Where λ is equal to the average number of occurrences of events in the same independent tests, i.e. λ = n × p, where p is the probability of an event in one trial, e = 2.71828 .
The distribution series of Poisson's law has the form:
Service assignment. The online calculator is used to build 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.
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 Moivre-Laplace theorems are used to calculate P n (m).
Numerical characteristics of a random variable X
The mathematical expectation of the Poisson distributionM[X] = λ
Poisson distribution variance
D[X] = λ
Example #1. The seeds contain 0.1% weeds. What is the probability of finding 5 weed seeds in a random selection of 2000 seeds?
Solution.
The probability p is small, and 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 #2. There are 0.4% weed seeds among rye seeds. Draw up the law of distribution of the number of weeds with a random selection of 5000 seeds. Find the mathematical expectation and variance of this random variable.
Solution. Expectation: M[X] = λ = 0.004*5000 = 20. Variance: D[X] = λ = 20
Distribution law:
| X | 0 | 1 | 2 | … | m | … |
| P | e-20 | 20e-20 | 200e-20 | … | 20 meters -20 / meters! | … |
Example #3. At the telephone exchange, an incorrect connection occurs with a probability of 1/200. Find the probability that among 200 connections there will be:
a) exactly one wrong connection;
b) less than three incorrect connections;
c) more than two incorrect connections.
Solution. According to the condition of the problem, the probability of an event is small, so we use the Poisson formula (15).
a) Given: n = 200, p = 1/200, k = 1. 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: to 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 large enough and p is small enough; we 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 in a time of 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 #4. The probability that a part is defective is 0.005. 400 parts are checked. Specify the formula for calculating the probability that more than 3 parts are defective.
Example number 5. The probability of the appearance of defective parts in their mass production is equal to 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, and the number n is large. np = 4.5< 10. Значит случайная величина Х – распределена по Пуассоновскому распределению. Составим закон.
The random variable X has the range (0,1,2,...,m). The probabilities of these values can be found by the formula:
Let's find the distribution series 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 is:
P(x<3) = P(0) + P(1) + P(2) = 0,01111 + 0,04999 + 0,1125 = 0,1736
Example number 6. An automatic telephone exchange receives, on average, N calls 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 ATS 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, by the formula we find the desired probability
Let's find the distribution series 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 number 7. We consider two elements that work independently of each other. The duration of uptime 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 flawlessly; b) only Probability that element #1 will not fail in 10 hours:
Solution.
P 1 (0) \u003d e -λ1 * t \u003d e -0.02 * 10 \u003d 0.8187
The probability that element #2 will not fail in 10 hours is:
P 2 (0) \u003d e -λ2 * t \u003d e -0.05 * 10 \u003d 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 number 7. Production gives 1% of the marriage. 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
How requests started coming in: "Where is Poisson? Where are the tasks on 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 euphoric familiar:
And the following 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:
By the condition, , and here everything is simple: the event consists of three incompatible outcomes:
The probability that a random variable will take on 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 the given random variable takes a positive value.
Solution and answer at the end of the lesson.
Apart from approximationbinomial distribution(Examples 1-3), the Poisson distribution has found wide application in queuing theory for a probabilistic characteristic the simplest event stream. I'll try to be concise:
Let some system receive requests (phone calls, incoming customers, etc.). The application flow is called the simplest if it satisfies the conditions stationarity, lack of consequences and ordinary. Stationarity implies that the intensity of applications 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 is no "dead hour". The absence of consequences means that the probability of the appearance of new applications does not depend on the “prehistory”, i.e. there is no such thing that “one grandmother told” and others “ran in” (or vice versa, fled). And, finally, the property of ordinaryness is characterized by the fact that for small enough time interval almost impossible appearance of two or more applications. "Two old ladies at the door?" - no, thank you, it’s more convenient to cut in order.
So, let some system receive the simplest flow of requests with medium intensity applications in a certain unit of time (minute, hour, day or any other). 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 dispatcher represent the simplest Poisson flow with an average intensity of 30 calls per hour. Find the probability that: a) in 1 min. 2-3 calls will be received, b) there will be at least one call within five minutes.
Solution: use the Poisson formula:
a) Given the stationarity of the flow, we calculate the average number of calls per 1 minute:
calls - an average of one minute.
According to the theorem of addition of probabilities of incompatible events:
- the probability that 2-3 calls will be received in the control room in 1 minute.
b) Calculate the average number of calls per five minutes: 
9. Poisson and Gauss distribution law
Poisson's law. Another name for it is the law of ra-determination of rare events. Poisson's Law (P.P.) is applied in cases where it is unlikely, and therefore the application of P/C/R is not practical.
The advantages of the law are: convenience in the calculation, the ability to calculate the probability in a given period of time, the ability to replace time with another continuous value, for example, linear dimensions.
Poisson's law has the following form:
and reads as follows: the probability of the occurrence of the event A in m times in n independent trials is expressed by a formula of the form (59), where a = pr is the average value of p(A), and a is the only parameter in Poisson's law.
The law of normal distribution (Gauss's law). Practice steadily confirms that the laws of error distribution obey the Gauss law with a sufficient approximation when measuring a wide variety of parameters: from linear and angular dimensions to the characteristics of the main mechanical properties of steel.
The probability density of the normal distribution law (hereinafter N. R.) has the form
where x 0 is the average value of a random variable;
? is the standard deviation of the same random variable;
e \u003d 2.1783 ... - the base of the natural logarithm;
W is a parameter that satisfies the condition.
The reason for the widespread use of the normal distribution law is theoretically determined by Lyapunov's theorem.
With known X 0 and? the ordinates of the curve of the function f(x) can be calculated by the formula
where t is a normalized variable,
(t) probability density z. If we substitute z and (t) into the formula, then it follows:
Curve Z.N.R. often called the Gaussian curve, this law describes very many phenomena in nature.
6. The law of transition to the supersystem Having exhausted the possibilities of development, the system is included in the supersystem as one of the parts; wherein further development takes place at the supersystem level. We have already spoken about this law. Let's move on to dynamics. It includes laws that
From the book Interface: New Directions in Design computer systems author Ruskin Jeff From the book Instrumentation author Babaev M A4.4.1. Fitts' Law Let's imagine that you move the cursor to a button shown on the screen. The button is the target of this move. The length of the straight line connecting the cursor's start position and the nearest point of the target object is defined in Fitts' law as a distance. On the
From the book Heat Engineering author Burkhanova Natalia4.4.2. Hick's Law Before moving the cursor to a target or performing any other action from a set of options, the user must select that object or action. Hick's Law states that when there are n options to choose from, the time to choose is
From the book Computational Linguistics for All: Myths. Algorithms. Language author Anisimov Anatoly Vasilievich6. Statistics of the distribution of random variables Main characteristics of random variables.1. Measures of position. These are called (considered) points around which the characteristics of quantities fluctuate. The sum of the products of empirical values of a random variable xi by
From the book Phenomenon of Science [Cybernetic Approach to Evolution] author Turchin Valentin Fedorovich10. Binomial and polynomial distribution laws. Uniprobable distribution. Eccentricity distribution law 1. Binomial distribution law. This law is mathematically expressed by the expansion formula for the binomial (q + p)2 in the following form where n! - read
From the book Nanotechnology [Science, Innovation and Opportunity] by Foster Lynn11. Other distribution laws In the technical industry, including instrument making, some other types of distribution laws are used, in addition to those discussed above. In this case, the distribution of random variables is already according to their most diverse parameters.
From the book History of Electrical Engineering author Team of authors22. Boyle-Mariotte Law One of the ideal gas laws is the Boyle-Mariotte law, which states: the product of pressure P and volume V of a gas is constant at constant gas mass and temperature. This equality is called the isotherm equation. The isotherm is displayed on
From the book History of outstanding discoveries and inventions (electrical engineering, electric power industry, radio electronics) author Shneiberg Jan Abramovich23. Gay-Lussac's law Gay-Lussac's law says: the ratio of the volume of a gas to its temperature at constant gas pressure and its mass is constant. V / T = m / MO R / P = const at P = const, m = const. the name of the isobar equation. An isobar is depicted on a PV diagram by a straight line,
From the author's book24. Charles' law Charles' law states that the ratio of gas pressure to its temperature is constant if the volume and mass of the gas are unchanged: P / T = m / MО R / V = const at V = const, m = const. .The isochore is depicted on a PV-diagram of a straight line parallel to the P axis, and
From the author's book30. The law of conservation and transformation of energy The first law of thermodynamics is based on the universal law of conservation and transformation of energy, which establishes that energy is neither created nor disappears. Bodies participating in a thermodynamic process interact with each other
From the author's bookTHE FROG PRINCESS AND THE LAW OF STABILITY As already emphasized earlier (the law of abstraction), primitive thinking was able to analyze concrete phenomena and synthesize new abstract systems. Since any object constructed by consciousness was perceived as living, and living
From the author's book1.1. The Basic Law of Evolution In the process of evolution of life, as far as we know, there has always been and is now an increase in the total mass of living matter and the complication of its organization. Complicating the organization biological formations, nature works by trial and error
From the author's book4.2. Moore's Law In its simplest form, Moore's Law is the statement that transistor circuit density doubles every 18 months. The authorship of the law is attributed to one of the founders of the well-known company Intel, Gordon Moore. Strictly speaking, in
Poisson distribution - a case of the binomial distribution when the number of trials n large enough, and the probability p developments A small().
The Poisson distribution is also called the distribution of rare events. For example, birth year three or four twins, the same distribution law applies to the number of radioactive atoms decaying per unit time, etc.
The probability of occurrence of rare events is calculated by the Poisson formula :
,
where m the number of occurrence of the event A;
Mean value of the Poisson distribution;
e\u003d 2.7183 - the base of the natural logarithm.
Poisson's law depends on one parameter - λ (lambda), the meaning of which is as follows: it is both the mathematical expectation and the variance of a random variable distributed according to the Poisson law.
Conditions for the occurrence of the Poisson distribution
Consider the conditions under which the Poisson distribution arises.
Firstly, the Poisson distribution is the limit for the binomial distribution when the number of experiments n increases indefinitely (tends to infinity) and at the same time the probability p success in one experiment decreases indefinitely (tends to zero), but in such a way that their product np remains in the limit constant and equal to λ (lambde):
In mathematical analysis, it is proved that the Poisson distribution with the parameter λ = np can be approximately applied instead of the binomial, when the number of experiments n very high, and the probability p is very small, that is, in each individual experience, the event A appears extremely rarely.
Secondly, Poisson distribution occurs when there is a stream of events called the simplest (or stationary Poisson stream) . A flow of events is a sequence of events such as the arrival of calls to the communication node, the arrival of visitors to the store, the arrival of trains on the hump, and the like. Poisson flow has the following properties:
- stationarity: probability of occurrence m events in a certain period of time is constant and does not depend on the origin of time, but depends only on the length of the time interval;
- ordinary: the probability of two or more events hitting a small time interval is negligible compared to the probability of one event hitting it;
- no consequence: likelihood of occurrence m events in a certain period of time does not depend on how many events occurred in the previous period.
Characteristics of a random variable distributed according to Poisson's law
Characteristics of a random variable distributed according to Poisson's law:
expected value ;
standard deviation ;
variance .
Poisson distribution and calculations in MS Excel
Poisson distribution probability P(m) and the value of the integral function F(m) can be calculated using the MS Excel function POISSON.DIST. The window for the corresponding calculation is shown below (click the left mouse button to enlarge).
MS Excel requires you to enter the following data:
- x- number of events m;
- average;
- integral - logical value: 0 - if you need to calculate the probability P(m) and 1 - if the probability F(m).
Solving examples with Poisson distribution
Example 1 The manager of a telecommunications company decided to calculate the probability that 0, 1, 2, ... calls will arrive in a small town within five minutes. Random intervals of five minutes were selected, the number of calls in each of their intervals was counted, and the average number of calls was calculated: .
Calculate the probability that 6 calls will arrive within five minutes.
Solution. According to the Poisson formula, we get:
We get the same result using the MS Excel function POISSON.DIST (the value of the integral value is 0):
P(6 ) = POISSON.DIST(6, 4.8, 0) = 0.1398.
Let's calculate the probability that no more than 6 calls will arrive within five minutes (the value of the integral value is 1):
P(≤6 ) = POISSON.DIST(6; 4.8; 1) = 0.7908.
Solve the example yourself and then see the solution
Example 2 The manufacturer sent 1000 tested, that is, serviceable TVs to a certain city. The probability that the TV will fail during transportation is 0.003. That is, in this case, the Poisson distribution law applies. Find the probability that out of all delivered televisions the following will be faulty: 1) two televisions; 2) less than two TVs.
We continue to solve examples together
Example 3 The customer call center receives a flow of calls with an intensity of 0.8 calls per minute. Find the probability that in 2 minutes: a) no calls will come; b) exactly one call will come; c) at least one call will come.
In many practical problems, one has to deal with random variables distributed according to a peculiar law, which is called Poisson's law.
Consider a discontinuous random variable , which can take only integer, non-negative values:
and the sequence of these values is theoretically unlimited.
A random variable is said to be distributed according to Poisson's law if the probability that it takes on a certain value is expressed by the formula
where a is some positive value, called the Poisson law parameter.
The distribution series of a random variable , distributed according to Poisson's law, has the form:
Let us first of all make sure that the sequence of probabilities given by formula (5.9.1) can be a distribution series, i.e. that the sum of all probabilities is equal to one. We have:
.
On fig. 5.9.1 shows the distribution polygons of a random variable distributed according to Poisson's law, corresponding to different values of the parameter . Table 8 of the appendix lists the values for various .
Let's define the main characteristics - mathematical expectation and variance - of a random variable distributed according to the Poisson law. By definition of mathematical expectation
.
The first term of the sum (corresponding to ) zero, therefore, the summation can be started with:
Let's denote ; then
. (5.9.2)
Thus, the parameter is nothing more than the mathematical expectation of a random variable.
To determine the dispersion, we first find the second initial moment of the quantity :
According to the previously proven
Moreover,
Thus, the dispersion of a random variable distributed according to the Poisson law is equal to its mathematical expectation.
This property of the Poisson distribution is often used in practice to decide whether the hypothesis that a random variable is distributed according to Poisson's law is plausible. To do this, determine from experience the statistical characteristics - the mathematical expectation and variance - of a random variable. If their values are close, then this can serve as an argument in favor of the Poisson distribution hypothesis; a sharp difference in these characteristics, on the contrary, testifies against the hypothesis.
For a random variable distributed according to Poisson's law, let's determine the probability that it will take a value not less than a given one. Let's denote this probability:
Obviously, the probability can be calculated as the sum
However, it is much easier to determine it from the probability of the opposite event:
(5.9.4)
In particular, the probability that the value will take a positive value is expressed by the formula
(5.9.5)
We have already mentioned that many practical tasks lead to a Poisson distribution. Consider one of the typical problems of this kind.
Let points be randomly distributed on the x-axis Ox (Fig. 5.9.2). Let's assume that random distribution points satisfies the following conditions:
1. The probability of hitting a given number of points on a segment depends only on the length of this segment, but does not depend on its position on the x-axis. In other words, the points are distributed on the x-axis with the same average density. Let's denote this density (i.e. the mathematical expectation of the number of points per unit length) as .
2. The points are distributed on the x-axis independently of each other, i.e. the probability that one or another number of points falls on a given segment does not depend on how many of them fall on any other segment that does not overlap with it.
3. The probability of hitting a small area of two or more points is negligible compared to the probability of hitting one point (this condition means the practical impossibility of coincidence of two or more points).
Let's single out a certain length segment on the abscissa axis and consider a discrete random variable - the number of points falling on this segment. Possible values of the quantity will be
Since the points fall on the segment independently of each other, it is theoretically possible that there will be an arbitrarily large number of them, i.e. series (5.9.6) continues indefinitely.
Let us prove that the random variable has the Poisson distribution law. To do this, we calculate the probability that exactly points fall on the segment.
Let's solve more first a simple task. Consider a small section on the Ox axis and calculate the probability that at least one point will fall on this section. We will argue as follows. The mathematical expectation of the number of points that fall on this section is obviously equal (because there are points on average per unit length). According to condition 3, for a small segment, the possibility of two or more points falling on it can be neglected. Therefore, the mathematical expectation of the number of points falling on the site will be approximately equal to the probability of one point falling on it (or, which is equivalent in our conditions, at least one).
Thus, up to infinitesimals of a higher order, at , we can assume that the probability that one (at least one) point will fall on the site is equal to , and the probability that none will fall is equal to .
