Let's not think about the lofty things for a long time - let's start right away with the definition.
Bernoulli's scheme is when n independent experiments of the same type are performed, in each of which the event of interest to us may appear A, and the probability of this event P (A) = p is known. We need to determine the probability that, after n trials, event A will occur exactly k times.
The problems that can be solved using the Bernoulli scheme are extremely varied: from simple ones (such as “find the probability that the shooter will hit 1 time in 10”) to very severe ones (for example, problems involving percentages or playing cards). In reality, this scheme is often used to solve problems related to product quality and reliability control various mechanisms, all characteristics of which must be known before starting work.
Let's return to the definition. Since we are talking about independent trials, and in each trial the probability of event A is the same, only two outcomes are possible:
- A is the occurrence of event A with probability p;
- “not A” - event A did not appear, which happens with probability q = 1 − p.
The most important condition, without which Bernoulli’s scheme loses its meaning, is constancy. No matter how many experiments we conduct, we are interested in the same event A, which occurs with the same probability p.
By the way, not all problems in probability theory are reduced to constant conditions. Any competent higher mathematics tutor will tell you about this. Even something as simple as taking colorful balls out of a box is not an experience with constant conditions. They took out another ball - the ratio of colors in the box changed. Consequently, the probabilities have also changed.
If the conditions are constant, we can accurately determine the probability that event A will occur exactly k times out of n possible. Let us formulate this fact in the form of a theorem:
Bernoulli's theorem. Let the probability of occurrence of event A in each experiment be constant and equal to p. Then the probability that in n independent trials event A will appear exactly k times is calculated by the formula:
where C n k is the number of combinations, q = 1 − p.
This formula is called Bernoulli's formula. It is interesting to note that the problems given below can be completely solved without using this formula. For example, you can apply the formulas for adding probabilities. However, the amount of computation will be simply unrealistic.
Task. The probability of producing a defective product on a machine is 0.2. Determine the probability that in a batch of ten parts produced on this machine exactly k parts will be without defects. Solve the problem for k = 0, 1, 10.
By condition, we are interested in the event A of releasing products without defects, which happens each time with probability p = 1 − 0.2 = 0.8. We need to determine the probability that this event will occur k times. Event A is contrasted with the event “not A”, i.e. release of a defective product.
Thus, we have: n = 10; p = 0.8; q = 0.2.
So, we find the probability that all parts in a batch are defective (k = 0), that only one part is without defects (k = 1), and that there are no defective parts at all (k = 10):
Task. The coin is tossed 6 times. Landing a coat of arms and heads is equally likely. Find the probability that:
- the coat of arms will appear three times;
- the coat of arms will appear once;
- the coat of arms will appear at least twice.
So, we are interested in the event A, when the coat of arms falls out. The probability of this event is p = 0.5. Event A is contrasted with the event “not A”, when the result is heads, which happens with probability q = 1 − 0.5 = 0.5. We need to determine the probability that the coat of arms will appear k times.
Thus, we have: n = 6; p = 0.5; q = 0.5.
Let us determine the probability that the coat of arms is drawn three times, i.e. k = 3:
Now let’s determine the probability that the coat of arms came up only once, i.e. k = 1:
It remains to determine with what probability the coat of arms will appear at least twice. The main catch is in the phrase “no less.” It turns out that we will be satisfied with any k except 0 and 1, i.e. we need to find the value of the sum X = P 6 (2) + P 6 (3) + ... + P 6 (6).
Note that this sum is also equal to (1 − P 6 (0) − P 6 (1)), i.e. enough of all possible options“cut out” those when the coat of arms fell out 1 time (k = 1) or did not fall out at all (k = 0). Since we already know P 6 (1), it remains to find P 6 (0):
Task. The probability that the TV has hidden defects is 0.2. 20 TVs arrived at the warehouse. Which event is more likely: that in this batch there are two TV sets with hidden defects or three?
Event of interest A is the presence of a latent defect. There are n = 20 TVs in total, the probability of a hidden defect is p = 0.2. Accordingly, the probability of receiving a TV without a hidden defect is q = 1 − 0.2 = 0.8.
We obtain the starting conditions for the Bernoulli scheme: n = 20; p = 0.2; q = 0.8.
Let’s find the probability of getting two “defective” TVs (k = 2) and three (k = 3):
\[\begin(array)(l)(P_(20))\left(2 \right) = C_(20)^2(p^2)(q^(18)) = \frac((20}{{2!18!}} \cdot {0,2^2} \cdot {0,8^{18}} \approx 0,137\\{P_{20}}\left(3 \right) = C_{20}^3{p^3}{q^{17}} = \frac{{20!}}{{3!17!}} \cdot {0,2^3} \cdot {0,8^{17}} \approx 0,41\end{array}\]!}
Obviously, P 20 (3) > P 20 (2), i.e. the probability of receiving three televisions with hidden defects is greater than the probability of receiving only two such televisions. Moreover, the difference is not weak.
A quick note about factorials. Many people experience a vague feeling of discomfort when they see the entry “0!” (read “zero factorial”). So, 0! = 1 by definition.
P. S. And the biggest probability in the last task is to get four TVs with hidden defects. Calculate for yourself and see for yourself.
Let independent tests be carried out, in each of which the probability of the event occurring is A equal to r . In other words, let Bernoulli's scheme hold. Is it possible to predict what the approximate relative frequency of occurrence of an event will be? A positive answer to this question is given by a theorem proven by J. Bernoulli 1, which is called the “law large numbers"and laid the foundation for probability theory as a science 2.
Bernoulli's theorem: If in each of 
independent tests conducted under identical conditions, the probability r
occurrence of an event A
is constant, then the relative frequency of occurrence of the event A
converges in probability to probability r
– the appearance of a given event in a separate experience, that is
.
Proof
. So, Bernoulli's scheme holds, 
. Let us denote by 
discrete random variable – the number of occurrences of the event A
V 
-th test. It is clear that each of the random variables can take only two values: 1
(event A
occurred) with probability r
And 0
(event A
did not occur) with probability 
, that is
|
|
|
|
|
R |
r |
|
(
)
Not hard to find
Is it possible to apply Chebyshev’s theorem to the quantities under consideration? It is possible if the random variables are pairwise independent and their variances are uniformly bounded. Both conditions are met. Indeed, pairwise independence of quantities 
follows from the fact that the tests are independent. Next 3 
at 
and, therefore, the variances of all quantities are limited, for example by the number 
. In addition, note that each of the random variables 
when an event occurs A
in the corresponding test takes a value equal to one. Therefore, the amount 
equal to the number 
- event occurrences A
V 
tests, which means
,
that is, a fraction 
equal to relative frequency 
occurrences of the event A
V 
tests.
Then, applying Chebyshev’s theorem to the quantities under consideration, we obtain:
Q.E.D.
Comment 1 : Bernoulli's theorem is the simplest special case of Chebyshev's theorem.
Comment 2 : In practice, unknown probabilities often have to be approximately determined from experience; a large number of experiments were carried out to check the agreement of Bernoulli’s theorem with experience. For example, the 18th century French naturalist Buffon tossed a coin 4040 times. The coat of arms fell out 2048 times. The frequency of appearance of the coat of arms in Buffon's experiment is approximately 0.507. The English statistician K. Pearson tossed a coin 12,000 times and observed 6,019 coins. The frequency of the coat of arms in this Pearson experiment is 0.5016. Another time he tossed a coin 24,000 times and the coat of arms came up 12,012 times; the frequency of loss of the coat of arms in this case turned out to be equal to 0.5005. As we can see, in all the above experiments, the frequency only slightly deviated from the probability of 0.5 - the appearance of a coat of arms as a result of one toss of a coin.
Comment
3
: It would be incorrect to conclude from Bernoulli’s theorem that as the number of trials increases, the relative frequency steadily approaches probability r
; in other words, Bernoulli's theorem does not imply the equality
. In the theorem it's just a matter of probability that with enough large number trials, the relative frequency will differ as little as desired from the constant probability of the event occurring in each trial. Thus, the convergence of the relative frequency 
to probability r
differs from convergence in the sense of ordinary analysis. To highlight this difference, introduce the concept of “convergence in probability”. More precisely, the difference between these types of convergence is as follows: if 
tends at 
To r
as much as possible in the sense of ordinary analysis, then, starting from some 
and for all subsequent values 
, the inequality is steadily satisfied 
;if 
tends according to probability To r
at 
, then for individual values 
the inequality may not hold.
Poisson and Markov theorems
Noticed if experimental conditions change, then the property of stability of the relative frequency of occurrence of an event A is saved. This circumstance was proven by Poisson.
Poisson's theorem: With an unlimited increase in the number of independent tests conducted under variable conditions, the relative frequency of occurrence of the event A converges in probability to the arithmetic mean of the probabilities of the occurrence of a given event in each of the experiments, that is
.
Comment 4 : It is easy to see that Poisson’s theorem is a special case of Chebyshev’s theorem.
Markov's theorem: If a sequence of random variables 
(however dependent) is such that when 
,
That, 
the condition is met: 
.
Comment
5
: Obviously, if random variables 
are pairwise independent, then the Markov condition takes the form: when 
.
This shows that Chebyshev’s theorem is a special case of Markov’s theorem.
Central limit theorem (Lyapunov's theorem)
The considered theorems of the law of large numbers concern the issues of approximation of certain random variables to certain limiting values, regardless of their distribution law. In probability theory, as already noted, there is another group of theorems concerning the limit laws of distribution of a sum of random variables. Common name this group of theorems - central limit chamber. Its various forms differ in the conditions imposed on the sum of the components of random variables. For the first time, one of the forms of the central limit theorem was proved by the outstanding Russian mathematician A.M. Lyapunov in 1900 using the method of characteristic functions specially developed by him.
Lyapunov's theorem: Law of distribution of the sum of independent random variables 
approaches the normal distribution law with unlimited increase 
(that is, when 
), if the following conditions are met:
,
It should be noted that the central limit theorem is valid not only for continuous, but also for discrete random variables. The practical significance of Lyapunov's theorem is enormous. Experience shows that the law of distribution of the sum of independent random variables comparable in their dispersion quickly approaches normal. Already with a number of terms of the order of ten, the distribution law of the sum can be replaced by a normal one (in particular, an example of such a sum can be the arithmetic mean of the observed values of random variables, that is 
).
A special case of the central limit theorem is Laplace's theorem. In it, as you remember, the case is considered when random variables 
are discrete, identically distributed and take only two possible values: 0 and 1.
Next, the probability that 
contained in the interval 
can be calculated using the formula
.
Using the Laplace function, the last formula can be written in a form convenient for calculations:
Where 
.
EXAMPLE. Let us measure some physical quantity. Any measurement gives only an approximate value of the measured value, since the measurement result is influenced by many independent random factors (temperature, instrument fluctuations, humidity, etc.). Each of these factors generates a negligible “partial error.” However, since the number of these factors is very large, their combined effect gives rise to a noticeable “total error”.
Considering the total error as the sum of a very large number of mutually independent partial errors, we have the right to conclude that the total error has a distribution close to normal. Experience confirms the validity of this conclusion.
2 The proof proposed by J. Bernoulli was complex; a simpler proof was given by P. Chebyshev in 1846.
3 It is known that the product of two factors, the sum of which is a constant value, has the greatest value when the factors are equal.
Definition of repeated independent tests. Bernoulli formulas for calculating probability and the most probable number. Asymptotic formulas for Bernoulli's formula (local and integral, Laplace's theorems). Using the integral theorem. Poisson's formula for unlikely random events.
Repeated independent tests
In practice, we have to deal with tasks that can be represented in the form of repeatedly repeated tests, as a result of each of which the event A may or may not appear. In this case, the outcome of interest is not the outcome of each individual test, but total quantity occurrences of event A as a result of a certain number of trials. In such problems, you need to be able to determine the probability of any number m of occurrences of event A as a result of n trials. Consider the case when the trials are independent and the probability of the occurrence of event A in each trial is constant. Such tests are called repeated independent.
An example of independent testing is checking the suitability of products taken one from a number of batches. If the percentage of defects in these lots is the same, then the probability that the selected product will be defective is a constant number in each case.
Bernoulli's formula
Let's use the concept complex event, which means the combination of several elementary events consisting of the appearance or non-occurrence of event A in the i-th trial. Let n independent trials be carried out, in each of which event A can either appear with probability p or not appear with probability q=1-p. Consider the event B_m, which is that event A will occur exactly m times in these n trials and, therefore, will not occur exactly (n-m) times. Let's denote A_i~(i=1,2,\ldots,(n)) occurrence of event A, a \overline(A)_i - non-occurrence of event A in the i-th trial. Due to the constancy of the test conditions, we have
Event A can appear m times in different sequences or combinations, alternating with the opposite event \overline(A) . The number of possible combinations of this kind is equal to the number of combinations of n elements by m, i.e. C_n^m. Consequently, the event B_m can be represented as a sum of complex events that are inconsistent with each other, and the number of terms is equal to C_n^m:
B_m=A_1A_2\cdots(A_m)\overline(A)_(m+1)\cdots\overline(A)_n+\cdots+\overline(A)_1\overline(A)_2\cdots\overline(A)_( n-m)A_(n-m+1)\cdots(A_n),
where each product contains the event A m times, and \overline(A) - (n-m) times.
