Find the probability that as a result of the test. Independent re-testing and the Bernoulli formula. Discrete random variables
Let it be carried out n independent trials, each of which contains some event A It may or may not appear. Let the probability of an event occur A in one test is constant and equal p(probability of non-occurrence of an event A equal to q = 1–p).
Under these conditions, the probability that an event A when conducting n tests will come exactly k times determined Bernoulli's formula:
Event occurrence number A in independent tests it is called most likely, if the probability that the event A occurs in these tests once, exceeds (or is not less than) the probabilities of other test outcomes. The number is determined using the double inequality:
If is a fractional number, then there is one most probable number.
If is an integer, then there are two most probable numbers and .
If is an integer, then .
Task. The probability that a product will not pass inspection is 0.125. Find the probability that among 12 products there will not be a single defective one.
Solution. Let's denote the event A- “the product will not pass inspection.” Held n= 12 independent trials. We need to find the probability that the event A will happen k= 0 times (there will not be a single product that fails inspection). Probability of event occurrence A p= 0.125=1/8, non-appearance – q= 0.875=7/8. Using the Bernoulli formula (17.1) we obtain:
Poisson's formula
In the case when, with increasing n probability p occurrence of the event of interest decreases, and is a constant number (we assume that a£ 10), then the probability that the event A when conducting n tests will come exactly k times can be calculated by Poisson's formula:
Poisson's formula is a good approximation of Bernoulli's formula in the case where the probability of an event is small ( p® 0, ), and the number of tests n great. Poisson's formula is called the law of rare events.
The flow of events is a sequence of events that occur at random times.
Flow intensity l is the average number of events that occur per unit time.
Probability of appearance k events of the simplest flow over time t is determined by the Poisson formula:
Task. 6 defective parts leave the assembly line per day. The conveyor operates in three shifts. Determine the probability that there will be no defective parts during a shift.
Solution. The intensity of the appearance of defects is l = 6/24 = 0.25. Period of time t = 8 (a.m.) – shift. Let's find the probability that there will be no defects during a shift:
Discrete random variables.
Numerical characteristics of discrete random variables.
Distribution function
Random variable is a quantity that, as a result of testing, takes on any previously unknown value from some number set. The value of a random variable depends on many random factors that cannot be taken into account before experimentation.
The random variable is called discrete, if it takes values from some fixed finite or countable set. In this case, the values of the random variable can be numbered.
Distribution law of a discrete random variable is called the correspondence between its possible values and their probabilities. The distribution law can be specified analytically, graphically and tabularly. The distribution law in tabular form has the form:
| X | X 1 | X 2 | … | x n |
| R | R 1 | R 2 | … | p n |
The first row of the table contains possible values of the random variable X, in the second - the probabilities of these values. On every test random value X can only take one value, so events X = x 1 , X = x 2 , …, X = x n form a complete group of pairwise incompatible events, and, therefore, .
Polygon(polygon)distribution a discrete random variable is a graphical representation of the law of its distribution. To construct a distribution polygon in a rectangular Cartesian coordinate system, it is necessary to sequentially connect points with coordinates , where are the possible values of the random variable X, - the corresponding probabilities ( i = 1, 2, …, n).
Mathematical expectation M(X X is the sum of the products of all its possible values and their probabilities:
Variance(scattering) D(X) discrete random variable X is the mathematical expectation of the squared deviation of a random variable from its mathematical expectation:
It is convenient to calculate the variance using the formula:
Standard deviation s( X)discrete random variable X is called the square root of the variance:
Distribution function (integral function) random variable X called function F(x), which determines the probability that the random variable X as a result of the test the value will be less X:
Properties of the distribution function
1. The values of the distribution function belong to the segment:
2. - non-decreasing function, i.e. , If .