Let's use this to calculate the probability of hitting exactly points on the segment. Divide the segment into equal parts of length . Let us agree to call an elementary segment "empty" if it does not contain a single point, and "occupied" if at least one has fallen into it. According to the above, the probability that the segment will be "occupied" is approximately equal to; the probability that it will be "empty" is . Since, according to condition 2, the hits of points in non-overlapping segments are independent, then our n segments can be considered as independent "experiments", in each of which the segment can be "occupied" with probability . Find the probability that among the segments there will be exactly "busy". According to the repetition theorem, this probability is equal to
or, denoting
(5.9.7)
For a sufficiently large value, this probability is approximately equal to the probability of hitting exactly points on the segment, since the hit of two or more points on the segment has a negligible probability. In order to find the exact value of , it is necessary in expression (5.9.7) to go to the limit at :
(5.9.8)
Let's transform the expression under the limit sign:
(5.9.9)
The first fraction and the denominator of the last fraction in expression (5.9.9) at obviously tend to unity. The expression does not depend on. The numerator of the last fraction can be converted as follows:
(5.9.10)
When and expression (5.9.10) tends to . Thus, it has been proved that the probability of exactly points falling into a segment is expressed by the formula
where , i.e. the quantity X is distributed according to the Poisson law with the parameter .
Note that the meaning of the value is the average number of points per segment .
The magnitude (probability that X will be positive) in this case expresses the probability that at least one point will fall on the segment:
Thus, we made sure that the Poisson distribution occurs where some points (or other elements) occupy a random position independently of each other, and the number of these points that fall into some area is counted. In our case, such an "area" was a segment on the x-axis. However, our conclusion can be easily extended to the case of distribution of points in the plane (random flat field of points) and in space (random spatial field of points). It is easy to prove that if the following conditions are met:
1) the points are distributed statistically uniformly in the field with an average density ;
2) points fall into non-overlapping regions independently;
3) points appear singly, and not in pairs, triples, etc., then the number of points falling into any area (flat or spatial) are distributed according to Poisson's law:
where is the average number of points falling into the area .
For the flat case
where is the area of the region; for spatial
where is the volume of the region.
Note that for the Poisson distribution of the number of points falling into a segment or region, the condition of constant density () is not essential. If the other two conditions are met, then Poisson's law still holds, only the parameter a in it acquires a different expression: it turns out not simple multiplication density on the length, area, or volume of a region, but by integrating a variable density over a segment, area, or volume. (For more on this, see n° 19.4)
The presence of random points scattered on a line, on a plane or on a volume is not the only condition under which the Poisson distribution occurs. One can, for example, prove that Poisson's law is limiting for the binomial distribution:
, (5.9.12)
if we simultaneously direct the number of experiments to infinity, and the probability to zero, and their product remains constant:
Indeed, this limiting property of the binomial distribution can be written as:
. (5.9.14)
But from condition (5.9.13) it follows that
Substituting (5.9.15) into (5.9.14), we obtain the equality
, (5.9.16)
which has just been proved by us on another occasion.
This limiting property of the binomial law is often used in practice. Let us assume that a large number of independent experiments are being made, in each of which an event has a very small probability . Then, to calculate the probability that an event will occur exactly once, you can use the approximate formula:
, (5.9.17)
where is the parameter of that Poisson's law, which approximately replaces the binomial distribution.
From this property of Poisson's law - to express the binomial distribution with a large number of experiments and a small probability of an event - comes its name, often used in statistics textbooks: the law of rare phenomena.
Let's look at a few examples related to the Poisson distribution from various fields of practice.
Example 1: An automatic telephone exchange receives calls with an average density of calls per hour. Assuming that the number of calls in any period of time is distributed according to the Poisson law, find the probability that exactly three calls will arrive at the station in two minutes.
Solution. The average number of calls per two minutes is:
sq.m. To hit the target, at least one fragment is enough to hit it. Find the probability of hitting the target for a given position of the discontinuity point.
Solution. . Using formula (5.9.4), we find the probability of hitting at least one fragment:
(To calculate the value of the exponential function, we use Table 2 of the Appendix).
Example 7. The average density of pathogenic microbes in one cubic meter air is 100. 2 cubic meters are taken for a sample. dm air. Find the probability that at least one microbe will be found in it.
Solution. Accepting the hypothesis of the Poisson distribution of the number of microbes in a volume, we find:
Example 8. 50 independent shots are fired at some target. The probability of hitting the target with one shot is 0.04. Using the limiting property of the binomial distribution (formula (5.9.17)), find approximately the probability that the target will hit: no projectile, one projectile, two projectiles.
Solution. We have . According to table 8 of the application, we find the probabilities.
