The product, or intersection, events l and in the event consisting in the simultaneous occurrence of events and l, and IN. Designation of the work AU or L and V.

For example, a double hit in the goal is a product of two events, the answer to both questions of the ticket on the exam is a product of two events.

Events L I. IN Call non-confidential if their work is an impossible event, i.e. LV \u003d V.

For example, the events l - the deposition of the coat of arms and IN - Loss of the numbers with a single throwing of the coin to step simultaneously cannot, their product is an event impossible, the events l and in incomplete.

The concepts of the amount and work of events have a visual geometric interpretation (Fig. 6.4).

Fig. 6.4. Geometric interpretation of the work (but)and sum (b) Two joint events

Let the event l be the set of points of the region L, the event B - the set of points of the region. The shaded region corresponds to the EFF event in Fig. 6. La and the event l + in fig. 6.46.

For incomplete events l and in having LV \u003d V.(Fig. 6.5a). The event l + B corresponds to the shaded area in Fig. 6.56.

Fig. 6.5. Geometric interpretation of the work ( but) and sum (b) Two incomplete events

Events BUT and BUT refer to the opposite if they are incomprehensible and in total make up a reliable event, i.e.

A \u003d v; A + a \u003d u.

For example, we will produce one shot of the target: Event BUT- the shooter hit the target BUT- missed; The coin is added:

event BUT- Eagle falling BUT - loss of numbers; Schoolchildren write testing: Event BUT- none

errors in control work, BUT- There are errors in the control work; The student came to rent a test: event BUT- passed

offset BUT- Did not pass off.

In the class there are boys and girls, excellent students, horoshists and triens, studying English and German. Let the event M - boy, o - excellent, and - studying English. Can a student who happened from the class and excellent student and studying English? This will be a work or intersection of Moa's events.

Example 6.15. Throw a playing cube - a cube made of homogeneous material, the faces of which are amiced. Watch the number (number of points) falling on the top face. Let the event BUT - The appearance of an odd number, event IN - The appearance of the number, multiple of three. Find outcomes that make up each of the events (? /, A, A. + In av) And specify their meaning.

Decision. Exodus - the appearance on the upper face of any of the numbers 1, 2, 3, 4, 5, 6. The set of all outcomes is the space of elementary events U. \u003d (1, 2, 3, 4, 5, 6). It is clear that the event A \u003d. (1, 3, 5), event In \u003d. {3, 6}.

Event BUT + In \u003d. (1, 3, 5, 6) - the appearance of either an odd number or number, multiple of three. When listed the outcomes, it is taken into account that each outcome in the set can be contained only once.

Event AV \u003d. (3) - the appearance and odd number, and numbers, multiple of three.

Example 6.16. Checked homework in three students. Let the event BUT ( - Making the task of the i-m student g. = 1, 2, 3.

What is the meaning of events: A \u003d A T + A 2. + L 3, BUT and B \u003d a t a 2 a 3?

Decision. Event BUT = A H. + A 2. + 3 - Performing a task at least one student, i.e. or any one student (or first, or second, or third), or any two, or all three.

Event A \u003d a x -a 2 -a 3 - The task is not fulfilled by any student - neither first nor the second nor the third. Event B \u003d A (2 A 3 -performing a task in three students - both first and second, and third.

When considering the joint onset of several events, there are cases when the appearance of one of them affects the possibility of the appearance of another. For example, if autumn is sunny day, it is less likely that the weather is spoiled (the rain starts). If the sun is not visible, then more chances that rain will go.

Event L. called independent of the event IN, If the probability of an event BUT does not change depending on whether or not an event occurred IN. Otherwise event BUT called event dependent IN. Two events AIIN They are called independent if the probability of one of them does not depend on the appearance or fault of the other dependent - otherwise. Events are called pairs independent if every two of them are independent of each other.

The probability multiplication theorem is formulated as follows. The probability of the work of two independent events is equal to the product of the probabilities of these events:

This theorem is valid for any finite number of events, unless they are independent in the aggregate, i.e. The probability of any of them does not depend on whether other of these events occurred.

Example 6.17. The student gives three exams. The probability of successful delivery of the first exam 0.9, the second - 0.65, the third - 0.35. Find the likelihood that he will not pass at least one exam.

