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3.3. Probability multiplication theorem

Lecture



Before setting out the theorem of multiplication of probabilities, we introduce another important concept: the concept of independent and dependent events.

Event   3.3.  Probability multiplication theorem called independent of the event   3.3.  Probability multiplication theorem if the probability of an event   3.3.  Probability multiplication theorem does not depend on what happened   3.3.  Probability multiplication theorem or not.

Event   3.3.  Probability multiplication theorem called event dependent   3.3.  Probability multiplication theorem if the probability of event A changes depending on whether the event has occurred   3.3.  Probability multiplication theorem or not.

Consider examples.

1) The experience consists in throwing two coins; deals with events:

  3.3.  Probability multiplication theorem - the appearance of the emblem on the first coin,

  3.3.  Probability multiplication theorem - The appearance of the emblem on the second coin.

In this case, the probability of an event   3.3.  Probability multiplication theorem does not depend on what happened   3.3.  Probability multiplication theorem or not; event   3.3.  Probability multiplication theorem whatever event   3.3.  Probability multiplication theorem .

2) In the urn two white balls and one black; two faces are removed from the urn one bowl at a time; deals with events:

  3.3.  Probability multiplication theorem - the appearance of a white ball in the 1st person,

  3.3.  Probability multiplication theorem - the appearance of a white ball in the 2nd person.

Event probability   3.3.  Probability multiplication theorem before something is known about the event   3.3.  Probability multiplication theorem equal to 2/3. If it became known that the event   3.3.  Probability multiplication theorem happened, then the probability of an event   3.3.  Probability multiplication theorem becomes equal to ½, from which we conclude that the event   3.3.  Probability multiplication theorem depends on the event   3.3.  Probability multiplication theorem .

Event probability   3.3.  Probability multiplication theorem calculated on condition that another event has occurred   3.3.  Probability multiplication theorem is called the conditional probability of an event   3.3.  Probability multiplication theorem and is denoted by

  3.3.  Probability multiplication theorem .

For the conditions of the last example

  3.3.  Probability multiplication theorem ;   3.3.  Probability multiplication theorem .

Condition independence event   3.3.  Probability multiplication theorem from the event   3.3.  Probability multiplication theorem can be written as:

  3.3.  Probability multiplication theorem ,

and the condition of dependence is in the form:

  3.3.  Probability multiplication theorem .

We proceed to the formulation and proof of the probability multiplication theorem.

The probability multiplication theorem is formulated as follows.

The probability of the product of two events is equal to the product of the probabilities of one of them by the conditional probability of the other, calculated under the condition that the first took place:

  3.3.  Probability multiplication theorem . (3.3.1)

Let us prove the multiplication theorem for the scheme of cases. Let possible outcomes of experience be reduced to   3.3.  Probability multiplication theorem For the sake of clarity, we will again depict them in the form   3.3.  Probability multiplication theorem points:

  3.3.  Probability multiplication theorem

Suppose event   3.3.  Probability multiplication theorem favorable   3.3.  Probability multiplication theorem events   3.3.  Probability multiplication theorem favorable   3.3.  Probability multiplication theorem cases. Since we did not anticipate events   3.3.  Probability multiplication theorem and   3.3.  Probability multiplication theorem incompatible, then generally there are cases favorable and eventful   3.3.  Probability multiplication theorem and event   3.3.  Probability multiplication theorem at the same time. Let the number of such cases   3.3.  Probability multiplication theorem . Then

  3.3.  Probability multiplication theorem .

Calculate   3.3.  Probability multiplication theorem i.e. conditional probability of an event   3.3.  Probability multiplication theorem under the assumption that   3.3.  Probability multiplication theorem took place. If it is known that the event   3.3.  Probability multiplication theorem happened then from the previously possible   3.3.  Probability multiplication theorem cases remain possible only those   3.3.  Probability multiplication theorem that favored the event   3.3.  Probability multiplication theorem . Of them   3.3.  Probability multiplication theorem cases are favorable event   3.3.  Probability multiplication theorem . Consequently,

  3.3.  Probability multiplication theorem .

Substituting expressions   3.3.  Probability multiplication theorem and   3.3.  Probability multiplication theorem in formula (3.3.1), we obtain the identity. The theorem is proved.