3. If possible values of a random variable belong to the interval ( a, b), then at , at .
4. Probability that a random variable X will take a value belonging to the interval [ a, b), is equal to the increment of the distribution function on this interval:
Task. Discrete random variable X is given by the distribution law:
| X | ||||
| R | 0,2 | 0,1 | 0,4 | 0,3 |
1. Construct a distribution polygon.
2. Find the mathematical expectation, variance and standard deviation.
3. Find the distribution function of the random variable X and plot it.
4. Find the probability that as a result of the test the random variable X will take a value from the interval . Numerical characteristics X:
Hence, 
. Deciding this system, we get two pairs of values: . Since according to the conditions of the problem, we finally have: 
.
Answer: .
Example 2.11. On average, under 10% of contracts, the insurance company pays insurance amounts in connection with the occurrence of an insured event. Calculate the mathematical expectation and dispersion of the number of such contracts among four randomly selected ones.
Solution: The mathematical expectation and variance can be found using the formulas:
.
Possible values of SV (number of contracts (out of four) with the occurrence of an insured event): 0, 1, 2, 3, 4.
We use Bernoulli's formula to calculate the probabilities of different numbers of contracts (out of four) for which the insurance amounts were paid:
.
The IC distribution series (the number of contracts with the occurrence of an insured event) has the form:
| 0,6561 | 0,2916 | 0,0486 | 0,0036 | 0,0001 |
Answer: , .
Example 2.12. Of the five roses, two are white. Draw up a law of distribution of a random variable expressing the number of white roses among two simultaneously taken.
Solution: In a selection of two roses, there may either be no white rose, or there may be one or two white roses. Therefore, the random variable X can take values: 0, 1, 2. Probabilities that X takes these values, we find it using the formula:
Where -- number of roses;
-- number of white roses;
– number of roses taken at the same time;
-- the number of white roses among those taken.
.
.
.
Then the distribution law of the random variable will be as follows:
Example 2.13. Among the 15 assembled units, 6 require additional lubrication. Draw up a distribution law for the number of units that need additional lubrication among five randomly selected from the total number.
Solution: Random value X– the number of units that require additional lubrication among the five selected – can take the following values: 0, 1, 2, 3, 4, 5 and has a hypergeometric distribution. Probabilities that X takes these values, we find it using the formula:
Where -- number of assembled units;
-- the number of units that require additional lubrication;
– number of selected units;
-- the number of units that require additional lubrication among those selected.
.
.
.
.
.
Then the distribution law of the random variable will be as follows:
Example 2.14. Of the 10 watches received for repair, 7 require general cleaning of the mechanism. The watches are not sorted by type of repair. The master, wanting to find watches that need cleaning, examines them one by one and, having found such watches, stops further viewing. Find the mathematical expectation and variance of the number of hours watched.
Solution: Random value X– the number of units that need additional lubrication among the five selected – can take the following values: 1, 2, 3, 4. Probabilities that X takes these values, we find it using the formula:
.
.
.
.
Then the distribution law of the random variable will be as follows:
Now let's calculate numerical characteristics quantities:
Answer: , .
Example 2.15. The subscriber has forgotten the last digit of the phone number he needs, but remembers that it is odd. Find the mathematical expectation and variance of the number of times he dials a phone number before reaching desired number, if he dials the last digit at random, and does not dial the dialed digit in the future.
Solution: The random variable can take the following values: . Since the subscriber does not dial the dialed digit in the future, the probabilities of these values are equal.
Let's compile a distribution series of a random variable:
| 0,2 |
Let's calculate the mathematical expectation and variance of the number of dialing attempts:
Answer: , .
Example 2.16. The probability of failure during reliability testing for each device in the series is equal to p. Determine the mathematical expectation of the number of devices that failed if they were tested N devices.
Solution: Discrete random variable X is the number of failed devices in N independent tests, in each of which the probability of failure is equal p, distributed according to the binomial law. The mathematical expectation of a binomial distribution is equal to the number of trials multiplied by the probability of an event occurring in one trial:
Example 2.17. Discrete random variable X takes 3 possible values: with probability ; with probability and with probability. Find and , knowing that M( X) = 8.
Solution: We use the definitions of mathematical expectation and the distribution law of a discrete random variable:
We find: .
Example 2.18. Department technical control checks products for standardness. The probability that the product is standard is 0.9. Each batch contains 5 products. Find the mathematical expectation of a random variable X– the number of batches, each of which contains exactly 4 standard products, if 50 batches are subject to inspection.
Solution: In this case, all experiments conducted are independent, and the probabilities that each batch contains exactly 4 standard products are the same, therefore, the mathematical expectation can be determined by the formula:
,
where is the number of parties;
The probability that a batch contains exactly 4 standard products.
We find the probability using Bernoulli's formula:
Answer: 
.
Example 2.19. Find the variance of a random variable X– number of occurrences of the event A in two independent trials, if the probabilities of the occurrence of an event in these trials are the same and it is known that M(X) = 0,9.
Solution: The problem can be solved in two ways.
1) Possible values of SV X: 0, 1, 2. Using the Bernoulli formula, we determine the probabilities of these events:
, , .
Then the distribution law X has the form:
From the definition of mathematical expectation, we determine the probability:
Let's find the dispersion of SV X:
.
2) You can use the formula:
.
Answer: 
.
Example 2.20. Expectation and standard deviation of a normally distributed random variable X respectively equal to 20 and 5. Find the probability that as a result of the test X will take the value contained in the interval (15; 25).
Solution: Probability of hitting a normal random variable X on the section from to is expressed through the Laplace function:
Example 2.21. Given function: 
At what parameter value C this function is the distribution density of some continuous random variable X? Find the mathematical expectation and variance of a random variable X.
Solution: In order for a function to be the distribution density of some random variable, it must be non-negative, and it must satisfy the property:
.
Hence:
Let's calculate the mathematical expectation using the formula:
.
Let's calculate the variance using the formula:
T is equal p. It is necessary to find the mathematical expectation and variance of this random variable.
Solution: The distribution law of a discrete random variable X - the number of occurrences of an event in independent trials, in each of which the probability of the event occurring is equal to , is called binomial. The mathematical expectation of the binomial distribution is equal to the product of the number of trials and the probability of occurrence of event A in one trial:
.
Example 2.25. Three independent shots are fired at the target. The probability of hitting each shot is 0.25. Determine the standard deviation of the number of hits with three shots.
Solution: Since three independent trials are performed, and the probability of the occurrence of event A (a hit) in each trial is the same, we will assume that the discrete random variable X - the number of hits on the target - is distributed according to the binomial law.
The variance of the binomial distribution is equal to the product of the number of trials and the probability of the occurrence and non-occurrence of an event in one trial:
Example 2.26. Average number of clients visiting insurance company in 10 minutes, equals three. Find the probability that at least one client will arrive in the next 5 minutes.
Average number of clients arriving in 5 minutes: 
. .
Example 2.29. The waiting time for an application in the processor queue obeys an exponential distribution law with an average value of 20 seconds. Find the probability that the next (random) request will wait on the processor for more than 35 seconds.
Solution: In this example, the mathematical expectation 
, and the failure rate is equal to .
Then the desired probability:
Example 2.30. A group of 15 students holds a meeting in a hall with 20 rows of 10 seats each. Each student takes a place in the hall randomly. What is the probability that no more than three people will be in the seventh place of the row?
Solution:
Example 2.31.
Then, according to the classical definition of probability:
Where -- number of parts in the batch;
-- number of non-standard parts in the batch;
– number of selected parts;
-- number of non-standard parts among those selected.
Then the distribution law of the random variable will be as follows.
In this lesson we will find the probability of an event occurring in independent trials when repeating trials . Trials are called independent if the probability of one or another outcome of each trial does not depend on what outcomes other trials had. . Independent tests can be carried out both under the same conditions and under different conditions. In the first case, the probability of the occurrence of some event is the same in all trials, in the second case it varies from trial to trial.
Examples of independent retests :
- one of the device nodes or two or three nodes will fail, and the failure of each node does not depend on the other node, and the probability of failure of one node is constant in all tests;
- a part, or three, four, five parts produced under certain constant technological conditions, will turn out to be non-standard, and one part may turn out to be non-standard regardless of any other part and the probability that the part will turn out to be non-standard is constant in all tests;
- out of several shots at a target, one, three or four shots hit the target regardless of the outcome of the other shots and the probability of hitting the target is constant in all trials;
- when dropping a coin, the machine will operate correctly one, two, or other number of times, regardless of the outcome of other coin drops, and the probability that the machine will operate correctly is constant across all trials.
These events can be described in one diagram. Each event occurs in each trial with the same probability, which does not change if the results of previous trials become known. Such tests are called independent, and the circuit is called Bernoulli scheme . It is assumed that such tests can be repeated as desired a large number of once.
If the probability p occurrence of an event A is constant in each trial, then the probability that in n independent testing event A will come m times, is located by Bernoulli's formula :
(Where q= 1 – p- the probability that the event will not occur)
Let us set the task - to find the probability that an event of this type in n independent tests will come m once.
Bernoulli's formula: examples of problem solving
Example 1. Find the probability that among five parts taken at random, two are standard, if the probability that each part turns out to be standard is 0.9.
Solution. Probability of event A, consisting in the fact that a part taken at random is standard, there is p=0.9 , and there is a probability that it is non-standard q=1–p=0.1. The event designated in the problem statement (we denote it by IN) will occur if, for example, the first two parts turn out to be standard, and the next three are non-standard. But the event IN will also occur if the first and third parts turn out to be standard and the rest are non-standard, or if the second and fifth parts are standard and the rest are non-standard. There are other possibilities for the event to occur IN. Any of them is characterized by the fact that out of five parts taken, two, occupying any places out of five, will turn out to be standard. Hence, total number various possibilities for the occurrence of an event IN is equal to the number of possibilities for placing two standard parts in five places, i.e. is equal to the number of combinations of five elements by two, and .
The probability of each possibility according to the probability multiplication theorem is equal to the product of five factors, of which two, corresponding to the appearance of standard parts, are equal to 0.9, and the remaining three, corresponding to the appearance of non-standard parts, are equal to 0.1, i.e. this probability is . Since these ten possibilities are incompatible events, by the addition theorem the probability of an event IN, which we denote
Example 2. The probability that the machine will require the attention of a worker within an hour is 0.6. Assuming that the problems on the machines are independent, find the probability that within an hour a worker’s attention will require any one machine out of the four he operates.
Solution. Using Bernoulli's formula at n=4 , m=1 , p=0.6 and q=1–p=0.4, we get
Example 3. For normal operation of the carpool, there must be at least eight vehicles on the line, and there are ten of them. The probability of each vehicle not entering the line is 0.1. Find the probability of normal operation of the car depot in the next day.
Solution. The carpool will work normally (event F), if eight or eight come on line (event A), or nine (event IN), or all ten cars event (event C). According to the theorem of addition of probabilities,
We find each term according to Bernoulli's formula. Here n=10 , m=8; 10 and p=1-0.1=0.9, since p should indicate the probability of the vehicle entering the line; Then q=0.1. As a result we get
Example 4. Let the probability that a customer needs size 41 men's shoes be 0.25. Find the probability that out of six buyers, at least two need shoes of size 41.
| Find the average score of students who received the following marks during the exam: 5; 3; 4; 5; 3; 2; 3; 5; 4; 3 | 3,7 |
| Discrete random variable X has a probability distribution law: (x=5;7 p=0.3;0.7): | 6,4 |
| the appearance of a jack and a queen when one card is taken from the deck once; | |
| The urn contains 5 white and 7 black balls. Two balls are drawn from the urn at the same time. The probability that both balls are white is: | 5/33 |
| The dice are tossed once. Event A – “the number of points rolled is greater than two”; event B – “the number of points rolled is less than five.” The following statement is true: | events A and B are joint |
| The dice are tossed once. The probability that a coin will appear on the top edge even number points is equal to: | 1/2 |
| The probability of some event occurring may be equal to: | 0,6 |
| Given the probability density of a continuous random variable X: Find the probability that as a result of the test X will take values belonging to the interval (0.3;1) | 0,91 |
| The mathematical expectation M(Y) of the random variable Y = 2X + 4 with M(X) = 3 is equal to: | |
| The first student will successfully answer this test option with a probability of 0.5, and the second - with a probability of 0.4. The probability that both students pass the test is: | 0,2 |
| The mathematical expectation of the difference between two random variables is: | differences in mathematical expectations of these random variables |
| If events A and B are incompatible, then the formula is valid: | P(A+B)=P(A)+P(B) |
| A continuous random variable X is given by an integral probability distribution function. Then the value of C is... | C=1/2, a=1 |
| Constant multiplier from under the dispersion sign... | Can be squared and taken out |
| The variance of a random variable characterizes... | dispersion of a random variable about the mean |
| The formula expresses | Markov's inequality |
| In a batch of 10 products, 8 products are defective. The probability that during random inspection, out of 5 selected products, 3 products will be defective (C is the symbol for the number of combinations): | 2/9 |
| The formula expresses | Chebyshev's inequality |
| The mathematical expectation of a random variable has the dimension | the most random value |
| The formula expresses | Bernoulli's theorem |
| The random variable is uniformly distributed over the interval [-2,2]. Then its probability density takes the value equal to | 1/4 |
| A discrete random variable X has a distribution law: (X=7;14;21;28 P=0.1;0.2Pз=0.4): Probability Pз is equal to: | 0,3 |
| A continuous random variable X is given by a differential probability distribution function. Then the value of C is... | 1/3 |
| The first student will successfully answer this test option with a probability of 0.5, and the second - with a probability of 0.7. The probability that both students pass the test is: | 0,35 |
| The urn contains a white and b black balls. Two balls are drawn (simultaneously or sequentially) from the urn. The probability that both balls are white is: | a*(a-1)/(a+b)*(a+b-1) |
| The following events are incompatible | the appearance of a coat of arms and numbers when one coin is tossed once; |
| The first shooter hits the target with probability 0.9, and the second shooter hits the target with probability 0.5. Each shooter fires one shot. The probability that both shooters will hit the target is: | 0,45 |
| Quantity in various ways choosing (the order does not matter) 3 volumes from an 8-volume collected works equals: | |
| The number of combinations that can be obtained by rearranging the letters included in the word “number” is equal to: | |
| If events A and B are simultaneous, then the formula is valid: | P(A+B)<=P(A)+P(B) |
| The number of five-digit numbers that can be read equally from left to right and from right to left is... | |
| There are 10 quality and 4 defective products. One product is removed. Event A – “a quality product was retrieved”, event B – “a defective product was retrieved”. For these events the following statement is incorrect: | the probability of event A is equal to the probability of event B; |
In a batch of N products, M products are defective. The probability that during random inspection out of n selected products m products will be defective (m | upper right term of the numerator (C(N-M))^n-m |
|
| The dice are tossed once. Event A – “the number of points rolled is greater than three”; event B – “the number of points rolled is less than three.” The following statement is true: | events A and B are incompatible |
| The probability for a student to pass the first exam is 0.6, the second - 0.4. The probability of passing either the first, or the second, or both exams is: | 0,76 |
| The dice are tossed once. The probability that a number of points equal to two or four will appear on the top side is: | 1/3 |
| The probability of some event occurring cannot be equal: | |
| The probability of producing a non-standard part is 0.11. Using Bernoulli's formula, find the probability that out of five parts taken at random there will be four standard ones. | 0,345 |
| In the test questions, there are 75% of questions to which students know the answers. The teacher selects two questions from them and asks them to the student. Determine the probability that among the questions a student receives there is at least one to which he knows the answer | 0,937 |
End of work -
This topic belongs to the section:
Given a differential function of a random variable x: find the probability that as a result of the test x will take values belonging to the interval 0.5; 1
What is a hypothesis containing only one assumption called a simple hypothesis?
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Therefore, your immediate pastime will be extremely useful. In addition, I will tell you what is wrong overwhelming majority participants of lotteries and gambling. ...Noooo, faith or a faint hope of “hitting the jackpot” has absolutely nothing to do with it ;-) Without even having time to blink an eye, we are immersed in the topic:
What's happened independent tests ? Almost everything is clear from the name itself. Let several tests be performed. If the probability of a certain event occurring in each of them does not depend from the results of the remaining tests, then... we finish the sentence in unison =) Well done. Moreover, the phrase “independent tests” often means repeated independent tests – when they are carried out one after another.
The simplest examples:
– the coin is tossed 10 times;
– the die is tossed 20 times.
It is absolutely clear that the probability of getting heads or tails in any test does not depend on the results of other throws. A similar statement, naturally, is true for the cube.
But the sequential removal of cards from the deck is not a series of independent tests - as you remember, this is a chain dependent events. However, if you return the card every time, the situation will become “as it should be.”
I hasten to please you - our guest is another Terminator, who is absolutely indifferent to his successes/failures, and therefore his shooting is an example of stability =):
Problem 1
The shooter fires 4 shots at the target. The probability of a hit with each shot is constant and equal. Find the probability that:
a) the shooter will hit only once;
b) the shooter will hit 2 times.
Solution: the condition is formulated in general and the probability of hitting the target with each shot considered famous. It is equal (if it’s really hard, assign the parameter some specific value, for example,) .
Once we know , it is easy to find the probability of a miss in each shot:
, that is, “ku” is also quantity known to us.
a) Consider the event "The shooter will only hit once" and denote its probability by (indices are understood as “one hit out of four”). This event consists of 4 incompatible outcomes: the shooter will hit the 1st or in the 2nd or in the 3rd or in the 4th attempt.
Find the probability that when throwing 10 coins, 3 coins will come up heads.
Here the tests are not repeated, but rather are carried out simultaneously, but, nevertheless, the same formula works: .
The solution will differ in meaning and some comments, in particular:
using these methods, you can choose 3 coins on which heads will appear.
– probability of getting heads on each of 10 coins
etc.
However, in practice, such problems do not occur so often, and, apparently, for this reason, Bernoulli’s formula is almost stereotypically associated only with repeated tests. Although, as just shown, repeatability is not at all necessary.
The following task is for you to solve on your own:
Problem 3
The dice are thrown 6 times. Find the probability that 5 points:
a) will not fall out (will appear 0 times);
b) will appear 2 times;
c) will appear 5 times.
Round the results to 4 decimal places.
A short solution and answer at the end of the lesson.
It is obvious that in the examples under consideration, some events are more likely, and some are less likely. So, for example, with 6 dice rolls, even without any calculations, it is intuitively clear that the probabilities of the events of points “a” and “be” are much greater than the probability that “five” will roll up 5 times. Now let's set the task to find
MOST LIKELY number of occurrences of an event in independent trials
Again, at the level of intuition in Problem No. 3, we can conclude that the most likely number of appearances of the “five” is equal to one - after all, there are six faces in total, and with 6 dice rolls, each of them should appear on average once. Those interested can calculate the probability and see if it is greater than the “competing” values and .
Let us formulate a strict criterion: to find the most likely number of occurrences of a random event in independent trials (with probability in each trial) are guided by the following double inequality:
1) if the value is fractional, then there is a single most probable number;
in particular, if is an integer, then it is the most probable number: ;
2) if it is a whole, then there are two most probable numbers: and .
The most likely number of occurrences of a “five” with 6 dice rolls falls under a special case of the first point:
In order to consolidate the material, we will solve a couple of problems:
Problem 4
The probability that a basketball player will hit the basket when throwing the ball is 0.3. Find the most probable number of hits with 8 throws and the corresponding probability.
And this is, if not the Terminator, then at least a cold-blooded athlete =)
Solution: to estimate the most likely number of hits we use the double inequality. In this case:
– total throws;
– the probability of hitting the basket with each throw;
– the probability of a miss with each throw.
Thus, the most likely number of hits with 8 throws is within the following limits:
Since the left border is a fractional number (point No. 1), then there is a single most probable value, and, obviously, it is equal to .
Using Bernoulli's formula, we calculate the probability that with 8 throws there will be exactly 2 hits:
Answer: – most likely number of hits with 8 throws,
– the corresponding probability.
A similar task for independent solution:
Problem 5
The coin is tossed 9 times. Find the probability of the most likely number of occurrences of an eagle
Sample solution and answer at the end of the lesson.
After a fascinating digression, we’ll look at a few more problems, and then I’ll share the secret to playing gambling and lotteries correctly.
Problem 6
Among the products produced on an automatic machine, on average, 60% of products are first grade. What is the probability that among 6 randomly selected items there will be:
a) from 2 to 4 first-class products;
b) at least 5 first-class products;
c) at least one product of a lower grade.
The likelihood of producing a first-class product does not depend on the quality of other products produced, so we are talking about independent testing here. Try not to neglect the analysis of the condition, otherwise it may turn out to be an event dependent or the task is about something else altogether.
Solution: the probability is encoded as a percentage, which, I remind you, needs to be divided by one hundred: - the probability that the selected product will be of the 1st grade.
Then: – the probability that it will not be first-class.
a) Event “Among 6 randomly selected products there will be from 2 to 4 first-class products” consists of three incompatible outcomes:
among the products there will be 2 first-class ones or 3 first-class or 4 first-class.
It is more convenient to deal with the outcomes separately. We use Bernoulli's formula three times:
– the probability that at least 5 out of six computers will work without failure during the day.
This value will also not suit us, since it is less than the required reliability of the computer center:
Thus, six computers are also not enough. Let's add one more:
3) Let there be computers in the computer center. Then 5, 6 or 7 computers should work flawlessly. Using Bernoulli's formula and theorem for adding probabilities of incompatible events, let’s find the probability that at least 5 out of seven computers will work flawlessly during the day:
Eat! The required level of reliability has been achieved.
You can, of course, supply more computers, but why overpay? =)
Answer: to ensure normal operation of the computer center during the day with a probability no less than , you need to install at least seven computers.
Bernoulli's formula is very convenient, but on the other hand, it also has a number of disadvantages. So, for example, with sufficiently large values of “en” and “em” its use is difficult due to the huge values of the factorials. In this case use Laplace's theorem, which we will look at in the next lesson. Another common situation in practice is when the probability of some event in a single trial is quite small, but the number of trials is large. The issue is resolved with Poisson formulas.
And finally, the promised secret:
...So, after all, how to play gambling and lotteries correctly?
Probably, many expected to hear from me something like: “It’s better not to play at all,” “Open your own casino,” “Organize a lottery,” etc.
Well, why not play? Game is one of the entertainments, and for entertainment, as you know, you need... absolutely right! Therefore, the money you play with should be considered a payment for entertainment, but in no case a tragic loss.
However, every gambler wants to win. And win a good amount. What tactics (there is no talk about strategy at all) it is most profitable to stick to a game with a known losing mathematical expectation, for example, in roulette? It's best to put it right away All chips, as an option, for “red” or “black”. You're likely to double (both quickly and a lot!), and if this happens, be sure to spend your winnings on other entertainment =)
It makes no sense to play according to some “system” (at least because it's stupid) and spend hours/days/weeks on this - in the same roulette, the house has a minimal advantage, and you can lose for a very long time. If in an offline casino this can somehow be understood (communication, drinking, girls, etc.), then the online game will leave you with red eyes and a feeling of deep annoyance.
As for lotteries, it is better to buy a ticket again for fun and... at random. Or "on a whim." True, for some reason I personally have never heard of psychics and fortune tellers who win the lottery =) No other way than they are encrypted.
Naturally, the listed tips do not apply to chronic gambling addicts and for them it’s just “It’s better not to play at all.” Well, for those visitors who dream of getting rich through gambling, I strongly recommend reading or re-reading the introductory article on
According to the theorem of addition of probabilities of incompatible events:
– the probability that in a series of 8 shots there will be none or 1 hit.
Let's find the probability of the opposite event:
– the probability that the target will be hit at least twice.
Answer
: