Poisson distribution.

Let's consider the most typical situation in which the Poisson distribution arises. Let the event A appears a certain number of times in a fixed area of ​​space (interval, area, volume) or a period of time with constant intensity. To be specific, consider the sequential occurrence of events over time, called a stream of events. Graphically, the flow of events can be illustrated by many points located on the time axis.

This could be a flow of calls in the service sector (repair of household appliances, calling an ambulance, etc.), a flow of calls to a telephone exchange, failure of some parts of the system, radioactive decay, pieces of fabric or metal sheets and the number of defects on each of them, etc. The Poisson distribution is most useful in those problems where it is necessary to determine only the number of positive outcomes (“successes”).

Let's imagine a bun with raisins, divided into small pieces of equal size. Due to the random distribution of raisins, all pieces cannot be expected to contain the same number of raisins. When the average number of raisins contained in these pieces is known, then the Poisson distribution gives the probability that any given piece contains X=k(k= 0,1,2,...,)number of raisins.

In other words, the Poisson distribution determines which part of a long series of pieces will contain equal to 0, or 1, or 2, or etc. number of highlights.

Let's make the following assumptions.

1. The probability of a certain number of events occurring in a given time interval depends only on the length of this interval, and not on its position on the time axis. This is the property of stationarity.

2. The occurrence of more than one event in a sufficiently short period of time is practically impossible, i.e. the conditional probability of the occurrence of another event in the same interval tends to zero as ® 0. This is the property of ordinaryness.

3. The probability of a given number of events occurring in a fixed period of time does not depend on the number of events appearing in other periods of time. This is the property of lack of aftereffect.

A flow of events that satisfies the above propositions is called the simplest.

Let's consider a fairly short period of time. Based on property 2, the event may appear once in this interval or not appear at all. Let us denote the probability of an event occurring by R, and non-appearance – through q = 1-p. Probability R is constant (property 3) and depends only on the value (property 1). The mathematical expectation of the number of occurrences of an event in the interval will be equal to 0× q+ 1× p = p. Then the average number of occurrences of events per unit time is called the flow intensity and is denoted by a, those. a = .

Consider a finite period of time t and divide it by n parts = . The occurrences of events in each of these intervals are independent (property 2). Let us determine the probability that in a period of time t at constant flow intensity A the event will appear exactly X = k won't appear again n–k. Since an event can in each of n gaps appear no more than 1 time, then for its appearance k once in a segment of duration t it should appear in any k intervals from the total n. There are total such combinations, and the probability of each is equal. Consequently, by the addition theorem of probabilities we obtain the well-known Bernoulli formula for the desired probability

This equality is written as an approximate one, since the initial premise for its derivation was property 2, which is fulfilled more accurately the smaller . To obtain exact equality, let us pass to the limit at ® 0 or, what is the same, n® . We will get it after replacement

P = a= and q = 1 – .

Let's introduce a new parameter = at, meaning the average number of occurrences of an event in a segment t. After simple transformations and passing to the limit in the factors, we obtain.

= 1, = ,

Finally we get

, k = 0, 1, 2, ...

e = 2.718... is the base of the natural logarithm.

Definition. Random value X, which takes only integer, positive values ​​0, 1, 2, ... has a Poisson distribution law with parameter if

For k = 0, 1, 2, ...

The Poisson distribution was proposed by the French mathematician S.D. Poisson (1781-1840). It is used to solve problems of calculating the probabilities of relatively rare, random, mutually independent events per unit of time, length, area and volume.

For the case when a) is large and b) k= , the Stirling formula is valid:

To calculate subsequent values, a recurrent formula is used

P(k + 1) = P(k).

Example 1. What is the probability that out of 1000 people on a given day: a) none, b) one, c) two, d) three people were born?

Solution. Because p= 1/365, then q= 1 – 1/365 = 364/365 "1.

Then

A) ,

b) ,

V) ,

G) .

Therefore, if there are samples of 1000 people, then the average number of people who were born on a particular day will accordingly be 65; 178; 244; 223.

Example 2. Determine the value at which with probability R the event appeared at least once.

Solution. Event A= (appear at least once) and = (not appear even once). Hence .

From here And .

For example, for R= 0.5, for R= 0,95 .