Obviously, when applying the multiplication theorem, it makes no difference which of the events   3.3.  Probability multiplication theorem and   3.3.  Probability multiplication theorem be considered the first, and which is the second, and the multiplication theorem can be written in this form:

  3.3.  Probability multiplication theorem .

We point out the corollaries arising from the multiplication theorem.

Corollary 1. If an event   3.3.  Probability multiplication theorem independent of event   3.3.  Probability multiplication theorem , then the event   3.3.  Probability multiplication theorem independent of event   3.3.  Probability multiplication theorem .

Evidence. Given that event   3.3.  Probability multiplication theorem does not depend on   3.3.  Probability multiplication theorem i.e.

  3.3.  Probability multiplication theorem . (3.3.2)

It is required to prove that the event   3.3.  Probability multiplication theorem does not depend on   3.3.  Probability multiplication theorem i.e.

  3.3.  Probability multiplication theorem .

In the proof, we will assume that   3.3.  Probability multiplication theorem .

We write the probability theorem in two forms:

  3.3.  Probability multiplication theorem ,

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem

or, according to condition (3.3.2),

  3.3.  Probability multiplication theorem . (3.3.3)

We divide both sides of equality (3.3.3) into   3.3.  Probability multiplication theorem . We get:

  3.3.  Probability multiplication theorem ,

Q.E.D.

Corollary 1 implies that the dependence or independence of events is always reciprocal. In this regard, it is fashionable to give the following new definition of independent events.

Two events are called independent if the appearance of one of them does not change the probability of the appearance of the other.

The notion of independence of events can be extended to the case of an arbitrary number of events. Several events are called independent if any of them does not depend on any aggregate of the others.

Corollary 2. The probability of the product of two independent events is equal to the product of the probabilities of these events.

The corollary follows directly from the definition of independent events.

The probability multiplication theorem can be generalized to the case of an arbitrary number of events. In general terms, it is formulated as follows.

The probability of producing several events is equal to the product of the probabilities of these events, and the probability of each subsequent event in order is calculated under the condition that all previous events occurred:

  3.3.  Probability multiplication theorem . (3.3.4)

The proof can be given by the same method of complete induction.

In the case of independent events, the theorem is simplified and takes the form:

  3.3.  Probability multiplication theorem , (3.3.5)

those. the probability of the product of independent events is equal to the product of the probabilities of these events.

Applying the sign of the product, the theorem can be written in the form:

  3.3.  Probability multiplication theorem . (3.3.6)

Consider examples on the application of the probability multiplication theorem.

Example 1. In the urn 2 white and 3 black balls. Two balls are taken out of the urn in a row. Find the probability that both balls are white.

Decision. Denote:

  3.3.  Probability multiplication theorem - the appearance of two white balls.

Event   3.3.  Probability multiplication theorem is a product of two events:

  3.3.  Probability multiplication theorem ,

Where   3.3.  Probability multiplication theorem - the appearance of a white ball when first taken out,   3.3.  Probability multiplication theorem - the appearance of a white ball during the second removal.

By the probability multiplication theorem

  3.3.  Probability multiplication theorem .

Example 2. The same conditions, but after first removing the ball returns to the urn, and the balls in the urn are mixed.

Decision. In this case, the event   3.3.  Probability multiplication theorem and   3.3.  Probability multiplication theorem independent and

  3.3.  Probability multiplication theorem .

Example 3. Instrument operating over time.   3.3.  Probability multiplication theorem , consists of three nodes, each of which, independently of the others, can over time   3.3.  Probability multiplication theorem refuse (fail). Failure of at least one node leads to failure of the device as a whole. During   3.3.  Probability multiplication theorem reliability (probability of failure-free operation) of the first node is equal to   3.3.  Probability multiplication theorem ; second   3.3.  Probability multiplication theorem ; the third   3.3.  Probability multiplication theorem . Find the reliability of the device as a whole.