Decision. Denote BUT Event - the student did not pass at least one exam. Then P (A.) \u003d 1 - / - '(1/1), where BUT - The opposite event - the student passed all exams. Since the surrender of each exam does not depend on other exams, R (a) \u003d 1 - p (1/1) \u003d \u003d 1 - 0.9 0.65 0.35 \u003d 0.7953.

Probability of an event BUT, calculated provided that the event occurs IN, called conditional probabilityevents BUT subject to appearance IN And denotes P in (a) or P (A / B).

Theorem.The probability of the appearance of a work of two events is equal to the probability of one of them on the conditional probability of the second calculated, provided that the first event happened:

Example 6.18. The student retrieves twice one ticket from 34. What is the likelihood that he will pass the exam, if they are prepared 30 tickets and for the first time a bad ticket will be reset?

Decision. Let the event BUT It is that the first time I got an unsuccessful ticket, an event IN- The second time is removed a good ticket. Then BUT?IN- The student will pass the exam (under the circumstances). Events BUT and IN Dependent, since the probability of choosing a successful ticket with a second attempt depends on the outcome of the first choice. Therefore, we use formula (6.6):

Note that the probability obtained in solving "0.107. Why was the probability of passing the exam, if 30 tickets from 34 were learning and two attempts are given?!

Expanded addition theoremformulated as follows. The probability of the sum of two events is equal to the sum of the probabilities of these events without the likelihood of their joint appearance (works):

Example 6.19. Two students solve the task. The likelihood that the first student will solve the task (event BUT), equal to 0.9; the likelihood that the second student will solve the task (event IN), equal to 0.8. What is the likelihood that the task will be solved?

Decision. We are interested in an event with, which is that the task will be solved, i.e. The first or second student, or two students at the same time. Thus, interests the pass event C \u003d a +IN. Events BUT and IN Joint, then the probability addition theorem is applicable for the case of joint events: P (A. + IN) = R (a) + P (B) - P (AB). For our case P (A. + C) \u003d \u003d 0.9 + 0.8 + 0.9 0.8 \u003d 0.98 (events BUT and IN jointly, but independent).

Example 6.20. The student knows 20 questions out of 25. What is the probability of answering three questions from 25?

Decision. We introduce the event L, - the student knows the answer to i.- ask the proposed question i. \u003d 1,2,3. Events L, L 2, L 3 - dependent. therefore

When finding the probabilities of events, a classic probability definition was used.

Study of the theory of probability begins with solving problems of addition and multiplication of probabilities. It is necessary to immediately mention that a student when mastering this area of \u200b\u200bknowledge may face a problem: if physical or chemical processes can be imagined to visually and understand empirically, the level of mathematical abstraction is very high, and understanding here comes only with experience.

However, the game is worth the candle, because the formula is both more complex - used today everywhere and may well come in handy in work.

Origin

Oddly enough, the impetus for the development of this section of mathematics has become ... gambling. Indeed, a game of bones, throwing coins, poker, roulette is typical examples in which the addition and multiplication of probabilities are used. On the example of tasks in any textbook it can be seen clearly. People were interested to know how to increase their chances of winning, and, I must say, some succeeded.

For example, in the 21st century, one person whose behalf of us will not open, used these knowledge accumulated in centuries to literally "clean" a casino, winning several tens of millions of dollars.

However, despite the increased interest in the subject, only by the 20th century, the theoretical base was developed, which makes the "theorer" of a full-fledged today, in almost any science, you can find calculations using probabilistic methods.

Applicability

An important point when using the formulas for addition and multiplication of probabilities, the conditional probability is the feasibility of the central limit theorem. Otherwise, even though it can not be realized by a student, all the calculations, whatever they seem like, will not be incorrect.

Yes, at a highlyotized student there is a temptation to use new knowledge with each convenient case. But in this case, you should slow down and strictly delineate the applicability framework.

The theory of probability deals with random events that are the results of experiments in an empirical plan: we can throw a cube with six faces, pull out a card from the deck, predicted the number of defective parts in the party. However, in some issues, the formulas from this section of mathematics are categorically impossible. Features of consideration of the probabilities of the event, the theorems of addition and multiplying events we will discuss at the end of the article, but as long as we turn to the examples.

