Poisson distribution series. Poisson distribution

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 .

Expectation 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 special case 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 , in the form

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.

$X$ has a Poisson distribution with parameter $\lambda$ ($\lambda$$>$0) if this value takes non-negative integer values $k=0, 1, 2,\dots$ with probabilities $pk$=$\frac (\lambda ^(:) )(: \cdot 5^{-\lambda } .$ (Это распределение впервые было рассмотрено французским математиком и физиком !} Simeon Denis Poisson in 1837)

Poisson distribution also called the law of rare events, because the probabilities pk give an approximate distribution of the number of occurrences of some rare event over a large number of independent trials. In this case, we assume $\lambda =n \cdot р$, where $n$ is the number of Bernoulli trials, $р$ is the probability of the event occurring in one trial.

The validity of using Poisson's law instead of the binomial distribution when large number tests is given by the following theorem.

Theorem 1

Poisson's theorem.

If in the Bernoulli scheme n$\rightarrow$$\infty$, p$\rightarrow$0, so that $n \cdot p$$\rightarrow$$\lambda$ (to a finite number), then

$!_(n)^(k) p^(k) (1-p)^(n-k) \to \frac(\lambda ^(k) )(k e^{-\lambda } $ при любых $k=0, 1, 2,... $!}

No proof.

Note 1

The Poisson formula becomes more accurate for small $p$ and large numbers $n$, and $n \cdot p $

Expectation random variable having a Poisson distribution with parameter $\lambda$:

$М(Х)$=$\sum \limits _(k=0)^(\infty )k\cdot \frac(\lambda ^(k) )(k e^{-\lambda } =\lambda \cdot e^{-\lambda } \sum \limits _{k=1}^{\infty }\frac{\lambda ^{k} }{k!} =\lambda \cdot e^{-\lambda } \cdot e^{\lambda } = $$\lambda$.!}

Dispersion random variable having a Poisson distribution with parameter $\lambda$:

$D(X)$=$\lambda$ .

Application of Poisson's formula in solving problems

Example 1

The probability of a defective product appearing during mass production is $0.002$. Find the probability that in a batch of $1500 $ products there will be no more than 3 defective ones. Find the average number of defective products.

Let $A$ be the number of defective products in a batch of $1500$ products. Then the desired probability is the probability that $A$ $\leq$ $3$. In this problem we have a Bernoulli scheme with $n=1500$ and $p=0.002$. To apply Poisson's theorem, let's set $\lambda=1500 \cdot 0.002=3$. Then the desired probability

The average number of defective products is $M(A)$=$\lambda$=3.

Example 2

The institution's switchboard serves $100$ subscribers. The probability that a subscriber will call within $1$ minute is $0.01$. Find the probability that no one will call within $1$ minute.

Let $A$ be the number of callers to the switchboard during $1$ minute. Then the desired probability is the probability that $A=0$. In this problem, the Bernoulli scheme with $n=100$, $p=0.01$ is applicable. To use Poisson's theorem, we set

$\lambda=100 \cdot 0.01=1$.

Then the desired probability

$P = e^-1$ $\approx0.37$.

Example 3

The factory sent $500$ of products to the base. The probability of damage to the product in transit is $0.002$. Find the probabilities that damage will occur along the way

exactly three products;
less than three products.

Having considered the remark to the Poisson formula, since the probability $p=0.002$ of damage to the product is small, and the number of products $n=500$ is large, and $a=n\cdot p=1

To solve the second problem, the formula is applicable, where $k1=0$ and $k2=2$. We have:

Example 4

The textbook was published in a circulation of $100,000 copies. The probability that one textbook is bound incorrectly is $0.0001$. What is the probability that the circulation contains $5$ of defective books?

According to the conditions of the problem, $n = 100000$, $p = 0.0001$.

The events “out of $n$ books exactly $m$ books are stitched incorrectly”, where $m = 0,1,2, \dots ,100000$, are independent. Since the number $n$ is large and the probability $p$ is small, the probability $P_n (m)$ can be calculated using the Poisson formula: $P_n$(m)$\approx \frac((\lambda )^m\cdot e^ (-\lambda ))(m$ , где $\lambda = np$.!}

In the problem under consideration

$\lambda = 100000 \cdot 0.0001 = $10.