Decision. Denoting:

  3.3.  Probability multiplication theorem - trouble-free operation of devices,

  3.3.  Probability multiplication theorem - trouble-free operation of the first node,

  3.3.  Probability multiplication theorem - trouble-free operation of the second node,

  3.3.  Probability multiplication theorem - trouble-free operation of the third node,

we have:

  3.3.  Probability multiplication theorem ,

whence by multiplication theorem for independent events

  3.3.  Probability multiplication theorem .

In practice, tasks are relatively rare in which you need to apply only the addition theorem or only the probability multiplication theorem. Usually both theorems have to be applied together. In this case, as a rule, the event whose probability is required to be determined is represented as the sum of several incompatible events (variants of this event), each of which in turn is a product of events.

Example 4. Three shots are fired at the same target. The probability of hitting the first, second and third shots are equal, respectively

  3.3.  Probability multiplication theorem

Find the probability that as a result of these three shots in the target will be exactly one hole.

Decision. Consider an event   3.3.  Probability multiplication theorem - exactly one hit in the target. This event can be realized in several ways, i.e. splits into several incompatible variants: there may be a hit on the first shot, a miss on the second and third; or hit on the second shot, misses on the first and third; or, at last, misses at the first and second shots and hit at the third. Consequently,

  3.3.  Probability multiplication theorem ,

Where   3.3.  Probability multiplication theorem - hit on the first, second, third shots,   3.3.  Probability multiplication theorem - a miss at the first, second, third shots.

Applying the theorems of addition and multiplication of probabilities and using the property of opposite events, we find:

  3.3.  Probability multiplication theorem .

Example 5. In the conditions of the previous example, find the probability that there will be at least one hole in the target.

Decision. Consider an event   3.3.  Probability multiplication theorem - at least one hit in the target. Using the same technique that was used in the previous example, the same notation, you can present the event   3.3.  Probability multiplication theorem as a sum of incompatible options:

  3.3.  Probability multiplication theorem ,

find the probability of each variant by the multiplication theorem and add all these probabilities. However, this way of solving the problem is too complicated; here it is advisable from a direct event   3.3.  Probability multiplication theorem go to the opposite:

  3.3.  Probability multiplication theorem - not a single hit in the target.

Obviously

  3.3.  Probability multiplication theorem .

By the multiplication theorem

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem .

The last example illustrates the principle of the expediency of using opposite events in probability theory. It can be formulated as follows.

If the opposite event falls into a smaller number of variants than a direct event, then it makes sense, when calculating the probabilities, to switch to the opposite event.

Example 6. There is a battle ("duel") between two participants (aircraft, tanks, ships)   3.3.  Probability multiplication theorem and   3.3.  Probability multiplication theorem . By the side   3.3.  Probability multiplication theorem there are two shots left, on the side   3.3.  Probability multiplication theorem - one. Starts shooting   3.3.  Probability multiplication theorem : he does by   3.3.  Probability multiplication theorem one shot and hits it with a probability of 0.2. If a   3.3.  Probability multiplication theorem not struck, he answers the enemy with a shot and hits him with a probability of 0.3. If a   3.3.  Probability multiplication theorem this shot is not struck, then he does by   3.3.  Probability multiplication theorem his last shot, which hits him with a probability of 0.4. Find the probability that the battle will be hit: a) party   3.3.  Probability multiplication theorem b) participant   3.3.  Probability multiplication theorem .

