Basic Concepts
Events are called incompatible if the occurrence of one of them excludes the occurrence of other events in the same trial. Otherwise they are called joint.
A complete group is a set of events, the combination of which is a reliable event.
The only two possible events that form a complete group are called opposite.
Events are called dependent if the probability of the occurrence of one of them depends on the occurrence or non-occurrence of other events.
Events are called independent if the probability of one of them does not depend on the occurrence or non-occurrence of others.
Theorem for adding probabilities of incompatible events
P(A+B)=P(A)+P(B),
where A, B are incompatible events.
Theorem for adding probabilities of joint events
P(A+B)=P(A)+P(B)-P(AB), where A and B are joint events.
Theorem for multiplying the probabilities of independent events
,
where A and B are independent events.
Theorem for multiplying probabilities of dependent events
P(AB)=P(A)P A (B),
where P A (B) is the probability of the occurrence of event B, provided that event A has occurred; A and B are dependent events.
Task 1.
The shooter fires two shots at the target. The probability of hitting each shot is 0.8. Compose a complete group of events and find their probabilities. Solution.
Test - Two shots are fired at a target.
Event A- missed both times.
Event IN- hit once.
Event WITH- hit both times. 
.
Control: P(A) +P(B) +P(C) = 1.
Task 2.
According to meteorologists' forecast, P(rain)=0.4; P(wind)=0.7; R(rain and wind)=0.2. What is the probability that it will rain or wind? Solution. By the theorem of addition of probabilities and due to the compatibility of the proposed events, we have:
P(rain or wind or both)=P(rain) +P(wind) –P(rain and wind)=0.4+0.7-0.2=0.9.
Task 3.
At the departure station there are 8 orders for goods to be sent: five for domestic shipments and three for export. What is the probability that two orders chosen at random will be for domestic consumption? Solution. Event A– the first order taken at random is within the country. Event IN– the second is also intended for domestic consumption. We need to find the probability. Then, by the theorem on the multiplication of probabilities of dependent events, we have 
Task 4.
From a batch of products, the merchandiser randomly selects the highest grade products. The probability that the selected item will be of the highest quality is 0.8; first grade – 0.7; second grade – 0.5. Find the probability that out of three randomly selected products there will be:
a) only two premium grades;
b) everyone is different. Solution. Let the event be a product of the highest quality; event - first-class product; the event is a second-class product.
According to the conditions of the problem; ; Events are independent.
a) Event A– only two top-grade products will look like this then
b) Event IN– all three products are different - let’s put it this way: 
, Then .
Task 5.
The probabilities of hitting the target when firing from three guns are as follows: p1= 0,8; p2=0,7; p3=0.9. Find the probability of at least one hit (event A) with one salvo from all guns. Solution. The probability of each gun hitting the target does not depend on the results of firing from other guns, therefore the events under consideration (hit by the first gun), (hit by the second gun) and (hit by the third gun) are independent in the aggregate.
The probabilities of events opposite to events (i.e., the probability of misses) are respectively equal to: 
Required probability
Task 6.
The printing house has 4 printing machines. For each machine, the probability that it is currently running is 0.9. Find the probability that at least one machine is currently working (event A).
Solution. The events “the machine is working” and “the machine is not working” (at the moment) are opposite, therefore the sum of their probabilities is equal to one:
Hence the probability that the machine is currently not working is equal to
The required probability. Problem 7. In the reading room there are 6 textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that both textbooks will be bound.
Solution. Consider the following events:
A1 - the first bound textbook taken;
A2 is the second bound textbook taken.
An event consisting in the fact that both taken textbooks are bound. Events A1 and A2 are dependent, since the probability of the occurrence of event A2 depends on the occurrence of event A1. To solve this problem, we use the theorem for multiplying the probabilities of dependent events: .
The probability of the occurrence of event A1 p(A1) in accordance with the classical definition of probability:
P(A1)=m/n=3/6=0.5.
The probability of the occurrence of event A2 is determined by the conditional probability of the occurrence of event A2 subject to the occurrence of event A1, i.e. (A2)==0.4.
Then the desired probability of the event occurring:
P(A)=0.5*0.4=0.2.
The study of probability theory begins with solving problems involving addition and multiplication of probabilities. It is worth mentioning right away that a student may encounter a problem when mastering this area of knowledge: if physical or chemical processes can be represented visually and understood empirically, then the level of mathematical abstraction is very high, and understanding here comes only with experience.