Basic concepts

Under a random event, it means some process or result that can manifest itself, and may not be manifested as a result of the experiment. For example, we throw the sandwich - it can fall up with oil up or oil down. Any of two outcomes will be random, and we do not know in advance which one will have a place.

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

The joints are called such events, the emergence of one of which does not exclude the appearance of another. Let's say two people simultaneously shoot targets. If one of them works successful in any way will not affect the possibility of the second to get into the "apple" or miss.

There will be such events that will be the appearance of which is simultaneously impossible. For example, pulling out only one ball out of the box, it is impossible to get both blue and red.

Designation

The concept of probability is indicated by the Latin capital letter P. Further, the arguments indicate some events follow in brackets.

In the formulas of the addition theorem, the conditioned probability, the multiplication theorems will see in the brackets of the expression, for example: a + b, ab or a | b. They will be calculated in various ways, we will now turn to them.

Addition

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

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

Suppose there is a box with 2 blue, 3 red and 5 yellow balls. Total there are 10 items in the box. What is the proportion of the approval that we pull out a blue or red ball? It will be equal to 2/10 + 3/10, i.e. fifty percent.

In the case of incomplete events, the formula is complicated, since an additional term is added. Let's return to him through one paragraph, after considering another formula.

Multiplication

Adjustment and multiplication of probabilities of independent events are used in different cases. If, by the experimental condition, we are satisfied with any of the two possible outcomes, we consider the amount; If we want to get two some outcome for each other, we will resort to the use of another formula.

Returning to example from the previous section, we want to pull the blue ball first, and then red. We know the first number - this is 2/10. What happens next? The balls remain 9, the red among them are all the same - three pieces. According to calculations, it will turn out 3/9 or 1/3. But what to do with two numbers now? The correct answer is to multiply to work out 2/30.

Joint events

Now you can again refer to the amount of the sum for joint events. Why are we distracted from the topic? To find out how the probabilities are prolonged. Now this knowledge is useful to us.

We already know what the first two components will be (the same as in the formula previously reviewed by the formula), now it will be necessary to subtract the work of the probabilities that we just learned to count on. For clarity, we write the formula: P (a + b) \u003d p (a) + p (b) - P (AB). It turns out that in one expression, both addition and multiplication of probabilities are used.

Suppose we have to solve any of the two tasks to get off. We can solve the first with a probability of 0.3, and the second - 0.6. Solution: 0.3 + 0.6 - 0.18 \u003d 0.72. Note, just summarize numbers here is not enough.

Conditional probability

Finally, there is a concept of a conditioned probability, the arguments of which are designated in brackets and are separated by a vertical feature. The P (A | B) record reads as follows: "The probability of event A under the condition of the B".

Let's see example: a friend gives you some device, let it be a phone. It can be broken (20%) or in charge (80%). Any device that fell into hand is able to fix with a probability of 0.4 or not able to do this (0.6). Finally, if the device is in working condition, you can call the right person with a probability of 0.7.

It is easy to see how in this case a conditional probability is manifested: you will not be able to call a person if the phone is broken, and if it is working, you do not need to repair it. Thus, to get any results at the "second level", you need to know which event was performed on the first.

Calculations

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

To begin with, we find the likelihood that you reproduce the device given to you. For this, firstly, it must be faulty, and secondly, you must cope with the repair. This is a typical task using multiplication: we get 0.2 * 0.4 \u003d 0.08.

What is the likelihood that you will immediately call the right person? Easier simple: 0.8 * 0.7 \u003d 0.56. In this case, you found that the phone is better and successfully made a call.

Finally, consider this option: you received a broken phone, repaired it, after which they scored a number, and the person at the opposite end took the phone. Here it is already required to multiply three components: 0.2 * 0.4 * 0.7 \u003d 0.056.

And what if you have two non-working phone at once? What is probability, do you fix at least one of them? For addition and multiplication of probabilities, since joint events are used. Solution: 0.4 + 0.4 - 0.4 * 0.4 \u003d 0.8 - 0.16 \u003d 0.64. Thus, if two broken apparatus fall into your hands, you will cope with the repair in 64% of cases.

