Pavel Ivanovich makes a walk from point a along the paths of the park at each fork, he randomly chooses. opposite event. Probability. What is this

MBOU Ostankino secondary school

Preparation for the exam

Solving problems in probability theory

A - coffee will end in the first machine; B - coffee will end in the second machine.

According to the task,

note that these events are not independent, otherwise

The probability of the opposite event "coffee will remain in both machines" is equal to

There are two types of weather in Fairyland: good and excellent, and the weather, having settled in the morning, remains unchanged all day. It is known that with a probability of 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in Fairyland is fine. Find the probability that there will be great weather in Magicland on July 6th.

4 options: XXO, XOO, OXO, OOO

P(ХХО) + P(ХОО) + P(ОХО) + P(ООО)=0.8∙0.8∙0.2+0.8∙0.2∙0.8+

0,2∙0,2∙0,2+0,2∙0,8∙0,8=0,128+0,128+0,008+0,128=0,392

Answer: 0.392

Egg bought from 1 farm

Egg bought from 2 farms

P∙0.4+(1-p)∙0.2=0.35

Two factories of the same firm produce the same mobile phones. The first factory produces 30% of all phones of this brand, and the second - the rest of the phones. It is known that of all the phones produced by the first factory, 1% have hidden defects, and those produced by the second factory have 1.5%. Find the probability that the purchased in the store, the phone of this brand has a hidden defect.

Phone released

at 1 factory

Phone released

at 2 factories

D-phone has a defect

0,3∙0,01+0,7∙0,015=0,003+0,0105=0,0135

Answer: 0.0135

Glasses released

1 factory

glasses released

2 factory

D-glasses are defective

0,45∙0,03+0,55∙0,01=0,0135+0,0055=0,019

Answer: 0.019

Pavel Ivanovich takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next path without going back. The track diagram is shown in the figure. Find the probability that Pavel Ivanovich will hit point G

Answer: 0.125

Pavel Ivanovich takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next path without going back. The track diagram is shown in the figure. Some of the routes lead to the village S, others - to the field F or to the swamp M. Find the probability that Pavel Ivanovich wanders into the swamp.

Event A - there are fewer than 15 passengers on the bus

Event B - in a bus from 15 to 19 passengers

Event A + B - there are fewer than 20 passengers on the bus

Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events:

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

P (B) \u003d 0.94 - 0.56 \u003d 0.38.

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

P(A)=0.97-0.89=0.08

Event A - the student will solve 11 problems

Event B - the student will solve more than 11 problems

Event A + B - the student will solve more than 10 problems

Answer: 0.035

Event A – John will take

shot revolver

Event B – John will take

unfired revolver

p(A)=0.4 p(B)=0.6

0,4∙0,1+0,6∙0,8=0,52

Event A-patient has hepatitis

Event B - the patient does not have hepatitis

0,05∙0,9+0,95∙0,01=0,0545

Answer: 0.0545

0,02∙0,99+0,98∙0,01=0,0296

Answer: 0.0296

Before the start of a football match, the referee tosses a coin to determine which team will start the ball. The Physicist team plays three matches with different teams. Find the probability that in these games "Physicist" wins the lot exactly twice

Convert to coins Since there are 3 matches, a coin is tossed three times.

Event A - the eagle will fall out 2 times (in the games "Physicist" will win the lot exactly twice)

Cases LLC, ORO, ROO

Answer: 0.375

Thank you for your attention

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Two identical vending machines sell coffee in the mall. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Find the probability that by the end of the day there will be coffee left in both vending machines. A - coffee will end in the first machine; B - coffee will end in the second machine. By the condition of the problem, we note that these events are not independent, otherwise the Probability of the opposite event “coffee will remain in both machines” is equal to Answer: 0.52

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Egg bought from farm 1 - egg bought from farm 2 P∙0.4+(1-p)∙0.2=0.35 0.2p=0.15 p=0.75 Answer: 0.75 D-egg The highest category Agrofirm buys chicken eggs in two households. 40% of the eggs from the first farm are eggs of the highest category, and from the second farm 20% of the eggs of the highest category. In total, 35% of eggs receive the highest category. Find the probability that an egg purchased from this farm will be from the first farm.

