The study of probability theory begins with solving problems for addition and multiplication of probabilities. It is worth mentioning right away that a student, when mastering this field of knowledge, may encounter a problem: if physical or chemical processes can be visualized 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 considered in this article and more complex ones - are used everywhere today and may well come in handy in work.
Origin
Oddly enough, the impetus for the development of this section of mathematics was ... gambling. Indeed, dice, coin toss, poker, roulette are typical examples that use addition and multiplication of probabilities. On the example of tasks in any textbook, this can be seen clearly. People were interested in learning how to increase their chances of winning, and I must say, some succeeded in this.
For example, already in the 21st century, one person, whose name we will not disclose, used this knowledge accumulated over the centuries to literally “cleanse” the casino, winning several tens of millions of dollars at roulette.
However, despite the increased interest in the subject, only by the 20th century a theoretical base was developed that made the "theorver" full-fledged. Today, in almost any science, one can find calculations using probabilistic methods.
Applicability
An important point when using the formulas for addition and multiplication of probabilities, conditional probability is the satisfiability of the central limit theorem. Otherwise, although it may not be realized by the student, all calculations, no matter how plausible they may seem, will be incorrect.
Yes, the highly motivated learner 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 are 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 categorically impossible 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 is some process or result that may or may not appear as a result of an experiment. For example, we toss a sandwich - it can fall butter up or butter 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 need two more concepts.
Joint events are such events, 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 affect the ability of the second to hit the bull's-eye or miss.
Inconsistent events will be such events, the occurrence of which is simultaneously impossible. For example, by pulling out only one ball from the 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 parentheses, there are arguments denoting some events.
In the formulas of the addition theorem, conditional probability, multiplication theorem, you will see expressions in brackets, for example: A+B, AB or A|B. They will count different ways, we will now turn to them.
Addition
Consider the cases in which the formulas for addition and multiplication of 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 10 items in total in the box. What is the percentage of the truth of the statement that we will draw a blue or 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. We will return to it in one paragraph, after considering one more formula.
Multiplication
Addition and multiplication of probabilities are not dependent events used in different occasions. If, according to the condition of the experiment, we are satisfied with either of the two possible outcomes, we will calculate the sum; if we want to get two certain outcomes one after the other, 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. The first number we know is 2/10. What happens next? There are 9 balls left, there are still the same number of red ones - three pieces. According to the calculations, you get 3/9 or 1/3. But what to do with two numbers now? 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 are we digressing from the topic? To learn how probabilities are multiplied. Now we need this knowledge.
We already know what the first two terms will be (the same as in the addition formula considered earlier), but now we need to subtract the product of probabilities, which we just learned how to calculate. For clarity, we write the formula: P(A+B) = P(A) + P(B) - P(AB). It turns out that in one expression both addition and multiplication of probabilities are used.
Let's say we have to solve either of two problems in order to get credit. We can solve the first one with a probability of 0.3, and the second - 0.6. Solution: 0.3 + 0.6 - 0.18 = 0.72. Note that simply summing the numbers here will not be enough.
Conditional Probability
Finally, there is the concept of conditional probability, the arguments of which are indicated in brackets and separated by a vertical bar. The entry P(A|B) reads as follows: "probability of event A given event B".
Let's look at an example: a friend gives you some device, let it be a phone. It can be broken (20%) or good (80%). You are able to repair any device that falls into your hands with a probability of 0.4 or you are not able to do this (0.6). Finally, if the device is in working condition, you can call right person with a probability of 0.7.
It's easy to see how conditional probability works in this case: you can't get through to the person if the phone is broken, and if it's good, you don't need to fix it. Thus, in order to get any results at the "second level", you need to know which event was executed at the first.
Calculations
Consider examples of solving problems for 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 cope with the repair. This is a typical multiplication problem: we get 0.2 * 0.4 = 0.08.
What is the probability that you will immediately get through to the right person? Easier than simple: 0.8 * 0.7 \u003d 0.56. In this case, you found that the phone is working and successfully made a call.
Finally, consider this scenario: you received a broken phone, fixed it, then dialed the number, and the person on the opposite end picked up the phone. Here, the multiplication of three components is already required: 0.2 * 0.4 * 0.7 \u003d 0.056.
But what 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 two broken devices fall into your hands, you will be able to fix it in 64% of cases.