Attentive use

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

The greater the series of experiments, the closer suits theoretically predicted importance to the in practice. For example, we throw a coin. Theoretically, knowing the existence of the formulas of addition and multiplication of probabilities, we can predict how many times "Eagle" and "Rushka" fall, if we carry out an experiment 10 times. We conducted an experiment, and by coincidence, the ratio of the fallen sides was 3 to 7. But if you hold a series of 100, 1000 or more attempts, it turns out that the distribution schedule is closer closer to theoretical: 44 K 56, 482 K 518 and so on.

And now imagine that this experiment is not carried out with a coin, but with the production of some new chemical, the likelihood of which we do not know. We would spend 10 experiments and, without having received a successful result, could generally: "Substance cannot be obtained." But who knows, we will spend the eleventh attempt - would we reach the goal or not?

Thus, if you apply to the unexplored area, the probability theory may not be applicable. Each subsequent attempt in this case may be successful and generalizations of the "X does not exist" or "X is impossible" will be premature.

Final word

So, we reviewed two types of addition, multiplication and conditional probabilities. With further study of this area, it is necessary to learn to distinguish the situations when each specific formula is used. In addition, it is necessary to submit whether the probabilistic methods are applicable in solving your task.

If you practice, after a while, you will begin to implement all the required operations exclusively in your mind. For those who are interested in card games, this skill can be considered extremely valuable - you will significantly increase your chances of winning, just counting the likelihood of a loss of one or another card or suit. However, the knowledge gained you can easily find the application in other areas of activity.

Event A called independent From Events B, if the probability of event A does not depend on whether an event B occurred or not. Event A called dependentfrom the event B, if the probability of event A varies depending on whether an event B occurred or not.

The probability of event A, computed, provided that the event B has already occurred, is called a conditional probability of event A and is indicated.

The condition of independence of the event A from the event b can be written as
.

Probability multiplication theorem. The probability of the work of two events is equal to the product of the likelihood of one of them on the conditional probability of the other, calculated, provided that the first took place:

If the event A does not depend on the event b, the event B does not depend on the event A.. \u200b\u200bIn this case, the probability of the work of events is equal to the product of their probability:

.

Example 14. There are 3 drawers containing 10 parts. In the first drawer 8, in the second - 7 and in the third 9 standard details. From each drawer, the muddy takes out one piece. Find the likelihood that all three details will be standard.

The likelihood that the standard part is removed from the first box (Event A) is equal to
. The likelihood that the standard part is removed from the second box (event) is equal to
. The probability that the third box is removed the standard part (eventc) is equal to
.

Since events A, B and C independent in aggregate, then by the multiplication theorem, the desired probability is equal

Let us give an example of sharing the addition and multiplication by theorems.

Example 15. The probability of appearance of independent events A 1 and a 2 is equal to respectively P 1 and P 2. Find the likelihood of only one of these events (Event A). Find the likelihood of at least one of these events (Event B).

Denote the likelihood of opposing events and it is 1 \u003d 1-p 1 and q 2 \u003d 1-P 2, respectively.

Event A will occur if an event A 1 occurs and event A 2 will occur, or if an event A 2 occurs and event A 1 will occur. Hence,

Event B will occur if an event A occurs, or events A 1 and A 2 will occur at the same time. Hence,

The probability of event B can be determined otherwise. Event The opposite event is that both events A 1 and A 2 will occur. Therefore, by the theorem of multiplication of probabilities for independent events we will get

which coincides with the expression obtained earlier, since the identity takes place

7. Formula of full probability. Bayes formula.

Theorem 1.. Suppose that events
form a complete group in pairs of incomplete events (such events are called hypotheses). Let be an arbitrary event. Then the probability of event A can be calculated by the formula

Evidence. Since hypothesis form a complete group, then, and, therefore.

Due to the fact that hypotheses are in pairs of incomplete events, the events are also in pairs are inconspicuous. By probability addition theorem

Using now the theorem of multiplication of probabilities, we get

Formula (1) is called a full probability formula. In abbreviated form, it can be written as follows.

.

The formula is useful if the conditional probabilities of the event A are calculated easier than the unconditional probability.

Example 16.. There are 3 decks of 36 cards and 2 decks of 52 cards. Minds choose one deck and bring one card out of it. Find the chance that the removed card is ace.

Let a be an event consisting in the fact that the removal map is ace. We introduce two hypotheses to consideration:

- the card is removed from the deck in 36 cards,

- The card is removed from the deck in 52 cards.

