Addition theorem for joint and inconsistent events. Probability addition formulas
Addition and multiplication of probabilities. This article will focus on solving problems in probability theory. Earlier, we have already analyzed some of the simplest tasks, to solve them it is enough to know and understand the formula (I advise you to repeat it).
There are types of tasks a little more complicated, to solve them you need to know and understand: the rule of addition of probabilities, the rule of multiplication of probabilities, the concepts of dependent and independent events, opposite events, joint and incompatible events. Do not be intimidated by the definitions, everything is simple)).In this article, we will consider just such tasks.
Some important and simple theory:
inconsistent if the appearance of one of them excludes the appearance of others. That is, only one specific event can occur, or another.
Classic example: when throwing dice(dice) can only be one, or only two, or only three, etc. Each of these events is incompatible with the others and the commission of one of them excludes the commission of the other (in one trial). It is the same with a coin - falling "heads" excludes the possibility of falling "tails".
This also applies to more complex combinations. For example, two lights are on. Each of them may or may not burn out for some time. There are options:
- The first burns out and the second burns out
- The first burns out and the second does not burn out
- The first does not burn out and the second burns out
- The first does not burn out and the second burns out.
All these 4 variants of events are incompatible - they simply cannot happen together and none of them with any other ...
Definition: Events are called joint if the appearance of one of them does not exclude the appearance of the other.
Example: a queen will be taken from the deck of cards and a spades card will be taken from the deck of cards. Two events are considered. These events are not mutually exclusive - you can pull out the Queen of Spades and thus both events will occur.
On the sum of probabilities
The sum of two events A and B is called event A + B, which consists in the fact that either event A or event B or both will occur simultaneously.
If happen inconsistent events A and B, then the probability of the sum of these events is equal to the sum of the probabilities of events:
Dice example:
We throw the dice. What is the probability of getting a number less than four?
Numbers less than four are 1,2,3. We know that the probability of getting one is 1/6, two is 1/6, three is 1/6. These are incompatible events. We can apply the addition rule. The probability of getting a number less than four is:
Indeed, if we proceed from the concept of classical probability: then the number of all possible outcomes is 6 (the number of all sides of the cube), the number of favorable outcomes is 3 (the fallout of one, two or three). The desired probability is 3 to 6 or 3/6 = 0.5.
* The probability of the sum of two joint events is equal to the sum of the probabilities of these events without taking into account their joint occurrence: P (A + B) = P (A) + P (B) -P (AB)
On the multiplication of probabilities
Let there be two inconsistent events A and B, their probabilities are respectively equal to P (A) and P (B). The product of two events A and B is called such an event A · B, which consists in the fact that these events will occur together, that is, both event A and event B will occur.The probability of such an event is equal to the product of the probabilities of events A and B.Calculated by the formula:
As you have already noticed, the logical connective "AND" means multiplication.
Example with the same die:Throw the dice two times. What is the probability of getting two sixes?
The probability of getting a six for the first time is 1/6. The second time it is also equal to 1/6. The probability of getting a six both for the first time and for the second time is equal to the product of the probabilities:
Speaking simple language: when a certain event occurs in one test, and then another (others) occurs (s), then the probability that they will occur together is equal to the product of the probabilities of these events.
We solved the problems with the dice, but we used only logical reasoning, we did not use the product formula. In the problems considered below, you cannot do without formulas, or rather, it will be easier and faster to get a result with them.
There is one more nuance worth mentioning. When reasoning in solving problems, the concept of SIMULTANITY of the occurrence of events is used. Events happen SIMULTANEOUSLY - this does not mean that they happen in one second (at the same time). This means that they occur in a certain period of time (in one test).
For instance:
Two lamps burn out within a year (it can be said - simultaneously within a year)
Two machines break down within a month (it can be said - simultaneously within a month)
The dice are rolled three times (points drop out at the same time, this means in one test)
The biathlete fires five shots. Events (shots) occur during one trial.
Events A and B are NOT dependent if the probability of any of them does not depend on the occurrence or non-occurrence of another event.
Consider the tasks:
Two factories produce the same glass for car headlights. The first factory produces 35% of these glasses, the second - 65%. The first factory produces 4% of defective glasses, and the second - 2%. Find the likelihood that a glass you accidentally buy in a store turns out to be defective.
The first factory produces 0.35 products (glasses). The probability of buying defective glass from the first factory is 0.04.
The second factory produces 0.65 glasses. The probability of buying defective glass from the second factory is 0.02.
The probability that the glass was bought at the first factory And at the same time it turns out to be defective is 0.35 ∙ 0.04 = 0.0140.
The probability that the glass was bought at the second factory And at the same time it turns out to be defective is 0.65 ∙ 0.02 = 0.0130.