Considerate use
As mentioned 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 approaches the value obtained in practice. For example, we are tossing a coin. Theoretically, knowing about the existence of formulas for addition and multiplication of probabilities, we can predict how many times heads and tails will fall out if we conduct the experiment 10 times. We conducted an experiment, and by coincidence, the ratio of the sides that fell out was 3 to 7. But if you 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, the probability of which we do not know. We would run 10 experiments and, without getting a successful result, we could generalize: "the substance cannot be obtained." But who knows, if we made the eleventh attempt, would we have reached the goal or not?
Thus, if you are going into the unknown, into an unexplored area, the theory of probability may not be applicable. Each subsequent attempt in this case may succeed, and generalizations like "X does not exist" or "X is impossible" will be premature.
Final word
So, we have considered 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 understand whether probabilistic methods are generally applicable in 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 addicted 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, the acquired knowledge can easily be applied in other areas of activity.
Theorems of addition and multiplication of probabilities.
Dependent and independent events
The title looks scary, but it's actually very simple. On this lesson we will get acquainted with the theorems of addition and multiplication of the probabilities of events, and also analyze typical problems that, along with task for the classical definition of probability will definitely meet or, more likely, have already met on your way. For effective learning materials of this article, you need to know and understand the basic terms probability theory and be able to perform simple arithmetic operations. As you can see, very little is required, and therefore a fat plus in the asset is almost guaranteed. But on the other hand, I again warn against a superficial attitude to practical examples - there are also enough subtleties. Good luck:
The addition theorem for the probabilities of incompatible events: the probability of occurrence of one of the two incompatible events or (no matter what), is equal to the sum of the probabilities of these events:
A similar fact is also true for a larger number of incompatible events, for example, for three incompatible events and :
Dream theorem =) However, such a dream is also subject to proof, which can be found, for example, in study guide V.E. Gmurman.
Let's get acquainted with new, hitherto unseen concepts:
Dependent and independent events
Let's start with independent events. Events are independent if the probability of occurrence any of them does not depend from the appearance/non-appearance of other events of the considered set (in all possible combinations). ... But what is there to grind out common phrases:
The theorem of multiplication of probabilities of independent events: the probability of joint occurrence of independent events and is equal to the product of the probabilities of these events:
Let's return to the simplest example of the 1st lesson, in which two coins are tossed and the following events:
- heads will fall on the 1st coin;
- Heads on the 2nd coin.
Let's find the probability of the event (heads will appear on the 1st coin And Eagle will appear on the 2nd coin - remember how to read product of events!)
. The probability of getting heads on one coin does not depend on the result of tossing another coin, therefore, the events and are independent. 
Similarly: 
is the probability that the 1st coin will land heads And on the 2nd tail; 
is the probability that heads appear on the 1st coin And on the 2nd tail; 
is the probability that the 1st coin will land on tails And on the 2nd eagle.
Note that events form full group and the sum of their probabilities is equal to one: .
The multiplication theorem obviously extends to a larger number of independent events, so, for example, if the events are independent, then the probability of their joint occurrence is: . Let's practice on concrete examples:
Task 3
Each of the three boxes contains 10 parts. In the first box there are 8 standard parts, in the second - 7, in the third - 9. One part is randomly removed from each box. Find the probability that all parts are standard.
Solution: the probability of extracting a standard or non-standard part from any box does not depend on which parts will be extracted from other boxes, so the problem is about independent events. Consider the following independent events:
– a standard part is removed from the 1st box;
– a standard part is removed from the 2nd box;
– A standard part has been removed from the 3rd drawer.
According to the classical definition:
are the corresponding probabilities.
Event we are interested in (Standard part will be taken from the 1st drawer And from the 2nd standard And from the 3rd standard) is expressed by the product.
According to the theorem of multiplication of probabilities of independent events:
is the probability that one standard part will be extracted from three boxes.
Answer: 0,504
After invigorating exercises with boxes, no less interesting urns await us:
Task 4
Three urns contain 6 white and 4 black balls. One ball is drawn at random from each urn. Find the probability that: a) all three balls will be white; b) all three balls will be the same color.
Based on the information received, guess how to deal with the “be” item ;-) Sample Sample decision is designed in an academic style with a detailed description of all events.