Therefore, the desired probability $P_(100000)$(5) is determined by the equality:

$P_(100000)$ (5)$\approx \frac(e^(-10)\cdot (10)^5)(5\approx $ ${10}^5$ $\frac{0,000045}{120}$ = $0,0375$.!}

Answer: $0.0375$.

Example 5

The plant sent $5,000 of good-quality products to the base. The probability that a product will be damaged during transit is $0.0002$. Find the probability that three unusable products will arrive at the base.

By condition $n=5000$; $p = 0.0002$; $k = 3$. Let's find $\lambda$:

$\lambda = n \cdot p = 5000 \cdot 0.0002 = 1$.

The required probability according to the Poisson formula is equal to:

Example 6

The probability that one subscriber will call a telephone exchange within one hour is 0.01. Within an hour, 200 subscribers called. Find the probability that 3 subscribers will call within an hour.

Having considered the condition of the problem, we see that:

Let's find $\lambda $ for Poisson's formula:

\[\lambda =np=200\cdot 0.01=2.\]

Substitute the values into the Poisson formula and get the value:

Example 7

There are 500 students at the faculty. What is the probability that September 1st is the birthday of 2 students at the same time?

We have $n=500$; $p=1/365 \approx 0.0027$, $q=0.9973$. Since the number of tests is large, and the probability of execution is very small and $npq=1.35\

In many practical problems one has to deal with random variables distributed according to a peculiar law called Poisson's law.

Consider a discontinuous random variable X, which can only take non-negative integer values:

and the sequence of these values is theoretically unlimited. They say that random variable X distributed according to Poisson's law if the probability that it will take a certain value T, expressed by the formula

Where A- some positive value, called parameter Poisson's law.

Distribution series of a random variable X, distributed according to Poisson's law, has the form:

Let us first make sure that the sequence of probabilities given by formula (5.9.1) can be a distribution series, i.e. sum of all probabilities R t equal to one. We have:

But

Figure 5.9.1 shows the polygons of the random variable distribution X, distributed according to Poisson's law, corresponding to different values of the parameter A. Appendix Table 8 shows the values R t for various A.

Let's define the main characteristics - mathematical expectation and variance - of a random variable X, distributed according to Poisson's law. By definition mathematical expectation

Rice. 5.9.1.

The first term of the sum (corresponding t = 0) is equal to zero, therefore, the summation can start from t = 1:

Let's denote t - 1 = k; Then

So the parameter A is nothing more than the mathematical expectation of a random variable X.

To determine the dispersion, we first find the second initial moment of the quantity X:

According to previously proven Besides, hence,

Thus, random variable variance, distributed according to Poisson's law, equal to its mathematical expectation a.

This property of the Poisson distribution is often used in practice to decide whether the hypothesis that a random variable X distributed according to Poisson's law. To do this, the statistical characteristics - mathematical expectation and variance - of a random variable are determined from experience. If their values are close, then this can serve as an argument in favor of the Poisson distribution hypothesis; the sharp difference in these characteristics, on the contrary, argues against the hypothesis.

Let us define for the random variable X, distributed according to Poisson's law, the probability that it will take a value no less than a given one To. Let's denote this probability Rk:

Obviously the probability Rk can be calculated as the sum

However, it is much easier to determine it from the probability of the opposite event:

In particular, the probability that the quantity X will take a positive value, expressed by the formula

We have already mentioned that many practice problems result in a Poisson distribution. Let's consider one of the typical problems of this kind.

Let points be randomly distributed on the x-axis Ox (Fig. 5.9.2). Let us assume that the random distribution of points satisfies the following conditions:

Rice. 5.9.2

1. The probability of a particular number of points falling 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 us denote this density (i.e., the mathematical expectation of the number of points per unit length) by X.
2. The points are distributed on the x-axis independently of each other, i.e. the probability of one or another number of points falling 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 two or more points hitting a small area Ax is negligible compared to the probability of one point hitting (this condition means the practical impossibility of two or more points coinciding).

Let us select a certain segment of length / on the abscissa axis and consider a discrete random variable X- the number of points falling on this segment. Possible values will be

Since the points fall on the segment independently of each other, it is theoretically possible that there will be as many of them there as desired, i.e. series (5.9.6) continues indefinitely.