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Glasses produced by 1 glass factory produced by 2 factories D-glasses are defective 0.45∙0.03+0.55∙0.01=0.0135+0.0055=0.019 Answer: 0.019 Two factories produce the same glass for car headlights. The first factory produces 45% of these glasses, the second - 55%. The first factory produces 3% of defective glasses, the author's - 1%. Find the probability that a glass accidentally bought in a store will be defective.

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Event A - there are less than 15 passengers on the bus Event B - there are from 15 to 19 passengers on the bus Event A + B - there are less than 20 passengers on the bus Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B). P (B) \u003d 0.94 - 0.56 \u003d 0.38. Answer: 0.38 A bus runs daily from the district center to the village. The probability that on Monday there will be less than 20 passengers on the bus is 0.94. The probability that there will be less than 15 passengers is 0.56. Find the probability that the number of passengers will be from 15 to 19.

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P(A + B + C) \u003d P (A) + P (B) + P (C) \u003d P (A) + P (B) P (A) \u003d 0.97-0.89 \u003d 0.08 Answer: 0.08 The probability that a new electric kettle will last more than a year is 0.97. The probability that it will last more than two years is 0.89. Find the probability that it will last less than two years but more than a year. Event A - the kettle will last more than a year, but less than two years Event B - the kettle will last more than two years Event C - the kettle will last exactly two years A + B + C - the kettle will last more than a year Events A, B and C are not combined, the probability of their sum is equal to the sum of the probabilities of these events. two years - strictly the same day, an hour and a second - is equal to zero.

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Event A - the student will solve 11 problems Event B - the student will solve more than 11 problems Event A + B - the student will solve more than 10 problems Р(А)=0.74-0.67=0.07 test in biology studentO. solves more than 11 problems correctly is 0.67. The probability that O. correctly solves more than 10 problems is 0.74. Find the probability that O. correctly solves exactly 11 problems. Events A and Outside, the probability of their sum is equal to the sum of the probabilities of these events: P(A + B) = P(A) + P(B).

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1-0.965 = 0.035 Answer: 0.035 When manufacturing bearings with a diameter of 67 mm, the probability that the diameter will differ from the specified one by no more than 0.01 mm is 0.965. Find the probability that a random bearing will have a diameter less than 66.99 mm or more than 67.01 mm.

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Event A – John will take a shot revolver Event B – John will take a shot revolver p(A)=0.4 p(B)=0.6 0.4∙0.1+0.6∙0.8=0.52 Answer: 0.52 Cowboy John hits a fly on the wall with a probability of 0.9 if he shoots with a shot revolver. If John shoots from the outside of a shot revolver, then he hits the fly with a probability of 0.2. There are 10 revolvers on the table, of which only 4 are shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots at the fly. Find the probability that John will miss.

An event consisting of those and only those elementary outcomes of experience that are not included in A is called the opposite of event A.

Incompatible events- events that do not occur in one experience. For example, opposite events are incompatible.

Probabilities of opposite events:

; .
Probability addition formula for joint events: The probability of the occurrence of at least one of the two joint events A and B is equal to the sum of their probabilities without the probability of their joint occurrence. .
The formula for adding probabilities for incompatible events: The probability of occurrence of at least one of two incompatible events A and B is equal to the sum of their probabilities.

Probability multiplication formula for independent events: The probability of the joint occurrence of two independent events A and B is equal to the product of the probabilities of events A and B.

Probability multiplication formula for dependent events: The probability of the joint occurrence of two dependent events A and B is equal to the product of the probability of one of them by the conditional probability of the other.

Here is a diagram that facilitates the use of formulas in solving problems:

The probability that a new ballpoint pen writes poorly (or does not write) is 0.1. The buyer in the store chooses one such pen. Find the probability that this pen writes well.

Solution.
Let's define the event A= (the selected pen writes well).
Then the opposite event = (the chosen pen writes badly).
From the condition, we know the probability of the opposite event: .
We use the formula for the probability of the opposite event: .
Answer: 0.9.

10. On the geometry exam, the student gets one question from the list of exam questions. The probability that this is an inscribed circle question is 0.2. The probability that this is a Parallelogram question is 0.15. There are no questions related to these two topics at the same time. Find the probability that the student will get a question on one of these two topics on the exam.

Solution.
Let's define events:
A= (question on the topic "Inscribed circle"),
B= (question on the topic "Parallelogram").
Events A and B are incompatible, since by condition there are no questions in the list related to these two topics at the same time.
Event C= (question on one of these two topics) is their union: .
We apply the formula for adding the probabilities of incompatible events: .
Answer: 0.35.