Buying defective glass in a store means that it (defective glass) was purchased EITHER from the first factory OR from the second. These are inconsistent events, that is, we add up the probabilities obtained:
0,0140 + 0,0130 = 0,027
Answer: 0.027
If grandmaster A. plays white, then he wins against grandmaster B. with a probability of 0.62. If A. plays black, then A. wins against B. with a probability of 0.2. Grandmasters A. and B. play two games, and in the second game they change the color of the pieces. Find the probability that A. wins both times.
The chances of winning the first and second games are independent of each other. It is said that a grandmaster must win both times, that is, win the first time AND, at the same time, win a second time. In the case when independent events must occur together, the probabilities of these events are multiplied, that is, the multiplication rule is used.
The probability of the occurrence of these events will be equal to 0.62 ∙ 0.2 = 0.124.
Answer: 0.124
In the geometry exam, the student gets one question from the list of exam questions. The probability that this is an inscribed circle question is 0.3. The probability that this is a Parallelogram question is 0.25. There are no questions that simultaneously relate to these two topics. Find the probability that a student will get a question on one of these two topics on the exam.
That is, it is necessary to find the probability that the student will get the question EITHER on the topic "Inscribed circle" OR on the topic "Parallelogram". In this case, the probabilities are summed up, since these events are inconsistent and any of these events can occur: 0.3 + 0.25 = 0.55.
* Incompatible events are events that cannot happen at the same time.
Answer: 0.55
The biathlete shoots targets five times. The probability of hitting a target with one shot is 0.9. Find the probability that the biathlete hits the targets the first four times and the last one misses. Round the result to the nearest hundredth.
Since the biathlete hits the target with a probability of 0.9, he misses with a probability of 1 - 0.9 = 0.1
* A miss and a hit are events that cannot occur simultaneously with one shot, the sum of the probabilities of these events is 1.
We are talking about the commission of several (independent) events. If an event occurs and another (subsequent) occurs at the same time (test), then the probabilities of these events are multiplied.
The probability of producing independent events is equal to the product of their probabilities.
Thus, the probability of the event “hit, hit, hit, hit, missed” is 0.9 ∙ 0.9 ∙ 0.9 ∙ 0.9 ∙ 0.1 = 0.06561.
Round to hundredths, we get 0.07
Answer: 0.07
There are two payment machines in the store. Each of them can be faulty with a probability of 0.07, regardless of the other machine. Find the probability that at least one machine is operational.
Let us find the probability that both automata are faulty.
These events are independent, which means the probability will be equal to the product of the probabilities of these events: 0.07 ∙ 0.07 = 0.0049.
This means that the probability that both automata are in good working order or one of them will be equal to 1 - 0.0049 = 0.9951.
* Both are in good working order and one completely - meets the condition "at least one".
One can imagine the probabilities of all (independent) events to be checked:
1. "faulty-faulty" 0.07 ∙ 0.07 = 0.0049
2. "good-bad" 0.93 ∙ 0.07 = 0.0651
3. "faulty-working" 0.07 ∙ 0.93 = 0.0651
4. "good-good" 0.93 ∙ 0.93 = 0.8649
To determine the probability that at least one machine is operational, it is necessary to add the probabilities of independent events 2, 3 and 4: A credible event is called an event that is likely to occur as a result of experience. The event is called impossible, if it never happens as a result of experience.
For example, if one ball is taken out at random from a box containing only red and green balls, then the appearance of a white ball among the taken out balls is an impossible event. The appearance of red and the appearance of green balls form a complete group of events.
Definition: Events are called equally possible if there is no reason to believe that one of them will appear as a result of the experiment with a higher probability.
In the above example, the appearance of red and green balls are equally possible events if the box contains the same number of red and green balls. If there are more red balls in the box than green ones, then the appearance of a green ball is an event less likely than the appearance of a red one.
In we will consider more problems where the sum and the product of the probabilities of events are used, do not miss it!
That's all. I wish you success!
Best regards, Alexander Krutitskikh.
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- So what?
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It can be difficult to count directly the cases favorable to a given event. Therefore, in order to determine the probability of an event, it can be advantageous to represent the given event as a combination of some other, simpler events. In this case, however, one must know the rules that govern the probabilities for a combination of events. It is to these rules that the theorems mentioned in the title of this section 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 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, favors, obviously, 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 of which 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 inconsistent and uniquely possible, then
Indeed, the event is either or or by assumption is reliable and its probability, as indicated in § 1, is equal to one. In particular, if they mean two mutually opposite events, then
Let us illustrate the addition theorem with examples.
Example 1. When shooting at a target, the probability of making a perfect shot is 0.3, and the probability of making a shot with a rating of "good" is 0.4. What is the probability of getting a rating of 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. In an urn containing balls of white, red and black, there are white balls and I red. What is the probability of taking out a ball that is not black?