Example 3. On looms operated by one weaver, 90 thread breaks occur within an hour. Find the probability that at least one thread break will occur in 4 minutes.

Solution. By condition t = 4 min. and the average number of breaks per minute, from where . The required probability is .

Properties. The mathematical expectation and variance of a random variable having a Poisson distribution with parameter are equal to:

M(X) = D(X) = .

These expressions are obtained by direct calculations:

This is where the replacement was made n = k– 1 and the fact that .

By performing transformations similar to those used in the output M(X), we get

The Poisson distribution is used to approximate the binomial distribution at large n

Brief theory

Let independent tests be carried out, in each of which the probability of the event occurring is equal to . To determine the probability of an event occurring in these tests, Bernoulli's formula is used. If it is large, then use or. However, this formula is not suitable if it is small. In these cases (great, small) they resort to asymptotic Poisson's formula.

Let us set ourselves the task of finding the probability that, given a very large number of trials, in each of which the probability of an event is very small, the event will occur exactly once. Let us make an important assumption: the product retains a constant value, namely . This means that the average number of occurrences of an event in different series of trials, i.e. for different values, remains unchanged.

Example of problem solution

Problem 1

The base received 10,000 electric lamps. The probability that the lamp will break during travel is 0.0003. Find the probability that among the lamps received, five lamps will be broken.

Solution

Condition for the applicability of the Poisson formula:

If the probability of an event occurring in a single trial is quite close to zero, then even for large values ​​of the number of trials, the probability calculated using Laplace’s local theorem turns out to be insufficiently accurate. In such cases, use the formula derived by Poisson.

Let the event - 5 lamps be broken

Let's use Poisson's formula:

In our case:

Answer

Problem 2

The enterprise has 1000 units of equipment of a certain type. The probability of a piece of equipment failing within an hour is 0.001. Draw up a distribution law for the number of equipment failures per hour. Find numerical characteristics.

Solution

Random variable - the number of equipment failures, can take values

Let's use Poisson's law:

Let's find these probabilities:

.

The mathematical expectation and variance of a random variable distributed according to Poisson’s law is equal to the parameter of this distribution:

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The most common case of various types of probability distributions is the binomial distribution. Let us use its versatility to determine the most common particular types of distributions encountered in practice.

Binomial distribution

Let there be some event A. The probability of occurrence of event A is equal to p, the probability of non-occurrence of event A is 1 p, sometimes it is designated as q. Let n number of tests, m frequency of occurrence of event A in these n tests.

It is known that the total probability of all possible combinations of outcomes is equal to one, that is:

1 = p n + n · p n 1 (1 p) + C n n 2 · p n 2 (1 p) 2 + + C n m · p m· (1 p) n – m+ + (1 p) n .

p n probability that in nn once; n · p n 1 (1 p) probability that in nn 1) once and will not happen 1 time; C n n 2 · p n 2 (1 p) 2 probability that in n tests, event A will occur ( n 2) times and will not happen 2 times; P m = C n m · p m· (1 p) n – m probability that in n tests, event A will occur m will never happen ( n – m) once; (1 p) n probability that in n in trials, event A will not occur even once; number of combinations of n By m .

Expected value M binomial distribution is equal to:

M = n · p ,

Where n number of tests, p probability of occurrence of event A.

Standard deviation σ :

σ = sqrt( n · p· (1 p)) .

Example 1. Calculate the probability that an event that has a probability p= 0.5, in n= 10 trials will happen m= 1 time. We have: C 10 1 = 10, and further: P 1 = 10 0.5 1 (1 0.5) 10 1 = 10 0.5 10 = 0.0098. As we can see, the probability of this event occurring is quite low. This is explained, firstly, by the fact that it is absolutely not clear whether the event will happen or not, since the probability is 0.5 and the chances here are “50 to 50”; and secondly, it is required to calculate that the event will occur exactly once (no more and no less) out of ten.

Example 2. Calculate the probability that an event that has a probability p= 0.5, in n= 10 trials will happen m= 2 times. We have: C 10 2 = 45, and further: P 2 = 45 0.5 2 (1 0.5) 10 2 = 45 0.5 10 = 0.044. The likelihood of this event occurring has increased!