Dependent events. The event is called dependent if its probability depends from one or more events that have already happened. You don’t have to go far for examples - just go to the nearest store:
- tomorrow at 19.00 will be on sale fresh bread.
The probability of this event depends on many other events: whether fresh bread will be delivered tomorrow, whether it will be sold out before 7 pm or not, etc. Depending on various circumstances, this event can be both reliable and impossible. So the event is dependent.
Bread ... and, as the Romans demanded, circuses:
- at the exam, the student will get a simple ticket.
If you go not the very first, then the event will be dependent, since its probability will depend on which tickets the classmates have already drawn.
How to determine dependency/independence of events?
Sometimes this is directly stated in the condition of the problem, but most often you have to conduct an independent analysis. There is no unambiguous guideline here, and the fact of dependence or independence of events follows from natural logical reasoning.
In order not to throw everything in one heap, tasks for dependent events I will highlight the next lesson, but for now we will consider the most common bunch of theorems in practice:
Problems on addition theorems for inconsistent probabilities
and multiplying the probabilities of independent events
This tandem, according to my subjective assessment, works in about 80% of the tasks on the topic under consideration. A hit of hits and a real classic of probability theory:
Task 5
Two shooters fired one shot each at the target. The probability of hitting for the first shooter is 0.8, for the second - 0.6. Find the probability that:
a) only one shooter will hit the target;
b) at least one of the shooters will hit the target.
Solution: The hit/miss probability of one shooter is obviously independent of the other shooter's performance.
Consider the events:
– 1st shooter will hit the target;
- The 2nd shooter will hit the target.
By condition: .
Let's find the probabilities of opposite events - that the corresponding arrows will miss: 
a) Consider the event: - only one shooter hits the target. This event consists of two incompatible outcomes:
1st shooter will hit And 2nd misses
or
1st will miss And 2nd will hit.
On the tongue event algebras this fact can be written as: 
First, we use the theorem of addition of probabilities of incompatible events, then - the theorem of multiplication of probabilities of independent events:
is the probability that there will be only one hit.
b) Consider the event: - at least one of the shooters will hit the target.
First of all, LET'S THINK - what does the condition "AT LEAST ONE" mean? In this case, this means that either the 1st shooter will hit (the 2nd will miss) or 2nd (1st misses) or both arrows at once - a total of 3 incompatible outcomes.
Method one: given the prepared probability of the previous item, it is convenient to represent the event as the sum of the following disjoint events:
one will get (an event consisting in turn of 2 incompatible outcomes) or
If both arrows hit, we denote this event by the letter .
Thus:
According to the theorem of multiplication of probabilities of independent events:
is the probability that the 1st shooter will hit And 2nd shooter will hit.
According to the theorem of addition of probabilities of incompatible events:
is the probability of at least one hit on the target.
Method two: consider the opposite event: – both shooters will miss.
According to the theorem of multiplication of probabilities of independent events:
As a result:
Special attention pay attention to the second method - in the general case it is more rational.
In addition, there is an alternative, third way of solving, based on the theorem of summing joint events, which was silent above.
! If you are reading the material for the first time, then in order to avoid confusion, it is better to skip the next paragraph.
Method three
: the events are joint, which means that their sum expresses the event “at least one shooter hits the target” (see Fig. event algebra). By theorem of addition of probabilities of joint events and the theorem of multiplication of probabilities of independent events:
Let's check: events and (0, 1 and 2 hits respectively) form a complete group, so the sum of their probabilities must be equal to one:
, which was to be verified.
Answer: 
With a thorough study of the theory of probability, you will come across dozens of tasks of a militaristic content, and, which is typical, after that you will not want to shoot anyone - the tasks are almost gift. Why not make the template even simpler? Let's shorten the entry:
Solution: according to the condition: , is the probability of hitting the corresponding shooters. Then their miss probabilities are: 
a) According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:
is the probability that only one shooter will hit the target.
b) According to the theorem of multiplication of probabilities of independent events:
is the probability that both shooters will miss.
Then: is the probability that at least one of the shooters will hit the target.
Answer: 
In practice, you can use any design option. Of course, much more often they go the short way, but one should not forget the 1st method - although it is longer, it is more meaningful - it is clearer in it, what, why and why adds up and multiplies. In some cases, a hybrid style is appropriate, when it is convenient to indicate only some events in capital letters.