A bus runs daily from the district center to the village. The probability that on Monday there will be less than 20 passengers on the bus is 0.94. The probability that there will be less than 15 passengers is 0.56. Find the probability that the number of passengers will be between 15 and 19.

Solution.
Consider the events A = “there are less than 15 passengers on the bus” and B = “there are between 15 and 19 passengers on the bus”. Their sum is the event A + B = "less than 20 passengers on the bus". Events A and B are incompatible, the probability of their sum is equal to the sum of the probabilities of these events:
P(A + B) = P(A) + P(B).

Then, using the data of the problem, we obtain: 0.94 = 0.56 + P(B), whence P(B) = 0.94 − 0.56 = 0.38.

Answer: 0.38.

Solution.
Let's define events
A= (coffee will end in the first machine),
B= (coffee will end in the second machine).
By the condition of the problem and .
Using the formula for adding probabilities, we find the probability of an event
= (coffee will end in at least one of the machines): .
Therefore, the probability of the opposite event (coffee will remain in both machines) is P=1-0.48=0.52.
Answer: 0.52.

The biathlete shoots five times at the targets. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hit the targets the first three times and missed the last two. Round the result to the nearest hundredth.

Solution.
In this problem, it is assumed that the result of each next shot does not depend on the previous ones. Therefore, the events “hit on the first shot”, “hit on the second shot”, etc. independent.
The probability of each hit is 0.8. So the probability of each miss is 1-0.8=0.2. We use the formula for multiplying the probabilities of independent events. We get that the sequence
A= (hit, hit, hit, miss, miss) has probability .
Answer: 0.02.

There are two payment machines in the store. Each of them can be faulty with a probability of 0.05, regardless of the other automaton. Find the probability that at least one automaton is serviceable.

Solution.
This problem also assumes the independence of the operation of automata.
Find the probability of the opposite event
= (both machines are faulty).
To do this, we use the formula for multiplying the probabilities of independent events: .
Hence, the probability of the event A= (at least one automaton is operational) is equal to .
Answer: 0.9975.

15. During artillery firing, the automatic system fires at the target. If the target is not destroyed, the system fires again. The shots are repeated until the target is destroyed. The probability of destroying a certain target with the first shot is 0.4, and with each subsequent shot - 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98?

Solution.
Let us find the probability of the opposite event, which is that the target will not be destroyed within n shots. The probability of missing on the first shot is 0.6, and on each subsequent shot is 0.4. These events are independent, the probability of their product is equal to the product of the probability of these events. Therefore, the probability of missing n shots is equal to: . It remains to find the smallest natural solution of the inequality ; . Consistently checking values n, equal to 1, 2, 3, etc. we find that the desired solution is n=5. Therefore, it is necessary to make 5 shots.

You can solve the problem "by actions", calculating the probability of surviving after a series of successive misses:

P(1) = 0.6.
P(2) = P(1) 0.4 = 0.24.
P(3) = P(2) 0.4 = 0.096.
P(4) = P(3) 0.4 = 0.0384;
P(5) = P(4) 0.4 = 0.015536.
The last probability is less than 0.02, so five shots on the target are sufficient.

16. Before the start of a volleyball match, the team captains draw fair lots to determine which team will start the ball game. The Stator team takes turns playing with the Rotor, Motor and Starter teams. Find the probability that Stator will start only the first and last games.

Solution.
It is required to find the probability of product of three events: "Stator" starts the first game, does not start the second game, starts the third game. The probability of producing independent events is equal to the product of the probabilities of these events. The probability of each of them is 0.5, whence we find: Р=0.5·0.5·0.5 = 0.125.

Another way to solve:

Because If the draw can be considered as a coin toss, then the problem can be solved using the technology of solving problems with coins. The draw was held three times, so N=2 3 =8. Let us assign the elementary event “Stator starts the game” the value “Eagle”. Then a favorable outcome corresponds only to the combination "ORI", i.e. N(A)=1. That's why

Answer: 0.125.

17. There are 21 students in the class. Among them are two friends: Anya and Nina. The class is randomly divided into 3 groups of 7 people each. Find the probability that Anya and Nina are in the same group.