Solution. If event A consists in the appearance of a white ball, and event B - a red ball, then the appearance of a ball is not black
means the appearance of either a white or a red ball. Since by definition of probability
then, by the addition theorem, the probability of the appearance of a ball that is not black is equal to;
This problem can be solved like this. Let event C consist in the appearance of a black ball. The number of black balls is equal so that P (C) The appearance of a ball that is not black is the opposite event C, therefore, based on the above corollary from the addition theorem, we have:
as before.
Example 3. In the cash lottery for a series of 1000 tickets, there are 120 cash and 80 item winnings. What is the probability of any winnings per lottery ticket?
Solution. If we denote by A an event consisting in the loss of a monetary gain and by B - a thing, then the definition of probability implies
The event of interest to us represents (A or B), therefore the addition theorem implies
Thus, the probability of any winning is 0.2.
Before moving on to the next theorem, you need to familiarize yourself with an important new 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 bulbs, and the second producing 25%. Suppose that among the bulbs manufactured by the first plant, 83% meet the conditions of a certain standard, and for the products of the second plant, this percentage is 63. Let us determine the probability that a light bulb accidentally taken from the warehouse will meet the conditions of the standard.
Note that the total number of standard bulbs available consists of the first manufactured bulbs.
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 light bulb we have chosen is standard.
In this calculation, no assumptions were made about which plant the selected light bulb 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 made at the first plant (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 event B, provided that event A takes place, is called the conditional probability of event B, provided that event A occurs and is denoted
If in the previous example we denote by A the event that the selected light bulb is made at the first factory, then we can write
Now we can formulate an important theorem related to calculating the probability of coincidence of events.
Multiplication theorem.
The probability of coinciding 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 means the onset of each of them, that is, the onset of both events A and 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 of the events that favor both event A and event B; to the second and third groups we will attribute such events that favor one of the two events of interest to us and do not favor the other, for example, to the second group - those that favor A, but do not favor B, and to the third - those that favor B, but do not A; finally to
the fourth group will include those events that do not favor either A or B.
Since the numbering of events does not matter, it can be assumed that this division into four groups looks like this:
Group I:
Group II:
IV group:
Thus, among the equally possible and pairwise incompatible events, there are events favorable to both event A and event B, I events, favorable to event A, but not favorable to the event of events favorable to B, but not favorable to A, and, finally, events that are not favorable neither A nor B.
Note, by the way, that any of the four groups we have considered (and not even one) may not contain a single event. In this case corresponding number, indicating the number of events in such a group, will be equal to zero.
The division into groups made by us allows you to immediately write
for the combination of events A and B is favored by the events of the first group and only they. The total number of events favorable to A is equal to the total number of events in the first and second groups, and those favorable to B is equal to the total number of events in the first and third groups.
Let's calculate now 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 onset of event A, and the number of possible cases turns out to be no longer equal. 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 asserted in the theorem:
It is clear that the identity we have written above makes sense only if it is always true, unless A is an impossible event.
Since events A and B are equal, then, 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 notice that we use 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 a certain enterprise, 96% of products are recognized as suitable (event A). The first grade (event B) turns out to belong to 75 items out of every hundred good ones. Determine the probability that a randomly taken product will be suitable and belong to the first grade.
Solution. The desired probability is the probability of coinciding events A and B. By the condition we have:. Therefore, 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 fuses fail (i.e., in 2% of cases the shot is not
Solution. Let event B mean that the shot will occur, and B means the opposite event. Then by hypothesis and by the corollary to the addition theorem. Further, by condition.
The defeat of the target means the combination of events A and B (a shot will occur and give a hit), therefore, according to the multiplication theorem
Important special case multiplication theorems can be obtained using the concept of independence of events.
Two events are called independent if the probability of one of them does not change as a result of the occurrence of the other.
Examples of independent events are the falling out of a different number of points when the dice or one or another side of the coins are thrown again when the coin is thrown again, since it is obvious that the probability of the coat of arms falling on the second throw is equal regardless of whether the coat of arms fell or did not fall in the first.
Likewise, the probability of removing a white ball from an urn with white and black balls a second time if the ball removed first is previously returned does not depend on whether the white or black ball was removed the first time. Therefore, the results of the first and second withdrawal are independent of each other. On the contrary, if the ball taken out 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 the conditional probabilities, we can write the condition for the independence of events A and B in the form
Using these equalities, we can reduce 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 is enough to assume in the original expression of the multiplication theorem what 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 the 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 independent events in the aggregate is equal to the product of the probabilities of these events:
Example 6. A worker is servicing three automatic machines, each of which must be approached to eliminate the malfunction 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 it is 0.7. Determine the probability that within an hour the worker will not need to approach any of the machines he maintains.