Similar tasks for independent decision:
Task 6
Two independently operating sensors are installed for fire alarm. The probabilities that the sensor will operate during a fire are 0.5 and 0.7 for the first and second sensors, respectively. Find the probability that in a fire:
a) both sensors will fail;
b) both sensors will work.
c) using addition theorem for the probabilities of events forming a complete group, find the probability that only one sensor will operate during a fire. Check the result by direct calculation of this probability (using addition and multiplication theorems).
Here, the independence of the operation of devices is directly spelled out in the condition, which, by the way, is an important clarification. The sample solution is designed in an academic style.
What if, in a similar problem, the same probabilities are given, for example, 0.9 and 0.9? You need to decide exactly the same! (which, in fact, has already been demonstrated in the example with two coins)
Task 7
The probability of hitting the target by the first shooter with one shot is 0.8. The probability that the target is not hit after the first and second shooters shoot one shot is 0.08. What is the probability of hitting the target by the second shooter with one shot?
And this is a small puzzle, which is framed in a short way. The condition can be reformulated more concisely, but I will not remake the original - in practice, I have to delve into more ornate fabrications.
Meet him - he is the one who cut an unmeasured amount of details for you =):
Task 8
A worker operates three machines. The probability that during the shift the first machine will require adjustment is 0.3, the second - 0.75, the third - 0.4. Find the probability that during the shift:
a) all machines will require adjustment;
b) only one machine will require adjustment;
c) at least one machine will require adjustment.
Solution: since the condition does not say anything about a single technological process, then the operation of each machine should be considered independent of the operation of other machines.
By analogy with Task No. 5, here you can enter into consideration events consisting in the fact that the corresponding machines will require adjustment during the shift, write down the probabilities , find the probabilities of opposite events, etc. But with three objects, I don’t really want to draw up the task like that - it will turn out long and tedious. Therefore, it is noticeably more profitable to use the “quick” style here:
By condition: - the probability that during the shift the corresponding machines will require tuning. Then the probabilities that they will not require attention are: 
One of the readers found a cool typo here, I won’t even correct it =)
a) According to the theorem of multiplication of probabilities of independent events:
is the probability that during the shift all three machines will require adjustment.
b) The event "During the shift, only one machine will require adjustment" consists of three incompatible outcomes:
1) 1st machine will require attention And 2nd machine will not require And 3rd machine will not require
or:
2) 1st machine will not require attention And 2nd machine will require And 3rd machine will not require
or:
3) 1st machine will not require attention And 2nd machine will not require And 3rd machine will require.
According to the theorems of addition of probabilities of incompatible and multiplication of probabilities of independent events:
- the probability that during the shift only one machine will require adjustment.
I think by now it should be clear to you where the expression came from
c) Calculate the probability that the machines will not require adjustment, and then the probability of the opposite event:
– the fact that at least one machine will require adjustment.
Answer:
Item "ve" can also be solved through the sum , where is the probability that during the shift only two machines will require adjustment. This event, in turn, includes 3 incompatible outcomes, which are signed by analogy with the "be" item. Try to find the probability yourself to check the whole problem with the help of equality.
Task 9
Three guns fired a volley at the target. The probability of hitting with one shot only from the first gun is 0.7, from the second - 0.6, from the third - 0.8. Find the probability that: 1) at least one projectile hits the target; 2) only two projectiles will hit the target; 3) the target will be hit at least twice.
Solution and answer at the end of the lesson.
And again about coincidences: in the event that, by condition, two or even all values of the initial probabilities coincide (for example, 0.7; 0.7 and 0.7), then exactly the same solution algorithm should be followed.
In conclusion of the article, we will analyze another common puzzle:
Task 10
The shooter hits the target with the same probability with each shot. What is this probability if the probability of at least one hit in three shots is 0.973.
Solution: denote by - the probability of hitting the target with each shot.
and through - the probability of a miss with each shot.
Let's write down the events:
- with 3 shots, the shooter will hit the target at least once;
- the shooter will miss 3 times.
According to the condition, then the probability of the opposite event:
On the other hand, according to the theorem of multiplication of probabilities of independent events: 
Thus: 
- the probability of a miss with each shot.
As a result:
is the probability of hitting each shot.