Let us prove that the random variable X has a Poisson distribution law. To do this, we calculate the probability R t that the segment / will fall exactly T points.

First let's solve more simple task. Let us consider a small section Ax on the Ox axis and calculate the probability that at least one point will fall on this section. We will reason as follows. The mathematical expectation of the number of points falling on this area is obviously equal to HAH(since per unit length falls on average X points). According to condition 3, for a small segment Ax we can neglect the possibility of two or more points falling on it. Therefore the mathematical expectation HAH the number of points falling on the area Ax will be approximately equal to the probability of one point falling on it (or, which in our conditions is equivalent, at least one).

Thus, up to infinitesimal higher order for Ax -» 0, we can consider the probability that one (at least one) point will fall on the section Ax equal to HAH, and the probability that none will hit is 1 - HAH.

Let's use this to calculate the probability R t hitting the segment / exactly T points. Divide the segment / into n equal parts length . Let us agree to call the elementary segment Ax “empty”,

if no points hit it, and “busy” if at least one hit it. According to the above, the probability that the segment Ax will be “busy” is approximately equal to ; probability

the fact that it turns out to be “empty” is equal to

Since, according to condition 2, points falling into non-overlapping segments are independent, then our n segments can be considered as n independent “experiments”, in each of which the segment can be “occupied” with probability. Let us find the probability that among n the segments will be exactly

T"busy". According to the theorem on repetition of experiments, this probability is equal to

or, denoting XI = a,

When large enough n this probability is approximately equal to the probability of hitting the segment / exactly T points, since the probability of two or more points falling on the segment Ax is negligible. To find the exact value R t, in expression (5.9.7) you need to go to the limit at n-> oo:

Let's transform the expression under the limit sign:

The first fraction and the denominator of the last fraction in expression (5.9.9) with n -> oo obviously tend to unity. Expression from n doesn't depend. The numerator of the last fraction can be transformed as follows:

At and expression (5.9.10) tends to e~ a.

Thus, it has been proven that the probability of hitting is exactly T points in the segment / is expressed by the formula

Where a = XI, those. magnitude X distributed according to Poisson's law with parameter A = XI.

Note that the value A in meaning it is the average number of points per segment I.

Magnitude R,(the probability that the value X will take a positive value) in this case expresses the likelihood that, that segment I contains at least one point:

Thus, we are convinced 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 falling into some area is counted. In our case, this “area” was the segment / on the abscissa axis. However, our conclusion can be easily extended to the case of distribution of points on the plane (random flat field of points) and in space (random spatial field of points). It is not difficult to prove that if the conditions are met:

1) points are distributed statistically evenly in the field with an average density X
2) the points fall into non-overlapping regions independently;
3) dots appear singly, and not in pairs, triplets, etc., then the number of dots X, falling into any area D(flat or spatial), distributed according to Poisson’s law:

Where A- average number of points falling into the area D.

For flat case

Where S D- area of the region D for spatial

Where V D- volume of the area D.

Note that for the presence of a Poisson distribution of the number of points falling into a segment or region, the condition of constant density (X = const) is unimportant. If the other two conditions are met, then Poisson’s law still holds, only the parameter A it takes on a different expression: it is not obtained by simply multiplying the density X by the length, area or volume of a region, but by integrating the variable density over a segment, area or volume (for more details, see subsection 19.4).

The presence of random points scattered on a line, plane, or volume is not the only condition under which a Poisson distribution occurs. One can, for example, prove that Poisson's law is limiting for the binomial distribution:

if you simultaneously direct the number of experiments p to infinity, and the probability r - to zero, and their product pr keeps a constant value:

Indeed, this limiting property of the binomial distribution can be written as:

But from condition (5.9.13) it follows that

Substituting (5.9.15) into (5.9.14), we obtain the equality

which we have just proved 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 carried out p, in each of which there is an event A has a very low probability r. Then to calculate the probability R t „ that the event A will appear exactly T times, you can use the approximate formula

Where pr = a- parameter of the Poisson law, which approximately replaces the binomial distribution.

From this property of Poisson’s law - to express a binomial distribution with a large number of experiments and a low probability of an event - its name, often used in statistics textbooks, comes from: law of rare phenomena.

Let's look at several examples related to the Poisson distribution from various areas of practice.

Example 1. The automatic telephone exchange receives calls with an average density TO calls per hour. Assuming that the number of calls in any period of time is distributed according to Poisson’s 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:

According to formula (5.9.1), the probability of receiving exactly three calls

Example 2. Under the conditions of the previous example, find the probability that at least one call will come in two minutes.

Solution. According to formula (5.9.4) we have:

Example 3. Under the same conditions, find the probability that at least three calls will arrive in two minutes.

Solution. According to formula (5.9.4) we have:

Example 4. On a weaving loom, the thread breaks on average 0.375 times per hour of loom operation. Find the probability that during a shift (8 hours) the number of thread breaks will be between 2 and 4 (no less than 2 and no more than 4 breaks).

Solution. Obviously,

we have:

According to table 8 of the appendix when A = 3

Example 5. On average, it escapes from a heated cathode per unit time q(t) electrons, where t- time elapsed since the beginning of the experiment. Find the probability that during a time period of duration t, starting at the moment t0, will fly straight out of the cathode T electrons.

Solution. We find the average number of electrons a emitted from the cathode over a given period of time. We have:

Based on the calculated a, we determine the required probability:

Example 6. The number of fragments hitting a small target at a given position of the breaking point is distributed according to Poisson's law. The average density of the fragmentation field in which the target finds itself at a given position of the breaking point is 3 fragments. /m2. The target area is S= 0.5 m2. To hit a target, at least one fragment is enough to hit it. Find the probability of hitting the target at a given position of the break point.

Solution, a = XS= 1.5. Using formula (5.9.4) we find the probability of hitting at least one fragment:

(To calculate the value of the exponential function e~a we use the table. 2 applications.)

Example 7. The average density of pathogenic microbes in one cubic meter of air is 100. A sample of 2 dm 3 of air is taken. Find the probability that at least one microbe will be found in it. Solution. Accepting the hypothesis of a Poisson distribution of the number of microbes in a volume, we find:

Example 8. 50 independent shots are fired at a certain 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 not a single projectile will hit the target, one projectile will hit, or two projectiles will hit.

Solution. We have a = pr = 50 0.04 = 2. Using table 8 of the appendix we find the probabilities:

For methods of experimentally determining these characteristics, see below, Chapters 7 and 14.

As soon as requests started coming in: “Where is Poisson? Where are the problems using the Poisson formula? etc.. 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 applications ( 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 period of time almost impossible the appearance of two or more applications. “Two old ladies at the door?” - No, excuse me, it’s more convenient to chop in order.

So, let some system receive the simplest flow of applications 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 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:

In many practically important applications, the Poisson distribution plays an important role. Many of the numerical discrete quantities are implementations of a Poisson process, which has the following properties:

We are interested in how many times a certain event occurs in a given range of possible outcomes of a random experiment. The area of possible outcomes can be a time interval, a segment, a surface, etc.
The probability of a given event is the same for all areas of possible outcomes.
The number of events occurring in one area of possible outcomes is independent of the number of events occurring in other areas.
The probability that a given event occurs more than once in the same area of possible outcomes tends to zero as the area of possible outcomes decreases.

To further understand the meaning of the Poisson process, suppose we examine the number of customers visiting a bank branch located in the central business district during lunch, i.e. from 12 to 13 o'clock. Suppose you want to determine the number of clients arriving in one minute. Does this situation have the features listed above? Firstly, the event that interests us is the arrival of a client, and the range of possible outcomes is a one-minute interval. How many clients will come to the bank in a minute - none, one, two or more? Secondly, it is reasonable to assume that the probability of a customer arriving within a minute is the same for all one-minute intervals. Third, the arrival of one customer during any one-minute interval is independent of the arrival of any other customer during any other one-minute interval. And finally, the probability that more than one client will come to the bank tends to zero if the time interval tends to zero, for example, becomes less than 0.1 s. So, the number of customers coming to the bank during lunch within one minute is described by the Poisson distribution.

The Poisson distribution has one parameter, denoted by the symbol λ (Greek letter "lambda") - the average number of successful trials in a given range of possible outcomes. The variance of the Poisson distribution is also λ, and its standard deviation is . Number of successful trials X Poisson random variable varies from 0 to infinity. The Poisson distribution is described by the formula:

Where P(X)- probability X successful trials, λ - expected number of successes, e- natural logarithm base equal to 2.71828, X- number of successes per unit of time.

Let's return to our example. Let's say that during the lunch break, on average, three customers come to the bank per minute. What is the probability that two customers will come to the bank at a given moment? What is the probability that more than two clients will come to the bank?

Let us apply formula (1) with the parameter λ = 3. Then the probability that two clients will come to the bank within a given minute is equal to

The probability that more than two clients will come to the bank is equal to P(X > 2) = P(X = 3) + P(X = 4) + … + P(X = ∞) . Since the sum of all probabilities must be equal to 1, the terms of the series on the right side of the formula represent the probability of addition to the event X ≤ 2. In other words, the sum of this series is equal to 1 – P(X ≤ 2). Thus, P(X>2) = 1 – P(X≤2) = 1 – [P(X = 0) + P(X = 1) + P(X = 2)]. Now, using formula (1), we get:

Thus, the probability that no more than two clients will come to the bank within a minute is 0.423 (or 42.3%), and the probability that more than two clients will come to the bank within a minute is 0.577 (or 57.7 %).

Such calculations may seem tedious, especially if the parameter λ is large enough. To avoid complex calculations, many Poisson probabilities can be found in special tables (Fig. 1). For example, the probability that two clients will come to the bank at a given minute, if on average three clients come to the bank per minute, is at the intersection of the line X= 2 and column λ = 3. Thus, it is equal to 0.2240 or 22.4%.

Rice. 1. Poisson probability at λ = 3

Nowadays, it is unlikely that anyone will use tables if they have Excel with its =POISSON.DIST() function at hand (Fig. 2). This function has three parameters: number of successful trials X, average expected number of successful trials λ, parameter Integral, taking two values: FALSE – in this case the probability of the number of successful trials is calculated X(X only), TRUE – in this case the probability of the number of successful trials from 0 to X.

Rice. 2. Calculation in Excel of the probabilities of the Poisson distribution at λ = 3

Approximation of the binomial distribution using the Poisson distribution

If the number n is large and the number r- small, the binomial distribution can be approximated using the Poisson distribution. The higher the number n and less number r, the higher the approximation accuracy. The following Poisson model is used to approximate the binomial distribution.

Where P(X)- probability X success with given parameters n And r, n- sample size, r- true probability of success, e- the base of the natural logarithm, X- number of successes in the sample (X = 0, 1, 2, …, n).

Theoretically, a random variable with a Poisson distribution takes values from 0 to ∞. However, in situations where the Poisson distribution is used to approximate the binomial distribution, the Poisson random variable is the number of successes among n observations - cannot exceed the number n. From formula (2) it follows that with increasing number n and a decrease in the number r the probability of detecting a large number of successes decreases and tends to zero.

As mentioned above, the expectation µ and the variance σ 2 of the Poisson distribution are equal to λ. Therefore, when approximating the binomial distribution using the Poisson distribution, formula (3) should be used to approximate the mathematical expectation.

(3) µ = E(X) = λ =n.p.

To approximate the standard deviation, formula (4) is used.

Note that the standard deviation calculated using formula (4) tends to the standard deviation in the binomial model - when the probability of success p tends to zero, and, accordingly, the probability of failure 1 – p tends to unity.

Let's assume that 8% of the tires produced at a certain plant are defective. To illustrate the use of the Poisson distribution to approximate the binomial distribution, let's calculate the probability of finding one defective tire in a sample of 20 tires. Let us apply formula (2), we obtain

If we were to calculate the true binomial distribution rather than its approximation, we would get the following result:

However, these calculations are quite tedious. However, if you use Excel to calculate probabilities, then using the Poisson distribution approximation becomes redundant. In Fig. Figure 3 shows that the complexity of calculations in Excel is the same. However, this section, in my opinion, is useful to understand that under some conditions the binomial distribution and the Poisson distribution give similar results.

Rice. 3. Comparison of the complexity of calculations in Excel: (a) Poisson distribution; (b) binomial distribution

So, in this and two previous notes, three discrete numerical distributions were considered: , and Poisson. To better understand how these distributions relate to each other, we present small tree questions (Fig. 4).