Example 7. Probability of shooting down an aircraft with a rifle shot What is the probability of destroying an enemy aircraft while firing 250 rifles at the same time?
Solution. The probability that the plane will not be shot down with a single shot is equal by the addition theorem.Then, using the multiplication theorem, we can calculate the probability that the plane will not be shot down after 250 shots, as the probability of coincidence of events. It is equal After that, we can again use the addition theorem and find the probability that the plane will be hit as the probability of the opposite event
It can be seen from this 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 quite 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 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 the occurrence of an even
the number of points when throwing a dice, and event B is the falling out of the number of points that is a multiple of three, then the event (A or B) favors the falling of 2, 3, 4 and 6 points, that is
On the other hand, that is. So in this case
Hence it is clear 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 previously considered addition theorem turns out to be a special case of a new one.
Events that A are not favorable.
All elementary events that favor the event (A or B) should favor either only A, or only B, or both A and B. Thus, the total number of such events is
and the probability
Q.E.D.
Applying formula (9) to the above example of the dropout of the number of points when throwing a dice, we get:
which is the same as the result of direct counting.
It is obvious that formula (1) is a special case of (9). Indeed, if events A and B are inconsistent, then the probability of coincidence
An example. Two fuses are connected in series in the electrical circuit. The probability of failure of the first fuse is 0.6, and the second 0.2. Let us determine the probability of a power outage as a result of 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 the formula (9):
Exercises
The addition theorem for the probabilities of two events. The probability of the sum of two 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).
The addition theorem for the probabilities of two incompatible events. The probability of the sum of two incompatible events is equal to the sum of the probabilities of these:
P (A + B) = P (A) + P (B).
Example 2.16. The shooter shoots at a target divided into 3 areas. The probability of hitting the first area is 0.45, the second - 0.35. Find the probability that the shooter will hit either the first or the second area in one shot.
Solution.
Events A- "the shooter hit the first area" and V- “the shooter hit the second area” - they are incompatible (hitting one area excludes hitting another), so the addition theorem is applicable.
The sought probability is:
P (A + B) = P (A) + P (B) = 0,45+ 0,35 = 0,8.
Probability addition theorem P inconsistent events. The probability of the sum of n inconsistent events is equal to the sum of the probabilities of these:
P (A 1 + A 2 + ... + A p) = P (A 1) + P (A 2) + ... + P (A p).
The sum of the probabilities of opposite events is equal to one:
Event probability V provided that an event has occurred A, is called the conditional probability of the event V and denoted as follows: P (B / A), or P A (B).
. The probability of the product of two events is equal to the product of the probability of one of them by the conditional probability of the other, provided that the first event occurred:
P (AB) = P (A) P A (B).
Event V does not depend on the event A, if
P A (B) = P (B),
those. probability of event V does not depend on whether the event happened A.
A theorem for the multiplication of the probabilities of two independent events.The probability of the product of two independent events is equal to the product of their probabilities:
P (AB) = P (A) P (B).
Example 2.17. 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 hitting at least one of the guns with one volley (from both 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, therefore the events A- "hit of the first gun" and V- "hit of the second weapon" are independent.
Event probability AB- "both guns hit":
Seeking probability
P (A + B) = P (A) + P (B) - P (AB)= 0,7 + 0,8 – 0,56 = 0,94.
Probability multiplication theorem P events.The probability of the product of n events is equal to the product of one of them by the conditional probabilities of all the others, calculated under the assumption that all previous events have occurred:
Example 2.18... The urn contains 5 white, 4 black and 3 blue balls. Each trial consists in removing one ball at random without returning it back. Find the probability that at the first test a white ball will appear (event A), at the second - a black (event B) and at the third - a blue (event C).
Solution.
The probability of a white ball appearing in the first trial:
The probability of a black ball appearing in the second trial, calculated on the assumption that a white ball appeared in the first trial, i.e. the conditional probability:
The probability of a blue ball appearing in the third trial, calculated on the assumption that a white ball appeared in the first trial, and a black ball in the second, i.e., the conditional probability:
The sought probability is:
Probability multiplication theorem P independent events.The probability of the product of n independent events is equal to the product of their probabilities:
P (A 1 A 2 ... A p) = P (A 1) P (A 2) ... P (A p).
The probability of the occurrence of at least one of the events. The probability of occurrence of at least one of the events А 1, А 2, ..., А п, independent in the aggregate, is equal to the difference between unity and the product of the probabilities of opposite events:
.
Example 2.19. The probabilities of hitting the target when firing three guns are as follows: p 1 = 0,8; p 2 = 0,7;p 3= 0.9. Find the probability of at least one hit (event A) with one volley from all guns.
Solution.
