Preparation for the Unified State Exam "solving problems in probability theory." Presentation "Preparation for the Unified State Exam"

MBOU Ostankino Secondary School

Preparation for the Unified State Exam

Solving problems in probability theory

A – coffee will run out in the first machine; B – coffee will run out in the second machine.

According to the conditions of the problem,

Note that these events are not independent, otherwise

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

4 options: ХХО, ХОО, ОХО, LLC

P(ХХО) + P(ХОО) + P(ХХО) + 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 purchased from 1 farm

Egg purchased from 2 farms

P∙0.4+(1-p)∙0.2=0.35

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

Phone released

at 1 factory

Phone released

at factory 2

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

glass is out

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 layout 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 layout is shown in the figure. Some routes lead to the village S, others to field F or swamp M. Find the probability that Pavel Ivanovich will wander into the swamp.

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) = 0.94 − 0.56 = 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 solves 11 problems

Event B - the student solves more than 11 problems

Event A + B - the student solves more than 10 problems

Answer:0.035

Event A – John will take

sighted revolver

Event B – John will take

unshot 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 with the ball. The Fizik team plays three matches with different teams. Find the probability that in these games “Physicist” will win the lot exactly twice

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

Event A - heads will appear 2 times (in games “Physicist” will win the lot exactly twice)

Cases of LLC, ORO, ROO

Answer:0.375

Thank you for your attention

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In a shopping center, two identical machines sell coffee. The probability that the machine will run out of coffee by the end of the day is 0.3. The probability that both machines will run out of coffee is 0.12. Find the probability that at the end of the day there will be coffee left in both machines. A – coffee will run out in the first machine; B – coffee will run out in the second machine. According to the conditions of the problem, 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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In the Magic Land there are two types of weather: good and excellent, and the weather, once established in the morning, remains unchanged all day. It is known that with probability 0.8 the weather tomorrow will be the same as today. Today is July 3rd, the weather in the Magic Land is good. Find the probability that the weather will be great in Fairyland on July 6th. 4 options: ХХО, ХОО, ОХО, LLC 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

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

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Glass produced by 1 factory Glass produced by 2 factory 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 glasses for car headlights. The first factory produces 45% of these glasses, the second - 55%. The first factory produces 3% of defective glasses, the second - 1%. Find the probability that glass accidentally bought in a store turns out to 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) = 0.94 − 0.56 = 0.38. Answer: 0.38 A bus runs from the district center to the village every day. 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) = P(A) + P(B)+ P(C)= P(A) + P(B) P(A) = 0.97-0.89 = 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 for more than a year, but less than two years Event B - the kettle will last for more than two years Event C - the kettle will last for exactly two years A + B + C - the kettle will last for more than a year Events A, B and C Incompatible, the probability of their sum is equal to the sum of the probabilities of these events. The probability of event C, consisting in the fact that the kettle will fail exactly in two years - strictly the same day, hour and 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 P(A) = 0.74-0.67 = 0.07 Answer: 0.07 Probability that Biology test student O. will solve more than 11 problems correctly is 0.67. The probability that O. will correctly solve more than 10 problems is 0.74. Find the probability that O. will correctly solve exactly 11 problems. Events A and Extras, 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 zeroed revolver Event B – John will take an unshot 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 fires a shot revolver. If John shoots from the outside of a zeroed revolver, then he hits the fly with probability 0.2. There are 10 revolvers on the table, only 4 of them have been shot. Cowboy John sees a fly on the wall, at random 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 the event A.

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

Probabilities of opposite events:

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

Formula for multiplying probabilities 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.

Formula for multiplying probabilities 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 makes it easier to use formulas when solving problems:

The probability that a new ballpoint pen writes poorly (or does not write) is 0.1. A buyer in a 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 selected pen writes poorly).
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. At 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 question on the topic “Parallelogram” is 0.15. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics in 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 that relate to these two topics at the same time.
Event C= (a question about one of these two topics) is their union: .
Let's 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 there will be fewer than 20 passengers on the bus on Monday is 0.94. The probability that there will be fewer 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 from 15 to 19 passengers on the bus.” Their sum is 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).