Answer: 0,7
Simple and elegant.
In the considered problem, additional questions can be raised about the probability of only one hit, only two hits, and the probability of three hits on the target. The solution scheme will be exactly the same as in the two previous examples: 
However, the fundamental substantive difference is that there are repeated independent tests, which are performed sequentially, independently of each other and with the same probability of outcomes.
It can be difficult to directly count the cases that favor a given event. Therefore, to determine the probability of an event, it is advantageous to represent a given event as a combination of some other, more simple events. In this case, however, one must know the rules that the probabilities obey when a combination of events occurs. It is to these rules that the theorems mentioned in the title of the paragraph refer.
The first of these relates to calculating the probability that at least one of several events will occur.
Addition theorem.
Let A and B be two incompatible events. Then the probability that at least one of these two events will occur is equal to the sum of their probabilities:
Proof. Let be a complete group of pairwise incompatible events. If then among these elementary events there are exactly events favorable to A, and exactly events favorable to B. Since the events A and B are incompatible, then none of the events can favor both of these events. The event (A or B), consisting in the fact that at least one of these two events occurs, is obviously favored by both each of the events favorable to A, and each of the events
Favorable V. Therefore total number events favorable to the event (A or B) is equal to the sum from which it follows:
Q.E.D.
It is easy to see that the addition theorem formulated above for the case of two events can be easily carried over to the case of any finite number of them. Namely, if pairwise incompatible events, then
For the case of three events, for example, one can write
An important consequence of the addition theorem is the statement: if events are pairwise incompatible and uniquely possible, then
Indeed, the event is either or or by assumption certain and its probability, as was indicated in § 1, is equal to one. In particular, if two mutually opposite events mean, then
Let us illustrate the addition theorem with examples.
Example 1. When shooting at a target, the probability of making an excellent shot is 0.3, and the probability of making a good shot is 0.4. What is the probability of getting at least "good" for a shot?
Solution. If event A means getting an excellent grade, and event B means getting a good grade, then
Example 2. An urn containing white, red and black balls contains white balls and I red ones. What is the probability of drawing a non-black ball?
Solution. If event A is the appearance of a white ball, and event B is a red ball, then the appearance of a ball is not black
means the appearance of either a white or red ball. Since by definition of probability
then, by the addition theorem, the probability of the appearance of a non-black ball is equal to;
This problem can be solved in this way. Let the event C consist in the appearance of a black ball. The number of black balls is equal to so that P (C) The appearance of a non-black ball is the opposite event C, therefore, based on the above corollary from the addition theorem, we have:
as before.
Example 3. In a cash and clothing lottery, for a series of 1000 tickets, there are 120 cash and 80 clothing winnings. What is the probability of any win per lottery ticket?
Solution. If we designate through A an event consisting in the loss of a monetary gain and through B - a clothing one, then from the definition of probability it follows
The event of interest to us is (A or B), so the addition theorem implies
Thus, the probability of any win is 0.2.
Before moving on to the next theorem, it is necessary to familiarize yourself with a new important concept - the concept of conditional probability. For this purpose, we will start by looking at the following example.
Suppose there are 400 light bulbs in the warehouse, made in two different factories, with the first producing 75% of all light bulbs, and the second - 25%. Suppose that among the bulbs manufactured by the first factory, 83% meet the conditions of a certain standard, and for the products of the second factory this percentage is 63. Let's determine the probability that a light bulb randomly taken from the warehouse will meet the conditions of the standard.
Note that the total number of standard bulbs available consists of bulbs made first.
factory, and 63 bulbs made by the second factory, that is, equal to 312. Since the choice of any bulb should be considered equally possible, we have 312 favorable cases out of 400, so that
where event B is that the bulb we have chosen is standard.
In this calculation, no assumptions were made about the production of which factory the light bulb we chose belongs to. If any assumptions of this kind are made, then it is obvious that the probability of interest to us may change. So, for example, if it is known that the selected light bulb was manufactured at the first factory (event A), then the probability that it is standard will no longer be 0.78, but 0.83.
This kind of probability, that is, the probability of an event B, provided that the event A takes place, is called the conditional probability of the event B, provided that the event A occurs and denote
If, in the previous example, we denote by A the event that the selected bulb is made in the first factory, then we can write
Now we can formulate an important theorem related to the calculation of the probability of coincidence of events.