The probability of hitting the target by each of the guns does not depend on the results of firing from other guns, therefore the events in question A 1(hit by the first gun), A 2(hit by the second gun) and A 3(hit by the third weapon) are independent in the aggregate.
Probabilities of events opposite to events A 1, A 2 and A 3(i.e. the probabilities of misses) are respectively equal to:
, , 
.
The sought probability is:
If independent events A 1, A 2, ..., A p have the same probability equal to R, then the probability of occurrence of at least one of these events is expressed by the formula:
P (A) = 1 - q n,
where q = 1- p
2.7. Formula of total probability. Bayes' formula.
Let the event A can occur if one of the inconsistent events occurs H 1, H 2, ..., H p forming a complete group of events. Since it is not known in advance which of these events will occur, they are called hypotheses.
Event probability A calculated by total probability formula:
P (A) = P (H 1) P (A / H 1) + P (H 2) P (A / H 2) + ... + P (H p) P (A / H p).
Assume that an experiment has been made, as a result of which the event A happened. Conditional probabilities of events H 1, H 2, ..., H p regarding the event A are determined Bayes' formulas:
,
Example 2.20... In a group of 20 students who came for the exam, 6 are excellent, 8 are good, 4 are satisfactory and 2 are poor. The exam tickets contain 30 questions. A well-prepared student can answer all 30 questions, a well-prepared student can answer 24, a satisfactory student can answer 15, and a poorly prepared student can answer 7.
The student called at random answered three at random questions asked... Find the probability that this student is prepared: a) excellent; b) bad.
Solution.
Hypotheses - “the student is well prepared”;
- “the student is well prepared”;
- "the student is prepared satisfactorily";
- "the student is poorly prepared."
Before experience:
; ; ; ;
7. What is called a complete group of events?
8. What events are called equally possible? Give examples of such events.
9. What is called an elementary outcome?
10. What outcomes do I call favorable for this event?
11. What operations can be performed on events? Give them definitions. How are they indicated? Give examples.
12. What is called probability?
13. What is the probability of a certain event?
14. What is the probability of an impossible event?
15. What are the limits of the probability?
16. How is the geometric probability on a plane determined?
17. How is probability in space determined?
18. How is the probability on a straight line determined?
19. What is the probability of the sum of two events?
20. What is the probability of the sum of two incompatible events?
21. What is the probability of the sum of n inconsistent events?
22. What probability is called conditional? Give an example.
23. Formulate the probability multiplication theorem.
24. How to find the probability of occurrence of at least one of the events?
25. What events are called hypotheses?
26. When are the total probability formula and Bayesian formulas applied?
In cases where the event of interest is the sum of other events, the addition formula is used to find its probability.
The addition formula has two main varieties - for joint events and for inconsistent events. These formulas can be substantiated using Venn diagrams (Fig. 21). Recall that in these diagrams the probabilities of events are numerically equal to the areas of the zones corresponding to these events.
For two incompatible events :
P (A + B) = P (A) + P (B).(8, a)
For N inconsistent events , the probability of their sum is equal to the sum of the probabilities of these events:
= . (8b)
There are two important consequences from the formula for adding incompatible events .
Corollary 1.For events forming a complete group, the sum of their probabilities is equal to one:
= 1.
This is explained as follows. For events that form a complete group, on the left side of expression (8b) is the probability that one of the events will occur A i, but since the full group exhausts the entire list of possible events, one of these events will surely occur. Thus, on the left side, the probability of an event that will definitely occur - a reliable event is recorded. Its probability is equal to one.
Corollary 2.The sum of the probabilities of two opposite events is equal to one:
P (A) + P (Ā)= 1.
This consequence follows from the previous one, since opposite events always form a complete group.
Example 15
V The probability of an operational state of a technical device is 0.8. Find the probability of failure of this device over the same observation period.
R solution.
Important note... In the theory of reliability, it is customary to denote the probability of a working state by the letterR, and the probability of failure - by the letter q. In what follows, we will use these notation. Both probabilities are functions of time. So, for long periods time, the probability of an efficient state of any object approaches zero. The probability of failure of any object is close to zero for short periods of time. In cases where the observation period is not specified in the tasks, it is assumed that it is the same for all objects under consideration.
Finding a device in operable and failure states are opposite events. Using Corollary 2, we get the probability of device failure:
q = 1 - p = 1 - 0.8 = 0.2.
For two joint events addition formula for probabilities looks like:
P (A + B) = P (A) + P (B) - P (AB), (9)
which is illustrated by the Venn diagram (Fig. 22).
Indeed, in order to find the entire shaded area (it corresponds to the sum of events A + B), it is necessary to subtract the area of the common zone from the sum of the areas of figures A and B (it corresponds to the product of events AB), since otherwise it will be counted twice.
