Many, faced with the concept of "probability theory", are frightened, thinking that this is something overwhelming, very complex. But it's really not all that tragic. Today we will consider the basic concept of probability theory, learn how to solve problems using specific examples.

The science

What does such a branch of mathematics as “probability theory” study? She notes patterns and magnitudes. For the first time, scientists became interested in this issue back in the eighteenth century, when they studied gambling. The basic concept of probability theory is an event. It is any fact that is ascertained by experience or observation. But what is experience? Another basic concept of probability theory. It means that this composition of circumstances was not created by chance, but for a specific purpose. As for observation, here the researcher himself does not participate in the experiment, but simply is a witness to these events, he does not influence what is happening in any way.

Events

We learned that the basic concept of probability theory is an event, but did not consider the classification. All of them fall into the following categories:

• Reliable.
• Impossible.
• Random.

No matter what kind of events are observed or created in the course of experience, they are all subject to this classification. We offer to get acquainted with each of the species separately.

Credible Event

This is a circumstance before which the necessary set of measures has been taken. In order to better understand the essence, it is better to give a few examples. Physics, chemistry, economics, and higher mathematics are subject to this law. Probability theory includes such an important concept as a certain event. Here are some examples:

• We work and receive remuneration in the form of wages.
• We passed the exams well, passed the competition, for this we receive a reward in the form of admission to an educational institution.
• We invested money in the bank, if necessary, we will get it back.

Such events are reliable. If we have fulfilled all the necessary conditions, then we will definitely get the expected result.

Impossible events

We now consider elements of probability theory. We propose to move on to an explanation of the next type of event, namely, the impossible. To begin with, we will stipulate the most important rule - the probability of an impossible event is zero.

It is impossible to deviate from this formulation when solving problems. To clarify, here are examples of such events:

• The water froze at a temperature of plus ten (this is impossible).
• The lack of electricity does not affect production in any way (just as impossible as in the previous example).

More examples should not be given, since the ones described above very clearly reflect the essence of this category. The impossible event will never happen during the experience under any circumstances.

random events

When studying the elements, special attention should be paid to this particular type of event. That is what science is studying. As a result of experience, something may or may not happen. In addition, the test can be repeated an unlimited number of times. Prominent examples are:

• Tossing a coin is an experience, or a test, heading is an event.
• Pulling the ball out of the bag blindly is a test, a red ball is caught is an event, and so on.

There can be an unlimited number of such examples, but, in general, the essence should be clear. To summarize and systematize the knowledge gained about events, a table is given. Probability theory studies only the last type of all presented.

Laws

Probability theory is a science that studies the possibility of an event occurring. Like the others, it has some rules. There are the following laws of probability theory:

• Convergence of sequences of random variables.
• The law of large numbers.

When calculating the possibility of the complex, a complex of simple events can be used to achieve the result in an easier and faster way. Note that the laws of probability theory are easily proved with the help of some theorems. Let's start with the first law.

Convergence of sequences of random variables

Note that there are several types of convergence:

• The sequence of random variables is convergent in probability.
• Almost impossible.
• RMS convergence.
• Distribution Convergence.

So, on the fly, it's very hard to get to the bottom of it. Here are some definitions to help you understand this topic. Let's start with the first look. The sequence is called convergent in probability, if the following condition is met: n tends to infinity, the number to which the sequence tends is greater than zero and close to one.

Let's move on to the next one, almost certainly. The sequence is said to converge almost certainly to a random variable with n tending to infinity, and P tending to a value close to unity.

The next type is RMS convergence. When using SC-convergence, the study of vector random processes is reduced to the study of their coordinate random processes.

The last type remains, let's briefly analyze it in order to proceed directly to solving problems. Distribution convergence has another name - “weak”, we will explain why below. Weak convergence is the convergence of distribution functions at all points of continuity of the limiting distribution function.

We will definitely fulfill the promise: weak convergence differs from all of the above in that the random variable is not defined on the probability space. This is possible because the condition is formed exclusively using distribution functions.

Law of Large Numbers

Excellent assistants in proving this law will be theorems of probability theory, such as:

• Chebyshev's inequality.
• Chebyshev's theorem.
• Generalized Chebyshev's theorem.
• Markov's theorem.

If we consider all these theorems, then this question can drag on for several tens of sheets. Our main task is to apply the theory of probability in practice. We invite you to do this right now. But before that, let's consider the axioms of probability theory, they will be the main assistants in solving problems.

Axioms

We already met the first one when we talked about the impossible event. Let's remember: the probability of an impossible event is zero. We gave a very vivid and memorable example: snow fell at an air temperature of thirty degrees Celsius.

The second is as follows: a certain event occurs with a probability equal to one. Now let's show how to write it down using the mathematical language: P(B)=1.

Third: A random event may or may not occur, but the possibility always ranges from zero to one. The closer the value is to one, the greater the chance; if the value approaches zero, the probability is very low. Let's write it in mathematical language: 0<Р(С)<1.

Consider the last, fourth axiom, which sounds like this: the probability of the sum of two events is equal to the sum of their probabilities. We write in mathematical language: P (A + B) \u003d P (A) + P (B).

The axioms of probability theory are the simplest rules that are easy to remember. Let's try to solve some problems, based on the knowledge already gained.

Lottery ticket

To begin with, consider the simplest example - the lottery. Imagine that you bought one lottery ticket for good luck. What is the probability that you will win at least twenty rubles? In total, a thousand tickets participate in the circulation, one of which has a prize of five hundred rubles, ten of one hundred rubles, fifty of twenty rubles, and one hundred of five. Problems in probability theory are based on finding the possibility of luck. Let's take a look at the solution to the above problem together.

If we denote by the letter A a win of five hundred rubles, then the probability of getting A will be 0.001. How did we get it? You just need to divide the number of "happy" tickets by their total number (in this case: 1/1000).

B is a win of one hundred rubles, the probability will be equal to 0.01. Now we acted on the same principle as in the previous action (10/1000)

C - the winnings are equal to twenty rubles. We find the probability, it is equal to 0.05.

The remaining tickets are of no interest to us, since their prize fund is less than that specified in the condition. Let's apply the fourth axiom: The probability of winning at least twenty rubles is P(A)+P(B)+P(C). The letter P denotes the probability of the occurrence of this event, we have already found them in the previous steps. It remains only to add the necessary data, in the answer we get 0.061. This number will be the answer to the question of the assignment.

card deck

Problems in the theory of probability are also more complex, for example, take the following task. Before you is a deck of thirty-six cards. Your task is to draw two cards in a row without mixing the pile, the first and second cards must be aces, the suit does not matter.

To begin with, we find the probability that the first card will be an ace, for this we divide four by thirty-six. They put it aside. We take out the second card, it will be an ace with a probability of three thirty-fifths. The probability of the second event depends on which card we drew first, we are interested in whether it was an ace or not. It follows that event B depends on event A.

The next step is to find the probability of simultaneous implementation, that is, we multiply A and B. Their product is found as follows: we multiply the probability of one event by the conditional probability of another, which we calculate, assuming that the first event happened, that is, we drew an ace with the first card.

In order to make everything clear, let's give a designation to such an element as events. It is calculated assuming that event A has occurred. Calculated as follows: P(B/A).

Let's continue the solution of our problem: P (A * B) \u003d P (A) * P (B / A) or P (A * B) \u003d P (B) * P (A / B). The probability is (4/36) * ((3/35)/(4/36). Calculate by rounding to hundredths. We have: 0.11 * (0.09/0.11)=0.11 * 0, 82 = 0.09 The probability that we will draw two aces in a row is nine hundredths.The value is very small, it follows that the probability of the occurrence of the event is extremely small.

Forgotten number

We propose to analyze a few more options for tasks that are studied by probability theory. You have already seen examples of solving some of them in this article, let's try to solve the following problem: the boy forgot the last digit of his friend's phone number, but since the call was very important, he began to dial everything in turn. We need to calculate the probability that he will call no more than three times. The solution of the problem is the simplest if the rules, laws and axioms of probability theory are known.

Before looking at the solution, try to solve it yourself. We know that the last digit can be from zero to nine, that is, there are ten values ​​in total. The probability of getting the right one is 1/10.

Next, we need to consider options for the origin of the event, suppose that the boy guessed right and immediately scored the right one, the probability of such an event is 1/10. The second option: the first call is a miss, and the second is on target. We calculate the probability of such an event: multiply 9/10 by 1/9, as a result we also get 1/10. The third option: the first and second calls turned out to be at the wrong address, only from the third the boy got where he wanted. We calculate the probability of such an event: we multiply 9/10 by 8/9 and by 1/8, we get 1/10 as a result. According to the condition of the problem, we are not interested in other options, so it remains for us to add up the results, as a result we have 3/10. Answer: The probability that the boy calls no more than three times is 0.3.

Cards with numbers

There are nine cards in front of you, each of which contains a number from one to nine, the numbers are not repeated. They were placed in a box and mixed thoroughly. You need to calculate the probability that

• an even number will come up;
• two-digit.

Before moving on to the solution, let's stipulate that m is the number of successful cases, and n is the total number of options. Find the probability that the number is even. It will not be difficult to calculate that there are four even numbers, this will be our m, there are nine options in total, that is, m = 9. Then the probability is 0.44 or 4/9.

We consider the second case: the number of options is nine, and there can be no successful outcomes at all, that is, m equals zero. The probability that the drawn card will contain a two-digit number is also zero.

Probability shows the possibility of an event with a certain number of repetitions. This is the number of possible outcomes with one or more outcomes divided by the total number of possible events. The probability of several events is calculated by dividing the problem into separate probabilities and then multiplying these probabilities.

Steps

Probability of a single random event

1. Choose an event with mutually exclusive results. Probability can only be calculated if the event in question either occurs or does not occur. It is impossible to get any event and the opposite result at the same time. An example of such events is rolling a 5 on a game die or winning a certain horse in a race. Five will either come up or not; a certain horse will either come first or not.

   • For example, it is impossible to calculate the probability of such an event: in one roll of the die, 5 and 6 will roll at the same time.

2. Identify all possible events and outcomes that could occur. Suppose we need to determine the probability that a three of a kind will come up when a dice with 6 numbers is rolled. "Three of a kind" is an event, and since we know that any of the 6 numbers can come up, the number of possible outcomes is six. Thus, we know that in this case there are 6 possible outcomes and one event whose probability we want to determine. Below are two more examples.

   • Example 1. In this case, the event is "selecting a day that falls on the weekend", and the number of possible outcomes is equal to the number of days of the week, that is, seven.
   • Example 2. The event is "draw the red ball", and the number of possible outcomes is equal to the total number of balls, that is, twenty.

3. Divide the number of events by the number of possible outcomes. This way you determine the probability of a single event. If we consider the case of a 3 on a die roll, the number of events is 1 (three is only on one side of the die), and the total number of outcomes is 6. The result is a ratio of 1/6, 0.166, or 16.6%. The probability of an event for the two examples above is found as follows:

   • Example 1. What is the probability that you will randomly choose a day that falls on a weekend? The number of events is 2, since there are two days off in one week, and the total number of outcomes is 7. Thus, the probability is 2/7. The result obtained can also be written as 0.285 or 28.5%.
   • Example 2. A box contains 4 blue, 5 red and 11 white balls. If you draw a random ball from the box, what is the probability that it will be red? The number of events is 5, since there are 5 red balls in the box, and the total number of outcomes is 20. Find the probability: 5/20 = 1/4. The result obtained can also be written as 0.25 or 25%.

4. Add up the probabilities of all possible events and see if the total is 1. The total probability of all possible events should be 1, or 100%. If you don't get 100%, chances are you made a mistake and missed one or more possible events. Check your calculations and make sure you account for all possible outcomes.

   • For example, the probability of rolling a 3 when rolling a die is 1/6. In this case, the probability of any other number falling out of the remaining five is also equal to 1/6. As a result, we get 1/6 + 1/6 + 1/6 + 1/6 + 1/6 + 1/6 = 6/6, that is, 100%.
   • If, for example, you forget about the number 4 on the die, adding the probabilities will give you only 5/6, or 83%, which is not equal to one and indicates an error.

5. Express the probability of an impossible outcome as 0. This means that the given event cannot happen and its probability is 0. This way you can account for impossible events.

   • For example, if you were calculating the probability that Easter falls on a Monday in 2020, you would get 0, since Easter is always celebrated on Sunday.

   Probability of several random events

   1. When considering independent events, calculate each probability separately. Once you determine what the probabilities of events are, they can be calculated separately. Let's say we want to know the probability of rolling a die twice in a row with a 5. We know that the probability of rolling one five is 1/6, and the probability of rolling the second five is also 1/6. The first outcome is not related to the second.

      • Several rolls of fives are called independent events, because what happens the first time does not affect the second event.

   2. Consider the influence of previous outcomes when calculating the probability for dependent events. If the first event affects the probability of the second outcome, it is said to calculate the probability dependent events. For example, if you choose two cards from a deck of 52 cards, after the first card is drawn, the composition of the deck changes, which affects the choice of the second card. To calculate the probability of the second of two dependent events, subtract 1 from the number of possible outcomes when calculating the probability of the second event.

      • Example 1. Consider the following event: Two cards are randomly drawn from the deck one after the other. What is the probability that both cards will have a club suit? The probability that the first card will have a club suit is 13/52, or 1/4, since there are 13 cards of the same suit in the deck.
        • After that, the probability that the second card will be a club suit is 12/51, since there is no longer one club card. This is because the first event affects the second. If you draw the 3 of clubs and don't put it back, there will be one less card in the deck (51 instead of 52).
      • Example 2. There are 4 blue, 5 red and 11 white balls in a box. If three balls are drawn at random, what is the probability that the first is red, the second is blue, and the third is white?
        • The probability that the first ball will be red is 5/20, or 1/4. The probability that the second ball will be blue is 4/19, since there is one less ball left in the box, but still 4 blue ball. Finally, the probability that the third ball is white is 11/18 since we have already drawn two balls.

   3. Multiply the probabilities of each individual event. Regardless of whether you are dealing with independent or dependent events, as well as the number of outcomes (there can be 2, 3 or even 10), you can calculate the overall probability by multiplying the probabilities of all the events in question by each other. As a result, you will get the probability of several events, the following one by one. For example, the task is Find the probability of rolling a dice twice in a row with a 5.. These are two independent events, the probability of each of which is 1/6. Thus, the probability of both events is 1/6 x 1/6 = 1/36, that is, 0.027, or 2.7%.

      • Example 1. Two cards are drawn at random from the deck, one after the other. What is the probability that both cards will have a club suit? The probability of the first event is 13/52. The probability of the second event is 12/51. We find the total probability: 13/52 x 12/51 = 12/204 = 1/17, that is, 0.058, or 5.8%.
      • Example 2. A box contains 4 blue, 5 red and 11 white balls. If three balls are randomly drawn from the box, one after the other, what is the probability that the first is red, the second is blue, and the third is white? The probability of the first event is 5/20. The probability of the second event is 4/19. The probability of the third event is 11/18. So the total probability is 5/20 x 4/19 x 11/18 = 44/1368 = 0.032, or 3.2%.

The following are the basic rules for determining the probability of a complex event occurring based on the known probabilities of its constituent simpler events.

1. Probability of certain event is equal to one:

2. Probability of association (sum) of incompatible events is equal to the sum of their probabilities:

These two equalities are axioms of the theory of probability, i.e., they are accepted as initial ones, but requiring proof of the properties of probabilities. On their basis, the whole theory of probability is built.

All other formulas given below without proofs can be derived from the accepted axioms.

3. Probability of an impossible event equals zero:

4. Probability of an event opposite event A is equal to

(4.5)

Formula (4.5) turns out to be useful in practice in cases where the calculation of the probability of the event itself A difficult, while the probability of the opposite event is easy to find (see p. 9 ).

5. Addition theorem. The probability of combining arbitrary events is equal to the sum of their probabilities minus the probability of producing events:

For incompatible events and formula (4.6) goes over into (4.3).

6. Conditional Probability. If you want to find the probability of an event IN assuming some other event has happened A, then such a situation is characterized by conditional probability. The conditional probability is equal to the ratio of the probability of the product of events A And IN to the probability of an event A:

(4.7)

In cases where events A And IN incompatible, and accordingly.

7. The definition of conditional probability in the form (4.7) makes it possible to write the following formula for calculating the probability of a product of events (probability multiplication theorem)

8. Since the probability of an event A(or IN) for independent events, by definition, does not change when another event occurs, then the conditional probability coincides with the probability of the event A, and the conditional probability is P(B). Probabilities P(A) And P(B) in contrast to conditional probabilities are called unconditional.

Probability multiplication theorem for independent events is written as follows:

i.e., the probability of the product of independent events is equal to the product of their probabilities.

9. Compute the probability of at least one event occurring in n trials

A- appearance in n trials at least once the event of interest to us.

- the event of interest to us did not appear in n trials never.

A 1 - the event of interest to us appeared in the first test.

A 2 - the event of interest to us appeared in the second test.

A n - the event of interest to us appeared in n th test.

10. Total Probability Formula.

If the event A can only occur when one of the incompatible events occurs H 1 , N 2 , …, N n, That

Example 4.3

An urn contains 5 white, 20 red and 10 black balls of the same size. The balls are thoroughly mixed and then 1 ball is taken out at random. What is the probability that the ball drawn is white or black?

Solution. Let the event A- the appearance of a white or black ball. Let's break this event into simpler ones. Let IN 1 - the appearance of a white ball, and IN 2 - black. Then, A=B 1 +V 2 P(A)=P(B 1 +V 2 ) . Because IN 1 And IN 2 are incompatible events, then according to the theorem on the probability of the sum of incompatible events (formula 4.3) P(B 1 +V 2 ) = P(B 1 )+P(B 2 ) .

Calculate the probabilities of events IN 1 And IN 2 . In this example, there are 35 equally possible (balls do not differ in size) outcomes of the experience, the event IN 1 (appearance of the white ball) is favored by 5 of them, so . Similarly,. Hence, .

Example 4.4

The search for two criminals is underway. Each of them, independently of the other, can be detected within a day with a probability of 0.5. What is the probability that at least one criminal will be found during the day?

Solution. Let the event A“At least one offender has been identified.” Let's break this event into simpler ones. Let IN 1 IN 2 The second perpetrator has been found. Then, A=B 1 +V 2 to determine the sum of events. Hence P(A)=P(B 1 +V 2 ) . Because IN 1 And IN 2 are joint events, then according to the theorem on the probability of the sum of events (formula 4.6)

P(B 1 +V 2 ) = P(B 1 )+P(B 2 )-P(B 1 IN 2 ) = 0,5+0,5 – 0,25=0,75 .

You can also solve through the reverse event: .

Example 4.5 a)

The criminal has 3 keys. In the dark, he opens the door by choosing a key at random. It takes 5 seconds to open each door. Find the probability that he will open all the doors in 15 seconds.

Solution. Let the event A“All doors are open.” Let's break this event into simpler ones. Let IN– “1st open”, WITH– “2nd open”, and D– “3rd is open”. Then, A=BCD P(A)=P(BCD). According to the theorem on the probability of the product of independent events (formula 4.10) P(BCD) = P(B)P(C) P(D).

Calculate the probabilities of events B, C And D. In this example, there are 3 equally possible (we choose each key from 3) outcomes of the experience. Each of the events B, C And D favors 1 of them, so ..

Example 4.5 b)

Let's change the problem: we believe that the criminal is a forgetful person. Let the criminal open the door and leave the key in it. What is then the probability that he will open all the doors in 15 seconds?

Solution. Event A“All doors are open.” Again, A=BCD by definition of the product of events. Hence P(A)=P(BCD). But now events B, C And D- dependent. By the theorem on the probability of product of dependent events P(BCD) = P(B)P(C|B) P(D|BC).

Calculate the probabilities: , (there are only two keys left and one of them is suitable!), and, therefore, .

Example 4.6

The search for two criminals is underway. Each of them, independently of the other, can be detected within a day with a probability of 0.5. After the capture of one of them, due to the increase in the number of employees involved in the search, the probability of finding the second one increases to 0.7. What is the probability that both criminals will be found during the day.

Solution. Let the event A“Two perpetrators found.” Let's break this event into simpler ones. Let IN 1 - The first offender was discovered, and IN 2 – the second criminal is found, after the first is caught. Then, A=B 1 IN 2 by definition of the product of events. Hence P(A)=P(B 1 IN 2 ) . Because IN 1 And IN 2 are dependent events, then according to the theorem on the probability of the product of dependent events (formula 4.8) P(B 1 IN 2 ) = P(B 1 )P(B 2 /IN 1 ) = 0,5 0,7=0,35 .

Example 4.7

Find the probability that when a coin is tossed 10 times, the coat of arms will fall at least once.

Solution. Let the event A– “the coat of arms will fall out at least 1 time". Consider the reverse event: - “the coat of arms will not fall out never". It is obvious that the reverse event is easier than the original one to be divided into simpler ones. Let A 1 - the coat of arms did not fall out on the first roll, A 2 - the coat of arms did not fall out on the second roll, ... A 10 - the coat of arms did not fall on the 10th roll. All events A 1 …A 10 are independent, therefore (formula 4.11)

Example 4.8

2 groups of snipers are participating in the operation to free the hostages: 10 people with an OP21 rifle and 20 people with AKM47. The probability of defeat from OP21 is 0.85, and AKM47 is 0.65. Find the probability that with one shot from an arbitrary sniper the criminal will be hit.

Solution. Let the event A- "The offender is defeated." Let's break this event into simpler ones. The criminal can be hit either from OP21 or from AKM47. The probability that an arbitrary sniper is armed with OP21 (an event H 1 ) is equal to 10/30. The probability that an arbitrary sniper is armed with an AKM47 (event H 2 ) is equal to 20/30.

The probability that the offender is hit is (formula 4.12)

In such problems, it is useful to draw a tree of all possible outcomes (indicating the probabilities of each outcome).

When a coin is tossed, it can be said that it will land heads up, or probability of this is 1/2. Of course, this does not mean that if a coin is tossed 10 times, it will necessarily land on heads 5 times. If the coin is "fair" and if it is tossed many times, then heads will come up very close half the time. Thus, there are two kinds of probabilities: experimental And theoretical .

Experimental and theoretical probability

If we toss a coin a large number of times - say 1000 - and count how many times it comes up heads, we can determine the probability that it will come up heads. If heads come up 503 times, we can calculate the probability of it coming up:
503/1000, or 0.503.

This experimental definition of probability. This definition of probability comes from observing and examining data and is quite common and very useful. For example, here are some probabilities that were determined experimentally:

1. The chance of a woman developing breast cancer is 1/11.

2. If you kiss someone who has a cold, then the probability that you will also get a cold is 0.07.

3. A person who has just been released from prison has an 80% chance of going back to prison.

If we consider the toss of a coin and taking into account that it is equally likely to come up heads or tails, we can calculate the probability of coming up heads: 1 / 2. This is the theoretical definition of probability. Here are some other probabilities that have been theoretically determined using mathematics:

1. If there are 30 people in a room, the probability that two of them have the same birthday (excluding the year) is 0.706.

2. During a trip, you meet someone and during the course of the conversation you discover that you have a mutual acquaintance. Typical reaction: "That can't be!" In fact, this phrase does not fit, because the probability of such an event is quite high - just over 22%.

Therefore, the experimental probability is determined by observation and data collection. Theoretical probabilities are determined by mathematical reasoning. Examples of experimental and theoretical probabilities, such as those discussed above, and especially those that we do not expect, lead us to the importance of studying probability. You may ask, "What is true probability?" Actually, there is none. It is experimentally possible to determine the probabilities within certain limits. They may or may not coincide with the probabilities that we obtain theoretically. There are situations in which it is much easier to define one type of probability than another. For example, it would be sufficient to find the probability of catching a cold using theoretical probability.

Calculation of experimental probabilities

Consider first the experimental definition of probability. The basic principle we use to calculate such probabilities is as follows.

Principle P (experimental)

If in an experiment in which n observations are made, the situation or event E occurs m times in n observations, then the experimental probability of the event is said to be P (E) = m/n.

Example 1 Sociological survey. An experimental study was conducted to determine the number of left-handers, right-handers and people in whom both hands are equally developed. The results are shown in the graph.

a) Determine the probability that the person is right-handed.

b) Determine the probability that the person is left-handed.

c) Determine the probability that the person is equally fluent in both hands.

d) Most PBA tournaments have 120 players. Based on this experiment, how many players can be left-handed?

Solution

a) The number of people who are right-handed is 82, the number of left-handers is 17, and the number of those who are equally fluent in both hands is 1. The total number of observations is 100. Thus, the probability that a person is right-handed is P
P = 82/100, or 0.82, or 82%.

b) The probability that a person is left-handed is P, where
P = 17/100 or 0.17 or 17%.

c) The probability that a person is equally fluent with both hands is P, where
P = 1/100 or 0.01 or 1%.

d) 120 bowlers and from (b) we can expect 17% to be left handed. From here
17% of 120 = 0.17.120 = 20.4,
that is, we can expect about 20 players to be left-handed.

Example 2 Quality control . It is very important for a manufacturer to keep the quality of their products at a high level. In fact, companies hire quality control inspectors to ensure this process. The goal is to release the minimum possible number of defective products. But since the company produces thousands of items every day, it cannot afford to inspect each item to determine if it is defective or not. To find out what percentage of products are defective, the company tests far fewer products.
The USDA requires that 80% of the seeds that growers sell germinate. To determine the quality of the seeds that the agricultural company produces, 500 seeds are planted from those that have been produced. After that, it was calculated that 417 seeds germinated.

a) What is the probability that the seed will germinate?

b) Do the seeds meet government standards?

Solution a) We know that out of 500 seeds that were planted, 417 sprouted. The probability of seed germination P, and
P = 417/500 = 0.834, or 83.4%.

b) Since the percentage of germinated seeds exceeded 80% on demand, the seeds meet the state standards.

Example 3 TV ratings. According to statistics, there are 105,500,000 TV households in the United States. Every week, information about viewing programs is collected and processed. Within one week, 7,815,000 households were tuned in to CBS' hit comedy series Everybody Loves Raymond and 8,302,000 households were tuned in to NBC's hit Law & Order (Source: Nielsen Media Research). What is the probability that one home's TV is tuned to "Everybody Loves Raymond" during a given week? to "Law & Order"?

Solution The probability that the TV in one household is set to "Everybody Loves Raymond" is P, and
P = 7.815.000/105.500.000 ≈ 0.074 ≈ 7.4%.
The possibility that the household TV was set to "Law & Order" is P, and
P = 8.302.000/105.500.000 ≈ 0.079 ≈ 7.9%.
These percentages are called ratings.

theoretical probability

Suppose we are doing an experiment, such as tossing a coin or dart, drawing a card from a deck, or checking products for quality on an assembly line. Each possible outcome of such an experiment is called Exodus . The set of all possible outcomes is called outcome space . Event it is a set of outcomes, that is, a subset of the space of outcomes.

Example 4 Throwing darts. Suppose that in the "throwing darts" experiment, the dart hits the target. Find each of the following:

b) Outcome space

Solution
a) Outcomes are: hitting black (H), hitting red (K) and hitting white (B).

b) There is an outcome space (hit black, hit red, hit white), which can be written simply as (B, R, B).

Example 5 Throwing dice. A die is a cube with six sides, each of which has one to six dots.


Suppose we are throwing a die. Find
a) Outcomes
b) Outcome space

Solution
a) Outcomes: 1, 2, 3, 4, 5, 6.
b) Outcome space (1, 2, 3, 4, 5, 6).

We denote the probability that an event E occurs as P(E). For example, "the coin will land on tails" can be denoted by H. Then P(H) is the probability that the coin will land on tails. When all outcomes of an experiment have the same probability of occurring, they are said to be equally likely. To see the difference between events that are equally likely and events that are not equally likely, consider the target shown below.

For target A, black, red, and white hit events are equally likely, since black, red, and white sectors are the same. However, for target B, the zones with these colors are not the same, that is, hitting them is not equally likely.

Principle P (Theoretical)

If an event E can happen in m ways out of n possible equiprobable outcomes from the outcome space S, then theoretical probability event, P(E) is
P(E) = m/n.

Example 6 What is the probability of rolling a 3 by rolling a die?

Solution There are 6 equally likely outcomes on the die and there is only one possibility of throwing the number 3. Then the probability P will be P(3) = 1/6.

Example 7 What is the probability of rolling an even number on the die?

Solution The event is the throwing of an even number. This can happen in 3 ways (if you roll 2, 4 or 6). The number of equiprobable outcomes is 6. Then the probability P(even) = 3/6, or 1/2.

We will be using a number of examples related to a standard 52-card deck. Such a deck consists of the cards shown in the figure below.

Example 8 What is the probability of drawing an ace from a well-shuffled deck of cards?

Solution There are 52 outcomes (the number of cards in the deck), they are equally likely (if the deck is well mixed), and there are 4 ways to draw an ace, so according to the P principle, the probability
P(drawing an ace) = 4/52, or 1/13.

Example 9 Suppose we choose without looking one marble from a bag of 3 red marbles and 4 green marbles. What is the probability of choosing a red ball?

Solution There are 7 equally likely outcomes to get any ball, and since the number of ways to draw a red ball is 3, we get
P(choosing a red ball) = 3/7.

The following statements are results from the P principle.

Probability Properties

a) If the event E cannot happen, then P(E) = 0.
b) If the event E is bound to happen then P(E) = 1.
c) The probability that event E will occur is a number between 0 and 1: 0 ≤ P(E) ≤ 1.

For example, in tossing a coin, the event that the coin lands on its edge has zero probability. The probability that a coin is either heads or tails has a probability of 1.

Example 10 Suppose that 2 cards are drawn from a deck with 52 cards. What is the probability that both of them are spades?

Solution The number of ways n of drawing 2 cards from a well-shuffled 52-card deck is 52 C 2 . Since 13 of the 52 cards are spades, the number m of ways to draw 2 spades is 13 C 2 . Then,
P(stretching 2 peaks) \u003d m / n \u003d 13 C 2 / 52 C 2 \u003d 78/1326 \u003d 1/17.

Example 11 Suppose 3 people are randomly selected from a group of 6 men and 4 women. What is the probability that 1 man and 2 women will be chosen?

Solution Number of ways to choose three people from a group of 10 people 10 C 3 . One man can be chosen in 6 C 1 ways and 2 women can be chosen in 4 C 2 ways. According to the fundamental principle of counting, the number of ways to choose the 1st man and 2 women is 6 C 1 . 4C2. Then, the probability that 1 man and 2 women will be chosen is
P = 6 C 1 . 4 C 2 / 10 C 3 \u003d 3/10.

Example 12 Throwing dice. What is the probability of throwing a total of 8 on two dice?

Solution There are 6 possible outcomes on each dice. The outcomes are doubled, that is, there are 6.6 or 36 possible ways in which the numbers on two dice can fall. (It's better if the cubes are different, say one is red and the other is blue - this will help visualize the result.)

Pairs of numbers that add up to 8 are shown in the figure below. There are 5 possible ways to get the sum equal to 8, hence the probability is 5/36.

Addition and multiplication of probabilities. This article will focus on solving problems in probability theory. Earlier, we have already analyzed some of the simplest tasks, to solve them it is enough to know and understand the formula (I advise you to repeat it).

There are tasks that are a little more complicated, for their solution you need to know and understand: the rule of addition of probabilities, the rule of multiplication of probabilities, the concepts of dependent and independent events, opposite events, joint and incompatible events. Do not be afraid of definitions, everything is simple)).In this article, we will consider just such tasks.

Some important and simple theory:

incompatible if the occurrence of one of them excludes the occurrence of the others. That is, only one particular event can occur, or another.

A classic example: when throwing a dice (dice), only one can fall out, or only two, or only three, etc. Each of these events is incompatible with the others, and the occurrence of one of them excludes the occurrence of the other (in one test). The same with the coin - the loss of "eagle" eliminates the possibility of loss of "tails".

This also applies to more complex combinations. For example, two lighting lamps are lit. Each of them may or may not burn out for some time. There are options:

1. The first burns out and the second burns out
2. The first burns out and the second does not burn out
3. The first does not burn out and the second burns out
4. The first does not burn out and the second burns out.

All these 4 variants of events are incompatible - they simply cannot happen together and none of them with any other ...

Definition: Events are called joint if the occurrence of one of them does not exclude the occurrence of the other.

Example: a queen will be taken from a deck of cards and a spade card will be taken from a deck of cards. Two events are considered. These events are not mutually exclusive - you can draw the Queen of Spades and thus both events will occur.

On the sum of probabilities

The sum of two events A and B is called the event A + B, which consists in the fact that either the event A or the event B or both will occur at the same time.

If occur incompatible events A and B, then the probability of the sum of these events is equal to the sum of the probabilities of the events:

Dice example:

We throw a dice. What is the probability of getting a number less than four?

Numbers less than four are 1,2,3. We know that the probability of getting a 1 is 1/6, a 2 is 1/6, and a 3 is 1/6. These are incompatible events. We can apply the addition rule. The probability of getting a number less than four is:

Indeed, if we proceed from the concept of classical probability: then the number of possible outcomes is 6 (the number of all faces of the cube), the number of favorable outcomes is 3 (one, two or three). The desired probability is 3 to 6 or 3/6 = 0.5.

* The probability of the sum of two joint events is equal to the sum of the probabilities of these events without taking into account their joint occurrence: P (A + B) \u003d P (A) + P (B) -P (AB)

On multiplication of probabilities

Let two incompatible events A and B occur, their probabilities are respectively P(A) and P(B). The product of two events A and B is called such an event A B, which consists in the fact that these events will occur together, that is, both event A and event B will occur. The probability of such an event is equal to the product of the probabilities of events A and B.Calculated according to the formula:

As you have already noticed, the logical connective "AND" means multiplication.

An example with the same dice:Throw a die twice. What is the probability of rolling two sixes?

The probability of rolling a six for the first time is 1/6. The second time is also equal to 1/6. The probability of getting a six both the first time and the second time is equal to the product of the probabilities:

In simple terms: when an event occurs in one test, AND then another (others) occurs, then the probability that they will occur together is equal to the product of the probabilities of these events.

We solved problems with dice, but we used only logical reasoning, we did not use the product formula. In the problems considered below, one cannot do without formulas, or rather, it will be easier and faster to get the result with them.

It is worth mentioning one more nuance. When reasoning in solving problems, the concept of SIMULTANEOUSITY of events is used. Events occur SIMULTANEOUSLY - this does not mean that they occur in one second (at one moment in time). This means that they occur in a certain period of time (with one test).

For example:

Two lamps burn out within a year (it can be said - simultaneously within a year)

Two automata break down within a month (it can be said - simultaneously within a month)

The dice is thrown three times (points fall out at the same time, which means in one test)

Biathlete makes five shots. Events (shots) occur during one test.

Events A and B are independent if the probability of either of them does not depend on the occurrence or non-occurrence of the other event.

Consider the tasks:

Two factories produce the same glass for car headlights. The first factory produces 35% of these glasses, the second - 65%. The first factory produces 4% of defective glasses, and the second - 2%. Find the probability that a glass accidentally bought in a store will be defective.

The first factory produces 0.35 products (glasses). The probability of buying defective glass from the first factory is 0.04.

The second factory produces 0.65 glasses. The probability of buying defective glass from the second factory is 0.02.

The probability that the glass was bought at the first factory AND at the same time it will be defective is 0.35∙0.04 = 0.0140.

The probability that the glass was bought at the second factory AND at the same time it will be defective is 0.65∙0.02 = 0.0130.

Buying defective glass in a store implies that it (defective glass) was purchased EITHER from the first factory OR from the second. These are incompatible events, that is, we add the resulting probabilities:

0,0140 + 0,0130 = 0,027

Answer: 0.027

If grandmaster A. plays white, then he wins grandmaster B. with a probability of 0.62. If A. plays black, then A. beats B. with a probability of 0.2. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.

The chances of winning the first and second games are independent of each other. It is said that a grandmaster must win both times, that is, win the first time AND at the same time win the second time. In the case when independent events must occur together, the probabilities of these events are multiplied, that is, the multiplication rule is used.

The probability of producing these events will be equal to 0.62∙0.2 = 0.124.

Answer: 0.124

In the geometry exam, the student gets one question from the list of exam questions. The probability that this is an inscribed circle question is 0.3. The probability that this is a Parallelogram question is 0.25. 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.

That is, it is necessary to find the probability that the student will get a question EITHER on the topic “Inscribed circle”, OR on the topic “Parallelogram”. In this case, the probabilities are summed up, since these events are incompatible and any of these events can occur: 0.3 + 0.25 = 0.55.

*Disjoint events are events that cannot happen at the same time.

Answer: 0.55

The biathlete shoots five times at the targets. The probability of hitting the target with one shot is 0.9. Find the probability that the biathlete hit the targets the first four times and missed the last one. Round the result to the nearest hundredth.

Since the biathlete hits the target with a probability of 0.9, he misses with a probability of 1 - 0.9 = 0.1

*A miss and a hit are events that cannot occur simultaneously with one shot, the sum of the probabilities of these events is 1.

We are talking about the commission of several (independent) events. If an event occurs and at the same time another (subsequent) occurs at the same time (test), then the probabilities of these events are multiplied.

The probability of producing independent events is equal to the product of their probabilities.

Thus, the probability of the event "hit, hit, hit, hit, missed" is equal to 0.9∙0.9∙0.9∙0.9∙0.1 = 0.06561.

Rounding up to hundredths, we get 0.07

Answer: 0.07

The store has two payment machines. Each of them can be faulty with a probability of 0.07, regardless of the other automaton. Find the probability that at least one automaton is serviceable.

Find the probability that both automata are faulty.

These events are independent, so the probability will be equal to the product of the probabilities of these events: 0.07∙0.07 = 0.0049.

This means that the probability that both automata are working or one of them will be equal to 1 - 0.0049 = 0.9951.

* Both are serviceable and some one is completely - meets the condition "at least one".

One can present the probabilities of all (independent) events to test:

1. “faulty-faulty” 0.07∙0.07 = 0.0049

2. “Good-Faulty” 0.93∙0.07 = 0.0651

3. "Faulty-Faulty" 0.07∙0.93 = 0.0651

4. “healthy-healthy” 0.93∙0.93 = 0.8649

To determine the probability that at least one automaton is in good condition, it is necessary to add the probabilities of independent events 2,3 and 4: a certain event An event is called an event that is certain to occur as a result of an experience. The event is called impossible if it never happens as a result of experience.

For example, if one ball is randomly drawn from a box containing only red and green balls, then the appearance of a white ball among the drawn balls is an impossible event. The appearance of the red and the appearance of the green balls form a complete group of events.

Definition: The events are called equally possible , if there is no reason to believe that one of them will appear as a result of the experiment with a greater probability.

In the above example, the appearance of red and green balls are equally likely events if the box contains the same number of red and green balls. If there are more red balls in the box than green ones, then the appearance of a green ball is less probable than the appearance of a red one.

In we will consider more problems where the sum and product of the probabilities of events are used, do not miss it!

That's all. I wish you success!

Sincerely, Alexander Krutitskikh.

Maria Ivanovna scolds Vasya:
Petrov, why weren't you at school yesterday?!
My mom washed my pants yesterday.
- So what?
- And I was walking past the house and saw that yours were hanging. I thought you weren't coming.

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