Then, using these problems, 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 run out in the first machine),
B= (coffee will run out in the second machine).
According to the conditions of the problem and .
Using the formula for adding probabilities, we find the probability of an event
= (coffee will run out in at least one of the machines): .
Therefore, the probability of the opposite event (coffee will remain in both machines) is equal to P = 1-0.48 = 0.52.
Answer: 0.52.

A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.8. Find the probability that the biathlete hits the targets the first three times and misses the last two times. 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. This means that the probability of each miss is 1-0.8=0.2. Let's use the formula for multiplying the probabilities of independent events. We find that the sequence
A= (hit, hit, hit, missed, missed) has a probability of .
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 machine. Find the probability that at least one machine is working.

Solution.
This problem also assumes that the automata operate independently.
Let's 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: .
This means that the probability of the event A= (at least one machine is operational) is equal to .
Answer: 0.9975.

15. During artillery fire, the automatic system fires a shot at the target. If the target is not destroyed, the system fires a second shot. 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 it is 0.6. How many shots will be required to ensure that the probability of destroying the target is at least 0.98?

Solution.
Let's find the probability of the opposite event, which is that the target will not be destroyed in n shots. The probability of missing the first shot is 0.6, and each subsequent shot is 0.4. These events are independent, the probability of their occurrence 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 to the inequality; . Consistently checking the values n, equal to 1, 2, 3, etc. we find that the desired solution is n=5. Therefore, 5 shots must be fired.

You can solve the problem “by action”, calculating the probability of surviving after a series of consecutive mistakes:

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 latter probability is less than 0.02, so five shots at the target is enough.

16. Before the start of a volleyball match, team captains draw fair lots to determine which team will start the game with the ball. 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.
You need to find the probability of three events happening: “Stator” starts the first game, does not start the second game, and starts the third game. The probability of a product of independent events is equal to the product of the probabilities of these events. The probability of each of them is 0.5, from which we find: P=0.5·0.5·0.5 = 0.125.

Another solution:

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

Answer: 0.125.

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

Solution.
Girlfriends can end up 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 equal to . Thus, the probability that both friends will be in this group is equal. The probability that Anya and Nina will be in the second group or third group will be the same. These events are inconsistent, then the required 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 some problems, the tree is built directly in the condition. In other problems 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 layout 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 follow the corresponding path. The choice of path at each fork is random, so the probability is equally divided among all possibilities. Let's assume that Pavel Ivanovich came to vertex C. Three edges CH, CK and CL come out of it. Therefore, the probability that Pavel Ivanovich will choose edge CH is 1/3. All probabilities can be arranged similarly.

Each route from the starting point A to any of the ending points is an elementary event in this experiment. The events here are not equally possible. The probability of each elementary event can be found using the multiplication rule.
We need to find the probability of an elementary event
G= (Pavel Ivanovich arrived at point G).

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

19. The picture shows a labyrinth. The spider crawls into the maze at the Entrance point. The spider cannot turn around and crawl back, so at each branch the spider chooses one of the paths along which it has not yet crawled. 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 probability 0.5. These are independent events, the probability of their occurrence (the spider reaches exit D) is equal to the product of the probabilities of these events. Therefore, the probability of arriving at exit D is (0.5) 4 = 0.0625.

Answer: 0.0625.

Let's consider a problem that generalizes the conditions of a number of probabilistic problems that can be 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, and 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 previous problems is that the probabilities on the edges are obtained not from equal possibility, but differently.

Let's denote the entire experiment with a letter (large omega) and put a dot near this letter - the root of the tree from which the branches-ribs grow down. From the point we draw an edge down to the left to point A. Event A has a probability of 0.3, so let’s label this edge with a probability of 0.3. The opposite event has a probability of 0.7. Let's draw the second edge to point .

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

To find the probability of event C, you need to select only those paths that lead from the root point to event C. In the figure, these paths are bright, and the paths that do not lead to C are depicted dimly. The highlighted paths are the elementary events that favor 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 obtain:

.
Answer: 0,34.

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

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

.
Answer: 0,0135

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

Solution.
This task is the reverse of the previous one. We will call the event “the egg has the highest category” H. We will call the events “the egg came from the first farm” and “the egg came from the second farm” A 1 and A 2, respectively. Let us denote the desired probability of event A 1 by the letter p and draw a tree.

We get: .
According to the condition, this value is 0.35.
Then ,
from where and, therefore, .
Answer: 0,75.

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

Based on the conditions of the problem, we will create a tree and find the necessary probabilities.

(A)

(IN)

John will miss if: A) he grabs a zeroed revolver and misses with it, or if B) he grabs an unshooted revolver and misses with it. According to the conditional probability formula, the probabilities of these events are equal, 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 equal to P = 0.04 + 0.48 = 0.52.

Answer: 0.52.

23. All patients with suspected hepatitis undergo a blood test. If the test reveals hepatitis, the test result is called positive. In patients with hepatitis, the test gives a positive result with a probability of 0.9. If the patient does not have hepatitis, 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 a patient admitted to the clinic with suspected hepatitis will test positive.

Based on the conditions of the problem, we will create a tree and find the necessary probabilities.

(A)

(IN)

A patient’s analysis 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.

“Point fluctuation” - Intermediate situation. The movement is damped and aperiodic. 5. Linear oscillations. 7. Free vibrations with viscous resistance. General solution = 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.

“Derivative of a function at a point” - What value does the derivative of the function y=f(x) take 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 make with the positive direction of the x-axis?

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

“Point coordinates” - The symmetry of the point relative to the abscissa axis (Ox). The lizard's body is symmetrical relative to 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 abscissa. Conclusion: Semirichnik is a rare plant, but the seven petals of the flower have bilateral symmetry.

“South Africa National Parks” - “Travel to the Republic of South Africa.” The famous Tugela waterfall (948 m) of five cascades is also located nearby. Third day National parks and reserves. First day The capital of South Africa. Hotel room rates start at $400. A rainbow glows in a cloud of water dust rising 100 meters.

“Four remarkable points of a triangle” - A perpendicular drawn from the vertex of a triangle to a straight line containing the opposite side is called. Height. Median of a triangle. Problem No. 1. Height of the triangle. Segment AN is a perpendicular dropped from point A to straight line a, if. The segment connecting a vertex to the middle of the opposite side is called.

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“How to Solve Probability Problems”

Mitrofanova Snezhana Viktorovna, MBOU "Verkhovskaya School" Vologda region

Subject: Workshop on solving problems in probability theory.

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 developed science is, it is impossible to make accurate predictions. We don't know what card we'll draw from the deck at random 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- this is a phenomenon whose behavior cannot be predicted, or it is an experiment whose result cannot be calculated in advance, etc. It is the probabilities of events that are calculated in standard Unified State Examination problems.

SLIDE 2 (AGAIN)

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

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

SLIDE 3

and then below I will illustrate its application with examples.

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

Select a formula (or several) to solve.

SLIDE 4

WHY DO WE READ OBJECTIVES 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 a 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?

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Task 1.

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Solution.

SLIDE 12

0.5 0.25= 0.125

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Task 2.

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Solution.

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

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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 diagram , according to which standard educational problems for calculating probability should be solved:

Find the main question of the task

A formula (or several) has been selected for the solution.

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 a 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 favorable to its occurrence to the number of all outcomes (incompatible, only possible and equally possible):

Problems solved by constructing a probability tree.

Task 1. 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 layout of the paths 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 follow the corresponding path. The choice of path at each fork is random, so the probability is equally divided among all possibilities.

This event is 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.

Solution. There are three routes into the swamp. Let us denote the vertices on these routes and write the corresponding probabilities on the edges along these routes. We will not consider other routes.

The probability of the event (Pavel Ivanovich getting into the swamp) is equal to

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

Answer: 0,125

Task 4. Two factories of the same company produce identical mobile phones.

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

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

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

Solution. Let's introduce the notations for events: A 1 = (the phone was released at the first factory), A 2 = (the phone was released at the second factory), D = (the phone has a hidden defect). Based on the conditions of the problem, we will create a tree and find the necessary probabilities.

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

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“How to Solve Probability Problems”

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