Directly counting cases favoring a given event may be difficult. Therefore, to determine the probability of an event, it can be advantageous to imagine this event as a combination of some other, simpler events. In this case, however, you need to know the rules that govern probabilities in combinations of events. It is to these rules that the theorems mentioned in the title of the paragraph relate.
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 no event can favor both of these events. An event (A or B), consisting in the occurrence of at least one of these two events, is obviously favored by both each of the events favoring A and each of the events
Favorable V. Therefore total number events favoring the event (A or B) is equal to the sum which follows:
Q.E.D.
It is easy to see that the addition theorem formulated above for the case of two events can easily be transferred to the case of any finite number of them. Precisely if there are 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 either or or is by assumption certain 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 an excellent shot is 0.3, and the probability of making a “good” shot is 0.4. What is the probability of receiving a score of at least “good” for a shot?
Solution. If event A means receiving an “excellent” rating, and event B means receiving a “good” rating, then
Example 2. In an urn containing white, red and black balls, there are white balls and I red balls. What is the probability of drawing a ball that is not black?
Solution. If event A consists of the appearance of a white ball, and event B of a red ball, then the appearance of the 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 a non-black ball appearing is equal;
This problem can be solved this way. 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 non-black ball is the opposite event of C, therefore, based on the above corollary from the addition theorem, we have:
as before.
Example 3. In a cash-material lottery, for a series of 1000 tickets there are 120 cash and 80 material winnings. What is the probability of winning anything on one lottery ticket?
Solution. If we denote by A an event consisting of a monetary gain and by B a material gain, then from the definition of probability it follows
The event of interest to us is represented by (A or B), therefore it follows from the addition theorem
Thus, the probability of any winning is 0.2.
Before moving on to the next theorem, it is necessary to become familiar with a new important concept - the concept of conditional probability. For this purpose, we will start by considering the following example.
Suppose there are 400 light bulbs in a warehouse, manufactured in two different factories, and the first one produces 75% of all light bulbs, and the second - 25%. Let us assume that among the light bulbs manufactured by the first plant, 83% satisfy 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 randomly taken from the warehouse will satisfy the conditions of the standard.
Note that the total number of standard light bulbs available consists of the light bulbs manufactured by the first
factory, and 63 light bulbs manufactured by the second plant, that is, equal to 312. Since the choice of any light bulb should be considered equally possible, we have 312 favorable cases out of 400, so
where event B is that the light bulb we have chosen is standard.
During this calculation, no assumptions were made about the product of which plant the light bulb we selected belonged to. If we make any assumptions of this kind, then it is obvious that the probability we are interested in may change. So, for example, if it is known that the selected light bulb was manufactured 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 given that event A occurs, is called the conditional probability of event B given the occurrence of event A and is denoted
If in the previous example we denote by A the event that the selected light bulb is manufactured at the first plant, then we can write
Now we can formulate an important theorem related to calculating the probability of combining events.
Multiplication theorem.
The probability of combining events A and B is equal to the product of the probability of one of the events and the conditional probability of the other, assuming that the first occurred:
In this case, the combination of events A and B means the occurrence of each of them, that is, the occurrence of both event A and event B.
Proof. Let us 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 us divide 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 those events that favor one of the two events of interest to us and do not favor the other, for example, the second group includes those that favor A but do not favor B, and the third group includes those that favor B but do not favor A; finally to
The fourth group includes those events that do not favor either A or B.
Since the numbering of events does not matter, we can assume that this division into four groups looks like this:
Group I:
Group II:
III group:
IV group:
Thus, among equally possible and pairwise incompatible events, there are events that favor both event A and event B, events that favor event A, but do not favor event A, events that favor B, but do not favor A, and, finally, events that do not favor neither A nor B.
Let us note, by the way, that any of the four groups we 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 breakdown into groups allows you 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 favoring A is equal to the total number of events in the first and second groups, and those favoring B is equal to the total number of events in the first and third groups.
Let us 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 occurrence 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 events of the first group, so we get:
To prove the theorem, it is enough now to write the obvious identity:
and replace all three fractions with the probabilities calculated above. We arrive at the equality stated in the theorem:
It is clear that the identity we wrote 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 you notice that using the identity
Comparing the right-hand sides of the two expressions for the probability P(A and B), we obtain a useful equality:
Let us now consider examples illustrating the multiplication theorem.
Example 4. In the products of a certain enterprise, 96% of the products are considered suitable (event A). 75 products out of every hundred suitable ones turn out to belong to the first grade (event B). Determine the probability that a randomly selected product will be suitable and belong to the first grade.
Solution. The desired probability is the probability of combining events A and B. By 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 the fuses fail (i.e., in 2% of cases the shot does not
Solution. Let the event B be that a shot will occur, and let B mean the opposite event. Then by condition 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 fire and hit), therefore, according to the multiplication theorem
Important special case multiplication theorems can be obtained by 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 whether the other occurs or does not occur.