Solution.
Friends can be together in any of the three groups. Let's consider some one group. The probability that Anya will be in it is equal to . If Anya is already in this group, then the probability that Nina will be in the same group is . Thus, the probability that both friends will be in this group is equal to . The same will be the probability that Anya and Nina will be in the second group or the third group. These events are incompatible, then the desired probability is equal to the sum of the probabilities of these events:

Answer: 0.3.

In the following problems, it is convenient to use probability tree. In terms of tasks, the tree is built directly in the condition. In other tasks, this tree should be built.

18. Pavel Ivanovich takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next path without going back.
The track diagram is shown in the figure. Find the probability that Pavel Ivanovich will hit point G.

Solution.
The track diagram is a graph, namely a tree. The edges (branches) of the tree correspond to the paths. Next to each edge we write the probability that Pavel Ivanovich will walk along the corresponding path. The choice of the path at each fork occurs at random, so the probability is equally divided among all possibilities. Let's assume that Pavel Ivanovich came to the vertex C. Three edges CH, CK and CL go out of it. Therefore, the probability that Pavel Ivanovich chooses the edge CH is 1/3. Similarly, you can arrange all the probabilities.

Each route from start point A to any of the end points is an elementary event in this experiment. Events here are not equally likely. The probability of each elementary event can be found by the multiplication rule.
We need to find the probability of an elementary event
G= (Pavel Ivanovich came to point G).

This event consists in the fact that Pavel Ivanovich passed the ABG route. The probability is found by multiplying the probabilities along the edges AB and BG: .
Answer: 0.125.

19. The figure shows a labyrinth. The spider crawls into the maze at the "Entrance" point. The spider cannot turn around and crawl back, therefore, at each fork, the spider chooses one of the paths that it has not crawled yet. Assuming that the choice of the further path is purely random, determine with what probability the spider will come to the exit.

Solution.

At each of the four marked forks, the spider can choose either the path leading to exit D or another path with a probability of 0.5. These are independent events, the probability of their product (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of reaching the exit D is (0.5) 4 = 0.0625.

Answer: 0.0625.

Consider a problem that generalizes the conditions of a number of probabilistic problems solved using a probability tree.

In some experiment, the probability of event A is 0.3. If event A occurs, then the probability of event C is 0.2, otherwise the probability of event C is 0.4. Find the probability of event C.

Solution.
In such problems it is convenient to depict the experiment graphically as a tree of probabilities. The difference from the previous problems is that the probabilities on the edges are obtained not from the equiprobability, but in a different way.

We denote the whole experiment with a letter (big omega) and put a dot near this letter - the root of the tree, from which the branches-ribs grow downward. Let's draw an edge down-left from a point to point A. Event A has a probability of 0.3, so we sign this edge with a probability of 0.3. The opposite event has a probability of 0.7. Draw the second edge to the point .

If the event A has taken place, then the event C by the condition has a probability of 0.2. Therefore, from point A we draw an edge down-left to point C and sign the probability. Proceeding in the same way further, we complete the entire tree (see Fig.).

To find the probability of the event C, we need to select only those paths that lead from the root point to the event C. In the figure, these paths are bright, and the paths that do not lead to C are shown pale. The distinguished paths are the elementary events favoring the event C.

Now we need to calculate the probabilities of the selected paths and add them. Using the rules of multiplication and addition of probabilities, we get:

.
Answer: 0,34.

20. Two factories of the same company produce the same mobile phones. The first factory produces 30% of all phones of this brand, and the second - the rest of the phones. It is known that of all phones produced by the first factory, 1% have hidden defects, and 1.5% of all phones manufactured by the second factory. Find the probability that a phone of this brand bought in a store has a hidden defect.

Solution.
Let's introduce notation for events:
A 1 = (phone released at the first factory),
A 2 = (phone made in the second factory),
D= (the phone has a hidden defect).

.
Answer: 0,0135

21. An agricultural firm buys chicken eggs from two households. 40% of the eggs from the first farm are eggs of the highest category, and from the second farm - 20% of the eggs of the highest category. In total, 35% of the eggs from these two farms receive the highest category. Find the probability that the egg purchased from this farm will be from the first farm.

Solution.
This problem is the reverse of the previous one. Let's call the event "egg has the highest category" H. Let's call the events "egg came from the first farm" and "the egg came from the second farm" A 1 and A 2 respectively. Let p denote the desired probability of the event A 1 and draw a tree.