However, the game is worth the candle, because the formulas - both those discussed in this article and more complex ones - are used everywhere today and may well be useful in work.
Origin
Oddly enough, the impetus for the development of this branch of mathematics was... gambling. Indeed, dice, coin toss, poker, roulette are typical examples that use addition and multiplication of probabilities. This can be seen clearly using the examples of problems in any textbook. People were interested in learning how to increase their chances of winning, and it must be said that some succeeded in this.
For example, already in the 21st century, one person, whose name we will not disclose, used this knowledge accumulated over centuries to literally “clean out” the casino, winning several tens of millions of dollars at roulette.
However, despite the increased interest in the subject, only by the 20th century was a theoretical framework developed that made the “theorem” complete. Today, in almost any science one can find calculations using probabilistic methods.
Applicability
An important point when using formulas for adding and multiplying probabilities and conditional probability is the satisfiability of the central limit theorem. Otherwise, although the student may not realize it, all calculations, no matter how plausible they may seem, will be incorrect.
Yes, a highly motivated student is tempted to use new knowledge at every opportunity. But in this case it is necessary to slow down a little and strictly outline the scope of applicability.
Probability theory deals with random events, which in empirical terms represent the results of experiments: we can roll a six-sided die, draw a card from a deck, predict the number of defective parts in a batch. However, in some questions it is strictly forbidden to use formulas from this section of mathematics. We will discuss the features of considering the probabilities of an event, the theorems of addition and multiplication of events at the end of the article, but for now let’s turn to examples.
Basic Concepts
A random event refers to some process or result that may or may not appear as a result of an experiment. For example, we toss a sandwich - it can land butter side up or butter side down. Either of the two outcomes will be random, and we do not know in advance which of them will take place.
When studying addition and multiplication of probabilities, we will need two more concepts.
Such events are called joint, the occurrence of one of which does not exclude the occurrence of the other. Let's say two people shoot at a target at the same time. If one of them produces a successful one, it will not in any way affect the ability of the second to hit the bull's eye or miss.
Incompatible events will be those events whose occurrence at the same time is impossible. For example, if you take out only one ball from a box, you cannot get both blue and red at once.
Designation
The concept of probability is denoted by the Latin capital letter P. Next in brackets are arguments denoting certain events.
In the formulas of the addition theorem, conditional probability, and multiplication theorem, you will see expressions in brackets, for example: A+B, AB or A|B. They will be calculated in various ways, and we will now turn to them.
Addition
Let's consider cases in which formulas for adding and multiplying probabilities are used.
For incompatible events, the simplest addition formula is relevant: the probability of any of the random outcomes will be equal to the sum of the probabilities of each of these outcomes.
Suppose there is a box with 2 blue, 3 red and 5 yellow marbles. There are a total of 10 items in the box. What is the truth of the statement that we will draw a blue or a red ball? It will be equal to 2/10 + 3/10, i.e. fifty percent.
In the case of incompatible events, the formula becomes more complicated, since an additional term is added. Let's return to it in one paragraph, after considering another formula.
Multiplication
Addition and multiplication of probabilities of independent events are used in different cases. If, according to the conditions of the experiment, we are satisfied with any of the two possible outcomes, we will calculate the sum; if we want to get two certain outcomes one after another, we will resort to using a different formula.
Returning to the example from the previous section, we want to draw the blue ball first and then the red one. We know the first number - it is 2/10. What happens next? There are 9 balls left, and there are still the same number of red ones - three. According to calculations, it will be 3/9 or 1/3. But what to do now with two numbers? The correct answer is to multiply to get 2/30.
Joint events
Now we can again turn to the sum formula for joint events. Why were we distracted from the topic? To find out how probabilities are multiplied. Now we will need this knowledge.
We already know what the first two terms will be (the same as in the addition formula discussed earlier), but now we need to subtract the product of probabilities, which we just learned to calculate. For clarity, let's write the formula: P(A+B) = P(A) + P(B) - P(AB). It turns out that both addition and multiplication of probabilities are used in one expression.
Let's say we have to solve any of two problems in order to get credit. We can solve the first with a probability of 0.3, and the second with a probability of 0.6. Solution: 0.3 + 0.6 - 0.18 = 0.72. Note that simply adding up the numbers here will not be enough.