Examples of independent events are dropout various numbers points when re-throwing a dice or one or another side of coins when re-throwing a coin, 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 out or not 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 is previously returned, does not depend on whether the ball was drawn the first time, white or black. Therefore, the results of the first and second removal 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, 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 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 put in the initial expression of the multiplication theorem, which follows from the independence of events, and we will obtain the required equality.
Let us now consider several events: We will call them collectively independent if the probability of the occurrence of any of them does not depend on whether any other events under consideration occurred or not
In the case of events that are jointly independent, the multiplication theorem can be extended to any final number them, thanks 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 operates three automatic machines, each of which must be approached to correct a 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 - 0.7. Determine the probability that within an hour the worker will not need to approach any of the machines he is servicing.
Example 7. Probability of shooting down a plane with a rifle shot. What is the probability of destroying an enemy plane if 250 rifles are fired at the same time?
Solution. The probability that the plane will not be shot down with a single shot is equal to the addition theorem. Then we can calculate, using the multiplication theorem, the probability that the plane will not be shot down with 250 shots, as the probability of combining events. It is equal to After this, we can again use the addition theorem and find the probability that the plane will be shot down as the probability of the opposite event
From this it can be seen that, although the probability of shooting down a plane with a single rifle shot is negligible, nevertheless, when firing from 250 rifles, the probability of shooting down a plane is already very noticeable. It increases significantly if the number of rifles is increased. So, when shooting from 500 rifles, the probability of shooting down a plane, as is easy to calculate, is equal to when shooting from 1000 rifles - even.
The multiplication theorem proved above allows us to somewhat expand the addition theorem, 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 die, and 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
From this it is clear that in the case of compatible events the theorem of addition of probabilities 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 the new one.
Events that are not favorable to 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 number of points appearing when throwing a dice, we obtain:
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 combination
For example. IN electrical circuit Two fuses are connected 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 power failure as a result of failure of at least one of these fuses.
Solution. Since events A and B, consisting of the failure of the first and second of the fuses, are compatible, the required probability will be determined by formula (9):
Exercises
Addition and multiplication of probabilities. This article will focus on solving problems in probability theory. Previously, 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 some problems that are a little more complicated; to solve them you need to know and understand: the rule of adding probabilities, the rule of multiplying probabilities, the concepts of dependent and independent events, opposite events, compatible and incompatible events. Don't be scared by the definitions, it's simple)).In this article we will consider just such tasks.
A little important and simple theory:
incompatible , if the appearance of one of them excludes the appearance of others. That is, only one specific event or another can happen.
A classic example: when throwing a dice, only a one can come up, or only a two, or only a three, etc. Each of these events is incompatible with the others, and the occurrence of one of them excludes the occurrence of the other (in one trial). It’s the same with a coin—when heads come up, it eliminates the possibility of tails coming up.
This also applies to more complex combinations. For example, two lighting lamps are on. Each of them may or may not burn out over 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 one does not burn out and the second one burns out
- The first one does not burn out and the second one burns out.
All these 4 options for 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 draw the queen of spades and thus both events will occur.
About the sum of probabilities
The sum of two events A and B is called the event A+B, which means that either event A or event B will occur, or both at the same time.
If there are incompatible events A and B, then the probability of the sum of these events is equal to the sum of the probabilities of the events:
Example with dice:
We throw the dice. What is the probability of rolling a number less than four?
Numbers less than four are 1,2,3. We know that the probability of getting a one is 1/6, a two is 1/6, and a three is 1/6. These are incompatible events. We can apply the addition rule. The probability of rolling a number less than four is:
Indeed, if we start from the concept classical probability: then the number of possible outcomes is 6 (the number of all sides of the cube), the number of favorable outcomes is 3 (rolling a 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)
About multiplying probabilities
Let two incompatible events A and B occur, their probabilities are respectively equal to P(A) and P(B). The product of two events A and B is 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:We throw the dice twice. What is the probability of getting two sixes?
The probability of rolling a six the first time is 1/6. The second time is also equal to 1/6. The probability of rolling a six the first time and the second time is equal to the product of the probabilities:
Speaking in simple language: when in one trial some event occurs, AND then another one (others) occurs, then the probability that they will occur together is equal to the product of the probabilities of these events.
We solved problems with dice, but we used only logical reasoning and did not use the product formula. In the tasks considered below, you cannot do without formulas; or rather, with them it will be easier and faster to get the result.
It is worth mentioning one more nuance. When reasoning in solving problems, the concept of SIMULTANEOUSNESS of events is used. Events occur SIMULTANEOUSLY - this does not mean that they occur in one second (at one point in time). This means that they occur over a certain period of time (during one test).
For example:
Two lamps burn out within a year (it can be said - simultaneously within a year)
Two machines break down within a month (one might say simultaneously within a month)
The dice are rolled three times (points appear at the same time, this means on one trial)
The biathlete fires five shots. Events (shots) occur during one trial.