For three joint events, the addition formula probabilities gets more complicated:
P (A + B + C) = P (A) + P (B) + P (C) - P (AB) - P (AC) - P (BC) + P (ABC).(10)
On the Venn diagram (Fig. 23), the desired probability is numerically equal to the total area of the zone formed by events A, B, and C (to simplify the figure, the unit square is not shown on it).
After the areas of zones AB, AC and CB were subtracted from the sum of the areas of zones A, B and C, it turned out that the area of zone ABC was summed up three times and subtracted three times. Therefore, to account for this area, it must be added to the final expression.
With an increase in the number of terms, the addition formula becomes more and more cumbersome, but the principle of its construction remains the same: first, the probabilities of events taken one by one are summed up, then the probabilities of all paired combinations of events are subtracted, the probabilities of events taken by triples are added, the probabilities of combinations of events taken by fours and etc.
As a result, it should be emphasized : formula for adding probabilities joint events with the number of terms from three or more is cumbersome and inconvenient to use, its use in solving problems is impractical.
Example 16
For the power supply diagram below (Fig. 24), determine the probability of failure of the system as a whole Q C according to the probability of failure q i individual elements (generator, transformers and lines).
Failure states individual elements of the power supply system, as well as and health states are always pairwise cooperative events, since there are no fundamental obstacles to the simultaneous repair of, for example, the line and the transformer. The failure of the system occurs when any of its elements fails: either the generator, or the 1st transformer, or the line, or the 2nd transformer, or when any pair, any three or all four elements fail. Consequently, the desired event - system failure is the sum of failures of individual elements. To solve the problem, the formula for adding joint events can be used:
Q c = q g + q t1 + q l + q t2 - q g q t1 - q g q l - q g q t2 - q t1 q l - q t1 q t2 - q l q t2 + q g q t1 q l + q g q l q t2 + q g q t1 q t2 + q t1 q t2 q l - q g q t1 q l q t2.
This solution once again convinces of the cumbersomeness of the addition formula for joint events. In the future, another more rational way of solving this problem will be considered.
The solution obtained above can be simplified taking into account the fact that the probabilities of failures of individual elements of the power supply system for the period of one year usually used in reliability calculations are rather small (about 10 -2). Therefore, all terms except the first four can be discarded, which practically does not affect the numerical result. Then you can write:
Q with≈q g + q t1 + q l + q t2.
However, such simplifications must be treated with caution, carefully studying their consequences, since the terms that are often discarded may turn out to be commensurate with the first.
Example 17
Determine the likelihood of a healthy state of the system P S consisting of three elements that reserve each other.
Solution... Reserving each other elements on the logic diagram of the reliability analysis are shown connected in parallel (Fig. 25):
The redundant system is operational when either the 1st, or the 2nd, or the 3rd element is operational, or any pair is operational, or all three elements together. Consequently, the operable state of the system is the sum of the operable states of individual elements. By the addition formula for joint events R c = R 1 + R 2 + R 3 - R 1 R 2 - R 1 R 3 - R 2 R 3 + R 1 R 2 R 3... , where R 1, R 2 and R 3- the probabilities of the operational state of elements 1, 2 and 3, respectively.
In this case, it is impossible to simplify the solution by discarding the paired products, since such an approximation will give a significant error (these products are usually numerically close to the first three terms). As in Example 16, this problem has a different, more compact solution.
Example 18
For a double-circuit power line (Fig. 26), the probability of failure of each circuit is known: q 1 = q 2= 0.001. Determine the probabilities that the line will have one hundred percent throughput - P (R 100), fifty percent throughput - P (R 50), and the probability that the system will fail - Q.
The line has one hundred percent throughput when both the 1st and 2nd circuits are operational:
P (100%) = p 1 p 2 = (1 - q 1) (1 - q 2) =
= (1 – 0,001)(1 – 0,001) = 0,998001.
The line fails when both the 1st and 2nd circuits fail:
P (0%) = q 1 q 2 = 0.001 ∙ 0.001 = 10 -6.
The line has fifty percent throughput when the 1st circuit is operational and the 2nd fails, or when the 2nd circuit is functional and the 1st fails:
P (50%) = p 1 q 2 + p 2 q 1 = 2 ∙ 0.999 ∙ 10 -3 = 0.001998.
The last expression uses the addition formula for inconsistent events, which they are.
The events considered in this problem make up a complete group, so the sum of their probabilities is one.
Educational institution "Belarusian State
agricultural Academy"
Department of Higher Mathematics
ADDITION AND MULTIPLICATION OF PROBABILITIES. REPEATED INDEPENDENT TESTS
Lecture for students of the Faculty of Land Management
correspondence courses
Gorki, 2012
Addition and multiplication of probabilities. Repeated
independent tests
Addition of probabilities
The sum of two joint events A and V called an event WITH, consisting in the onset of at least one of the events A or V... Similarly, the sum of several joint events is an event consisting in the occurrence of at least one of these events.