To calculate the probability of an event, we will use the formula of the full probability:

Theorem 2.. Suppose that events
form a complete group in pairs of incomplete events. Let be an arbitrary event. Conditional probability of hypothesis in the assumption that an event occurred, it can be calculated by the Bayes formula:

Evidence. Of the probability multiplication theorem for dependent events, it follows that.

.

Applying the full probability formula, we obtain (2).

Probability hypothesis
called a priori, and probabilities of hypotheses
Provided that the event A occurred is called a posteriori. Bayeys formulas themselves are called hypotheses probability formulas.

Example 17.. There are 2 urns. The first urn contains 2 white and 4 black bowls, and the second urn contains 7 white and 5 black balls. I chose the urn and I will remove one ball from it. It turned out to be black (event A happened). Find the likelihood that the ball was removed from the first urn (hypothesis
). Find the chance that the ball was removed from the second urn (hypothesis
).

Apply Bayes formulas:

,

.

Example 18.. At the factory, the bolts are produced by three machines, which are produced by 25%, respectively, 35% and 40% of all bolts. Marriage of these machines is respectively 5%, 4%, 2%. One bolt was selected from the products of all three cars. It turned out to be defective (Event a). Find the chance that the bolt was released first, second, third car.

Let be
- an event consisting in the fact that the bolt was released first car
- second car,
- Third car. These events are in pairs are incomprehensible and form a complete group. We use Bayes formulas

As a result, we get

,

,

.

Establishment of Education "Belarusian State

agricultural Academy"

Department of Higher Mathematics

Addition and multiplication of probabilities. Repeated independent tests

Lecture for students of the land management faculty

correspondence formation

Gorki, 2012.

Addition and multiplication of probabilities. Repeated

independent tests

1. Addition of probabilities

Sum of two joint events BUT and IN called event FROMconsisting in the occurrence of at least one of the events BUT or IN. Similar to the sum of several joint events is the event consisting in the occurrence of at least one of these events.

The sum of two inconsistent events BUT and IN called event FROMoccurrence or event BUT, or events IN. Similar to the sum of several incomplete events is an event consisting in the onset of any of these events.

Fair the theorem of the addition of probability of incomplete events: the probability of the sum of two inconsistent events is equal to the sum of the probabilities of these events . . This theorem can be extended to any finite number of incomplete events.

From this theorem follows:

The sum of the probabilities of events forming a complete group is equal to one;

The sum of the probabilities of opposing events is equal to one, i.e.
.

Example 1. . There are 2 whites in the box, 3 red and 5 blue balls. Balls are mixed and at random is removed one. What is the likelihood that the ball will be color?

Decision . Denote the events:

A.\u003d (removed colored ball);

B.\u003d (white ball extracted);

C.\u003d (red ball extracted);

D.\u003d (The blue ball is removed).

Then A.= C.+ D.. Since events C., D. Uncomfortable, we will use the theorem of the addition of probabilities of incomplete events :.

Example 2. . In the urn there are 4 white balls and 6 - black. From the urn at random, it takes 3 balls. What is the likelihood that all of them are one color?

Decision . Denote the events:

A.\u003d (taken out of the balls of the same color);

B.\u003d (taken out of white balls);

C.\u003d (cut black balls).

As A.= B.+ C. and events IN and FROM incomprehensible, then by the theorem of the addition of probabilities of incomplete events
. Probability of an event IN equal
where
4,

. Substitute k. and n. in the formula and get
Similarly, find the likelihood of an event. FROM:
where
,
.
. Then
.

Example 3. . From the deck in 36 cards at random takes out 4 cards. Find the chance that among them will be at least three aces.

Decision . Denote the events:

A.\u003d (among the removed cards of at least three aces);

B.\u003d (among the cuts of three aces);

C.\u003d (Among the filled cards four aces).

As A.= B.+ C., and events IN and FROM inconsistent, T.
. Find the probabilities of events IN and FROM:

,
. Therefore, the likelihood that among the removed cards of at least three aces is equal to

0.0022.

1. Multiplication of probabilities

Work Two events BUT and IN called event FROMconsisting in the joint occurrence of these events:
. This definition applies to any finite number of events.