Events A and B are INDEPENDENT if the probability of either of them does not depend on the occurrence or non-occurrence of the other event.
Let's consider the tasks:
Two factories produce identical glasses for car headlights. The first factory produces 35% of these glasses, the second – 65%. The first factory produces 4% of defective glass, and the second – 2%. Find the probability that glass accidentally purchased in a store will be defective.
The first factory produces 0.35 products (glass). 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 purchased at the first factory and that it turns out to be defective is 0.35∙0.04 = 0.0140.
The probability that the glass was purchased at the second factory and that it turns out to be defective is 0.65∙0.02 = 0.0130.
Buying defective glass in a store implies that it (the defective glass) was purchased EITHER from the first factory OR from the second. These are incompatible events, that is, we add up the resulting probabilities:
0,0140 + 0,0130 = 0,027
Answer: 0.027
If grandmaster A. plays white, then he wins against grandmaster B. with probability 0.62. If A. plays black, then A. wins against B. with probability 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 possibility of winning the first and second games does not depend on each other. It is said that a grandmaster must win both times, that is, win the first time AND at the same time win the second time. In the case when independent events must occur together, the probabilities of these events are multiplied, that is, the multiplication rule is used.
The probability of the occurrence of these events will be equal to 0.62∙0.2 = 0.124.
Answer: 0.124
At 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 question on the topic “Parallelogram” 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 in the exam.
That is, it is necessary to find the probability that the student will get a question EITHER on the topic “Inscribed Circle” OR on the topic “Parallelogram”. In this case, the probabilities are summed up, since these are incompatible events and any of these events can happen: 0.3 + 0.25 = 0.55.
*Incompatible events are events that cannot happen at the same time.
Answer: 0.55
A biathlete shoots at targets five times. The probability of hitting the target with one shot is 0.9. Find the probability that the biathlete hits the targets the first four times and misses the last one. Round the result to hundredths.
Since the biathlete hits the target with probability 0.9, he misses with probability 1 – 0.9 = 0.1
*Miss and hit are events that cannot occur simultaneously with one shot; the sum of the probabilities of these events is equal to 1.
We are talking about the occurrence of several (independent) events. If an event occurs and at the same time another (subsequent) event occurs at the same time (test), then the probabilities of these events are multiplied.
The probability of a product of 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 the nearest hundredth, we get 0.07
Answer: 0.07
There are two payment machines in the store. Each of them can be faulty with probability 0.07, regardless of the other machine. Find the probability that at least one machine is working.
Let's find the probability that both machines 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 machines or one of them is working will be equal to 1 – 0.0049 = 0.9951.
*Both are operational and one of them is fully operational – meets the “at least one” condition.
We can present the probabilities of all (independent) events to be tested:
1. “faulty-faulty” 0.07∙0.07 = 0.0049
2. “defective-defective” 0.93∙0.07 = 0.0651
3. “defective-defective” 0.07∙0.93 = 0.0651
4. “defective-defective” 0.93∙0.93 = 0.8649
To determine the probability that at least one machine is working, it is necessary to add the probabilities of independent events 2,3 and 4: A reliable event is an event that is certain to occur as a result of an experience. The event is called impossible, if it never occurs as a result of experience.
For example, if one ball is drawn at random from a box containing only red and green balls, then the appearance of a white one among the drawn balls is an impossible event. The appearance of the red and the appearance of the green balls form a complete group of events.
Definition: The events are called equally possible , unless there is reason to believe that one of them is more likely to appear as a result of experience.
In the example above, the appearance of red and green balls are equally likely events if there are the same number of red and green balls in the box. If there are more red balls in the box than green ones, then the appearance of a green ball is a less probable event than the appearance of a red one.
In we will look at more problems where the sum and product of the probabilities of events are used, don’t miss it!
That's all. Good luck to you!
Sincerely, Alexander Krutitskikh.
Marya Ivanovna scolds Vasya:
- Petrov, why weren’t you at school yesterday?!
“My mother washed my pants yesterday.”
- So what?
- And I walked past the house and saw that yours were hanging. I thought you wouldn't come.
P.S: I would be grateful if you tell me about the site on social networks.
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 compatible and incompatible events. These formulas can be justified using Venn diagrams (Fig. 21). Let us 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 incompatible 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 that form 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 complete group exhausts the entire list of possible events, then one of these events will necessarily occur. Thus, on the left side is written the probability of an event that will definitely happen - a reliable event. 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
IN the probability of a technical device being in working order is 0.8. Find the probability of failure of this device over the same observation period.
R decision.
Important Note. In reliability theory, it is customary to denote the probability of an operational state by the letterr, and the probability of failure is the letter q. In what follows we will use these notations. Both probabilities are functions of time. Thus, for long periods of time, the probability of an operational 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 operability and failure states are opposite events. Using Corollary 2, we obtain the probability of device failure:
q = 1 – p = 1 – 0.8 = 0.2.