Example 3. Let's increase the likelihood of the event itself occurring. Let's make it more likely. Calculate the probability that an event that has a probability p= 0.8, in n= 10 trials will happen m= 1 time. We have: C 10 1 = 10, and further: P 1 = 10 0.8 1 (1 0.8) 10 1 = 10 0.8 1 0.2 9 = 0.000004. The probability has become less than in the first example! The answer, at first glance, seems strange, but since the event has a fairly high probability, it is unlikely to happen only once. It is more likely that it will happen more than once. Indeed, counting P 0 , P 1 , P 2 , P 3, , P 10 (probability that an event in n= 10 trials will happen 0, 1, 2, 3, , 10 times), we will see:

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.8 0 (1 0.8) 10 0 = 1 1 0.2 10 = 0.0000;
P 1 = 10 0.8 1 (1 0.8) 10 1 = 10 0.8 1 0.2 9 = 0.0000;
P 2 = 45 0.8 2 (1 0.8) 10 2 = 45 0.8 2 0.2 8 = 0.0000;
P 3 = 120 0.8 3 (1 0.8) 10 3 = 120 0.8 3 0.2 7 = 0.0008;
P 4 = 210 0.8 4 (1 0.8) 10 4 = 210 0.8 4 0.2 6 = 0.0055;
P 5 = 252 0.8 5 (1 0.8) 10 5 = 252 0.8 5 0.2 5 = 0.0264;
P 6 = 210 0.8 6 (1 0.8) 10 6 = 210 0.8 6 0.2 4 = 0.0881;
P 7 = 120 0.8 7 (1 0.8) 10 7 = 120 0.8 7 0.2 3 = 0.2013;
P 8 = 45 0.8 8 (1 0.8) 10 8 = 45 0.8 8 0.2 2 = 0.3020(highest probability!);
P 9 = 10 0.8 9 (1 0.8) 10 9 = 10 0.8 9 0.2 1 = 0.2684;
P 10 = 1 0.8 10 (1 0.8) 10 10 = 1 0.8 10 0.2 0 = 0.1074

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

Normal distribution

If we depict the quantities P 0 , P 1 , P 2 , P 3, , P 10, which we calculated in example 3, on the graph, it turns out that their distribution has a form close to the normal distribution law (see Fig. 27.1) (see lecture 25. Modeling of normally distributed random variables).

Rice. 27.1. Type of binomial distribution
probabilities for different m at p = 0.8, n = 10

The binomial law becomes normal if the probabilities of occurrence and non-occurrence of event A are approximately the same, that is, we can conditionally write: p≈ (1 p) . For example, let's take n= 10 and p= 0.5 (that is p= 1 p = 0.5 ).

We will come to such a problem meaningfully if, for example, we want to theoretically calculate how many boys and how many girls there will be out of 10 children born in a maternity hospital on the same day. More precisely, we will count not boys and girls, but the probability that only boys will be born, that 1 boy and 9 girls will be born, that 2 boys and 8 girls will be born, and so on. Let us assume for simplicity that the probability of having a boy and a girl is the same and equal to 0.5 (but in fact, to be honest, this is not the case, see the course “Modeling Artificial Intelligence Systems”).

It is clear that the distribution will be symmetrical, since the probability of having 3 boys and 7 girls is equal to the probability of having 7 boys and 3 girls. The greatest likelihood of birth will be 5 boys and 5 girls. This probability is equal to 0.25, by the way, it is not that big in absolute value. Further, the probability that 10 or 9 boys will be born at once is much less than the probability that 5 ± 1 boy will be born out of 10 children. The binomial distribution will help us make this calculation. So.

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.5 0 (1 0.5) 10 0 = 1 1 0.5 10 = 0.000977;
P 1 = 10 0.5 1 (1 0.5) 10 1 = 10 0.5 10 = 0.009766;
P 2 = 45 0.5 2 (1 0.5) 10 2 = 45 0.5 10 = 0.043945;
P 3 = 120 0.5 3 (1 0.5) 10 3 = 120 0.5 10 = 0.117188;
P 4 = 210 0.5 4 (1 0.5) 10 4 = 210 0.5 10 = 0.205078;
P 5 = 252 0.5 5 (1 0.5) 10 5 = 252 0.5 10 = 0.246094;
P 6 = 210 0.5 6 (1 0.5) 10 6 = 210 0.5 10 = 0.205078;
P 7 = 120 0.5 7 (1 0.5) 10 7 = 120 0.5 10 = 0.117188;
P 8 = 45 0.5 8 (1 0.5) 10 8 = 45 0.5 10 = 0.043945;
P 9 = 10 0.5 9 (1 0.5) 10 9 = 10 0.5 10 = 0.009766;
P 10 = 1 0.5 10 (1 0.5) 10 10 = 1 0.5 10 = 0.000977

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

Let us display the quantities on the graph P 0 , P 1 , P 2 , P 3, , P 10 (see Fig. 27.2).

