Brief theory

Let independent trials be carried out, in each of which the probability of an 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, for very large number tests, in each of which the probability of the 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. at different meanings, 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 an individual trial is sufficiently close to zero, then even for large values of the number of trials, the probability calculated by local theorem Laplace 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 pieces of equipment 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:

Average solution cost test work 700 - 1200 rubles (but not less than 300 rubles for the entire order). The price is greatly influenced by the urgency of the decision (from a day to several hours). The cost of online help for an exam/test is from 1000 rubles. for solving the ticket.

You can leave a request directly in the chat, having previously sent the task conditions and informed you of the time frame for the solution you need. Response time is a few minutes.

Poisson distribution.

Let's consider the most typical situation, in which the Poisson distribution appears. 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”).

Imagine a raisin bun, divided into small pieces equal size. Due to random distribution raisins cannot be expected that all pieces will contain them same number. 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. conditional probability occurrence of another event in the same interval tends to zero at ® 0. This is the property of ordinaryness.

3. Probability of occurrence given number events 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 = .

Let's consider final segment 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 between total number n. There are total such combinations, and the probability of each is equal. Consequently, by the addition theorem of probabilities we obtain for the desired probability well-known formula Bernoulli

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'll get it after replacement.

P = a= and q = 1 – .

Let's introduce 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... – base natural logarithm.

Definition. Random variable X, which only accepts integers, positive values 0, 1, 2, ... has a Poisson distribution with parameter if

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

The Poisson distribution has been proposed French mathematician S.D. Poisson (1781-1840). It is used to solve problems of calculating probabilities of relatively rare, mutually random 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

Let us again recall the situation that was called the Bernoulli scheme: n independent trials, each of which contains some event A can appear with the same probability r. Then, to determine the probability that in these n testing event A will appear exactly k times (this probability was denoted P n (k) ) can be calculated exactly using Bernoulli's formula, where q=1− p. However, with a large number of tests n Calculations using Bernoulli's formula become very inconvenient, as they lead to operations with very large numbers. Therefore (if you remember − this was once covered when studying the Bernoulli scheme and formula when studying the first part of the theory of probability “Random events”) for large n much more convenient (albeit approximate) formulas were proposed, which turned out to be more accurate the more n(Poisson formula, local and integral Moivre-Laplace formula). If in the Bernoulli scheme the number of experiments n is high and the probability r occurrence of an event A is small in each test, then the mentioned Poisson formula gives a good approximation
, where the parameter a =n∙ p. This formula leads to the Poisson distribution. Let's give precise definitions

Discrete random variable X has Poisson distribution, if it takes values 0, 1, 2, ... with probabilities r 0 , p 1 , ... , which are calculated by the formula

and the number A is a parameter of the Poisson distribution. Please note that the possible values of r.v. X infinitely many − These are all non-negative integers. Thus, d.s.v X with the Poisson distribution has the following distribution law:

When calculating mathematical expectation(according to their definition for d.r.v. with a known distribution law) one will now have to consider not final amounts, and the sums of the corresponding infinite series (since the table of the distribution law has infinitely many columns). If we calculate the sums of these series, it turns out that both the mathematical expectation and the variance of the random variable X with Poisson distribution coincides with the parameter A of this distribution:

,
.

Let's find fashion d(X) Poisson distributed random variable X. Let us apply the same technique that was used to calculate the mode of a binomially distributed random variable. By definition of fashion d(X)= k, if probability
greatest among all probabilities r 0 , p 1 , ... . Let's find such a number k (this is a non-negative integer). With this k probability p k must be no less than its neighboring probabilities: p k −1 ≤ p k ≤ p k +1 . Substituting the corresponding formula for each probability, we obtain that the number k must satisfy the double inequality:

If we write down the formulas for factorials and carry out simple transformations, we can get that left inequality gives k≤ a, and the right k≥ a −1. So the number k satisfies the double inequality a −1 ≤k≤ a, i.e. belongs to the segment [ a −1, a] . Since the length of this segment is obviously equal to 1 , then it can contain either one or 2 integers. If the number A whole, then in the segment [ a −1, a] there are 2 integers lying at the ends of the segment. If the number A is not an integer, then there is only one integer in this segment.

Thus, if the number A integer, then the mode of the Poisson distributed random variable X takes 2 adjacent values: d(X)=a−1 And d(X)=a. If the number A not the whole, then fashion has one value d(X)= k, Where k is the only integer that satisfies the inequality a −1 ≤k≤ a, i.e. d(X)= [A] .

Example. The plant sent 5,000 products to the base. The probability that the product will be damaged in transit is 0.0002. What is the probability that 18 products will be damaged? What is the average value of damaged products? What is the most likely number of damaged products and what is its probability?

For example, the number of traffic accidents per week on a certain section of the road is recorded. This number is a random variable that can take the following values: ( upper limit No). The number of road accidents can be as large as you like. If we consider any short period of time during a week, say a minute, then an incident will either happen during that period or it will not. The probability of a traffic accident within a single minute is very small, and it is approximately the same for all minutes.

The probability distribution of the number of incidents is described by the formula:

where m is the average number of accidents per week on a certain section of the road; e is a constant equal to 2.718...

Characteristic features of data for which in the best possible way The appropriate Poisson distribution is as follows:

1. Each small interval of time can be considered as an experience, the result of which is one of two things: either an incident (“success”) or its absence (“failure”). The intervals are so small that there can only be one “success” in one interval, the probability of which is small and constant.

2. Number of “successes” in one large interval does not depend on their number in another, i.e. “successes” are randomly scattered across time periods.

3. The average number of “successes” is constant throughout the entire time. The Poisson probability distribution can be used not only when working with random variables over time intervals, but also when taking into account road surface defects per kilometer of travel or typos per page of text. General formula Poisson probability distributions:

where m is the average number of “successes” per unit.

In Poisson probability distribution tables, the values are tabulated for certain values of m and

Example 2.7. On average per telephone exchange order three telephone conversations within five minutes. What is the probability that 0, 1,2, 3, 4 or more than four calls will be ordered within five minutes?

We will apply the Poisson probability distribution, since:

1. There is an unlimited number of experiments, i.e. small periods of time when an order for a telephone conversation may appear, the probability of which is small and constant.

2. The demand for telephone conversations is assumed to be randomly distributed over time.

3. It is believed that the average telephone conversations in any minute period of time is the same.

In this example, the average number of orders is 3 in 5 minutes. Hence, the Poisson distribution:

With the Poisson probability distribution, knowing the average number of “successes” in a 5-minute period (for example, as in example 2.7), in order to find out the average number of “successes” in one hour, you simply need to multiply by 12. In example 2.7, the average number of orders in hour will be: 3 x 12 = 36. Similarly, if you want to determine the average number of orders per minute:

Example 2.8. On average in five days working week 3.4 malfunctions occur on the automatic line. What is the probability of two malfunctions on each day of operation? Solution.

You can apply the Poisson distribution:

1. There is an unlimited number of experiments, i.e. small periods of time, during each of which a malfunction may or may not occur on the automatic line. The probability of this for each period of time is small and constant.

2. It is assumed that the problems are randomly distributed in time.

3. The average number of failures over any five days is assumed to be constant.

The average number of problems is 3.4 in five days. Hence the number of problems per day:

Hence,

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

We are interested in how many times a certain event occurs in a given range of possible outcomes 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 in the same area of possible outcomes this event occurs more than once, tends to zero as the range 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 a 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 λ ( greek letter“lambda”) is the average number of successful trials in a given area 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

Now it’s 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 probabilities Poisson distribution at λ = 3

Approximation of the binomial distribution using the Poisson distribution

If the number n is large and the number r- few, binomial distribution can be approximated using the Poisson distribution. How larger 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 random variable, which has 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 probability of detection large number success rate 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.

Please note that the standard deviation calculated using formula (4) tends to 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: , and Poisson. To better understand how these distributions relate to each other, we present a small tree of questions (Fig. 4).