For two joint events probability addition formula has the form:
P(A+B) = P(A) + P(B) – P(AB), (9)
as 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 is probabilities gets more complicated:
P(A+B+C)=P(A) + P(B) + P(C) – P(AB) – P(AC) – P(BC) + P(ABC).(10)
In 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).
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 three times and subtracted three times. Therefore, to account for this area, it must be added to the final expression.
As the number of terms increases, the addition formula becomes more and more cumbersome, but the principle of its construction remains the same: first, the probabilities of events taken individually are summed up, then the probabilities of all paired combinations of events are subtracted, the probabilities of events taken in triplets are added, the probabilities of combinations of events taken in quadruples are subtracted. etc.
Finally, it should be emphasized : formula for adding probabilities joint events with a number of terms of 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 failure probabilities qi individual elements (generator, transformers and lines).
Failure States individual elements of the power supply system, as well as and health states are always pairwise joint events, since there are no fundamental obstacles to simultaneously carrying out repairs, for example, of a line and a transformer. A system failure occurs when any of its elements fails: either a generator, or the 1st transformer, or a 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 с = 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 us 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 failure probabilities of individual elements of the power supply system for the period of one year usually used in reliability calculations are quite small (about 10 -2). Therefore, all terms except the first four can be discarded, which will have virtually no effect on the numerical result. Then we 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 often discarded terms may turn out to be comparable with the first ones.
Example 17
Determine the probability of the system being operational R S, consisting of three elements reserving each other.
Solution. Elements that back up each other in the logical diagram of reliability analysis are shown connected in parallel (Fig. 25):
A redundant system is operational when either the 1st, or 2nd, or 3rd element is operational, or any pair is operational, or all three elements together. Consequently, the operational state of the system is the sum of the operational states of individual elements. Using the addition formula for joint events Р с = Р 1 + Р 2 + Р 3 – Р 1 Р 2 – Р 1 Р 3 – Р 2 Р 3 + Р 1 Р 2 Р 3. , Where R 1, R 2 And R 3– probabilities of the operable state of elements 1, 2 and 3, respectively.
In this case, it is impossible to simplify the solution by discarding 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 another, 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 capacity - P(R 100), fifty percent capacity - P(R 50), and the probability that the system will fail - Q.
The line has 100% throughput when both the 1st and 2nd circuits are operational:
Р(100%) = р 1 р 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 capacity when the 1st circuit is operational and the 2nd has failed, or when the 2nd circuit is operational and the 1st has failed:
P(50%)= p 1 q 2 + p 2 q 1 = 2∙0.999∙10 -3 = 0.001998.
IN last expression the addition formula was used for incompatible events, which they are.
The events considered in this problem form a complete group, so the sum of their probabilities is one.
Probability addition and multiplication theorems
Addition theorem
The probability of the occurrence of one of several incompatible events is equal to the sum of the probabilities of these events.
In the case of two incompatible events A and B we have:
P(A+B) = P(A) + P(B) (7)
The event opposite to event A is denoted by . The combination of events A gives a reliable event, and since events A are incompatible, then
P(A) + P() = 1 (8)
The probability of event A, calculated under the assumption that event B has occurred, is called conditional probability event A and is denoted by the symbol P B (A).
If events A and B are independent, then P(B) = P A (B).
Events A, B, C, ... are called collectively independent, if the probability of each of them does not change due to the occurrence or non-occurrence of other events separately or in any combination of them and in any number.
Multiplication theorem
The probability that events A, B, and C will occur... is equal to the product of their probabilities, calculated under the assumption that all events preceding each of them took place, i.e.
P(AB) = P(A)P A (B)(9)
The notation P A (B) denotes the probability of event B under the assumption that event A has already occurred.
If events A, B, C, ... are collectively independent, then the probability that they will all occur is equal to the product of their probabilities:
P(ABC) = P(A)P(B)P(C) (10)
Example 3.1. The bag contains balls: 10 white, 15 black, 20 blue and 25 red. One ball was taken out. Find the probability that the drawn ball will be white? black? And one more thing: white or black?
Solution.
The number of all possible trials n = 10 + 15 + 20 + 25 = 70;
Probability P(b) = 10/70 = 1/7, P(h) = 15/70 = 3/14.
We apply the probability addition theorem:
R(b + h) = R(b) + R(h) = 1/7 + 3/14 = 5/14.
Note: capital letters in brackets respectively indicate the color of each ball according to the conditions of the problem.
Example 3.2 The first box contains two white and ten black balls. The second box contains eight white and four black balls. A ball was taken from each box. Determine the probability that both balls will be white.
Solution.
Event A is the appearance of a white ball from the first box. Event B is the appearance of a white ball from the second box. Events A and B are independent.
Probabilities P(A) = 2/12 = 1/6, P(B) = 8/12 = 2/3.
We apply the probability multiplication theorem:
P(AB) = P(A)P(B) = 2/18 = 1/9.
Review questions
1 What is factorial?
2 List the main tasks of combinatorics.
3 What are permutations called?
4 What are movements called?
5 What are combinations called?
6 What events are called reliable?
7 What events are called incompatible?
8 What is the probability of an event?
9 What is called conditional probability?
10 Formulate theorems for addition and multiplication of probabilities.
11 pr.Placement from n elements by k (k ≤ p ) is any set consisting of To elements taken in a specific order from data n elements.
Thus, two placements from n elements by To are considered different if they differ in the elements themselves or in the order of their arrangement Number of placements from n elements by To denote A p k and calculated using the formula
A p k =
If placements from n elements by n differ from each other only in the order of the elements, then they represent permutations of n elements
Example1. Second grade students study 9 subjects. In how many ways can you make a schedule for one day so that it contains 4 different subjects?
Solution: Any schedule for one day, made up of 4 different subjects, differs from the other either in the set of subjects or in the order in which they are presented. This means that in this example we are talking about placements of 9 elements of 4. We have
A 9 4 = = 6 ∙ 7 ∙ 8 ∙ 9 = 3024
You can create a schedule in 3024 ways
Example 2. How many three-digit numbers(without repeating the numbers in the number) can you make up 0,1,2,3,4,5,6 from the numbers?
Solution If there is no zero among seven digits, then the number of three-digit numbers (without repeating digits) that can be made from these digits is equal to the number of placements
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of 7 elements of 3 each. However, among these numbers there is the number 0, with which a three-digit number cannot begin. Therefore, from arrangements of 7 elements by 3, it is necessary to exclude those whose first element is 0. Their number is equal to the number of arrangements of their 6 elements by 2. =
This means that the required number of three-digit numbers is
A 7 3 - A 6 2 = - = 5 ∙ 6 ∙ 7 - 5 ∙ 6 = 180.
3. Consolidation of acquired knowledge in the process of solving problems
№754 . In how many ways can a family of three sleep in a four-seater compartment if there are no other passengers in the compartment?
Solution. The number of ways is equal A 4 3 = = 1∙ 2 ∙ 3 ∙ 4 = 24
№755. Of the 30 meeting participants, a chairman and secretary must be selected. In how many ways can this be done?
Solution. Since any of the participants can be either a secretary or a chairman, the number of ways to elect them is equal
A 30 2 = = = 29 ∙ 30 = 870
№762 How many four-digit numbers that do not have identical digits can be made from the following digits: a) 1,3,5,7,9. b) 0,2,4,6,8?
Solution a) A 5 4 = = 1∙ 2 ∙ 3 ∙ 4 ∙ 5 = 120
b)) A 5 4 - A 4 3 = 5! – 4! = 120 – 24 = 96
Homework No. 756, No. 757, No. 758, No. 759.
Lesson 6 Topic: “Combinations”
Purpose: To give the concept of combinations, introduce the formula for calculating combinations, teach how to use this formula to count the number of combinations.
1 Checking homework.
№ 756 . There are 7 alternate tracks at the station. In how many ways can 4 trains be arranged on them?
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Solution : A 7 4 = = 4 ∙ 5 ∙ 6 ∙ 7 = 20 ∙ 42 = 840 ways
№757 In how many ways can a coach determine which of the 12 athletes ready to participate in the 4x100m relay will run in the first, second, third and fourth stages?
Solution: A 12 4 = = 9 ∙ 10 ∙ 11 ∙12 = 90 ∙132 = 11 880
№ 758. In a pie chart, the circle is divided into 5 sectors. We decided to paint over the sectors different colors, taken from a set containing 10 colors. In how many ways can this be done?
Solution: A 10 5 = = 6 ∙ 7 ∙ 8 ∙ 9∙ 10 = 30 240
№759. In how many ways can 6 students taking an exam take seats in a classroom with 20 single tables?
Solution: A 20 6 = = 15∙ 16 ∙17∙ 18∙19 ∙20 = 27 907 200
You can arrange for your homework to be checked in different ways: verbally check the solutions to home exercises, write down the solutions to some of them on the board, and while the solutions are being recorded, conduct a survey of students on the following questions:
1. What does the entry mean? p!
2.What is called a permutation from n elements?
3.What formula is used to calculate the number of permutations?
4. What is called placement from n elements by To?
5. n elements by To?
2 Explanation of new material
Let there be 5 carnations of different colors. Let's designate them with letters a, c, c, d, f. You need to make a bouquet of three carnations. Let's find out what bouquets can be composed.
If the bouquet includes carnations A , then you can make the following bouquets:
avs, avd, ave, asd, ase, ade.
If the bouquet does not include carnations A, but cloves come in V , then you can get the following bouquets:
all, all, everywhere.
Finally, if the bouquet does not include a carnation A, not a clove V, then only one option for composing a bouquet is possible:
sde.
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We have indicated everything possible ways making bouquets in which three out of 5 carnations are combined in different ways. They say that we have made all possible combinations out of 5 elements, 3 each, we found that C 5 3 = 10.
Let us derive the formula for the number of combinations from n elements in k, where k ≤ p.
Let us first find out how C 5 3 is expressed through A 5 3 and P 3 . We found that their 5 elements can be made into the following combinations of 3 elements:
avs, avd, ave, asd, ase, ade, vsd, all, vde, sde.
In each combination we will perform all permutations. The number of permutations of 3 elements is equal to P 3 . As a result, we get all possible combinations of 5 elements of 3, which differ either in the elements themselves or in the order of the elements, i.e. all placements of 5 elements are 3 each. In total we get A 5 3 placements.
Means , C 5 3 ∙ P 3 = A 5 3, hence C 5 3 = A 5 3: P 3
Reasoning in general case we get C p k = A p k: P k,
Using the fact that A p k = , where k ≤ p., we get C p k = .
This is the formula for calculating the number of combinations of n elements by To at any
k ≤ p.
Example1. From a set of 15 paints, you need to choose 3 colors to paint the box. In how many ways can this choice be made?
Solution: Each choice of three colors differs from the other in at least one color. This means that here we are talking about combinations of 15 elements of 3
From 15 3 = = (13∙ 14∙15) : ( 1∙ 2 ∙ 3) = 455
Prime2 There are 12 boys and 10 girls in the class. Three boys and two girls are required to clean the area near the school. In how many ways can this choice be made?
Solution: You can choose 3 boys out of 12 with 12 3 , and two girls out of 10 can be selected with 10 2 . Since for each choice of boys it is possible to choose girls in 10 2 ways, then you can make the choice of students, which is discussed in the problem
С 12 3 ∙ С 10 2 = ∙ = 220 ∙ 45 = 9900
3) Consolidation of new material in the process of solving problems
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Task
Sasha has 8 historical novels in her home library. Petya wants to take any 2 novels from him. In how many ways can this choice be made?
Solution: C 8 2 = = ( 7 ∙ 8) : ( 1∙ 2) = 56: 2 = 28
№779 a
There are 16 people in the chess club. In how many ways can a coach select a team of 4 people from them for the upcoming tournament?
Solution: C 16 4 = = ( 13∙ 14∙15 ∙16) : ( 1∙ 2 ∙ 3 ∙ 4) = 13 ∙ 7 ∙5∙ 4 = 91 ∙20 = 1820
№774 The team renovating the school consists of 12 painters and 5 carpenters. Of these, 4 painters and 2 carpenters need to be allocated to repair the sports hall. In how many ways can this be done?
С 12 4 ∙ С 5 2 = ∙ = 495 ∙ 10 = 4950
Homework No. 768, No. 769, No. 770, No. 775
Lesson 7 Topic: “Solving problems using formulas to calculate the number of movements, placements, combinations”
Goal: Consolidation of students' knowledge. Formation of skills for solving simple combinatorial problems
1 Checking homework
№768 There are 7 people in the class who are successfully doing mathematics. In how many ways can you choose two of them to participate in the Mathematical Olympiad?
Solution: C 7 2 = = (6∙ 7) : 2 = 21
№769 The Philately store sells 8 different stamp sets dedicated to sports theme. In how many ways can you choose 3 sets from them?
Solution: C 8 3 = = ( 6 ∙ 7 ∙ 8) : ( 1∙ 2 ∙ 3) = 56
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№770 The students were given a list of 10 books to read during the holidays. In how many ways can a student choose 6 books from them?
Solution: C 10 6 = = ( 7 ∙ 8 ∙ 9∙ 10) : ( 1∙ 2 ∙ 3 ∙ 4) = 210
№775 The library offered the reader a choice of 10 books and 4 magazines from new arrivals. In how many ways can he choose 3 books and 2 magazines from them?
Solution: C 10 3 ∙ C 4 2 = ∙ = 120 ∙ 6 = 720
Questions for the class
1.What is called a permutation from n elements?
2.What formula is used to calculate the number of permutations?
3. What is called placement from n elements by To?
4. What formula is used to calculate the number of placements from n elements by To?
5. What is called a combination of n elements by To?
6. What formula is used to calculate the number of combinations of n elements by To?
Problems for joint solution
When solving each problem, there is first a discussion: which of the three studied formulas will help get the answer and why
1. How many four-digit numbers can be made from the numbers 4,6,8,9, provided that all the numbers are different?
2. From 15 people in a group of students, you must choose a headman and his deputy. In how many ways can this be done?
3. Of the 10 best students at the school, two people should be sent to the leaders’ meeting.
In how many ways can this be done?
Comment: In problem No. 3, it doesn’t matter who to choose: any 2 people out of 10, so the formula for counting the number of combinations works here.
In problem No. 2, an ordered pair is chosen, because in the selected pair, if the surnames are swapped, it will be a different choice, so the formula for calculating the number of placements works here
Answers to problems for joint solution:
No. 1 on the 24th. No. 2 210 ways. No. 3 45 ways
Problems for joint discussion and independent calculations
No. 1 6 friends met and each shook hands with each other. How many handshakes were there?
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No. 2 In how many ways can you create a schedule for 1st grade students for one day if they have 7 subjects and there should be 4 lessons on that day?
(Number of placements from 7 to 4)
No. 3 There are 6 people in the family, and there are 6 chairs at the table in the kitchen. It was decided to sit on these 6 chairs every evening before dinner. in a new way. How many days can family members do this without repetition?
No. 4 They came to the owner of the house guests A, B, C, D. There are five different chairs at the round table. How many seating methods are there?
(4 people came to visit + the owner = 5 people sit on 5 chairs, you need to count the number of permutations)
5. In the coloring book, a non-intersecting triangle, square and circle are drawn. Each figure must be painted in one of the colors of the rainbow, different figures in different colors. How many ways of coloring are there?
(Count the number of placements from 7 to 3)
No. 6 There are 10 boys and 4 girls in the class. It is necessary to choose 3 people on duty so that among them there are 2 boys and 1 girl. In how many ways can this be done?
(The number of combinations of 10 by 2 multiplied by the number of combinations of 4 by 1)
Answers for self-calculation problems
№1 15 handshakes
№2,840 ways
№3 720 days
№5 120 ways
№6,180 ways
Homework No. 835, No. 841
Lesson 8 Topic: “Independent work”
Purpose: Testing students' knowledge
1.Check homework
№^ 835 How many even four-digit numbers in which the digits are not repeated can be written using the numbers a) 1,2,3,7. b) 1,2,3,4.
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a) Our numbers must end with an even digit, such a digit in condition one is digit 2, we will put it in last place, and we will rearrange the remaining 3 digits, the number of such permutations is 3! = 6. So you can make 6 even numbers
b) we reason as in example a) by putting the number 2 in the last place we get 6 even numbers, by putting the number 4 in the last place we get 6 more even numbers,
that means there are only 12 even numbers
№841 In how many ways can you choose from a class of 24 students: a) two attendants; b) the headman and his assistant?
a) because any 2 people out of 24 can be on duty, then the number of pairs is equal
C 24 2 = = 23 ∙ 24:2 = 276
b) here they tear out an ordered pair of elements from 24 elements, the number of such pairs is A 24 2 = = 23 ∙ 24 = 552
Option 1 solves tasks No. 1,2,3,4,5.
Option 2 solves tasks No. 6,7,8,9,10.
Solving the simplest combinatorial problems
(based on materials from K.R. in April 2010)
1 . In how many ways can five books by different authors be arranged on a shelf?
2. In how many ways can you make an afternoon snack from a drink and a pie if the menu includes: tea, coffee, cocoa and apple or cherry pies?
3. On Wednesday, according to the schedule, there should be 5 lessons in grade 9 “A”: chemistry, physics, algebra, biology and life safety. In how many ways can you create a schedule for this day?
4. There are 2 white horses and 4 bay horses. How many ways can you
make a pair of horses of different colors?
5. How many ways can you put 5 different coins into 5 different pockets?
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6. There are 3 hats of different styles and 4 scarves on the shelf in the closet. different colors. In how many ways can you make a set of one hat and one scarf?
7. 4 participants reached the finals of the beauty contest. In how many ways
Is it possible to establish the order of performance of the participants in the beauty final?
^ 8 .There are 4 ducks and 3 geese. In how many ways can you choose two different birds?
9. In how many ways can 5 different letters be divided into 5 different ones?
envelopes, if only one letter is placed in each envelope?
10. A box contains 5 red and 4 green balls. In how many ways can you make a pair of balls of different colors?
Answers for independent work tasks
An experiment is being considered E. It is assumed that it can be carried out repeatedly. As a result of the experiment, various events may appear, making up a certain set F. Observable events are divided into three types: reliable, impossible, random.
Reliable an event that is sure to occur as a result of an experiment is called E. Denoted by Ω.
Impossible an event that is known to not occur as a result of an experiment is called E. Denoted by .
Random an event that may or may not occur as a result of an experiment is called E.
Additional (opposite) event A is an event, denoted by , that occurs if and only if the event does not occur A.
Sum (combination) events is an event that occurs if and only if at least one of these events occurs (Figure 3.1). Notation.
Figure 3.1
Product (intersection) events is an event that occurs if and only if all these events occur together (simultaneously) (Figure 3.2). Notation. It is obvious that events A and B incompatible , If .
Figure 3.2
Full group of events is a set of events whose sum is a certain event:
Event IN called a special case of an event A, if with the occurrence of an event IN the event appears A. They also say that the event IN entails an event A(Figure 3.3). Designation
Figure 3.3
Events A And IN are called equivalent , if they occur or do not occur together during the experiment E. Designation Obviously, if...
A difficult event call an observed event expressed through other events observed in the same experiment using algebraic operations.
The probability of a particular complex event occurring is calculated using the formulas for adding and multiplying probabilities.
Probability addition theorem
Consequences:
1) if events A And IN are inconsistent, the addition theorem takes the form:
2) in the case of three terms, the addition theorem is written in the form
3) the sum of the probabilities of mutually opposite events is equal to 1:
The set of events ,, ..., is called full group of events , If
The sum of the probabilities of events forming a complete group is equal to 1:
Probability of event occurrence A provided that the event IN happened, they call it conditional probability and denote or.
A And IN – dependent events , If .
A And IN – independent events , If .
Probability multiplication theorem
Consequences:
1) for independent events A And IN
2) in the general case, for the product of three events, the probability multiplication theorem has the form:
Examples of problem solving
Example1 - Three elements are connected in series to the electrical circuit, operating independently of each other. The failure probabilities of the first, second and third elements are respectively equal to ,. Find the probability that there will be no current in the circuit.
Solution
First way.
Let us denote the following events: - failure of the first, second and third elements occurred in the circuit, respectively.
Event A– there will be no current in the circuit (at least one of the elements will fail, since they are connected in series).
Event - there is current in the circuit (three elements are working), . The probability of opposite events is related by formula (3.4). An event is the product of three events that are pairwise independent. Using the theorem for multiplying the probabilities of independent events, we obtain
Then the probability of the desired event is .
Second way.
Taking into account the previously accepted notation, we write down the desired event A– at least one of the elements will fail:
Since the terms included in the sum are compatible, one should apply the theorem of addition of probabilities in general view for the case of three terms (3.3):
Answer: 0,388.
Problems to solve independently
1 The reading room has six textbooks on probability theory, three of which are bound. The librarian took two textbooks at random. Find the probability that both textbooks will be bound.
2 There are threads mixed in the bag, 30% of which are white and the rest are red. Determine the probabilities that two threads drawn at random will be: the same color; different colors.
3 The device consists of three elements that work independently. The probabilities of failure-free operation for a certain period of time of the first, second and third elements, respectively, are 0.6; 0.7; 0.8. Find the probabilities that during this time only one element will work without failure; only two elements; all three elements; at least two elements.
4 Three dice are thrown. Find the probabilities of the following events:
a) five points will appear on each side drawn;
b) the same number of points will appear on all dropped sides;
c) one point will appear on two dropped sides, and another number of points will appear on the third side;
d) a different number of points will appear on all dropped faces.
5 The probability of a shooter hitting the target with one shot is 0.8. How many shots must the shooter fire so that with a probability of less than 0.4 it can be expected that there will be no misses?
6 From the numbers 1, 2, 3, 4, 5, one is first selected, and then the second digit is selected from the remaining four. It is assumed that all 20 possible outcomes are equally likely. Find the probability that an odd number will be chosen: for the first time; for the second time; both times.
7 The probability that a pair of size 46 shoes will be sold again in the men's shoe section of the store is 0.01. How many pairs of shoes must be sold in a store so that with a probability of at least 0.9 one can expect that at least one pair of size 46 shoes will be sold?
8 The box contains 10 parts, including two non-standard ones. Find the probability that out of six randomly selected parts there will be no more than one non-standard one.
9 The technical control department checks products for standardness. The probability that the product is non-standard is 0.1. Find the probability that:
a) out of three tested products, only two will turn out to be non-standard;
b) only the fourth product tested in order will turn out to be non-standard.
10 32 letters of the Russian alphabet are written on cut-out alphabet cards:
a) three cards are taken out at random one after another and placed on the table in the order of appearance. Find the probability that the word “world” will come out;
b) the three cards removed can be swapped in any way. What is the probability that they can be used to form the word “world”?
11 A fighter attacks a bomber and fires two independent bursts at it. The probability of shooting down a bomber with the first burst is 0.2, and the second - 0.3. If the bomber is not shot down, it fires at the fighter from its rear guns and shoots it down with a probability of 0.25. Find the probability that a bomber or fighter is shot down as a result of an air battle.
Homework
1 Total probability formula. Bayes' formula.
2 Solve problems
Task1 . A worker operates three machines that operate independently of each other. The probability that the first machine will not require the worker’s attention within an hour is 0.9, the second – 0.8, and the third – 0.85. Find the probability that within an hour at least one machine will require the attention of a worker.
Task2 . The computer center, which must continuously process incoming information, has two computing devices. It is known that each of them has a probability of failure over some time equal to 0.2. You need to determine the probability:
a) the fact that one of the devices will fail, and the second will be operational;
b) trouble-free operation of each device.
Task3 . Four hunters agreed to shoot at game in a certain sequence: the next hunter fires a shot only if the previous one misses. The probability of a hit for the first hunter is 0.6, for the second – 0.7, for the third – 0.8. Find the probability that shots will be fired:
d) four.
Task4 . The part goes through four processing operations. The probability of receiving a defect during the first operation is 0.01, during the second - 0.02, during the third - 0.03, and during the fourth - 0.04. Find the probability of receiving a part without defects after four operations, assuming that the events of receiving defects in individual operations are independent.