Rice. 27.2. Graph of binomial distribution with parameters
p = 0.5 and n = 10, bringing it closer to the normal law

So, under the conditions m ≈ n/2 and p≈ 1 p or p≈ 0.5 instead of the binomial distribution, you can use the normal one. For large values n the graph shifts to the right and becomes more and more flat, as the mathematical expectation and variance increase with increasing n : M = n · p , D = n · p· (1 p) .

By the way, the binomial law tends to normal and with increasing n, which is quite natural, according to the central limit theorem (see lecture 34. Recording and processing statistical results).

Now consider how the binomial law changes in the case when p ≠ q, that is p> 0 . In this case, the hypothesis of normal distribution cannot be applied, and the binomial distribution becomes a Poisson distribution.

Poisson distribution

The Poisson distribution is a special case of the binomial distribution (with n>> 0 and at p>0 (rare events)).

A formula is known from mathematics that allows you to approximately calculate the value of any member of the binomial distribution:

Where a = n · p Poisson parameter (mathematical expectation), and the variance is equal to the mathematical expectation. Let us present mathematical calculations that explain this transition. Binomial distribution law

P m = C n m · p m· (1 p) n – m

can be written if you put p = a/n , as

Because p is very small, then only the numbers should be taken into account m, small compared to n. Work

very close to unity. The same applies to the size

Magnitude

very close to e – a. From here we get the formula:

Example. The box contains n= 100 parts, both high-quality and defective. The probability of receiving a defective product is p= 0.01 . Let's say that we take out a product, determine whether it is defective or not, and put it back. By doing this, it turned out that out of 100 products that we went through, two turned out to be defective. What is the likelihood of this?

From the binomial distribution we get:

From the Poisson distribution we get:

As you can see, the values ​​turned out to be close, so in the case of rare events it is quite acceptable to apply Poisson’s law, especially since it requires less computational effort.

Let us show graphically the form of Poisson's law. Let's take the parameters as an example p = 0.05 , n= 10 . Then:

C 10 0 = 1 , C 10 1 = 10 , C 10 2 = 45 , C 10 3 = 120 , C 10 4 = 210 , C 10 5 = 252 ,
C 10 6 = 210 , C 10 7 = 120 , C 10 8 = 45 , C 10 9 = 10 , C 10 10 = 1 ;

P 0 = 1 0.05 0 (1 0.05) 10 0 = 1 1 0.95 10 = 0.5987;
P 1 = 10 0.05 1 (1 0.05) 10 1 = 10 0.05 1 0.95 9 = 0.3151;
P 2 = 45 0.05 2 (1 0.05) 10 2 = 45 0.05 2 0.95 8 = 0.0746;
P 3 = 120 0.05 3 (1 0.05) 10 3 = 120 0.05 3 0.95 7 = 0.0105;
P 4 = 210 0.05 4 (1 0.05) 10 4 = 210 0.05 4 0.95 6 = 0.00096;
P 5 = 252 0.05 5 (1 0.05) 10 5 = 252 0.05 5 0.95 5 = 0.00006;
P 6 = 210 0.05 6 (1 0.05) 10 6 = 210 0.05 6 0.95 4 = 0.0000;
P 7 = 120 0.05 7 (1 0.05) 10 7 = 120 0.05 7 0.95 3 = 0.0000;
P 8 = 45 0.05 8 (1 0.05) 10 8 = 45 0.05 8 0.95 2 = 0.0000;
P 9 = 10 0.05 9 (1 0.05) 10 9 = 10 0.05 9 0.95 1 = 0.0000;
P 10 = 1 0.05 10 (1 0.05) 10 10 = 1 0.05 10 0.95 0 = 0.0000

Of course P 0 + P 1 + P 2 + P 3 + P 4 + P 5 + P 6 + P 7 + P 8 + P 9 + P 10 = 1 .