We get: .
By condition, this value is equal to 0.35.
Then ,
whence and, hence, .
Answer: 0,75.

22. Cowboy John hits a fly on the wall with a probability of 0.9 if he shoots with a shot revolver. If John fires an unshot revolver, he hits a fly with a probability of 0.2. There are 10 revolvers on the table, of which only 4 are shot. Cowboy John sees a fly on the wall, randomly grabs the first revolver he comes across and shoots at the fly. Find the probability that John misses.

According to the condition of the problem, we will make a tree and find the necessary probabilities.

(A)

(IN)

John will miss if: A) grabs a fired revolver and misses with it, or if B) grabs an unfired revolver and misses with it. According to the conditional probability formula, the probabilities of these events are, respectively, P(A)=0.4 (1 - 0.9) = 0.04 and P(B)=0.6 (1 - 0.2) = 0, 48. These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events. Then the desired probability is P=0.04 + 0.48 = 0.52.

Answer: 0.52.

23. All patients with suspected hepatitis are given a blood test. If the analysis reveals hepatitis, then the result of the analysis is called positive. In hepatitis patients, the analysis gives a positive result with a probability of 0.9. If the patient does not have hepatitis, then the test may give a false positive result with a probability of 0.01. It is known that 5% of patients admitted with suspected hepatitis actually have hepatitis. Find the probability that the test result of a patient admitted to the clinic with suspected hepatitis will be positive.

According to the condition of the problem, we will make a tree and find the necessary probabilities.

(A)

(IN)

The analysis of the patient can be positive for two reasons: A) the patient has hepatitis, his analysis is correct; B) the patient does not have hepatitis, his analysis is false. These are incompatible events, the probability of their sum is equal to the sum of the probabilities of these events. We have: July 4 P(A) = 0.8 0.8 0.2 = 0.128;
P(B) = 0.8 0.2 0.8 = 0.128;
P(C) = 0.2 0.2 0.2 = 0.008;
P(D) = 0.2 0.8 0.8 = 0.128.

These events are incompatible, the probability of their sum is equal to the sum of the probabilities of these events:
P= 0.128 + 0.128 + 0.008 + 0.128 = 0.392.

SLIDE 4

View document content
"How to solve probability problems"

Mitrofanova Snezhana Viktorovna, MBOU "Verkhovskaya School" Vologda Region

Subject: Workshop on solving problems in the theory of probability.

SLIDE 1

How to solve probability problems?

Probability. What is this?

SLIDE 2

Probability theory, as the name suggests, deals with probabilities. We are surrounded by many things and phenomena about which, no matter how advanced science is, it is impossible to make accurate predictions. We don't know which card we will randomly draw from the deck or how many days it will rain in May, but with some additional information we can make predictions and calculate the probabilities of these random events.

Thus, we are faced with the basic concept random event- these are phenomena, the behavior of which cannot be predicted, or it is an experiment, the result of which cannot be calculated in advance, etc. It is the probabilities of events that are calculated in typical USE problems.

SLIDE 2 (ONE AGAIN)

Probability- this is some, strictly speaking, a function that takes values from 0 to 1 and characterizes a given random event.

Then we use sample diagram, which should be used to solve standard learning problems for calculating the probability of a random event,

SLIDE 3

and then below with examples I will illustrate its application.

Find the main question of the task (find what is the outcome of the task, find favorable outcomes.)

Select a formula (or several) for the solution.

SLIDE 4

WHY DO WE READ TASKS CAREFULLY?

Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will be able to answer the ticket chosen at random?

Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will not be able to answer the ticket chosen at random?

SLIDE 5,6,7

SLIDE 8.9

SLIDE 10

Task 1.

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Solution.

SLIDE 12

0.5 0.25= 0.125

SLIDE 13

Task 2.

SLIDE 14

Solution.

P(M)=P(ABD)+P(ABE)+P(ACF)

SLIDE 15

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SLIDE 19, 20

Task 4.

View presentation content
"Presentation"

How to solve problems

on probability?

Mitrofanova Snezhana Viktorovna,

mathematic teacher

MBOU "Verkhovskaya School"

Vologda region

Probability. What is it ?

Probability is a function that takes values from 0 to 1.

Approximate scheme , which should be used to solve standard educational problems for calculating the probability:

Find the main question of the task

The formula (or several) for the solution is selected.

Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will be able to answer the ticket chosen at random?

Of the 20 tickets offered in the exam, the student can answer only 17. What is the probability that the student will not be able to answer the ticket chosen at random?

Probability events is the ratio of the number of outcomes that favor its occurrence to the number of all outcomes (incompatible, the only possible and equally possible):

Problems solved by constructing a tree of probabilities.

Task 1. Pavel Ivanovich takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next track without going back. The track diagram is shown in the figure. Find the probability that Pavel Ivanovich will hit point G.

Solution.

Next to each edge we write the probability that Pavel Ivanovich will walk along the corresponding path. The choice of the path at each fork occurs at random, so the probability is equally divided among all possibilities.

This event consists in the fact that Pavel Ivanovich passed the ABG route. The probability is found by multiplying the probabilities along the edges AB and BG

0,5 0.25= 0.125

Task 2.

Pavel Ivanovich takes a walk from point A along the paths of the park. At each fork, he randomly chooses the next path without going back. The track diagram is shown in the figure. Some of the routes lead to the village S , others - to the field F or to the swamp M . Find the probability that Pavel Ivanovich wanders into the swamp.

Solution. Three routes lead to the swamp. We denote the vertices on these routes and write the corresponding probabilities on the edges along these routes. Other routes will not be considered.

The probability of an event (Pavel Ivanovich will fall into a swamp) is equal to

P(M)=P(ABD)+P(ABE)+P(ACF)

Answer: 0,125

Task 4. Two factories of the same firm produce the same mobile phones.

The first factory produces 30% of all phones of this brand, and the second - the rest of the phones.

It is known that of all phones produced by the first factory, 1% have hidden defects, and 1.5% of all phones manufactured by the second factory.

Find the probability that a phone of this brand bought in a store has a hidden defect.

Solution. Let's introduce the notation for the events: A 1 = (the phone is released at the first factory), A 2 = (the phone is released at the second factory), D = (the phone has a hidden defect). According to the condition of the problem, we will make a tree and find the necessary probabilities.

P(D)=0.3 *0.01+0.7 *0.015=0.003+0.0105=0.0135 .

"Point fluctuation" - Intermediate situation. The motion is damped and aperiodic. 5. Linear oscillations. 7. Free vibrations with viscous resistance. General solution \u003d general solution + particular solution of homogeneous y-i of inhomogeneous y-i. 1. Examples of oscillations. Harmonic driving force. Free vibrations caused by a driving force.

"The derivative of the function at a point" - What is the value of the derivative of the function y \u003d f (x) at point B? The figure shows a graph of the derivative y= f‘(x) of the function f(x) defined on the interval (-3;3). What value does the derivative of the functions y= f(x) take at point A? What angle does the tangent to the graph of the function form with the positive direction of the x-axis?

"Critical points of the function" - Among the critical points there are extremum points. A necessary condition for an extremum. Critical points of the function Extremum points. Definition. Extreme points (repetition). But, if f "(x0) = 0, then it is not necessary that the point x0 will be an extremum point. Examples. Critical points.

"Point coordinates" - Symmetry of the point about the abscissa axis (Ox). The body of the lizard is symmetrical about a straight line. The human body has an axis of symmetry. In nature, the structure of animal bodies also obeys the laws of symmetry. Point B (3; 6) is symmetrical to point B (3; - 6), located below the x-axis. Conclusion: Semirichnik is a rare plant, but the seven petals of the flower have bilateral symmetry.

"South Africa National Parks" - "Journey to the Republic of South Africa". Nearby is the famous Tugela waterfall (948 m) of five cascades. Third day National parks and reserves. First day The capital of South Africa. The cost of hotel rooms starts from $400. A rainbow glows in a cloud of water dust rising 100 meters.

"Four remarkable points of the triangle" - The perpendicular dropped from the vertex of the triangle to the line containing the opposite side is called. Height. The median of a triangle. Problem number 1. The height of the triangle. Segment AN is a perpendicular dropped from point A to line a, if. A line segment that connects a vertex to the midpoint of the opposite side is called.

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Pavel Ivanovich makes a walk from point a along the paths of the park at each fork, he randomly chooses. opposite event. Probability. What is this

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"How to solve probability problems"

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