Multiplication theorem.
The probability of combining events A and B is equal to the product of the probability of one of the events by the conditional probability of the other, assuming that the first took place:
In this case, the combination of events A and B is understood as the occurrence of each of them, that is, the occurrence of both events A and events B.
Proof. Consider a complete group of equally possible pairwise incompatible events, each of which can be favorable or unfavorable for both event A and event B.
Let's break all these events into four different groups as follows. The first group includes those events that favor both event A and event B; the second and third groups include such events that favor one of the two events of interest to us and do not favor the other, for example, the second group - those that favor A, but do not favor B, and the third - those that favor B, but do not favor A; finally to
The fourth group includes those of the events that do not favor either A or B.
Since the numbering of events does not play a role, we can assume that this division into four groups looks like this:
I group:
II group:
III group:
IV group:
Thus, among equally possible and pairwise incompatible events, there are events that favor both event A and event B, I events that favor event A, but do not favor the event, events that favor B, but do not favor A, and, finally, events that do not favor neither A nor B.
Note, by the way, that any of the four groups we have considered (and even more than one) may not contain a single event. In this case, the corresponding number, indicating the number of events in such a group, will be equal to zero.
Our grouping allows us to immediately write
for the combination of events A and B is favored by the events of the first group and only by them. The total number of events favorable to A is equal to the total number of events in the first and second groups, and favorable to B is equal to the total number of events in the first and third groups.
We now calculate the probability, that is, the probability of event B, provided that event A took place. Now the events included in the third and fourth groups disappear, since their appearance would contradict the occurrence of event A, and the number of possible cases is no longer equal to . Of these, event B is favored only by the events of the first group, so we get:
To prove the theorem, it suffices now to write the obvious identity:
and replace all three fractions in it with the probabilities calculated above. We arrive at the equality stated in the theorem:
It is clear that the identity we have written above makes sense only if A is always true, unless A is an impossible event.
Since the events A and B are equal, by swapping them, we get another form of the multiplication theorem:
However, this equality can be obtained in the same way as the previous one, if we note that using the identity
Comparing the right-hand sides of the two expressions for the probability P(A and B), we get a useful equality:
Let us now consider examples illustrating the multiplication theorem.
Example 4. In the products of some enterprise, 96% of the products are recognized as suitable (event A). The first grade (event B) is owned by 75 items out of every hundred fit. Determine the probability that an arbitrarily taken product will be suitable and belongs to the first grade.
Solution. The desired probability is the probability of combining events A and B. According to the condition, we have: . So the multiplication theorem gives
Example 5. The probability of hitting the target with a single shot (event A) is 0.2. What is the probability of hitting the target if 2% of the fuses fail (i.e., in 2% of cases the shot does not
Solution. Let the event B be that the shot will occur, and B be the opposite event. Then by assumption and according to the corollary of the addition theorem . Further, according to the condition
Hitting the target means the combination of events A and B (the shot will occur and give a hit), therefore, according to the multiplication theorem
An important special case of the multiplication theorem can be obtained by using the concept of independence of events.
Two events are said to be independent if the probability of one of them does not change as a result of whether or not the other occurs.
Examples of independent events are the loss of a different number of points when a dice is thrown again or one or another side of the coins when a coin is thrown again, since it is obvious that the probability of a coat of arms falling out on the second throw is equal regardless of whether the coat of arms fell or did not fall in the first.
Similarly, the probability of drawing a white ball a second time from an urn with white and black balls, if the first ball drawn first is previously returned, does not depend on whether the white or black ball was drawn the first time. Therefore, the results of the first and second taking out are independent of each other. Conversely, if the ball drawn first does not return to the urn, then the result of the second removal depends on the first one, because the composition of the balls in the urn after the first removal changes depending on its outcome. Here we have an example of dependent events.
Using the notation adopted for conditional probabilities, we can write the condition for the independence of events A and B in the form
Using these equalities, we can bring the multiplication theorem for independent events to the following form.
If events A and B are independent, then the probability of their combination is equal to the product of the probabilities of these events:
Indeed, it suffices to put in the original expression of the multiplication theorem, which follows from the independence of events, and we obtain the required equality.