Two events are called independent If the probability of the occurrence of one of them does not depend on whether another event occurred or not. Events , , … , called independent in aggregate If the probability of occurrence of each of them does not depend on whether other events occurred or did not occur.

Example 4. . Two shooters shoot target. Denote the events:

A.\u003d (the first arrows hit the target);

B.\u003d (The second arrows hit the target).

It is obvious that the likelihood of hitting the target the first shooter does not depend on whether he fell or did not hit the second arrows, and vice versa. Consequently, events BUT and IN Independent.

Fair the theorem of multiplication of probabilities of independent events: the probability of the work of two independent events is equal to the product of the probabilities of these events. : .

This theorem is valid for n. independent in aggregate events :.

Example 5. . Two arrows shoot one goal. The probability of hitting the first arrow is 0.9, and the second - 0.7. Both arrow simultaneously make one shot. Determine the likelihood that there will be two hits in the target.

Decision . Denote the events:

A.

B.

C.\u003d (both arrows will fall into the goal).

As
, and events BUT and IN Independent, T.
. .

Events BUT and IN called dependent If the probability of the occurrence of one of them depends on whether another event occurred or not. The likelihood of an event BUT provided that the event IN has already come, called conditional probability And denotes
or
.

Example 6. . There are 4 white and 7 black balls in the urn. Balls are extracted from the urn. Denote the events:

A.\u003d (white ball extracted);

B.\u003d (Black ball extracted).

Before the start of extracting balls from the urn
. One ball was removed from the urn and he turned out to be black. Then the likelihood of an event BUT After the events IN will be another equal . This means that the probability of an event BUT Depends on the event IN. These events will be dependent.

Fair the theorem of multiplication of probabilities of dependent events: the probability of the work of two dependent events is equal to the product of the likelihood of one of them on the conditional likelihood of another calculated in the assumption that the first event has already come. or .

Example 7. . In the urn there are 4 white balls and 8 red. From her at random, two balls are consistently removed. Find the chance that both balls will be black.

Decision . Denote the events:

A.\u003d (first extracted black ball);

B.\u003d (The second is extracted the black ball).

Events BUT and IN dependent because
, but
. Then
.

Example 8. . Three arrow shoot targets independently of each other. The probability of entering the target for the first arrow is 0.5, for the second - 0.6 and for the third - 0.8. Find the likelihood that two hitting goals occur if each shooter makes one shots.

Decision . Denote the events:

A.\u003d (there are two hits in the goal);

B.\u003d (the first shooter will fall into the target);

C.\u003d (the second shooter will fall into the goal);

D.\u003d (the third arrows will fall into the target);

\u003d (the first shooter will not fall into the goal);

\u003d (the second shooter will not fall into the goal);

\u003d (the third shooter will not fall into the target).

Under the condition of example
,
,
,

,
,
. Since, using the theorem of the addition of probabilities of incomplete events and the theorem of multiplying the probabilities of independent events, we obtain:

Let the events
Form a complete group of events of some test, and the event BUT May come only with one of these events. If probabilities and conditional probabilities are known BUT, The probability of event A is calculated by the formula:

Or
. This formula is called formula full probability , and events
hypothesis .

Example 9. . 700 parts from the first machine and 300 parts arrive at the assembly conveyor From the second. The first machine gives a 0.5% marriage, and the second is 0.7%. Find the chance that the detail taken will be defective.

Decision . Denote the events:

A.\u003d (taken the item will be defective);

\u003d (item is made on the first machine);

\u003d (The item is made on the second machine).

The probability that the item is made on the first machine is equal to
. For the second machine
. By condition, the likelihood of obtaining a defective part made on the first machine is equal to
. For the second machine, this probability is equal
. Then the likelihood that the detail taken will be defective, is calculated by the formula of the full probability.

If it is known that a certain event has come as a result of the test BUTthen the likelihood that this event came with a hypothesis
, equal
where
- full probability of an event BUT. This formula is called bayes formula and allows you to calculate the probabilities of events
After it became known that the event BUT has already come.

Example 10. . The same type of cars are produced at two factories and go to the store. The first plant produces 80% of the total number of parts, and the second is 20%. Products of the first plant contains 90% of standard parts, and the second is 95%. The buyer bought one detail and it turned out to be standard. Find the likelihood that this item is made in the second factory.

Decision . Denote the events:

A.\u003d (standard item purchased);

\u003d (item is made in the first factory);

\u003d (The item is made in the second factory).

Under the condition of example
,
,
and
. Calculate the full probability of an event BUT: 0.91. The probability that the item is made in the second factory, calculated by the Bayes formula:

.

Tasks for independent work

The probability of entering the target for the first arrow is 0.8, for the second - 0.7 and for the third - 0.9. The arrows made one shots. Find the likelihood that there are at least two hits in the target.

15 tractors came to the repair shop. It is known that 6 of them need to replace the engine, and the rest - in the replacement of individual nodes. Three tractors are randomly selected. Find the likelihood that the engine replacement is required no more than two selected tractors.

A panels are made on the reinforced concrete plant, 80% of which are of the highest quality. Find the likelihood that from three at random the selected panels of at least two will be the highest grade.

Three workers collect bearings. The likelihood that the bearing assembled by the first working, top quality is 0.7, second - 0.8 and third - 0.6. To control at random, it is taken by one bearing from the collected every workers. Find the chance that at least two of them will be the highest quality.

The probability of winning the first release lottery ticket is 0.2, the second - 0.3 and the third - 0.25. There are one ticket each release. Find the chance that you will win at least two tickets.

The accountant performs calculations, using three directories. The probability that its interests are in the first directory is 0.6, in the second - 0.7 yves third - 0.8. Find the likelihood that the accountant you are interested in are no more than two reference books.

Three machines make details. The first automatic makes the highest quality item with a probability of 0.9, the second - with a probability of 0.7 and the third - with a probability of 0.6. At rags take one detail from each machine. Find the likelihood that among them at least two top quality.

On two machines are processed by the same type. The probability of manufacturing a non-standard part for the first machine is 0.03, in for the second - 0.02. Processed parts are folded in one place. Among them are 67% from the first machine, and the rest - from the second. At random, the detail taken is standard. Find the chance that it is made on the first machine.

Two boxes of the same type of capacitors arrived in the workshop. In the first box there were 20 capacitors, of which 2 faulty. In the second box 10 capacitors, of which are 3 faulty. Capacitors were transferred to one box. Find the likelihood that the capacitor taken from the box will be serviceable.

On three windows, the same type of items are made, which come to the general conveyor. Among all the details are 20% from the first machine, 30% - from the second and 505 - from the third. The probability of manufacturing a standard part on the first machine is 0.8, on the second - 0.6 and on the third - 0.7. The detail turned out to be standard. Find the likelihood of this item is made on the third machine.

A complete component gets for the assembly of 40% of the details from the factory BUT, and the rest - from the factory IN. The probability that the item from the factory BUT - highest quality, equal to 0.8, and from the factory IN - 0.9. The pledge holder at random took one part and it was not the highest quality. Find the likelihood that this item from the factory IN.

For participation in student sports competitions, 10 students from the first group and 8 are allocated - from the second. The likelihood that a student from the first group will fall into the national team of the Academy, is 0.8, and from the second - 0.7. At rags, the selected student came to the national team. Find the likelihood that he is from the first group.

\\ (\\ blacktriangleright \\) If it is necessary to perform both joints (which may occur simultaneously) events \\ (a \\) and \\ (B \\) (\\ (C \u003d \\ (A \\) and \\ ((C \u003d \\ (A \\) B \\) \\)), then the probability of event \\ (C \\) is equal to the product of the probabilities of events \\ (a \\) and \\ (b \\).

Note that if the events are incomplete, the probability of their simultaneous origin is equal to \\ (0 \\).

\\ (\\ BLACKTRIANGLERIGHT \\) Each event can be designated in the form of a circle. Then if events are jointly, the circles must intersect. The probability of an event \\ (C \\) is the probability of getting into both circles at the same time.

\\ (\\ blacktriangleright \\) For example, when throwing a playing bone to find the probability \\ (C \u003d \\) (the loss of the number \\ (6 \\)).
The event \\ (C \\) can be formulated as \\ (a \u003d \\) (the loss of an even number) and \\ (b \u003d \\) (the loss of the number divided by three).
Then \\ (P \\, (c) \u003d p \\, (a) \\ cdot p \\, (b) \u003d \\ dfrac12 \\ cdot \\ dfrac13 \u003d \\ dfrac16 \\).

