State.. Financial and Economic Institute.. Department of Statistics and Econometrics..

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