Let us now consider several events: We will call them independent in the aggregate if the probability of occurrence of any of them does not depend on whether any other events in question have occurred or not
In the case of events that are independent in the aggregate, the multiplication theorem can be extended to any finite number of them, due to which it can be formulated as follows:
The probability of combining events that are independent in the aggregate is equal to the product of the probabilities of these events:
Example 6. A worker maintains three automatic machines, each of which must be approached to troubleshoot if the machine stops. The probability that the first machine will not stop within an hour is 0.9. The same probability for the second machine is 0.8 and for the third - 0.7. Determine the probability that within an hour the worker will not need to go to any of the machines he serves.
Example 7. Probability of shooting down an aircraft with a rifle shot What is the probability of destroying an enemy aircraft if 250 rifles are fired simultaneously?
Solution. The probability that the plane will not be shot down with a single shot, according to the addition theorem, is Then, using the multiplication theorem, the probability that the plane will not be shot down with 250 shots can be calculated as the probability of combining events. It equals After that, we can use the addition theorem again and find the probability that the plane will be shot down as the probability of the opposite event
This shows that although the probability of shooting down an aircraft with a single rifle shot is negligible, nevertheless, when firing from 250 rifles, the probability of shooting down an aircraft is already very tangible. It increases significantly if the number of rifles is increased. So, when firing from 500 rifles, the probability of shooting down an aircraft, as it is easy to calculate, is equal to when firing from 1000 rifles - even.
The multiplication theorem proved above allows us to somewhat extend the addition theorem by extending it to the case of compatible events. It is clear that if events A and B are compatible, then the probability of the occurrence of at least one of them is not equal to the sum of their probabilities. For example, if event A means an even number
the number of points when throwing a dice, and the event B is the loss of a number of points that is a multiple of three, then the event (A or B) is favored by the loss of 2, 3, 4 and 6 points, that is
On the other hand, that is. So in this case
This shows that in the case of compatible events, the probability addition theorem must be changed. As we will now see, it can be formulated in such a way that it is valid for both compatible and incompatible events, so that the addition theorem considered earlier turns out to be a special case of the new one.
Events that do not favor A.
All elementary events that favor an event (A or B) must favor either only A, or only B, or both A and B. Thus, the total number of such events is equal to
and the probability
Q.E.D.
Applying formula (9) to the above example of the loss of the number of points when throwing a dice, we get:
which coincides with the result of direct calculation.
Obviously, formula (1) is a special case of (9). Indeed, if events A and B are incompatible, then the probability of coincidence
example. IN electrical circuit two fuses in series. The probability of failure of the first fuse is 0.6, and the second is 0.2. Let us determine the probability of a power failure as a result of the failure of at least one of these fuses.
Solution. Since events A and B, consisting in the failure of the first and second of the fuses, are compatible, the desired probability is determined by formula (9):
Exercises
Lecture 7. Probability theory
CONSEQUENCES OF THEOREMS OF ADDITION AND MULTIPLICATION
Addition theorem for joint event probabilities
The addition theorem for incompatible events. Here we will present the addition theorem for joint events.
Two events are called joint, if the appearance of one of them does not exclude the appearance of the other in the same trial.
Example 1 . A - the appearance of four points when throwing a dice; B - the appearance of an even number of points. Events A and B are joint.
Let the events A and B be joint, and the probabilities of these events and the probability of their joint occurrence are given. How to find the probability of an event A + B, consisting in the fact that at least one of the events A and B will appear? The answer to this question is given by the addition theorem for the probabilities of joint events.
Theorem. The probability of occurrence of at least one of the two joint events is equal to the sum of the probabilities of these events without the probability of their joint occurrence: P(A + B) = P(A) + P(B) - P(AB).
Proof . Since the events A and B, by condition, are joint, the event A + B will occur if one of the following three incompatible events occurs: . According to the theorem of addition of probabilities of incompatible events, we have:
P (A + B) \u003d P (A) + P (B) + P (AB).(*)
Event A will occur if one of two incompatible events occurs: A
or AV. By the addition theorem for the probabilities of incompatible events, we have
P (A) \u003d P (A) + P (AB).
P (A) \u003d P (A) - P (AB).(**)
Similarly, we have
P(B) = P(ĀB) + P(AB).