Conditional probability
Finally, there is the concept of conditional probability, the arguments of which are indicated in parentheses and separated by a vertical bar. The entry P(A|B) reads as follows: “the probability of event A given event B.”
Let's look at an example: a friend gives you some device, let it be a telephone. It may be broken (20%) or intact (80%). You are able to repair any device that comes into your hands with a probability of 0.4, or you are unable to do so (0.6). Finally, if the device is in working order, you can reach the right person with a probability of 0.7.
It's easy to see how conditional probability plays out in this case: you won't be able to reach a person if the phone is broken, but if it's working, you don't need to fix it. Thus, in order to get any results at the “second level”, you need to find out what event was executed at the first.
Calculations
Let's look at examples of solving problems involving addition and multiplication of probabilities, using the data from the previous paragraph.
First, let's find the probability that you will repair the device given to you. To do this, firstly, it must be faulty, and secondly, you must be able to fix it. This is a typical problem using multiplication: we get 0.2 * 0.4 = 0.08.
What is the likelihood that you will immediately reach the right person? It's as simple as that: 0.8*0.7 = 0.56. In this case, you found that the phone is working and successfully made the call.
Finally, consider this scenario: you get a broken phone, fix it, then dial a number and the person on the other end picks up. Here we already need to multiply three components: 0.2*0.4*0.7 = 0.056.
What to do if you have two non-working phones at once? How likely are you to fix at least one of them? on addition and multiplication of probabilities, since joint events are used. Solution: 0.4 + 0.4 - 0.4*0.4 = 0.8 - 0.16 = 0.64. Thus, if you get two broken devices, you will be able to fix it in 64% of cases.
Careful Use
As stated at the beginning of the article, the use of probability theory should be deliberate and conscious.
The larger the series of experiments, the closer the theoretically predicted value comes to the one obtained in practice. For example, we throw a coin. Theoretically, knowing the existence of formulas for addition and multiplication of probabilities, we can predict how many times “heads” and “tails” will appear if we conduct the experiment 10 times. We conducted an experiment, and by coincidence the ratio of the sides drawn was 3 to 7. But if we conduct a series of 100, 1000 or more attempts, it turns out that the distribution graph is getting closer and closer to the theoretical one: 44 to 56, 482 to 518, and so on.
Now imagine that this experiment is carried out not with a coin, but with the production of some new chemical substance, the probability of which we do not know. We would conduct 10 experiments and, without obtaining a successful result, we could generalize: “it is impossible to obtain the substance.” But who knows, had we made the eleventh attempt, would we have achieved the goal or not?
So if you are going into the unknown, into an unexplored area, probability theory may not apply. Each subsequent attempt in this case may be successful and generalizations like “X does not exist” or “X is impossible” will be premature.
Final word
So, we looked at two types of addition, multiplication and conditional probabilities. With further study of this area, it is necessary to learn to distinguish situations when each specific formula is used. In addition, you need to imagine whether probabilistic methods are generally applicable to solving your problem.
If you practice, after a while you will begin to carry out all the required operations exclusively in your mind. For those who are interested in card games, this skill can be considered extremely valuable - you will significantly increase your chances of winning just by calculating the probability of a particular card or suit falling out. However, you can easily find application of the acquired knowledge in other areas of activity.
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.
The need to act on probabilities occurs when the probabilities of some events are known, and it is necessary to calculate the probabilities of other events that are associated with these events.
Addition of probabilities is used when you need to calculate the probability of a combination or logical sum of random events.
Sum of events A And B denote A + B or A ∪ B. The sum of two events is an event that occurs if and only if at least one of the events occurs. It means that A + B– an event that occurs if and only if the event occurred during observation A or event B, or simultaneously A And B.
If events A And B are mutually inconsistent and their probabilities are given, then the probability that one of these events will occur as a result of one trial is calculated using the addition of probabilities.
Probability addition theorem. The probability that one of two mutually incompatible events will occur is equal to the sum of the probabilities of these events:
For example, while hunting, two shots are fired. Event A– hitting a duck with the first shot, event IN– hit from the second shot, event ( A+ IN) – a hit from the first or second shot or from two shots. So, if two events A And IN– incompatible events, then A+ IN– the occurrence of at least one of these events or two events.
Example 1. There are 30 balls of the same size in a box: 10 red, 5 blue and 15 white. Calculate the probability that a colored (not white) ball will be picked up without looking.
Solution. Let us assume that the event A- “the red ball is taken”, and the event IN- “The blue ball was taken.” Then the event is “a colored (not white) ball is taken.” Let's find the probability of the event A:
and events IN:
Events A And IN– mutually incompatible, since if one ball is taken, then it is impossible to take balls of different colors. Therefore, we use the addition of probabilities:
The theorem for adding probabilities for several incompatible events. If events constitute a complete set of events, then the sum of their probabilities is equal to 1:
The sum of the probabilities of opposite events is also equal to 1:
Opposite events form a complete set of events, and the probability of a complete set of events is 1.
Probabilities of opposite events are usually indicated in small letters p And q. In particular,
from which the following formulas for the probability of opposite events follow:
Example 2. The target in the shooting range is divided into 3 zones. The probability that a certain shooter will shoot at the target in the first zone is 0.15, in the second zone – 0.23, in the third zone – 0.17. Find the probability that the shooter will hit the target and the probability that the shooter will miss the target.
Solution: Find the probability that the shooter will hit the target:
Let's find the probability that the shooter will miss the target:
For more complex problems, in which you need to use both addition and multiplication of probabilities, see the page "Various problems involving addition and multiplication of probabilities".
Addition of probabilities of mutually simultaneous events
Two random events are called joint if the occurrence of one event does not exclude the occurrence of a second event in the same observation. For example, when throwing a die the event A The number 4 is considered to be rolled out, and the event IN– rolling an even number. Since 4 is an even number, the two events are compatible. In practice, there are problems of calculating the probabilities of the occurrence of one of the mutually simultaneous events.
Probability addition theorem for joint events. The probability that one of the joint events will occur is equal to the sum of the probabilities of these events, from which the probability of the common occurrence of both events is subtracted, that is, the product of the probabilities. The formula for the probabilities of joint events has the following form:
Since events A And IN compatible, event A+ IN occurs if one of three possible events occurs: or AB. According to the theorem of addition of incompatible events, we calculate as follows:
Event A will occur if one of two incompatible events occurs: or AB. However, the probability of the occurrence of one event from several incompatible events is equal to the sum of the probabilities of all these events:
Likewise:
Substituting expressions (6) and (7) into expression (5), we obtain the probability formula for joint events:
When using formula (8), it should be taken into account that events A And IN can be:
- mutually independent;
- mutually dependent.
Probability formula for mutually independent events:
Probability formula for mutually dependent events:
If events A And IN are inconsistent, then their coincidence is an impossible case and, thus, P(AB) = 0. The fourth probability formula for incompatible events is:
Example 3. In auto racing, when you drive the first car, you have a better chance of winning, and when you drive the second car. Find:
- the probability that both cars will win;
- the probability that at least one car will win;
1) The probability that the first car will win does not depend on the result of the second car, so the events A(the first car wins) and IN(the second car will win) – independent events. Let's find the probability that both cars win:
2) Find the probability that one of the two cars will win:
For more complex problems, in which you need to use both addition and multiplication of probabilities, see the page "Various problems involving addition and multiplication of probabilities".
Solve the addition of probabilities problem yourself, and then look at the solution
Example 4. Two coins are tossed. Event A- loss of the coat of arms on the first coin. Event B- loss of the coat of arms on the second coin. Find the probability of an event C = A + B .
Multiplying Probabilities
Probability multiplication is used when the probability of a logical product of events must be calculated.
In this case, random events must be independent. Two events are said to be mutually independent if the occurrence of one event does not affect the probability of the occurrence of the second event.
Probability multiplication theorem for independent events. Probability of simultaneous occurrence of two independent events A And IN is equal to the product of the probabilities of these events and is calculated by the formula:
Example 5. The coin is tossed three times in a row. Find the probability that the coat of arms will appear all three times.
Solution. The probability that the coat of arms will appear on the first toss of a coin, the second time, and the third time. Let's find the probability that the coat of arms will appear all three times:
Solve probability multiplication problems on your own and then look at the solution
Example 6. There is a box of nine new tennis balls. To play, three balls are taken, and after the game they are put back. When choosing balls, played balls are not distinguished from unplayed balls. What is the probability that after three games there will be no unplayed balls left in the box?
Example 7. 32 letters of the Russian alphabet are written on cut-out alphabet cards. Five cards are drawn at random one after another and placed on the table in order of appearance. Find the probability that the letters will form the word "end".
Example 8. From a full deck of cards (52 sheets), four cards are taken out at once. Find the probability that all four of these cards will be of different suits.
Example 9. The same task as in example 8, but each card after being removed is returned to the deck.
More complex problems, in which you need to use both addition and multiplication of probabilities, as well as calculate the product of several events, can be found on the page "Various problems involving addition and multiplication of probabilities".
The probability that at least one of the mutually independent events will occur can be calculated by subtracting from 1 the product of the probabilities of opposite events, that is, using the formula:
Example 10. Cargo is delivered by three modes of transport: river, rail and road transport. The probability that the cargo will be delivered by river transport is 0.82, by rail 0.87, by road transport 0.90. Find the probability that the cargo will be delivered by at least one of the three modes of transport.
Important notes!
1. If you see gobbledygook instead of formulas, clear your cache. How to do this in your browser is written here:
2. Before you start reading the article, pay attention to our navigator for the most useful resources for
What is probability?
The first time I encountered this term, I would not have understood what it was. Therefore, I will try to explain clearly.
Probability is the chance that the event we want will happen.
For example, you decided to go to a friend’s house, you remember the entrance and even the floor on which he lives. But I forgot the number and location of the apartment. And now you are standing on the staircase, and in front of you there are doors to choose from.
What is the chance (probability) that if you ring the first doorbell, your friend will answer the door for you? There are only apartments, and a friend lives only behind one of them. With an equal chance we can choose any door.
But what is this chance?
The door, the right door. Probability of guessing by ringing the first doorbell: . That is, one time out of three you will accurately guess.
We want to know, having called once, how often will we guess the door? Let's look at all the options:
- You called 1st door
- You called 2nd door
- You called 3rd door
Now let’s look at all the options where a friend could be:
A. Behind 1st the door
b. Behind 2nd the door
V. Behind 3rd the door
Let's compare all the options in table form. A checkmark indicates options when your choice coincides with a friend's location, a cross - when it does not coincide.
How do you see everything Maybe options your friend's location and your choice of which door to ring.
A favorable outcomes of all . That is, you will guess once by ringing the doorbell once, i.e. .
This is probability - the ratio of a favorable outcome (when your choice coincides with your friend’s location) to the number of possible events.
The definition is the formula. Probability is usually denoted by p, so:
It is not very convenient to write such a formula, so we will take for - the number of favorable outcomes, and for - the total number of outcomes.
The probability can be written as a percentage; to do this, you need to multiply the resulting result by:
The word “outcomes” probably caught your eye. Since mathematicians call various actions (in our case, such an action is a doorbell) experiments, the result of such experiments is usually called the outcome.
Well, there are favorable and unfavorable outcomes.
Let's go back to our example. Let's say we rang one of the doors, but a stranger opened it for us. We didn't guess right. What is the probability that if we ring one of the remaining doors, our friend will open it for us?
If you thought that, then this is a mistake. Let's figure it out.
We have two doors left. So we have possible steps:
1) Call 1st door
2) Call 2nd door
The friend, despite all this, is definitely behind one of them (after all, he wasn’t behind the one we called):
a) Friend for 1st the door
b) Friend for 2nd the door
Let's draw the table again:
As you can see, there are only options, of which are favorable. That is, the probability is equal.
Why not?
The situation we considered is example of dependent events. The first event is the first doorbell, the second event is the second doorbell.
And they are called dependent because they influence the following actions. After all, if after the first ring the doorbell was answered by a friend, what would be the probability that he was behind one of the other two? Right, .
But if there are dependent events, then there must also be independent? That's right, they do happen.
A textbook example is tossing a coin.
- Toss a coin once. What is the probability of getting heads, for example? That's right - because there are all the options (either heads or tails, we will neglect the probability of the coin landing on its edge), but it only suits us.
- But it came up heads. Okay, let's throw it again. What is the probability of getting heads now? Nothing has changed, everything is the same. How many options? Two. How many are we happy with? One.
And let it come up heads at least a thousand times in a row. The probability of getting heads at once will be the same. There are always options, and favorable ones.
It is easy to distinguish dependent events from independent ones:
- If the experiment is carried out once (they throw a coin once, ring the doorbell once, etc.), then the events are always independent.
- If an experiment is carried out several times (a coin is thrown once, the doorbell is rung several times), then the first event is always independent. And then, if the number of favorable ones or the number of all outcomes changes, then the events are dependent, and if not, they are independent.
Let's practice determining probability a little.
Example 1.
The coin is tossed twice. What is the probability of getting heads twice in a row?
Solution:
Let's consider all possible options:
- Eagle-eagle
- Heads-tails
- Tails-Heads
- Tails-tails
As you can see, there are only options. Of these, we are only satisfied. That is, the probability:
If the condition simply asks you to find the probability, then the answer must be given in the form of a decimal fraction. If it were specified that the answer should be given as a percentage, then we would multiply by.
Answer:
Example 2.
In a box of chocolates, all the chocolates are packaged in the same wrapper. However, from sweets - with nuts, with cognac, with cherries, with caramel and with nougat.
What is the probability of taking one candy and getting a candy with nuts? Give your answer as a percentage.
Solution:
How many possible outcomes are there? .
That is, if you take one candy, it will be one of those available in the box.
How many favorable outcomes?
Because the box contains only chocolates with nuts.
Answer:
Example 3.
In a box of balloons. of which are white and black.
- What is the probability of drawing a white ball?
- We added more black balls to the box. What is now the probability of drawing a white ball?
Solution:
a) There are only balls in the box. Of them are white.
The probability is:
b) Now there are more balls in the box. And there are just as many whites left - .
Answer:
Total probability
| The probability of all possible events is equal to (). |
Let's say there are red and green balls in a box. What is the probability of drawing a red ball? Green ball? Red or green ball?
Probability of drawing a red ball
Green ball:
Red or green ball:
As you can see, the sum of all possible events is equal to (). Understanding this point will help you solve many problems.
Example 4.
There are markers in the box: green, red, blue, yellow, black.
What is the probability of drawing NOT a red marker?
Solution:
Let's count the number favorable outcomes.
NOT a red marker, that means green, blue, yellow or black.
| The probability that an event will not occur is equal to minus the probability that the event will occur. |
Rule for multiplying the probabilities of independent events
You already know what independent events are.
What if you need to find the probability that two (or more) independent events will occur in a row?
Let's say we want to know what is the probability that if we flip a coin once, we will see heads twice?
We have already considered - .
What if we toss a coin once? What is the probability of seeing an eagle twice in a row?
Total possible options:
- Eagle-eagle-eagle
- Heads-heads-tails
- Heads-tails-heads
- Heads-tails-tails
- Tails-heads-heads
- Tails-heads-tails
- Tails-tails-heads
- Tails-tails-tails
I don’t know about you, but I made mistakes several times when compiling this list. Wow! And only option (first) suits us.
For 5 throws, you can make a list of possible outcomes yourself. But mathematicians are not as hardworking as you.
Therefore, they first noticed and then proved that the probability of a certain sequence of independent events each time decreases by the probability of one event.
In other words,
Let's look at the example of the same ill-fated coin.
Probability of getting heads in a challenge? . Now we flip the coin once.
What is the probability of getting heads in a row?
This rule doesn't only work if we are asked to find the probability that the same event will happen several times in a row.
If we wanted to find the sequence TAILS-HEADS-TAILS for consecutive tosses, we would do the same.
The probability of getting tails is , heads - .
Probability of getting the sequence TAILS-HEADS-TAILS-TAILS:
You can check it yourself by making a table.
The rule for adding the probabilities of incompatible events.
So stop! New definition.
Let's figure it out. Let's take our worn-out coin and toss it once.
Possible options:
- Eagle-eagle-eagle
- Heads-heads-tails
- Heads-tails-heads
- Heads-tails-tails
- Tails-heads-heads
- Tails-heads-tails
- Tails-tails-heads
- Tails-tails-tails
So, incompatible events are a certain, given sequence of events. - these are incompatible events.
If we want to determine what the probability of two (or more) incompatible events is, then we add the probabilities of these events.
You need to understand that heads or tails are two independent events.
If we want to determine the probability of a sequence (or any other) occurring, then we use the rule of multiplying probabilities.
What is the probability of getting heads on the first toss, and tails on the second and third tosses?
But if we want to know what is the probability of getting one of several sequences, for example, when heads comes up exactly once, i.e. options and, then we must add up the probabilities of these sequences.
Total options suit us.
We can get the same thing by adding up the probabilities of occurrence of each sequence:
Thus, we add probabilities when we want to determine the probability of certain, inconsistent, sequences of events.
There is a great rule to help you avoid getting confused when to multiply and when to add:
Let's go back to the example where we tossed a coin once and wanted to know the probability of seeing heads once.
What is going to happen?
Should fall out:
(heads AND tails AND tails) OR (tails AND heads AND tails) OR (tails AND tails AND heads).
This is how it turns out:
Let's look at a few examples.
Example 5.
There are pencils in the box. red, green, orange and yellow and black. What is the probability of drawing red or green pencils?
Solution:
Example 6.
If a die is thrown twice, what is the probability of getting a total of 8?
Solution.
How can we get points?
(and) or (and) or (and) or (and) or (and).
The probability of getting one (any) face is .
We calculate the probability:
Training.
I think now you understand when you need to calculate probabilities, when to add them, and when to multiply them. Is not it? Let's practice a little.
Tasks:
Let's take a card deck containing cards including spades, hearts, 13 clubs and 13 diamonds. From to Ace of each suit.
- What is the probability of drawing clubs in a row (we put the first card pulled out back into the deck and shuffle it)?
- What is the probability of drawing a black card (spades or clubs)?
- What is the probability of drawing a picture (jack, queen, king or ace)?
- What is the probability of drawing two pictures in a row (we remove the first card drawn from the deck)?
- What is the probability, taking two cards, to collect a combination - (jack, queen or king) and an ace? The sequence in which the cards are drawn does not matter.
Answers:
If you were able to solve all the problems yourself, then you are great! Now you will crack probability theory problems in the Unified State Exam like nuts!
PROBABILITY THEORY. AVERAGE LEVEL
Let's look at an example. Let's say we throw a die. What kind of bone is this, do you know? This is what they call a cube with numbers on its faces. How many faces, so many numbers: from to how many? Before.
So we roll the dice and we want it to come up or. And we get it.
In probability theory they say what happened auspicious event(not to be confused with prosperous).
If it happened, the event would also be favorable. In total, only two favorable events can happen.
How many are unfavorable? Since there are total possible events, it means that the unfavorable ones are events (this is if or falls out).
Definition:
Probability is the ratio of the number of favorable events to the number of all possible events. That is, probability shows what proportion of all possible events are favorable.
They denote probability with a Latin letter (apparently from the English word probability - probability).
It is customary to measure probability as a percentage (see topics and). To do this, the probability value must be multiplied by. In the dice example, probability.
And in percentage: .
Examples (decide for yourself):
- What is the probability of getting heads when tossing a coin? What is the probability of landing heads?
- What is the probability of getting an even number when throwing a die? Which one is odd?
- In a box of simple, blue and red pencils. We draw one pencil at random. What is the probability of getting a simple one?
Solutions:
- How many options are there? Heads and tails - just two. How many of them are favorable? Only one is an eagle. So the probability
It's the same with tails: .
- Total options: (how many sides the cube has, so many different options). Favorable ones: (these are all even numbers:).
Probability. Of course, it’s the same with odd numbers. - Total: . Favorable: . Probability: .
Total probability
All pencils in the box are green. What is the probability of drawing a red pencil? There are no chances: probability (after all, favorable events -).
Such an event is called impossible.
What is the probability of drawing a green pencil? There are exactly the same number of favorable events as there are total events (all events are favorable). So the probability is equal to or.
Such an event is called reliable.
If a box contains green and red pencils, what is the probability of drawing green or red? Yet again. Let's note this: the probability of pulling out green is equal, and red is equal.
In sum, these probabilities are exactly equal. That is, the sum of the probabilities of all possible events is equal to or.
Example:
In a box of pencils, among them are blue, red, green, plain, yellow, and the rest are orange. What is the probability of not drawing green?
Solution:
We remember that all probabilities add up. And the probability of getting green is equal. This means that the probability of not drawing green is equal.
Remember this trick: The probability that an event will not occur is equal to minus the probability that the event will occur.
Independent events and the multiplication rule
You flip a coin once and want it to come up heads both times. What is the likelihood of this?
Let's go through all the possible options and determine how many there are:
Heads-Heads, Tails-Heads, Heads-Tails, Tails-Tails. What else?
Total options. Of these, only one suits us: Eagle-Eagle. In total, the probability is equal.
Fine. Now let's flip a coin once. Do the math yourself. Happened? (answer).
You may have noticed that with the addition of each subsequent throw, the probability decreases by half. The general rule is called multiplication rule:
The probabilities of independent events change.
What are independent events? Everything is logical: these are those that do not depend on each other. For example, when we throw a coin several times, each time a new throw is made, the result of which does not depend on all previous throws. We can just as easily throw two different coins at the same time.
More examples:
- The dice are thrown twice. What is the probability of getting it both times?
- The coin is tossed once. What is the probability that it will come up heads the first time, and then tails twice?
- The player rolls two dice. What is the probability that the sum of the numbers on them will be equal?
Answers:
- The events are independent, which means the multiplication rule works: .
- The probability of heads is equal. The probability of tails is the same. Multiply:
- 12 can only be obtained if two -ki are rolled: .
Incompatible events and the addition rule
Events that complement each other to the point of full probability are called incompatible. As the name suggests, they cannot happen simultaneously. For example, if we flip a coin, it can come up either heads or tails.
Example.
In a box of pencils, among them are blue, red, green, plain, yellow, and the rest are orange. What is the probability of drawing green or red?
Solution .
The probability of drawing a green pencil is equal. Red - .
Favorable events in all: green + red. This means that the probability of drawing green or red is equal.
The same probability can be represented in this form: .
This is the addition rule: the probabilities of incompatible events add up.
Mixed type problems
Example.
The coin is tossed twice. What is the probability that the results of the rolls will be different?
Solution .
This means that if the first result is heads, the second must be tails, and vice versa. It turns out that there are two pairs of independent events, and these pairs are incompatible with each other. How not to get confused about where to multiply and where to add.
There is a simple rule for such situations. Try to describe what is going to happen using the conjunctions “AND” or “OR”. For example, in this case:
It should come up (heads and tails) or (tails and heads).
Where there is a conjunction “and” there will be multiplication, and where there is “or” there will be addition:
Try it yourself:
- What is the probability that if a coin is tossed twice, the coin will land on the same side both times?
- The dice are thrown twice. What is the probability of getting a total of points?
Solutions:
Another example:
Toss a coin once. What is the probability that heads will appear at least once?
Solution:
PROBABILITY THEORY. BRIEFLY ABOUT THE MAIN THINGS
Probability is the ratio of the number of favorable events to the number of all possible events.
Independent events
Two events are independent if the occurrence of one does not change the probability of the other occurring.
Total probability
The probability of all possible events is equal to ().
The probability that an event will not occur is equal to minus the probability that the event will occur.
Rule for multiplying the probabilities of independent events
The probability of a certain sequence of independent events is equal to the product of the probabilities of each event
Incompatible events
Incompatible events are those that cannot possibly occur simultaneously as a result of an experiment. A number of incompatible events form a complete group of events.
The probabilities of incompatible events add up.
Having described what should happen, using the conjunctions “AND” or “OR”, instead of “AND” we put a multiplication sign, and instead of “OR” we put an addition sign.
Well, the topic is over. If you are reading these lines, it means you are very cool.
Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!
Now the most important thing.
You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.
The problem is that this may not be enough...
For what?
For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.
I won’t convince you of anything, I’ll just say one thing...
People who have received a good education earn much more than those who have not received it. This is statistics.
But this is not the main thing.
The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...
But think for yourself...
What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?
GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.
You won't be asked for theory during the exam.
You will need solve problems against time.
And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.
It's like in sports - you need to repeat it many times to win for sure.
Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!
You can use our tasks (optional) and we, of course, recommend them.
In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.
How? There are two options:
- Unlock all hidden tasks in this article -
- Unlock access to all hidden tasks in all 99 articles of the textbook - Buy a textbook - 499 RUR
Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.
Access to all hidden tasks is provided for the ENTIRE life of the site.
In conclusion...
If you don't like our tasks, find others. Just don't stop at theory.
“Understood” and “I can solve” are completely different skills. You need both.
Find problems and solve them!