Decision. Consider the events:

  3.3.  Probability multiplication theorem - member loss   3.3.  Probability multiplication theorem ,

  3.3.  Probability multiplication theorem - member loss   3.3.  Probability multiplication theorem .

To perform the event   3.3.  Probability multiplication theorem It is necessary to combine (product) two events: 1)   3.3.  Probability multiplication theorem didn't hit   3.3.  Probability multiplication theorem first shot and 2)   3.3.  Probability multiplication theorem struck and his return shot. By the probability multiplication theorem, we obtain

  3.3.  Probability multiplication theorem .

Go to the event   3.3.  Probability multiplication theorem . It obviously consists of two incompatible options:

  3.3.  Probability multiplication theorem

Where   3.3.  Probability multiplication theorem - member loss   3.3.  Probability multiplication theorem first shot   3.3.  Probability multiplication theorem ,   3.3.  Probability multiplication theorem - member loss   3.3.  Probability multiplication theorem second shot   3.3.  Probability multiplication theorem .

By the addition theorem

  3.3.  Probability multiplication theorem .

By condition   3.3.  Probability multiplication theorem = 0.2. As for the event   3.3.  Probability multiplication theorem then it is a combination (product) of three events, namely:

1) first side shot   3.3.  Probability multiplication theorem should not hit   3.3.  Probability multiplication theorem ;

2) side response shot   3.3.  Probability multiplication theorem should not hit   3.3.  Probability multiplication theorem ;

3) the last (second) side shot   3.3.  Probability multiplication theorem must hit   3.3.  Probability multiplication theorem .

By the probability multiplication theorem

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem .

Example 7. The target, which is being fired, consists of three parts of different vulnerability. To hit a target, one hit in the first part, or two hits in the second, or three hits in the third. If the projectile hit the target, then the probability of it getting into one or another part is proportional to the area of ​​this part. On the projection of the target on a plane perpendicular to the direction of shooting, the first, second and third parts occupy relative areas of 0.1, 0.2 and 0.7. It is known that exactly two shells hit the target. Find the probability that the target will be hit.

Decision. Denote   3.3.  Probability multiplication theorem - defeat the purpose;   3.3.  Probability multiplication theorem - conditional probability of hitting the target, provided that it hit exactly two shells. Two projectiles that hit the target can hit it in two ways: either at least one of them gets into the first part, or both shells hit the second. These options are incompatible, since only two shells hit the target; therefore, the addition theorem can be applied. The probability that at least one projectile falls into the first part can be calculated through the probability of the opposite event (none of the two projectiles will fall into the first part) and be equal to   3.3.  Probability multiplication theorem . The probability that both shells fall into the second part is equal to   3.3.  Probability multiplication theorem . Consequently,

  3.3.  Probability multiplication theorem .

Example 8. For the conditions of the previous example, find the probability of hitting the target, if it is known that three shells hit it.

Decision. We solve the problem in two ways: through a direct and opposite event.

A direct event — the defeat of a target with three hits — splits into four incompatible options:

  3.3.  Probability multiplication theorem - at least one hit in the first part,

  3.3.  Probability multiplication theorem - two hits in the second part and one - in the third,

  3.3.  Probability multiplication theorem - three hits in the second part,

  3.3.  Probability multiplication theorem - three hits in the third part.

The probability of the first option is found as in the previous example:

  3.3.  Probability multiplication theorem .

Find the probability of the second option. Three projectile hit can be distributed in the second and third parts as needed (two to the second and one to the third) in three ways (  3.3.  Probability multiplication theorem ). Consequently,

  3.3.  Probability multiplication theorem .

Next, we find:

  3.3.  Probability multiplication theorem ,

  3.3.  Probability multiplication theorem .

From here

  3.3.  Probability multiplication theorem .

However, it is easier to solve the problem, if we go to the opposite event - failure to hit the target with three hits. This event can be realized only in one way: if two projectiles out of three fall into the third part, and one into the second. There may be three such combinations (   3.3.  Probability multiplication theorem ), therefore

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem .

Example 9. A coin is thrown 6 times. Find the probability that more coats of arms will fall out than numbers.

Decision.To find the probability of an event of interest to us   3.3.  Probability multiplication theorem (more coats of arms will fall out than numbers) it would be possible to list all its possible variants, for example:

  3.3.  Probability multiplication theorem - six emblems and not a single number will appear,

  3.3.  Probability multiplication theorem - five coats of arms and one figure will be drawn

etc.

However, it will be easier to apply another technique. We list all possible outcomes of the experiment:

  3.3.  Probability multiplication theorem - more coats of arms will fall out than numbers

  3.3.  Probability multiplication theorem - there will be more figures than emblems,

  3.3.  Probability multiplication theorem - the same number of figures and coats of arms will appear.

Developments   3.3.  Probability multiplication theorem ,   3.3.  Probability multiplication theorem ,   3.3.  Probability multiplication theorem incompatible and form a complete group. Consequently,

  3.3.  Probability multiplication theorem .

Since the task is symmetrical with respect to the “coat of arms” and “numbers”,

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem

and

  3.3.  Probability multiplication theorem .

Let us find the probability of an event   3.3.  Probability multiplication theorem consisting in the fact that with six throwing of a coin, exactly three coats of arms will appear (and, therefore, exactly three digits). The probability of any of the variants of the event   3.3.  Probability multiplication theorem (for example, the sequence g, c, g, g, c, c with six throws) is the same and is equal to  3.3.  Probability multiplication theorem .The number of such combinations is equal to   3.3.  Probability multiplication theorem (the number of ways, from which it is possible to choose from six throws three, in which the coat of arms appeared). Consequently,

  3.3.  Probability multiplication theorem ;

from here

  3.3.  Probability multiplication theorem .

Пример 10. Прибор состоит из четырех узлов:   3.3.  Probability multiplication theorem , причем узел   3.3.  Probability multiplication theorem дублирует узел   3.3.  Probability multiplication theorem , а узел   3.3.  Probability multiplication theorem дублирует узел   3.3.  Probability multiplication theorem . При отказе (выходе из строя) любого из основных узлов (   3.3.  Probability multiplication theorem or   3.3.  Probability multiplication theorem ) происходит автоматическое переключение на дублирующий узел. Надежность (вероятность безотказной работы) в течение заданного времени каждого из узлов равна соответственно   3.3.  Probability multiplication theorem . Надежность каждого из переключающих устройств равна   3.3.  Probability multiplication theorem . Все элементы выходят из строя независимо друг от друга. Определить надежность прибора.

Decision. Рассмотрим совокупность узлов   3.3.  Probability multiplication theorem и соответствующего переключающего устройства как один «обобщенный узел»   3.3.  Probability multiplication theorem , а совокупность узлов   3.3.  Probability multiplication theorem и соответствующего обобщающего устройства – как обобщенный узел   3.3.  Probability multiplication theorem . Consider the events:

  3.3.  Probability multiplication theorem – безотказная работа прибора,

  3.3.  Probability multiplication theorem – безотказная работа обобщенного узла   3.3.  Probability multiplication theorem ,

  3.3.  Probability multiplication theorem – безотказная работа обобщенного узла   3.3.  Probability multiplication theorem .

Obviously

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem .

Find the probability of the event   3.3.  Probability multiplication theorem . It falls into two variants:

  3.3.  Probability multiplication theorem - the node worked properly   3.3.  Probability multiplication theorem

and

  3.3.  Probability multiplication theorem - the node   3.3.  Probability multiplication theorem failed, but the switching device and the node turned out to be serviceable  3.3.  Probability multiplication theorem .

We have:

  3.3.  Probability multiplication theorem ,

similarly

  3.3.  Probability multiplication theorem ,

from where

  3.3.  Probability multiplication theorem .


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Probability theory. Mathematical Statistics and Stochastic Analysis

Terms: Probability theory. Mathematical Statistics and Stochastic Analysis