Rice. 27.3. Poisson distribution plot at p = 0.05 and n = 10

At n> ∞ the Poisson distribution turns into a normal law, according to the central limit theorem (see.

Where λ is equal to the average number of occurrences of events in identical independent trials, i.e. λ = n × p, where p is the probability of an event in one trial, e = 2.71828.

The Poisson law distribution series has the form:

Purpose of the service. The online calculator is used to construct the Poisson distribution and calculate all the characteristics of the series: mathematical expectation, variance and standard deviation. The report with the decision is drawn up in Word format.

Number of tests: n= , Probability p =
Calculate the probability for: m =
will come once
less once
no less once
more once
no more once
no less and no more once
will happen at least once

In the case when n is large and λ = p n > 10, the Poisson formula gives a very rough approximation and the local and integral theorems of Moivre-Laplace are used to calculate P n (m).

Numerical characteristics of random variable X

Expectation of Poisson distribution
M[X] = λ

Variance of Poisson distribution
D[X] = λ

Example No. 1. Seeds contain 0.1% weeds. What is the probability of finding 5 weed seeds if you randomly select 2000 seeds?
Solution.
The probability p is small, but the number n is large. np = 2 P(5) = λ 5 e -5 /5! = 0.03609
Expected value: M[X] = λ = 2
Dispersion: D[X] = λ = 2

Example No. 2. Among rye seeds there are 0.4% weed seeds. Draw up a distribution law for the number of weeds with a random selection of 5000 seeds. Find the mathematical expectation and variance of this random variable.
Solution. Mathematical expectation: M[X] = λ = 0.004*5000 = 20. Dispersion: D[X] = λ = 20
Distribution law:

X	0	1	2	…	m	…
P	e -20	20e -20	200e -20	…	20 m e -20 /m!	…

Example No. 3. At a telephone exchange, an incorrect connection occurs with a probability of 1/200. Find the probability that among 200 connections the following will occur:
a) exactly one incorrect connection;
b) less than three incorrect connections;
c) more than two incorrect connections.
Solution. According to the conditions of the problem, the probability of the event is low, so we use the Poisson formula (15).
a) Given: n = 200, p = 1/200, k = 1. Let’s find P 200 (1).
We get: . Then P 200 (1) ≈ e -1 ≈ 0.3679.
b) Given: n = 200, p = 1/200, k< 3. Найдем P 200 (k < 3).
We have: a = 1.

c) Given: n = 200, p = 1/200, k > 2. Find P 200 (k > 2).
This problem can be solved more simply: find the probability of the opposite event, since in this case you need to calculate fewer terms. Taking into account the previous case, we have

Consider the case where n is sufficiently large and p sufficiently small; let's put np = a, where a is some number. In this case, the desired probability is determined by the Poisson formula:

The probability of occurrence of k events during a time duration t can also be found using the Poisson formula:
where λ is the intensity of the flow of events, that is, the average number of events that appear per unit time.

Example No. 4. The probability that the part is defective is 0.005. 400 parts are checked. Provide a formula for calculating the probability that more than 3 parts are defective.

Example No. 5. The probability of defective parts appearing during mass production is p. determine the probability that a batch of N parts contains a) exactly three parts; b) no more than three defective parts.
p=0.001; N = 4500
Solution.
The probability p is small, but the number n is large. np = 4.5< 10. Значит случайная величина Х – распределена по Пуассоновскому распределению. Составим закон.
The random variable X has a range of values ​​(0,1,2,...,m). The probabilities of these values ​​can be found using the formula:

Let's find the distribution series of X.
Here λ = np = 4500*0.001 = 4.5
P(0) = e - λ = e -4.5 = 0.01111
P(1) = λe -λ = 4.5e -4.5 = 0.04999

Then the probability that a batch of N parts contains exactly three parts is equal to:

Then the probability that a batch of N parts contains no more than three defective parts:
P(x<3) = P(0) + P(1) + P(2) = 0,01111 + 0,04999 + 0,1125 = 0,1736

