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14.10.2019

Problems on the classical determination of probability. Simple problems in probability theory. Basic formula

Events that happen in reality or in our imagination can be divided into 3 groups. These are certain events that will definitely happen, impossible events and random events. Probability theory studies random events, i.e. events that may or may not happen. This article will briefly present the theory of probability formulas and examples of solving problems in probability theory, which will be in task 4 of the Unified State Exam in mathematics (profile level).

Why do we need probability theory?

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

Playing cards, dice, and roulette created situations where any of a finite number of equally possible events could occur. There was a need to give numerical estimates of the possibility of the occurrence of a particular 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. The modern theory of probability was created.

Basic concepts of probability theory

The object of study of probability theory is events and their probabilities. If an 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 occurred simultaneously.

The product of events A and B is an event C, which means that both event A and event B occurred.

Events A and B are called incompatible if they cannot occur simultaneously.

An event A is called impossible if it cannot happen. Such an event is indicated by the symbol.

An event A is called certain if it is sure to happen. Such an event is indicated by the symbol.

Let each event A be associated with a number P(A). This number P(A) is called the probability of event A if the following conditions are met with this correspondence.

An important special 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 entered using the formula. Probability introduced in this way is called classical probability. It can be proven that in this case properties 1-4 are satisfied.

Probability theory problems that appear on the Unified State Examination in mathematics are mainly related to classical probability. Such tasks can be very simple. The probability theory problems in the demonstration versions are especially simple. It is easy to calculate the number of favorable outcomes; the number of all outcomes is written right in the condition.

We get the answer using the formula.

An example of a problem from the Unified State Examination in mathematics on determining probability

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

Solution.

There are 20 equally probable elementary outcomes, that is, Marina can take any of the 20 pies. But we need to estimate the probability that Marina will take the rice pie, that is, where A is the choice of the rice pie. This means that the number of favorable outcomes (choices of pies with rice) is only 8. Then the probability will be determined by the formula:

Independent, Opposite and Arbitrary Events

However, more complex tasks began to be found in the open task bank. Therefore, let us draw the reader’s attention to other issues studied in probability theory.

Events A and B are said to be independent if the probability of each does not depend on whether the other event occurs.

Event B is that event A did not happen, 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. .

Probability 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 occurrence 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.

Counting the number of outcomes is not always so simple. In some cases it is necessary to use combinatorics formulas. The most important thing is to count the number of events that satisfy certain conditions. Sometimes these kinds of 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 for the second student to take a place. There are 4 free places left for the third student, 3 for the fourth, 2 for the fifth, and 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 reads "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.

Let us now consider 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 for the second student to take a place. To find the number of all options, you need to find the product.

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

In our case .

And the last case in this series. In how many ways can you choose three students out of 6? The first student can be selected in 6 ways, the second - in 5 ways, the third - in four ways. But among these options, the same three students appear 6 times. To find the number of all options, you need to calculate the value: . In general, the answer to this question is given by the formula for the number of combinations of elements by element:

In our case .

Examples of solving problems from the Unified State Exam in mathematics to determine probability

Task 1. From the collection edited by. Yashchenko.

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

.

Answer: 0.3.

Task 2. From the collection edited by. Yashchenko.

In each batch of 1000 light bulbs, on average, 20 are defective. Find the probability that a light bulb taken at random from a batch will be working.

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

Answer: 0.98.

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

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

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

Answer: 0.06.

In a geometry exam, a student answers one question from a list of exam questions. The probability that this is a Trigonometry question is 0.2. The probability that this is a question on External Angles 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.

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 within a year is 0.29. Find the probability that at least one lamp will not burn out during the 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 options for such events. Let's use the following notations: - the light bulb is on, - the light bulb is burnt out. And immediately next we calculate the probability of the event. For example, the probability of an event in which three independent events “the light bulb is burned out”, “the light bulb is on”, “the light bulb is on” occurred: , where the probability of the event “the light bulb is on” is calculated as the probability of the event opposite to the event “the light bulb is not on”, namely: .

At When assessing the probability of the occurrence of any random event, it is very important to have a good understanding of whether the probability () of the occurrence of the event we are interested in depends on how other events develop.

In the case of the classical scheme, when all outcomes are equally probable, we can already estimate the probability values ​​of the individual event of interest to us independently. We can do this even if the event is a complex collection of several elementary outcomes. What if several random events occur simultaneously or sequentially? How does this affect the likelihood of the event we are interested in happening?

If I roll a die several times and want a six to come up, and I keep getting unlucky, does that mean I should increase my bet because, according to probability theory, I'm about to get lucky? Alas, probability theory does not state anything like this. No dice, no cards, no coins can't remember what they showed us last time. It doesn’t matter to them at all whether it’s the first time or the tenth time I’m testing my luck today. Every time I repeat the roll, I know only one thing: and this time the probability of getting a six is ​​again one sixth. Of course, this does not mean that the number I need will never come up. This only means that my loss after the first throw and after any other throw are independent events.

Events A and B are called independent, if the implementation of one of them does not in any way affect the probability of another event. For example, the probabilities of hitting a target with the first of two weapons do not depend on whether the target was hit by the other weapon, so the events “the first weapon hit the target” and “the second weapon hit the target” are independent.

If two events A and B are independent, and the probability of each of them is known, then the probability of the simultaneous occurrence of both event A and event B (denoted AB) can be calculated using the following theorem.

Probability multiplication theorem for independent events

P(AB) = P(A)*P(B)- probability simultaneous the onset of two independent events is equal to work the probabilities of these events.

Example.The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 =0.7; p 2 =0.8. Find the probability of a hit with one salvo by both guns simultaneously.

Solution: as we have already seen, events A (hit by the first gun) and B (hit by the second gun) are independent, i.e. P(AB)=P(A)*P(B)=p 1 *p 2 =0.56.


What happens to our estimates if the initial events are not independent? Let's change the previous example a little.

Example.Two shooters shoot at targets at a competition, and if one of them shoots accurately, the opponent begins to get nervous and his results worsen. How to turn this everyday situation into a mathematical problem and outline ways to solve it? It is intuitively clear that it is necessary to somehow separate the two options for the development of events, to essentially create two scenarios, two different tasks. In the first case, if the opponent missed, the scenario will be favorable for the nervous athlete and his accuracy will be higher. In the second case, if the opponent has taken his chance decently, the probability of hitting the target for the second athlete decreases.


To separate possible scenarios (often called hypotheses) for the development of events, we will often use a “probability tree” diagram. This diagram is similar in meaning to the decision tree that you have probably already dealt with. Each branch represents a separate scenario for the development of events, only now it has its own meaning of the so-called conditional probabilities (q 1, q 2, q 1 -1, q 2 -1).


This scheme is very convenient for analyzing sequential random events.

It remains to clarify one more important question: where do the initial values ​​of the probabilities come from? real situations ? After all, probability theory doesn’t work with just coins and dice? Usually these estimates are taken from statistics, and when statistical information is not available, we conduct our own research. And we often have to start it not with collecting data, but with the question of what information we actually need.

Example.Let's say we need to estimate in a city with a population of one hundred thousand inhabitants the market volume for a new product that is not an essential item, for example, for a balm for the care of colored hair. Let's consider the "probability tree" diagram. In this case, we need to approximately estimate the probability value on each “branch”. So, our estimates of market capacity:

1) of all city residents, 50% are women,

2) of all women, only 30% dye their hair often,

3) of them, only 10% use balms for colored hair,

4) of them, only 10% can muster the courage to try a new product,

5) 70% of them usually buy everything not from us, but from our competitors.




Solution: According to the law of multiplication of probabilities, we determine the probability of the event we are interested in A = (a city resident buys this new balm from us) = 0.00045.

Let's multiply this probability value by the number of city residents. As a result, we have only 45 potential customers, and considering that one bottle of this product lasts for several months, the trade is not very lively.

And yet there is some benefit from our assessments.

Firstly, we can compare forecasts of different business ideas; they will have different “forks” in the diagrams, and, of course, the probability values ​​will also be different.

Secondly, as we have already said, a random variable is not called random because it does not depend on anything at all. Just her exact the meaning is not known in advance. We know that the average number of buyers can be increased (for example, by advertising a new product). So it makes sense to focus our efforts on those “forks” where the probability distribution does not suit us particularly, on those factors that we are able to influence.

Let's look at another quantitative example of consumer behavior research.

Example. On average, 10,000 people visit the food market per day. The probability that a market visitor enters the dairy products pavilion is 1/2. It is known that this pavilion sells an average of 500 kg of various products per day.

Can we say that the average purchase in the pavilion weighs only 100 g?

Discussion. Of course not. It is clear that not everyone who entered the pavilion ended up buying something there.




As shown in the diagram, to answer the question about the average weight of a purchase, we must find an answer to the question, what is the probability that a person entering the pavilion will buy something there. If we do not have such data at our disposal, but we need it, we will have to obtain it ourselves by observing the visitors to the pavilion for some time. Let’s say our observations showed that only a fifth of pavilion visitors buy something.

Once we have obtained these estimates, the task becomes simple. Out of 10,000 people who come to the market, 5,000 will go to the dairy products pavilion; there will be only 1,000 purchases. The average purchase weight is 500 grams. It is interesting to note that in order to build a complete picture of what is happening, the logic of conditional “branching” must be defined at each stage of our reasoning as clearly as if we were working with a “specific” situation, and not with probabilities.

Self-test tasks

1. Let there be an electrical circuit consisting of n elements connected in series, each of which operates independently of the others.




The probability p of failure of each element is known. Determine the probability of proper operation of the entire section of the circuit (event A).

2. The student knows 20 out of 25 exam questions. Find the probability that the student knows the three questions given to him by the examiner.

3. Production consists of four successive stages, at each of which equipment operates, for which the probabilities of failure over the next month are equal to p 1, p 2, p 3 and p 4, respectively. Find the probability that there will be no production stoppages due to equipment failure in a month.

Knowing that probability can be measured, let's try to express it in numbers. There are three possible ways.

Rice. 1.1. Measuring Probability

PROBABILITY DETERMINED BY SYMMETRY

There are situations in which possible outcomes are equally probable. For example, when tossing a coin once, if the coin is standard, the probability of “heads” or “tails” appearing is the same, i.e. P("heads") = P("tails"). Since only two outcomes are possible, then P(“heads”) + P(“tails”) = 1, therefore, P(“heads”) = P(“tails”) = 0.5.

In experiments where outcomes have equal chances of occurrence, the probability of event E, P (E) is equal to:

Example 1.1. The coin is tossed three times. What is the probability of two heads and one tail?

First, let's find all possible outcomes: To make sure that we have found all possible options, we will use a tree diagram (see Chapter 1, Section 1.3.1).

So, there are 8 equally possible outcomes, therefore, the probability of them is 1/8. Event E - two heads and tails - three occurred. That's why:

Example 1.2. A standard die is rolled twice. What is the probability that the score is 9 or more?

Let's find all possible outcomes.

Table 1.2. The total number of points obtained by rolling a die twice

So, in 10 out of 36 possible outcomes the sum of points is 9 or therefore:

EMPIRICALLY DETERMINED PROBABILITY

Example with a coin from the table. 1.1 clearly illustrates the mechanism for determining probability.

Given the total number of experiments that are successful, the probability of the required result is calculated as follows:

A ratio is the relative frequency of occurrence of a certain result over a sufficiently long experiment. The probability is calculated either based on the data of the experiment performed, based on past data.

Example 1.3. Of the five hundred electric lamps tested, 415 worked for more than 1000 hours. Based on the data from this experiment, we can conclude that the probability of normal operation of a lamp of this type for more than 1000 hours is:

Note. Testing is destructive in nature, so not all lamps can be tested. If only one lamp were tested, the probability would be 1 or 0 (i.e. whether it can last 1000 hours or not). Hence the need to repeat the experiment.

Example 1.4. In table 1.3 shows data on the length of service of men working in the company:

Table 1.3. Men's work experience

What is the probability that the next person hired by the company will work for at least two years:

Solution.

The table shows that 38 out of 100 employees have been working in the company for more than two years. The empirical probability that the next employee will remain with the company for more than two years is:

At the same time, we assume that the new employee is “typical and the working conditions are unchanged.

SUBJECTIVE PROBABILITY ASSESSMENT

In business, situations often arise in which there is no symmetry, and there is no experimental data either. Therefore, determining the likelihood of a favorable outcome under the influence of the views and experience of the researcher is subjective.

Example 1.5.

1. An investment expert estimates that the probability of making a profit in the first two years is 0.6.

2. Marketing manager's forecast: the probability of selling 1000 units of a product in the first month after its appearance on the market is 0.4.

Let's talk about problems in which the phrase “at least one” appears. Surely you have encountered such problems in homework and tests, and now you will learn how to solve them. First, I will talk about the general rule, and then we will look at a special case and write down formulas and examples for each.

General methodology and examples

General technique to solve problems in which the phrase “at least one” occurs is as follows:

  • Write down the initial event $A$ = (Probability that... at least...).
  • Formulate opposite event $\bar(A)$.
  • Find the probability of the event $P(\bar(A))$.
  • Find the desired probability using the formula $P(A)=1-P(\bar(A))$.

    Now let's look at it with examples. Forward!

    Example 1. The box contains 25 standard and 6 defective parts of the same type. What is the probability that among three randomly selected parts, at least one will be defective?

    We act directly point by point.
    1. We write down an event whose probability must be found directly from the problem statement:
    $A$ =(From 3 selected parts at least one defective).

    2. Then the opposite event is formulated as follows: $\bar(A)$ = (From 3 selected details none defective) = (All 3 selected parts will be standard).

    3. Now we need to understand how to find the probability of the event $\bar(A)$, for which we will look at the problem again: we are talking about objects of two types (defective parts and not), from which a certain number of objects are taken out and studied (defective or not). This problem is solved using the classical definition of probability (more precisely, using the hypergeometric probability formula, read more about it in the article).

    For the first example, we will write down the solution in detail, then we will abbreviate it (and you will find complete instructions and calculators at the link above).

    First, let's find the total number of outcomes - this is the number of ways to choose any 3 parts from a batch of 25+6=31 parts in a box. Since the order of choice is unimportant, we apply the formula for the number of combinations of 31 objects of 3: $n=C_(31)^3$.

    Now let's move on to the number of outcomes favorable to the event. To do this, all 3 selected parts must be standard; they can be selected in $m = C_(25)^3$ ways (since there are exactly 25 standard parts in the box).

    The probability is:

    $$ P(\bar(A))=\frac(m)(n)=\frac(C_(25)^3 )(C_(31)^3) = \frac(23 \cdot 24\cdot 25) (29\cdot 30\cdot 31) =\frac(2300)(4495)= 0.512. $$

    4. Then the desired probability:

    $$ P(A)=1-P(\bar(A))=1- 0.512 = 0.488. $$

    Answer: 0.488.


    Example 2. From a deck of 36 cards, 6 cards are taken at random. Find the probability that among the cards taken there will be at least two spades.

    1. We record the event $A$ =(Of the 6 selected cards there will be at least two peaks).

    2. Then the opposite event is formulated as follows: $\bar(A)$ = (Out of 6 selected cards there will be less than 2 spades) = (Out of 6 selected cards there will be exactly 0 or 1 spades, the rest of a different suit).

    Comment. Here I will stop and make a small remark. Although in 90% of cases the "go to the opposite event" technique works perfectly, there are cases when it is easier to find the probability of the original event. In this case, if you directly look for the probability of event $A$, you will need to add up 5 probabilities, and for event $\bar(A)$ - only 2 probabilities. But if the problem was “out of 6 cards at least 5 are peaks,” the situation would be reversed and it would be easier to solve the original problem. If I try to give instructions again, I’ll say this. In tasks where you see “at least one”, feel free to move on to the opposite event. If we are talking about “at least 2, at least 4, etc.”, then you need to figure out what is easier to count.

    3. We return to our problem and find the probability of the event $\bar(A)$ using the classical definition of probability.

    The total number of outcomes (ways to choose any 6 cards out of 36) is $n=C_(36)^6$ (calculator).

    Let's find the number of outcomes favorable to the event. $m_0 = C_(27)^6$ - the number of ways to select all 6 cards of a non-peak suit (there are 36-9=27 in the deck), $m_1 = C_(9)^1\cdot C_(27)^5$ - number ways to choose 1 card of the spades suit (out of 9) and 5 other suits (out of 27).

    $$ P(\bar(A))=\frac(m_0+m_1)(n)=\frac(C_(27)^6+C_(9)^1\cdot C_(27)^5 )(C_( 36)^6) =\frac(85215)(162316)= 0.525. $$

    4. Then the desired probability:

    $$ P(A)=1-P(\bar(A))=1- 0.525 = 0.475. $$

    Answer: 0.475.


    Example 3. There are 2 white, 3 black and 5 red balls in the urn. Three balls are drawn at random. Find the probability that among the drawn balls at least two will be of different colors.

    1. We record the event $A$ =(Among 3 drawn balls at least two different color). That is, for example, “2 red balls and 1 white”, or “1 white, 1 black, 1 red”, or “2 black, 1 red” and so on, there are a lot of options. Let's try the rule of transition to the opposite event.

    2. Then the opposite event is formulated as follows: $\bar(A)$ = (All three balls are the same color) = (3 black balls or 3 red balls are chosen) - there are only 2 options, which means this method of solution simplifies the calculations. By the way, all white balls cannot be selected, since there are only 2 of them, and 3 balls are taken out.

    3. The total number of outcomes (ways to choose any 3 balls from 2+3+5=10 balls) is $n=C_(10)^3=120$.

    Let's find the number of outcomes favorable to the event. $m = C_(3)^3+C_(5)^3=1+10=11$ - the number of ways to choose either 3 black balls (out of 3) or 3 red balls (out of 5).

    $$ P(\bar(A))=\frac(m)(n)=\frac(11)(120). $$

    4. Required probability:

    $$ P(A)=1-P(\bar(A))=1- \frac(11)(120)=\frac(109)(120) = 0.908. $$

    Answer: 0.908.

    A special case. Independent events

    Let's go further and come to a class of problems where several independent events are considered (arrows hit, light bulbs burn out, cars start, workers get sick with different probabilities, etc.) and we need "find the probability of at least one event occurring". In variations, this may sound like this: “find the probability that at least one shooter out of three will hit the target”, “find the probability that at least one out of two buses arrive at the station on time”, “find the probability that at least one element in a device consisting of four elements will fail within a year,” etc.

    If in the examples above we were talking about the use of the classical probability formula, here we come to the algebra of events, we use the formulas for adding and multiplying probabilities (a small theory).

    So, several independent events $A_1, A_2,...,A_n$ are considered, the probabilities of each occurrence are known and equal to $P(A_i)=p_i$ ($q_i=1-p_i$). Then the probability that at least one of the events will occur as a result of the experiment is calculated by the formula

    $$ P=1-q_1\cdot q_2 \cdot ...\cdot q_n. \quad(1) $$

    Strictly speaking, this formula is also obtained by applying the basic technique "go to the opposite event". Indeed, let $A$=(At least one event from $A_1, A_2,...,A_n$ will occur), then $\bar(A)$ = (None of the events will occur), which means:

    $$ P(\bar(A))=P(\bar(A_1) \cdot \bar(A_2) \cdot ... \bar(A_n))=P(\bar(A_1)) \cdot P(\ bar(A_2)) \cdot ... P(\bar(A_n))=\\ =(1-P(A_1)) \cdot (1-P(A_2)) \cdot ... (1-P( A_n))=\\ =(1-p_1) \cdot (1-p_2) \cdot ... (1-p_n)=q_1\cdot q_2 \cdot ...\cdot q_n,\\ $$ from where we get our formula $$ P(A)=1-P(\bar(A))=1-q_1\cdot q_2 \cdot ...\cdot q_n. $$

    Example 4. The unit contains two independently operating parts. The probabilities of failure of parts are 0.05 and 0.08, respectively. Find the probability of a unit failure if it is enough for at least one part to fail.

    Event $A$ =(Node has failed) = (At least one of the two parts has failed). Let's introduce independent events: $A_1$ = (The first part failed) and $A_2$ = (The second part failed). By condition $p_1=P(A_1)=0.05$, $p_2=P(A_2)=0.08$, then $q_1=1-p_1=0.95$, $q_2=1-p_2=0, $92. Let's apply formula (1) and get:

    $$ P(A)=1-q_1\cdot q_2 = 1-0.95\cdot 0.92=0.126. $$

    Answer: 0,126.

    Example 5. The student looks for the formula he needs in three reference books. The probability that the formula is contained in the first directory is 0.8, in the second - 0.7, in the third - 0.6. Find the probability that the formula is contained in at least one reference book.

    We proceed in the same way. Consider the main event
    $A$ =(The formula is contained in at least one reference book). Let's introduce independent events:
    $A_1$ = (The formula is in the first reference book),
    $A_2$ = (The formula is in the second reference book),
    $A_3$ = (The formula is in the third reference book).

    By condition $p_1=P(A_1)=0.8$, $p_2=P(A_2)=0.7$, $p_3=P(A_3)=0.6$, then $q_1=1-p_1=0 ,2$, $q_2=1-p_2=0.3$, $q_3=1-p_3=0.4$. Let's apply formula (1) and get:

    $$ P(A)=1-q_1\cdot q_2\cdot q_3 = 1-0.2\cdot 0.3\cdot 0.4=0.976. $$

    Answer: 0,976.

    Example 6. A worker maintains 4 machines that operate independently of each other. The probability that during a shift the first machine will require a worker’s attention is 0.3, the second - 0.6, the third - 0.4 and the fourth - 0.25. Find the probability that during a shift at least one machine will not require the attention of a foreman.

    I think you have already grasped the principle of the solution, the only question is the number of events, but this does not affect the complexity of the solution (unlike general problems involving addition and multiplication of probabilities). Just be careful, the probabilities are indicated for “will require attention,” but the question of the problem is “at least one machine will NOT require attention.” You need to enter events that are the same as the main one (in this case, with NOT) in order to use the general formula (1).

    We get:
    $A$ = (During the shift at least one machine will NOT require the attention of a foreman),
    $A_i$ = ($i$-th machine will NOT require the attention of the master), $i=1,2,3,4$,
    $p_1 = 0.7$, $p_2 = 0.4$, $p_3 = 0.6$, $p_4 = 0.75$.

    Required probability:

    $$ P(A)=1-q_1\cdot q_2\cdot q_3 \cdot q_4= 1-(1-0.7)\cdot (1-0.4)\cdot (1-0.6)\cdot ( 1-0.75)=0.982. $$

    Answer: 0.982. Almost certainly the master will rest for the entire shift;)

    A special case. Repeated tests

    So, we have $n$ independent events (or repetitions of some experience), and the probabilities of the occurrence of these events (or the occurrence of an event in each of the experiments) are now the same and are equal to $p$. Then formula (1) simplifies to the form:

    $$ P=1-q_1\cdot q_2 \cdot ...\cdot q_n = 1-q^n. $$

    In fact, we are narrowing down to a class of problems called “repeated independent trials” or “Bernoulli scheme”, where $n$ experiments are carried out, the probability of an event occurring in each of them is equal to $p$. We need to find the probability that the event will occur at least once out of $n$ repetitions:

    $$ P=1-q^n. \quad(2) $$

    You can read more about Bernoulli's scheme in the online textbook, and also look at calculator articles about solving various subtypes of problems (about shots, lottery tickets, etc.). Below, only problems with “at least one” will be discussed.

    Example 7. Let the probability that the TV will not require repairs during the warranty period be equal to 0.9. Find the probability that during the warranty period at least one of 3 TVs will not require repair.

    In short, you haven't seen the solution yet.
    We simply write out from the condition: $n=3$, $p=0.9$, $q=1-p=0.1$.
    Then the probability that during the warranty period at least one of 3 TVs will not require repair, according to formula (2):

    $$ P=1-0.1^3=1-0.001=0.999 $$

    Answer: 0,999.

    Example 8. 5 independent shots are fired at a certain target. The probability of a hit with one shot is 0.8. Find the probability that there will be at least one hit.

    Again, we start by formalizing the problem, writing out known quantities. $n=5$ shots, $p=0.8$ - hit probability with one shot, $q=1-p=0.2$.
    And then the probability that there will be at least one hit out of five shots is equal to: $$ P=1-0.2^5=1-0.00032=0.99968 $$

    Answer: 0,99968.

    I think that using formula (2) everything is more than clear (do not forget to read about other problems solved within the framework of Bernoulli’s scheme, the links were above). And below I will give a slightly more complex problem. Such problems occur less frequently, but the method of solving them must also be learned. Go!

    Example 9. N independent experiments are performed, in each of which some event A appears with probability 0.7. How many experiments need to be done to guarantee at least one occurrence of event A with a probability of 0.95?

    We have a Bernoulli scheme, $n$ is the number of experiments, $p=0.7$ is the probability of occurrence of event A.

    Then the probability that at least one event A will occur in $n$ experiments is equal to formula (2): $$ P=1-q^n=1-(1-0.7)^n=1-0, 3^n $$ According to the condition, this probability must be no less than 0.95, therefore:

    $$ 1-0.3^n \ge 0.95,\\ 0.3^n \le 0.05,\\ n \ge \log_(0.3) 0.05 = 2.49. $$

    Rounding up, we get that you need to conduct at least 3 experiments.

    Answer: You need to do a minimum of 3 experiments.

    • If you are interested in the title question, you are probably a student or schoolboy who is faced with a new subject. Probability theory problems are now being solved by fifth-grade students at advanced schools, high school students before the Unified State Exam, and students of literally all specialties - from geographers to mathematicians. What kind of object is this, and how to approach it?

    Probability. What is this?

    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 which 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 probabilities these random events.

    Thus, we are faced with the basic concept random event- a phenomenon whose behavior cannot be predicted, an experiment whose result cannot be calculated in advance, etc. It is the probabilities of events that are calculated in typical problems.

    Probability- this is some, strictly speaking, function that takes values ​​from 0 to 1 and characterizes a given random event. 0 - the event is practically impossible, 1 - the event is almost certain, 0.5 (or "50 to 50") - with equal probability the event will occur or not.

    Algorithm for solving probability problems

    You can learn more about the basics of probability theory, for example, in the online textbook.

    Now let's not beat around the bush, and let's formulate diagram, according to which you should solve standard educational problems on calculating the probability of a random event, and then below Let's illustrate with examples its application.

    • Carefully read the task and understand what exactly is happening (what is being pulled out of which box, what was lying where, how many devices are working, etc.)
    • Find the main question of the problem like “calculate the probability that...” and write this ellipsis in the form of an event, the probability of which must be found.
    • The event is recorded. Now we need to understand which “scheme” of probability theory the problem belongs to in order to correctly select formulas for the solution. Answer test questions like:
      • there is one test (for example, throwing two dice) or several (for example, checking 10 devices);
      • if there are several tests, are the results of one dependent on the others (dependence or independence of events);
      • an event occurs in a single situation or the problem speaks of several possible hypotheses (for example, a ball is drawn from any of the three boxes, or from a specific one).
      The more experience you have in solving problems, the easier it will be to determine which formulas are appropriate.
    • A formula (or several) has been selected for the solution. We write down all the task data and substitute it into this formula.
    • Voila, the probability has been found.

    How to Solve Problems: Classical Probability

    Example 1. In a group of 30 students, on the test, 6 students received a “5”, 10 students received a “4”, 9 students received a “3”, the rest received a “2”. Find the probability that 3 students called to the board received a “2” on the test.

    We begin the solution according to the points described above.

    • The problem involves selecting 3 students from a group who satisfy certain conditions.
    • Enter the main event $X$ = (All 3 students called to the board received a “2” on the test).
    • Since only one test occurs in the task and it is associated with selection/choice under a certain condition, we are talking about the classical definition of probability. Let us write the formula: $P=m/n$, where $m$ is the number of outcomes favorable to the occurrence of event $X$, and $n$ is the number of all equally possible elementary outcomes.
    • Now we need to find the values ​​of $m$ and $n$ for this problem. First, let's find the number of all possible outcomes - the number of ways to choose 3 students out of 30. Since the order of choice does not matter, this is the number of combinations of 30 by 3: $$n=C_(30)^3=\frac(30{3!27!}=\frac{28\cdot 29 \cdot 30}{1\cdot 2 \cdot 3}=4060.$$ Найдем число способов вызвать только студентов, получивших "2". Всего таких студентов было $30-6-10-9=5$ человек, поэтому $$m=C_{5}^3=\frac{5!}{3!2!}=\frac{4 \cdot 5}{1\cdot 2}=10.$$!}
    • We get the probability: $$P(X)=\frac(m)(n)=\frac(10)(4060)=0.002.$$ Problem solved.

    No time to decide? Find a solved problem

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