P(ĀB) = P(B) - P(AB).(***)
Substituting (**) and (***) into (*), we finally get
P(A + B) = P(A) + P(B) - P(AB).(****)
Q.E.D.
Remark 1. When using the resulting formula, it should be borne in mind that events A and B can be both independent, and dependent.
For independent events
P (A + B) \u003d P (A) + P (B) - P (A) * P (B);
For dependent events
P (A + B) \u003d P (A) + P (B) - P (A) * P A (B).
Remark 2. If events A and B incompatible, then their combination is an impossible event and, therefore, P(AB) = 0.
The formula (****) for incompatible events takes the form
P(A + B) = P(A) + P(B).
We have again obtained the addition theorem for incompatible events. Thus, the formula (****) is valid for both joint and non-joint events.
Example 2
The probabilities of hitting the target when firing the first and second guns are respectively equal: p 1 = 0.7; p2 = 0.8. Find the probability of hitting with one volley
(from both guns) by at least one of the guns.
Solution . The probability of hitting the target by each of the guns does not depend on the result of firing from the other gun, so the events A (hit by the first gun) and B (hit by the second gun) are independent.
Probability of event AB (both guns hit)
P (AB) \u003d P (A) * P (B) \u003d 0.7 * 0.8 \u003d 0.56.
The desired probability P (A + B) \u003d P (A) + P (B) - P (AB) \u003d 0.7 + 0.8 - 0.56 \u003d 0.94.
Remark 3. Since in this example events A and B are independent, it was possible to use the formula Р = 1 - q 1 q 2
Indeed, the probabilities of events opposite events A and B, i.e. miss probabilities are:
q 1 \u003d 1 - p 1 \u003d 1 - 0.7 \u003d 0.3;
q 2 \u003d 1 - p 2 \u003d 1 - 0.8 \u003d 0.2;
The desired probability that in one volley at least one gun will hit is equal to
P \u003d 1 - q 1 q 2 \u003d 1 - 0.3 * 0.2 \u003d 1 - 0.06 \u003d 0.94.
As expected, the same result is obtained.
Let the events A And IN are incompatible, and the probabilities of these events are known. Question: how to find the probability that one of these disjoint events will occur? This question is answered by the addition theorem.
Theorem.The probability of occurrence of one of two incompatible events is equal to the sum of the probabilities of these events:
p(A + IN) = p(A) + p(IN) (1.6)
Proof. Indeed, let n- the total number of all equally possible and incompatible (i.e. elementary) outcomes. Let the event A favors m 1 outcomes, and the event IN – m 2 outcomes. Then, according to the classical definition, the probabilities of these events are: p(A) = m 1 / n, p(B) = m 2 / n .
Since the events A And IN inconsistent, then none of the outcomes favorable to the event A, does not favor the event IN(see diagram below).
Therefore, the event A+IN will favor m 1 + m 2 outcomes. Therefore, for the probability p(A+B) we get:
Consequence 1. The sum of the probabilities of events forming a complete group is equal to one:
p(A) + p(IN) + p(WITH) + … + p(D) = 1.
Indeed, let the events A,IN,WITH, … , D form a complete group. Because of this, they are incompatible and the only possible ones. Therefore the event A + B + C + …+D, which consists in the appearance (as a result of testing) of at least one of these events, is reliable, i.e. A+B+C+…+D = And p(A+B+C+ …+D) = 1.
Due to the incompatibility of events A,IN,WITH, … , D the correct formula is:
p(A+B+C+ …+D) = p(A) + p(IN) + p(WITH) + … + p(D) = 1.
Example. An urn contains 30 balls, of which 10 are red, 5 are blue and 15 are white. Find the probability of drawing a red or blue ball, provided that only one ball is drawn from the urn.
Solution. Let the event A 1 is the extraction of the red ball, and the event A 2 - extraction of the blue ball. These events are incompatible, and p(A 1) = 10 / 30 = 1 / 3; p(A 2) = 5/30 = 1/6. By the addition theorem, we get:
p(A 1 + A 2) = p(A 1) + p(A 2) = 1 / 3 + 1 / 6 = 1 / 2.
Remark 1. We emphasize that, according to the meaning of the problem, it is necessary first of all to establish the nature of the events under consideration - whether they are incompatible. If the above theorem is applied to joint events, then the result will be incorrect.