Example No. 6. An automatic telephone exchange receives N calls on average per hour. Determine the probability that in a given minute she will receive: a) exactly two calls; b) more than two calls.
N=18
Solution.
In one minute, the automatic telephone exchange receives on average λ = 18/60 min. = 0.3
Assuming that a random number X of calls received at the PBX in one minute,
obeys Poisson's law, using the formula we will find the desired probability

Let's find the distribution series of X.
Here λ = 0.3
P(0) = e - λ = e -0.3 = 0.7408
P(1) = λe -λ = 0.3e -0.3 = 0.2222

The probability that she will receive exactly two calls in a given minute is:
P(2) = 0.03334
The probability that she will receive more than two calls in a given minute is:
P(x>2) = 1 – 0.7408 – 0.2222 – 0.03334 = 0.00366

Example No. 7. Two elements operating independently of each other are considered. The duration of failure-free operation has an exponential distribution with the parameter λ1 = 0.02 for the first element and λ2 = 0.05 for the second element. Find the probability that in 10 hours: a) both elements will work without failure; b) only the Probability that element No. 1 will not fail in 10 hours:
Decision.
P 1 (0) = e -λ1*t = e -0.02*10 = 0.8187

Probability that element No. 2 will not fail in 10 hours:
P 2 (0) = e -λ2*t = e -0.05*10 = 0.6065

a) both elements will work flawlessly;
P(2) = P 1 (0)*P 2 (0) = 0.8187*0.6065 = 0.4966
b) only one element will fail.
P(1) = P 1 (0)*(1-P 2 (0)) + (1-P 1 (0))*P 2 (0) = 0.8187*(1-0.6065) + (1-0.8187) *0.6065 = 0.4321

Example No. 7. Production produces 1% defects. What is the probability that out of 1100 products taken for research, no more than 17 will be rejected?
Note: since here n*p =1100*0.01=11 > 10, it is necessary to use

As soon as requests started coming in: “Where is Poisson? Where are the problems using the Poisson formula? and so on. And so I'll start with private use Poisson distribution - due to the high demand for the material.

The task is painfully familiar:

And the next two tasks are fundamentally different from the previous ones:

Example 4

The random variable is subject to Poisson's law with mathematical expectation. Find the probability that a given random variable will take a value less than its mathematical expectation.

The difference is that here we are talking EXACTLY about the Poisson distribution.

Solution: random variable takes values with probabilities:

According to the condition, , and here everything is simple: the event consists of three inconsistent outcomes:

The probability that a random variable will take a value less than its mathematical expectation.

Answer:

A similar comprehension task:

Example 5

The random variable is subject to Poisson's law with mathematical expectation. Find the probability that a given random variable will take a positive value.

The solution and answer are at the end of the lesson.

Besides approachingbinomial distribution(Examples 1-3), the Poisson distribution has found wide application in queuing theory for probabilistic characteristics the simplest stream of events. I'll try to be concise:

Let some system receive requests (phone calls, incoming clients, etc.). The flow of applications is called the simplest, if it satisfies the conditions stationarity, no consequences And ordinariness. Stationarity implies that the intensity of requests constant and does not depend on the time of day, day of the week or other time frames. In other words, there is no “rush hour” and there are no “dead hours”. The absence of consequences means that the probability of new applications does not depend on the “prehistory”, i.e. there is no such thing as “one grandmother told” and others “ran up” (or, on the contrary, ran away). And finally, the property of ordinaryness is characterized by the fact that small enough time interval almost impossible the appearance of two or more applications. “Two old ladies at the door?” - no, excuse me.

So, let some system receive the simplest flow of applications with medium intensity applications per minute (per hour, per day or at an arbitrary period of time). Then the probability that for a given period of time, the system will receive exactly requests is equal to:

Example 6

Calls to the taxi dispatch center are a simple Poisson flow with an average intensity of 30 calls per hour. Find the probability that: a) in 1 min. 2-3 calls will arrive, b) there will be at least one call within five minutes.

Solution: we use the Poisson formula:

a) Taking into account the stationarity of the flow, we calculate the average number of calls per 1 minute:
call - on average in one minute.

According to the theorem of addition of probabilities of incompatible events:
– the probability that in 1 minute the control room will receive 2-3 calls.

b) Calculate the average number of calls per five minutes: