"the theory of probability in the tasks of the exam and oge". Simple Problems in Probability Theory

Presented to date in the open bank of USE problems in mathematics (mathege.ru), the solution of which is based on only one formula, which is a classical definition of probability.

The easiest way to understand the formula is with examples.
Example 1 There are 9 red balls and 3 blue ones in the basket. The balls differ only in color. At random (without looking) we get one of them. What is the probability that the ball chosen in this way will be blue?

Comment. In probability problems, something happens (in this case, our action of pulling the ball) that can have different result- outcome. It should be noted that the result can be viewed in different ways. "We pulled out a ball" is also a result. "We pulled out the blue ball" is the result. "We drew this particular ball out of all possible balls" - this least generalized view of the result is called the elementary outcome. It is the elementary outcomes that are meant in the formula for calculating the probability.

Decision. Now we calculate the probability of choosing a blue ball.
Event A: "the chosen ball turned out to be blue"
Total number of all possible outcomes: 9+3=12 (number of all balls we could draw)
Number of outcomes favorable for event A: 3 (the number of such outcomes in which event A occurred - that is, the number of blue balls)
P(A)=3/12=1/4=0.25
Answer: 0.25

Let us calculate for the same problem the probability of choosing a red ball.
The total number of possible outcomes will remain the same, 12. The number of favorable outcomes: 9. The desired probability: 9/12=3/4=0.75

The probability of any event always lies between 0 and 1.
Sometimes in everyday speech (but not in probability theory!) The probability of events is estimated as a percentage. The transition between mathematical and conversational assessment is done by multiplying (or dividing) by 100%.
So,
In this case, the probability is zero for events that cannot happen - improbable. For example, in our example, this would be the probability of drawing a green ball from the basket. (The number of favorable outcomes is 0, P(A)=0/12=0 if counted according to the formula)
Probability 1 has events that will absolutely definitely happen, without options. For example, the probability that "the chosen ball will be either red or blue" is for our problem. (Number of favorable outcomes: 12, P(A)=12/12=1)

We've looked at a classic example that illustrates the definition of probability. All similar USE tasks according to probability theory are solved by applying this formula.
Instead of red and blue balls, there can be apples and pears, boys and girls, learned and unlearned tickets, tickets containing and not containing a question on a certain topic (prototypes , ), defective and high-quality bags or garden pumps (prototypes , ) - the principle remains the same.

They differ slightly in the formulation of the problem of the USE probability theory, where you need to calculate the probability of an event occurring on a certain day. ( , ) As in the previous tasks, you need to determine what is an elementary outcome, and then apply the same formula.

Example 2 The conference lasts three days. On the first and second days, 15 speakers each, on the third day - 20. What is the probability that the report of Professor M. will fall on the third day, if the order of the reports is determined by lottery?

What is the elementary outcome here? - Assigning a professor's report to one of all possible serial numbers for a speech. 15+15+20=50 people participate in the draw. Thus, Professor M.'s report can receive one of 50 numbers. This means that there are only 50 elementary outcomes.
What are the favorable outcomes? - Those in which it turns out that the professor will speak on the third day. That is, the last 20 numbers.
According to the formula, the probability P(A)= 20/50=2/5=4/10=0.4
Answer: 0.4

The drawing of lots here is the establishment of a random correspondence between people and ordered places. In Example 2, matching was considered in terms of which of the places a particular person could take. You can approach the same situation from the other side: which of the people with what probability could get to a particular place (prototypes , , , ):

Example 3 5 Germans, 8 Frenchmen and 3 Estonians participate in the draw. What is the probability that the first (/second/seventh/last - it doesn't matter) will be a Frenchman.

The number of elementary outcomes is the number of all possible people who could, by lot, get into given place. 5+8+3=16 people.
Favorable outcomes - the French. 8 people.
Desired probability: 8/16=1/2=0.5
Answer: 0.5

The prototype is slightly different. There are tasks about coins () and dice () that are somewhat more creative. Solutions to these problems can be found on the prototype pages.

Here are some examples of coin tossing or dice tossing.

Example 4 When we toss a coin, what is the probability of getting tails?
Outcomes 2 - heads or tails. (it is believed that the coin never falls on the edge) Favorable outcome - tails, 1.
Probability 1/2=0.5
Answer: 0.5.

Example 5 What if we flip a coin twice? What is the probability that it will come up heads both times?
The main thing is to determine which elementary outcomes we will consider when tossing two coins. After tossing two coins, one of the following results can occur:
1) PP - both times it came up tails
2) PO - first time tails, second time heads
3) OP - the first time heads, the second time tails
4) OO - heads up both times
There are no other options. This means that there are 4 elementary outcomes. Only the first one is favorable, 1.
Probability: 1/4=0.25
Answer: 0.25

What is the probability that two tosses of a coin will land on tails?
The number of elementary outcomes is the same, 4. Favorable outcomes are the second and third, 2.
Probability of getting one tail: 2/4=0.5

In such problems, another formula may come in handy.
If with one toss of a coin options we have 2 results, then for two throws the results will be 2 2=2 2 =4 (as in example 5), for three throws 2 2 2=2 3 =8, for four: 2 2 2 2 =2 4 =16, … for N throws there are 2·2·...·2=2 N possible outcomes.

So, you can find the probability of getting 5 tails out of 5 coin tosses.
The total number of elementary outcomes: 2 5 =32.
Favorable outcomes: 1. (RRRRRR - all 5 times tails)
Probability: 1/32=0.03125

The same is true for the dice. With one throw, there are 6 possible results. So, for two throws: 6 6=36, for three 6 6 6=216, etc.

Example 6 We throw a dice. What is the probability of getting an even number?

Total outcomes: 6, according to the number of faces.
Favorable: 3 outcomes. (2, 4, 6)
Probability: 3/6=0.5

Example 7 Throw two dice. What is the probability that the total rolls 10? (round to hundredths)

There are 6 possible outcomes for one die. Hence, for two, according to the above rule, 6·6=36.
What outcomes will be favorable for a total of 10 to fall out?
10 must be decomposed into the sum of two numbers from 1 to 6. This can be done in two ways: 10=6+4 and 10=5+5. So, for cubes, options are possible:
(6 on the first and 4 on the second)
(4 on the first and 6 on the second)
(5 on the first and 5 on the second)
In total, 3 options. Desired probability: 3/36=1/12=0.08
Answer: 0.08

Other types of B6 problems will be discussed in one of the following "How to Solve" articles.

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Key tasks in probability theory Preparation for the OGE No. 9 MBOU "Gymnasium No. 4 named after. A.S. Pushkin” Compiled by: Sofina N.Yu.

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Basic verifiable requirements for mathematical preparation No. 9 OGE in mathematics Solve practical problems that require a systematic enumeration of options; compare the chances of occurrence of random events, evaluate the probabilities of a random event, compare and explore models of a real situation using the apparatus of probability and statistics. No. 9 - basic task. The maximum score for completing the task is 1.

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The probability of an event A is the ratio of the number m of outcomes favorable to this event to total number n all equally possible incompatible events that can occur as a result of one test or observation Classical definition of probability Recall the formula for calculating the classical probability of a random event Р = n m

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Classical definition of probability Example: The Parent Committee bought 40 coloring pages for graduation gifts for children school year. Of these, 14 are based on the fairy tales of A.S. Pushkin and 26 based on the fairy tales of G.Kh. Andersen. Gifts are distributed randomly. Find the probability that Nastya will get a coloring book based on the fairy tales of A.S. Pushkin. Solution: m= 14; n= 14 +26=40 Р= 14/40= 0.35 Answer: 0.35.

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Example: There were 60 questions for the exam. Ivan did not learn 3 of them. Find the probability that he will come across the learned question. Solution: Here n=60. Ivan did not learn 3, so he learned all the rest, i.e. m=60-3=57. P=57/60=0.95. Classic definition of probability Answer: 0.95.

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“The order is determined by a draw” Example: 20 athletes participate in the gymnastics championship: 8 from Russia, 7 from the USA, the rest from China. The order in which the gymnasts perform is determined by lot. Find the probability that the fifth athlete is from China. Solution: In the condition of the problem there is a “magic” word “lot”, which means we forget about the order of speaking. Thus, m= 20-8-7=5 (from China); n=20. P \u003d 5/20 \u003d 0.25. Answer: 0.25.

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Example: A scientific conference is held in 5 days. A total of 75 reports are planned - the first 3 days, 17 reports each, the rest are distributed equally between the 4th and 5th days. The order of reports is determined by a draw. What is the probability that Professor Ivanov's report will be scheduled for the last day of the conference? Solution: Let's put the data in the table. We got that m=12; n=75. P=12/75=0.16. Answer: 0.16. “Order determined by lottery” Day I II III IV V Total Number of presentations 17 17 17 12 12 75

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Event frequency In the same way as the probability, the frequency of the event is found, the tasks for which are also in the prototypes. What is the difference? Probability is a predictable value, and frequency is a statement of fact. Example: The probability that a new tablet will be repaired within a year is 0.045. In a certain city, out of 1000 tablets sold during the year, 51 pieces arrived at the warranty workshop. How different is the frequency of the “warranty repair” event from its probability in this city? Solution: Find the frequency of the event: 51/1000=0.051. And the probability is equal to 0.045 (by condition). This means that in this city the event “warranty repair” occurs more often than expected. Let's find the difference ∆= 0.051- 0.045= 0.006. At the same time, we must take into account that the sign of the difference is NOT important to us, but only its absolute value. Answer: 0.006.

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Problems with enumeration of options ("coins", "matches") Let k be the number of coin tosses, then the number of possible outcomes: n = 2k. Example: In a random experiment, a symmetrical coin is tossed twice. Find the probability that heads come up exactly once. Solution: Coin drop options: OO; OR; RR; RO. Thus, n=4. Favorable outcomes: RR and RR. That is, m = 2. P = 2/4 = 1/2 = 0.5. Answer: 0.5.

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Example: Before starting football match The referee tosses a coin to determine which team will have the ball first. The team "Mercury" plays in turn with the teams "Mars", "Jupiter", "Uranus". Find the probability that in all matches the right to own the ball will be won by the team "Mercury"? Problems with enumeration of options ("coins", "matches") Solution: Let's designate the right of possession of the first ball of the "Mercury" team in the match with one of the other three teams as "Tails". Then the right of possession of the second ball of this team is “Eagle”. So, let's write down all the possible outcomes of tossing a coin three times. "O" - heads, "P" - tails. ; i.e., n=8; m=1. P=1/8=0.125. Answer: 0.125 n = 23 "Mars" "Jupiter" "Uranus"

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Problems on "dice" (dice) Let k be the number of throws of the dice, then the number of possible outcomes: n = 6k. Example: Dasha rolls a dice twice. Find the probability that her total rolled 8. Round the result to the nearest hundredth. Answer: 0.14. Solution: The sum of the two dice must be 8 points. This is possible if there are the following combinations: 2 and 6 6 and 2 3 and 5 5 and 3 4 and 4 m= 5 (5 suitable combinations) n \u003d 36 P \u003d 5/36 \u003d 0.13 (8)

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Independent events and the law of multiplication The probability of finding both the 1st, and 2nd, and n-th events are found by the formula: Р= Р1*Р2*…*Рn Example: 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 hit the targets the first three times and missed the last two. Round the result to the nearest hundredth. Answer: 0.02. Solution: 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 a miss is 1 - 0.8 = 0.2. 1 shot: 0.8 2 shot: 0.8 3 shot: 0.8 4 shot: 0.2 5 shot: 0.2 .8 ∙ 0.2 ∙ 0.2 = 0.02048 ≈ 0.02.

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Combinations of "and" laws and "or" laws Example: An office buys stationery for employees of 3 different firms. Moreover, the products of the 1st company make up 40% of all deliveries, and the rest of the 2nd company is equally divided. It turned out that 2% of the pens of the 2nd company are defective. The percentage of marriage in the 1st and 3rd firms, respectively, is 1% and 3%. Employee A took a pen from a new shipment. Find the probability that it will be correct. Solution: The products of 2nd and 3rd firms are (100%-40%):2=30% of supplies. P(marriage) \u003d 0.4 0.01 + 0.3 0.02 + 0.3 0.03 \u003d 0.019. P (serviceable pens) \u003d 1 - 0.019 \u003d 0.981. Answer: 0.981.

Easy tasks

There are 25 pies on the table: 7 - with jam, 9 - with potatoes, the rest with cabbage. What is the probability that a randomly selected pie will be with cabbage?

0,36

The taxi employs 40 cars: 14 are Lada brands, 8 are Renault brands, 2 are Mercedes brands, and the rest are Skoda brands. What is the probability that a Mercedes will come to your call?

0,05

Determine the probability that a number of at least three will come up when a dice is thrown.

Ira, Dima, Vasya, Natasha and Andrey pass the standard in 60 meters. What is the probability that the girl runs the fastest?

The probability that a phone bought in an underpass is fake is 0.83. What is the probability that the phone bought in the transition will not be a fake?

0,17

20 teams take part in the basketball tournament, including the “Guys” team. All teams are divided into 4 groups: A, B, C, D. What is the probability that the “Guys” team will be in group A?

0,25

The lottery bag contains kegs numbered from 5 to 94 inclusive. What is the probability that the keg taken from the bag contains a two-digit number? Round your answer to the nearest hundredth.

0,94

Before the exam, Igor held out to the last and managed to learn only 5 tickets out of 80. Determine the probability that he will come across a learned ticket.

0,0625

Anya turns on the radio and randomly selects a radio wave. In total, her radio receiver catches 20 radio waves and only 7 of them in this moment music is playing. Find the probability that Anya will fall on a musical wave.

0,35

In every twentieth bottle of soda, a code with a win is hidden under the cap. Determine the probability that the purchased bottle will have a winning code under the cap.

0,05

Tasks are more difficult

What is the probability that a randomly chosen 3-digit number is divisible by 5?

0,2

The height (in cm) of five students is recorded: 166, 158, 132, 136, 170. How much does the arithmetic mean of this set of numbers differ from its median?

According to the statistics of one small country, it is known that the probability that the baby born will be a boy is 0.507. In 2017, there were an average of 486 girls per 1,000 babies born in this country. How different is the frequency of female births in 2017 in this country from the probability of this event?

0,007

A die is thrown twice. Find the probability that the sum of the two numbers drawn is 3 or 7. Round your answer to the nearest hundredth.

0,22

What is the probability that a randomly chosen three-digit number is divisible by 2?

0,5

Find the probability that two coin tosses come up tails exactly once.

0,5

A die is thrown twice, find the probability that a number greater than three will come up both times. Round your answer to the nearest hundredth.

0,31

According to the statistics of one small country, it is known that the probability that a baby born will be a boy is 0.594. In 2017, there were an average of 513 girls per 1,000 babies born in this country. How different is the frequency of female births in 2017 in this country from the probability of this event?

0,107

The height (in cm) of five students is recorded: 184, 145, 176, 192, 174. How much does the arithmetic mean of this set of numbers differ from its median?

1,8

The average height of the inhabitants of the village "Giants" is 194 cm. The height of Nikolai Petrovich is 195 cm. Which of the following statements is correct?

1) The height of one of the villagers must be 194 cm.

2) Nikolai Petrovich is the tallest resident of the village.

3) There will definitely be at least one man from this village below Nikolai Petrovich.

4) There will definitely be at least one resident from this village below Nikolai Petrovich.

4

Difficult tasks

The shooter shoots 4 times with a gun at the targets. The probability of its exact hit on the target with one shot is 0.5. Find the probability that the shooter hits the target the first two times and misses the last two.

0,0625

The probability that the battery is defective is 0.05. The customer in the store chooses a random package with two batteries. Find the probability that both batteries are good.

0,9025

The shooter shoots at the targets 5 times in a row. The probability of hitting the target when fired is 0.7. Find the probability that the shooter hit the target the first four times and missed the last time. Round the result to the nearest hundredth.

Events that occur in reality or in our imagination can be divided into 3 groups. These are certain events that are bound to happen, impossible events, and random events. Probability theory studies random events, i.e. events that may or may not occur. This article will be presented in summary theory of probability formulas and examples of solving problems in the theory of probability, which will be in the 4th task of the exam in mathematics (profile level).

Why do we need the theory of probability

Historically, the need to study these problems arose in the 17th century in connection with the development and professionalization of gambling and the advent of the casino. It was a real phenomenon that required its study and research.

Playing cards, dice, roulette created situations where any of a finite number of equally probable events could occur. There was a need to give numerical estimates of the possibility of the occurrence of an event.

In the 20th century, it became clear that this seemingly frivolous science plays an important role in understanding the fundamental processes occurring in the microcosm. Was created modern theory probabilities.

Basic concepts of probability theory

The object of study of probability theory is events and their probabilities. If the event is complex, then it can be broken down into simple components, the probabilities of which are easy to find.

The sum of events A and B is called event C, which consists in the fact that either event A, or event B, or events A and B happened at the same time.

The product of events A and B is the event C, which consists in the fact that both the event A and the event B happened.

Events A and B are said to be incompatible if they cannot happen at the same time.

An event A is said to be impossible if it cannot happen. Such an event is denoted by the symbol .

An event A is called certain if it will definitely occur. Such an event is denoted by the symbol .

Let each event A be assigned a number P(A). This number P(A) is called the probability of the event A if the following conditions are satisfied with such a correspondence.

An important particular case is the situation when there are equally probable elementary outcomes, and arbitrary of these outcomes form events A. In this case, the probability can be introduced by the formula . The probability introduced in this way is called classical probability. It can be proved that properties 1-4 hold in this case.

Problems in the theory of probability, which are found on the exam in mathematics, are mainly related to classical probability. Such tasks can be very simple. Particularly simple are problems in probability theory in demo versions. It is easy to calculate the number of favorable outcomes, the number of all outcomes is written directly in the condition.

We get the answer according to the formula.

An example of a task from the exam in mathematics to determine the probability

There are 20 pies on the table - 5 with cabbage, 7 with apples and 8 with rice. Marina wants to take a pie. What is the probability that she will take the rice cake?

Decision.

There are 20 equiprobable elementary outcomes in total, that is, Marina can take any of the 20 pies. But we need to estimate the probability that Marina will take the rice patty, that is, where A is the choice of the rice patty. This means that we have a total of 8 favorable outcomes (choosing rice pies). Then the probability will be determined by the formula:

Independent, Opposite, and Arbitrary Events

However, in the open bank of tasks, more than difficult tasks. Therefore, let us draw the reader's attention to other questions studied in probability theory.

Events A and B are called independent if the probability of each of them does not depend on whether the other event occurred.

Event B consists in the fact that event A did not occur, i.e. event B is opposite to event A. The probability of the opposite event is equal to one minus the probability of the direct event, i.e. .

Addition and multiplication theorems, formulas

For arbitrary events A and B, the probability of the sum of these events is equal to the sum of their probabilities without the probability of their joint event, i.e. .

For independent events A and B, the probability of the product of these events is equal to the product of their probabilities, i.e. in this case .

The last 2 statements are called the theorems of addition and multiplication of probabilities.

Not always counting the number of outcomes is so simple. In some cases, it is necessary to use combinatorics formulas. The most important thing is to count the number of events that meet certain conditions. Sometimes such calculations can become independent tasks.

In how many ways can 6 students be seated in 6 empty seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways to place the second student. For the third student there are 4 free places, for the fourth - 3, for the fifth - 2, the sixth will take the only remaining place. To find the number of all options, you need to find the product, which is denoted by the symbol 6! and read "six factorial".

In the general case, the answer to this question is given by the formula for the number of permutations of n elements. In our case, .

Consider now another case with our students. In how many ways can 2 students be seated in 6 empty seats? The first student will take any of the 6 places. Each of these options corresponds to 5 ways to place the second student. To find the number of all options, you need to find the product.

In the general case, the answer to this question is given by the formula for the number of placements of n elements by k elements

In our case .

And the last one in this series. How many ways are there to choose 3 students out of 6? The first student can be chosen in 6 ways, the second in 5 ways, and the third in 4 ways. But among these options, the same three students occur 6 times. To find the number of all options, you need to calculate the value: . In the general case, the answer to this question is given by the formula for the number of combinations of elements by elements:

In our case .

Examples of solving problems from the exam in mathematics to determine the probability

Task 1. From the collection, ed. Yashchenko.

There are 30 pies on a plate: 3 with meat, 18 with cabbage and 9 with cherries. Sasha randomly chooses one pie. Find the probability that he ends up with a cherry.

.

Answer: 0.3.

Problem 2. From the collection, ed. Yashchenko.

In each batch of 1000 light bulbs, an average of 20 defective ones. Find the probability that a light bulb chosen at random from a batch is good.

Solution: The number of serviceable light bulbs is 1000-20=980. Then the probability that a light bulb taken at random from the batch will be serviceable is:

Answer: 0.98.

The probability that student U. correctly solves more than 9 problems on a math test is 0.67. The probability that U. correctly solves more than 8 problems is 0.73. Find the probability that U. correctly solves exactly 9 problems.

If we imagine a number line and mark points 8 and 9 on it, then we will see that the condition "U. correctly solve exactly 9 problems” is included in the condition “U. correctly solve more than 8 problems", but does not apply to the condition "W. correctly solve more than 9 problems.

However, the condition "U. correctly solve more than 9 problems" is contained in the condition "U. correctly solve more than 8 problems. Thus, if we designate events: “W. correctly solve exactly 9 problems" - through A, "U. correctly solve more than 8 problems" - through B, "U. correctly solve more than 9 problems ”through C. Then the solution will look like this:

Answer: 0.06.

In the geometry exam, the student answers one question from the list of exam questions. The probability that this is a trigonometry question is 0.2. The probability that this is an Outer Corners 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.

Let's think about what events we have. We are given two incompatible events. That is, either the question will relate to the topic "Trigonometry", or to the topic "External angles". According to the probability theorem, the probability of incompatible events is equal to the sum of the probabilities of each event, we must find the sum of the probabilities of these events, that is:

Answer: 0.35.

The room is illuminated by a lantern with three lamps. The probability of one lamp burning out in a year is 0.29. Find the probability that at least one lamp does not burn out within a year.

Let's consider possible events. We have three light bulbs, each of which may or may not burn out independently of any other light bulb. These are independent events.

Then we will indicate the variants of such events. We accept the notation: - the light bulb is on, - the light bulb is burned out. And immediately next we calculate the probability of an event. For example, the probability of an event in which three independent events “light bulb burned out”, “light bulb on”, “light bulb on” occurred: where the probability of the event “light bulb on” is calculated as the probability of an event opposite to the event “light bulb off”, namely .

Note that there are only 7 incompatible events favorable to us. The probability of such events is equal to the sum of the probabilities of each of the events: .

Answer: 0.975608.

You can see another problem in the picture:

Thus, you and I understood what the theory of probability is, formulas and examples of problem solving for which you can meet in the version of the exam.

This presentation presents the most frequently encountered tasks on the probability theory exam. Basic level tasks. The presentation will help both teachers in the lessons of generalizing repetition, and students in self-training to the exam.

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PROBABILITY THEORY KEY TASKS Getting ready for the OGE

COIN THROW

1. A coin is tossed twice. What is the probability of getting one head and one tail? Decision: When tossing one coin, two outcomes are possible - “heads” or “tails”. When throwing two coins - 4 outcomes (2 * 2 \u003d 4): “eagle” - “tails” “tails” - “tails” “tails” - “eagles” “eagles” - “eagles” One “eagle” and one “ tails" will fall out in two cases out of four. P(A)=2:4=0.5. Answer: 0.5.

2. A coin is tossed three times. What is the probability of getting two heads and one tail? Solution: When thrown three coins 8 outcomes are possible (2*2*2=8): "eagle" - "tails" - "tails" "tails" - "tails" - "tails" "tails" - "heads" - "tails" "heads" - "eagle" - "tails" "tails" - "tails" - "heads" "tails" - "eagles" - "eagles" "eagles" - "tails" - "eagles" "eagles" - "eagles" - "eagles" » Two "eagles" and one "tails" will fall out in three cases out of eight. P(A)=3:8=0.375. Answer: 0.375.

3. In a random experiment, a symmetrical coin is tossed four times. Find the probability that heads will never come up. Solution: When throwing four coins, 16 outcomes are possible: (2*2*2*2=16): Favorable outcomes - 1 (four tails will fall out). P(A)=1:16=0.0625. Answer: 0.0625.

GAME OF DICE

4. Determine the probability that more than three points fell out when the die was rolled. Solution: There are 6 possible outcomes in total. Big numbers are 3 - 4, 5, 6. P(A)=3:6=0.5. Answer: 0.5.

5. A die is thrown. Find the probability of getting an even number of points. Solution: Total possible outcomes - 6. 1, 3, 5 - odd numbers; 2, 4, 6 are even numbers. The probability of getting an even number of points is 3:6=0.5. Answer: 0.5.

6. In a random experiment, two dice are thrown. Find the probability of getting 8 points in total. Round the result to the nearest hundredth. Solution: This action - throwing two dice has a total of 36 possible outcomes, since 6² = 36. Favorable outcomes: 2 6 3 5 4 4 5 3 6 2 The probability of getting eight points is 5:36 ≈ 0.14. Answer: 0.14.

7. Throw a dice twice. In total, 6 points fell out. Find the probability of getting 5 on one of the rolls. Decision: Total outcomes of 6 points - 5: 2 and 4; 4 and 2; 3 and 3; 1 and 5; 5 and 1. Favorable outcomes - 2. P(A)=2:5=0.4. Answer: 0.4.

8. There were 50 tickets in the exam, Timofey did not learn 5 of them. Find the probability that he will get the learned ticket. Solution: Timofey learned 45 tickets. P(A)=45:50=0.9. Answer: 0.9.

COMPETITIONS

9. 20 athletes participate in the gymnastics championship: 8 from Russia, 7 from the USA, the rest from China. The order of performance is determined by lot. Find the probability that the athlete who competes first is from China. Solution: Total outcomes 20. Favorable outcomes 20-(8+7)=5. P(A)=5:20=0.25. Answer: 0.25.

10. 4 athletes from France, 5 from England and 3 from Italy came to the shot throwing competition. The order of performances is determined by a draw. Find the probability that the fifth athlete is from Italy. Solution: The number of all possible outcomes is 12 (4 + 5 + 3 = 12). The number of favorable outcomes is 3. P(A)=3:12=0.25. Answer: 0.25.

11. Before the start of the first round of the badminton championship, the participants are randomly divided into game pairs by drawing lots. In total, 26 badminton players participate in the championship, including 12 participants from Russia, including Vladimir Orlov. Find the probability that in the first round Vladimir Orlov will play with any badminton player from Russia? Decision: Total outcomes - 25 (Vladimir Orlov with 25 badminton players). Favorable outcomes - (12-1) = 11. P(A)=11:25=0.44. Answer: 0.44.

12. The competition of performers is held in 5 days. A total of 75 performances were announced - one from each country. There are 27 performances on the first day, the rest are distributed equally among the remaining days. The order of performances is determined by a draw. What is the probability that the performance of the representative of Russia will take place on the third day of the competition? Decision: Total outcomes - 75. Performers from Russia perform on the third day. Favorable outcomes - (75-27): 4 = 12. P(A)=12: 75=0.16. Answer: 0.16.

13. Kolya chooses a two-digit number. Find the probability that it is divisible by 5. Solution: Two-digit numbers: 10;11;12;…;99. Total outcomes - 90. Numbers divisible by 5: 10; fifteen; 20; 25; …; 90; 95. Favorable outcomes - 18. P(A)=18:90=0.2. Answer: 0.2.

DIFFERENT TASKS FOR DETERMINING PROBABILITY

14. The factory produces bags. On average, for every 170 quality bags, there are six bags with hidden defects. Find the probability that the purchased bag will be of high quality. Round the result to the nearest hundredth. Solution: Total outcomes - 176. Favorable outcomes - 170. Р(А)=170:176 ≈ 0.97. Answer: 0.97.

15. On average, out of every 100 batteries sold, 94 batteries are charged. Find the probability that the purchased battery is not charged. Solution: Total outcomes - 100. Favorable outcomes - 100-94=6. P(A)=6:100=0.06. Answer: 0.06.

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