The sum of two incompatible events A and V called an event WITH consisting of an offensive or event A, or events V... Similarly, the sum of several incompatible events is an event consisting in the occurrence of any one of these events.
The theorem of addition of the probabilities of inconsistent events is valid: the probability of the sum of two incompatible events is equal to the sum of the probabilities of these events , i.e. ... This theorem can be extended to any finite number of inconsistent events.
This theorem implies:
the sum of the probabilities of events forming a complete group is equal to one;
the sum of the probabilities of opposite events is equal to one, i.e. 
.
Example 1 ... The box contains 2 white, 3 red and 5 blue balls. The balls are mixed and one is taken out at random. What is the probability that the ball will be colored?
Solution ... Let's designate events:
A= (colored ball removed);
B= (removed white ball);
C= (red ball removed);
D= (blue ball removed).
Then A= C+ D... Since events C, D are inconsistent, then we will use the theorem of addition of the probabilities of inconsistent events:.
Example 2 ... The urn contains 4 white balls and 6 black ones. 3 balls are taken out of the urn at random. What is the likelihood that they are all the same color?
Solution ... Let's designate events:
A= (balls of the same color are taken out);
B= (white balls are taken out);
C= (black balls are taken out).
Because A=
B+
C and events V and WITH are inconsistent, then by the addition theorem for the probabilities of inconsistent events 
... Event probability V is equal to 
, where 
4,
... Substitute k and n into the formula and get 
Similarly, we find the probability of the event WITH:
, where 
,
, i.e. 
... Then 
.
Example 3 ... 4 cards are drawn at random from a deck of 36 cards. Find the probability that there are at least three aces among them.
Solution ... Let's designate events:
A= (among the drawn cards, at least three aces);
B= (among the drawn cards there are three aces);
C= (among the drawn cards there are four aces).
Because A=
B+
C and events V and WITH inconsistent, then 
... Find the probabilities of events V and WITH:
,
... Therefore, the probability that among the drawn cards there are at least three aces is equal to
0.0022.
Multiplication of probabilities
By product
two events A and V called an event WITH, consisting in the joint occurrence of these events: 
... This definition applies to any finite number of events.
Two events are called independent
if the probability of one of them occurring does not depend on whether another event has occurred or not. Events 
,
,
… ,
are called collectively independent
if the probability of occurrence of each of them does not depend on whether other events have occurred or not.
Example 4 ... Two arrows shoot at the target. Let's designate events:
A= (the first shooter hits the target);
B= (the second shooter hits the target).
Obviously, the probability of hitting the target by the first shooter does not depend on whether the second shooter hit or missed, and vice versa. Hence the events A and V independent.
The theorem of multiplication of probabilities of independent events is valid: the probability of the product of two independent events is equal to the product of the probabilities of these events : .
This theorem is also valid for n independent events in the aggregate:.
Example 5 ... Two shooters shoot at one target. The probability of hitting the first shooter is 0.9, and the second - 0.7. Both shooters fire one shot at the same time. Determine the probability that there will be two hits on the target.
Solution ... Let's designate events:
A
B
C= (both arrows hit the target).
Because 
and events A and V independent, then 
, i.e. ...
Events A and V are called dependent
, if the probability of one of them occurring depends on whether another event occurred or not. The probability of an event occurring A provided that the event V has already arrived, it is called conditional probability
and denoted 
or 
.
Example 6 ... The urn contains 4 white and 7 black balls. Balls are removed from the urn. Let's designate events:
A= (removed white ball);
B= (black ball removed).
Before starting to remove the balls from the urn 
... One ball was removed from the urn and it turned out to be black. Then the probability of the event A after the event V will be already different, equal 
... This means that the probability of an event A depends on the event V, i.e. these events will be dependent.
The theorem of multiplication of probabilities of dependent events is valid: the probability of the product of two dependent events is equal to the product of the probability of one of them by the conditional probability of the other, calculated under the assumption that the first event has already occurred, i.e. or .
Example 7 ... The urn contains 4 white balls and 8 red ones. Two balls are taken from it at random in succession. Find the probability that both balls are black.
Solution ... Let's designate events:
A= (the black ball is drawn first);
B= (the black ball is removed second).
Events A and V dependent since 
, a 
... Then 
.
Example 8 ... Three shooters shoot at the target independently of each other. The probability of hitting the target for the first shooter is 0.5, for the second - 0.6 and for the third - 0.8. Find the probability that there will be two hits on the target if each shooter fires one shot.
Solution ... Let's designate events:
A= (there will be two hits on the target);
B= (the first shooter hits the target);
C= (the second shooter will hit the target);
D= (third shooter will hit the target);
= (the first shooter will miss the target);
= (second shooter will miss the target);
= (third shooter will miss the target).
By the condition of the example 
,
,
,
,
,
... Since, then using the theorem of addition of the probabilities of incompatible events and the theorem of multiplication of the probabilities of independent events, we get:
Let the events 
form a complete group of events of some test, and the event A can occur with only one of these events. If the probabilities and conditional probabilities of the event are known A, then the probability of event A is calculated by the formula:
Or 
... This formula is called total probability formula
and events 
hypotheses
.
Example 9
... The assembly line receives 700 parts from the first machine and 300 parts 
from the second. The first machine gives 0.5% of scrap, and the second - 0.7%. Find the probability that the taken part will be defective.
Solution ... Let's designate events:
A= (the taken part will be defective);
= (the part is made on the first machine);
= (the part is made on the second machine).
The probability that the part is made on the first machine is 
... For the second machine 
... According to the condition, the probability of obtaining a defective part made on the first machine is equal to 
... For the second machine, this probability is 
... Then the probability that the taken part will be defective is calculated using the formula for the total probability 
If it is known that as a result of the test some event occurred A, then the probability that this event occurred with the hypothesis 
, is equal to 
, where 
- full probability of the event A... This formula is called Bayes' formula
and allows you to calculate the probabilities of events 
after it became known that the event A has already arrived.
Example 10 ... The same type of parts for cars are produced at two factories and go to the store. The first plant produces 80% of the total number of parts, and the second - 20%. The products of the first plant contain 90% of standard parts, and the second one - 95%. The buyer bought one part and it turned out to be standard. Find the probability that this part was manufactured at a second plant.
Solution ... Let's designate events:
A= (standard part purchased);
= (part made at the first plant);
= (part manufactured at the second plant).
By the condition of the example 
,
,
and 
... We calculate the total probability of the event A: 0.91. The probability that the part is manufactured at the second plant is calculated using the Bayes formula: 
.
Self-study assignments
The probability of hitting the target for the first shooter is 0.8, for the second - 0.7 and for the third - 0.9. The shooters fired one shot at a time. Find the probability that there are at least two hits on the target.
The repair shop received 15 tractors. It is known that 6 of them need to replace the engine, and the rest need to replace individual units. Three tractors are randomly selected. Find the probability that engine replacement is necessary for no more than two selected tractors.
The reinforced concrete plant produces panels, 80% of which are of the highest quality. Find the probability that out of three randomly selected panels, at least two will be of the highest grade.
Three workers are assembling bearings. The probability that a bearing assembled by the first worker is of the highest quality is 0.7, the second is 0.8 and the third is 0.6. For control, one bearing was taken at random from those assembled by each worker. Find the probability that at least two of them will be of the highest quality.
The probability of winning a lottery ticket for the first issue is 0.2, the second - 0.3, and the third - 0.25. There are one ticket for each issue. Find the probability of winning at least two tickets.
The accountant performs calculations using three reference books. The probability that the data of interest is in the first reference book is 0.6, in the second - 0.7, and in the third - 0.8. Find the probability that the data of interest to the accountant is contained in no more than two reference books.
Three automata make parts. The first automaton makes a top quality part with a probability of 0.9, the second - with a probability of 0.7, and the third - with a probability of 0.6. One piece is taken at random from each machine. Find the probability that there are at least two of the highest quality among them.
Parts of the same type are processed on two machines. The probability of manufacturing a non-standard part for the first machine is 0.03, for the second - 0.02. The processed parts are stacked in one place. Among them, 67% from the first machine, and the rest from the second. The part taken at random turned out to be standard. Find the probability that it was made on the first machine.
The workshop received two boxes of the same type of capacitors. The first box had 20 capacitors, of which 2 were faulty. In the second box there are 10 capacitors, of which 3 are faulty. The capacitors were put in one box. Find the probability that a capacitor taken from a box at random turns out to be operational.
On three machines, parts of the same type are made, which are fed to a common conveyor. Among all parts, 20% from the first machine, 30% from the second and 505 from the third. The probability of making a standard part on the first machine is 0.8, on the second - 0.6 and on the third - 0.7. The part taken turned out to be standard. Find the probability that this part was made on the third machine.
The picker receives 40% of the parts from the factory for assembly A, and the rest - from the factory V... The likelihood that the part is from the factory A- of the highest quality, equal to 0.8, and from the factory V- 0.9. The picker took one piece at random and it was not of the highest quality. Find the probability that this part is from the factory V.
10 students from the first group and 8 from the second were allocated to participate in student sports competitions. The probability that a student from the first group will be included in the national team of the academy is 0.8, and from the second - 0.7. The student chosen at random was included in the national team. Find the probability that he is from the first group.