Task 1 # 3092

Task level: equal to ege

The store sells sneakers of two firms: Dike and Ananas. The likelihood that the randomly selected pair of sneakers will be Dike, is equal to \\ (0.6 \\). Each firm may be mistaken in writing its name on sneakers. The likelihood that the Dike company will be mistaken in writing the name is equal to \\ (0.05 \\); The likelihood that Ananas is mistaken in writing the name is equal to \\ (0.025 \\). Find the likelihood that the randomly purchased pair of sneakers will be with the right writing of the company name.

Event A: "A pair of sneakers will be with the correct name" equal to the amount of events B: "A pair of sneakers will be a Dike company and with the correct name" and C: "The pair of sneakers will be Ananas and with the correct name."
The probability of event B is equal to the product of the probability of events "sneakers will be the company Dike" and "The name of the company Dike wrote correctly": \\ Similarly for Event C: \ Hence, \

Answer: 0,96

Task 2 # 166

Task level: equal to ege

If Timur plays white checkers, he wins in Vani with a probability of 0.72. If Timur plays black checkers, he wins the Vanya with a probability of 0.63. Timur and Vanya play two parties, and in the second batch change the color of the checkers. Find the likelihood that Vanya wins both times.

Vanya wins white with a probability \\ (0.37 \\), and black with a probability \\ (0.28 \\). Events "From two batches of Vanya won white" \\ (\\ \\) and "from two batches Vanya won black" \\ (\\ \\) - independent, then the likelihood of their simultaneous occurrence is equal to

Answer: 0.1036.

Task 3 # 172

Task level: equal to ege

The entrance to the museum is guarded by two guards. The likelihood that the senior of them will forget the radio equal to \\ (0.2 \\), and the likelihood that the younger of them will forget the radio equal to \\ (0.1 \\). What is the likelihood that they will not have a single radio?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. Then the desired probability is equal to \\

Answer: 0.02.

Task 4 # 167

Task level: equal to ege

Jumping from a height of 1 meter, Kostya breaks his leg with a probability \\ (0.05 \\). Jumping from a height of 1 meter, Vanya breaks his leg with a probability \\ (0.01 \\). Jumping from a height of 1 meter, Anton breaks the leg with a probability \\ (0.01 \\). Kostya, Vanya and Anton simultaneously jump from a height of 1 meter. What is the likelihood that of them only Kostya break his leg? Answer round up to thousandths.

Events "When jumping from a height of 1 meter Kostya broke the leg" \\ (, \\) "When jumping from a height of 1 meter Vanya did not break the leg" \\ (\\) and "When jumping from a height of 1 meter, Anton did not break the leg" \\ ( \\ \\) - independent, therefore, the probability of their simultaneous occurrence is equal to the product of their probability: \ After rounding, we finally get \\ (0.049 \\).

Answer: 0,049

Task 5 # 170

Task level: equal to ege

Maxim and Vanya decided to play bowling. Maxim fairly figured out that on average he knocks the strike once in eight throws. Vanya fairly figured that on average he knocks the strike once every five shots. Maxim and Vanya make exactly one throw (regardless of the result). What is the likelihood that there will be no strike among them?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. In this case, the likelihood that Maxim will not choose a strike equal \ The likelihood that Vanya will not choose a strike equal to \\ (1 - 0.2 \u003d 0.8 \\). Then the desired probability is equal \\ [\\ DFRAC (7) (8) \\ CDot 0.8 \u003d 0.7. \\]

Answer: 0,7

Task 6 # 1646

Task level: equal to ege

Anton and Kostya play table tennis. The likelihood that Kostya will fall as a corona blow to the table \\ (0.9 \\). The likelihood that Anton will win the draw in which Kostya tried to apply a crown strike \\ (0.3 \\). Kostya tried to get to his corona blow to the table. What is the likelihood that Kostya really gets his coronal blow and eventually win this draw?

Since the events under consideration are independent, the probability of their simultaneous occurrence is equal to the product of their probabilities. At the same time, the probability that Anton will not win the draw, in which Kostya tried to apply his corona strike \\ (1 - 0.3 \u003d 0.7 \\). Then the desired probability is equal to \\