The probability of each complex event included in formula (3.1), according to the probability multiplication theorem for independent events is equal to p^(m)q^(n-m) . Since the total number of such events is equal to C_n^m, then, using the probability addition theorem for incompatible events, we obtain the probability of the event B_m (we denote it P_(m,n) )
P_(m,n)=C_n^mp^(m)q^(n-m)\quad \text(or)\quad P_(m,n)=\frac(n{m!(n-m)!}p^{m}q^{n-m}. !}
Formula (3.2) is called Bernoulli's formula, and repeated trials that satisfy the condition of independence and constancy of the probabilities of the occurrence of event A in each of them are called Bernoulli tests, or Bernoulli scheme.
Example 1. The probability of going beyond the tolerance zone when processing parts on lathe equal to 0.07. Determine the probability that out of five parts selected at random during a shift, one has diameter dimensions that do not correspond to the specified tolerance.
Solution. The condition of the problem satisfies the requirements of the Bernoulli scheme. Therefore, assuming n=5,\,m=1,\,p=0,\!07, using formula (3.2) we obtain
P_(1,5)=C_5^1(0,\!07)^(1)(0,\!93)^(5-1)\approx0,\!262.
Example 2. Observations have established that in a certain area there are 12 rainy days in September. What is the probability that out of 8 days chosen at random this month, 3 days will be rainy?
Solution.
P_(3;8)=C_8^3(\left(\frac(12)(30)\right)\^3{\left(1-\frac{12}{30}\right)\!}^{8-3}=\frac{8!}{3!(8-3)!}{\left(\frac{2}{5}\right)\!}^3{\left(\frac{3}{5}\right)\!}^5=56\cdot\frac{8}{125}\cdot\frac{243}{3125}=\frac{108\,864}{390\,625}\approx0,\!2787. !}
Most likely number of occurrences of an event
Most likely date of occurrence event A in n independent trials is called such a number m_0 for which the probability corresponding to this number exceeds or, at least, is not less than the probability of each of the other possible numbers of occurrence of event A. To determine the most probable number, it is not necessary to calculate the probabilities of the possible number of occurrences of an event; it is enough to know the number of trials n and the probability of occurrence of event A in separate test. Let us denote P_(m_0,n) the probability corresponding to the most probable number m_0. Using formula (3.2), we write
P_(m_0,n)=C_n^(m_0)p^(m_0)q^(n-m_0)=\frac(n{m_0!(n-m_0)!}p^{m_0}q^{n-m_0}. !}
According to the definition of the most probable number, the probabilities of the occurrence of event A, respectively m_0+1 and m_0-1 times, must at least not exceed the probability P_(m_0,n), i.e.
P_(m_0,n)\geqslant(P_(m_0+1,n));\quad P_(m_0,n)\geqslant(P_(m_0-1,n))
Substituting the value P_(m_0,n) and the probability expressions P_(m_0+1,n) and P_(m_0-1,n) into the inequalities, we obtain
Solving these inequalities for m_0, we obtain
M_0\geqslant(np-q),\quad m_0\leqslant(np+p)
Combining the last inequalities, we get a double inequality, which is used to determine the most probable number:
Np-q\leqslant(m_0)\leqslant(np+p).
Since the length of the interval defined by inequality (3.4) is equal to one, i.e.
(np+p)-(np-q)=p+q=1,
and the event can only occur in n trials an integer number of times, then it should be borne in mind that:
1) if np-q is an integer, then there are two values of the most probable number, namely: m_0=np-q and m"_0=np-q+1=np+p ;
2) if np-q is a fractional number, then there is one most probable number, namely: the only integer contained between the fractional numbers obtained from inequality (3.4);
3) if np is an integer, then there is one most probable number, namely: m_0=np.
For large values of n, it is inconvenient to use formula (3.3) to calculate the probability corresponding to the most probable number. If we substitute the Stirling formula into equality (3.3)
N!\approx(n^ne^(-n)\sqrt(2\pi(n))),
valid for sufficiently large n, and take the most probable number m_0=np, then we obtain a formula for approximate calculation of the probability corresponding to the most probable number:
P_(m_0,n)\approx\frac(n^ne^(-n)\sqrt(2\pi(n))\,p^(np)q^(nq))((np)^(np) e^(-np)\sqrt(2\pi(np))\,(nq)^(nq)e^(-nq)\sqrt(2\pi(nq)))=\frac(1)(\ sqrt(2\pi(npq)))=\frac(1)(\sqrt(2\pi)\sqrt(npq)).
Example 2. It is known that \frac(1)(15) part of the products supplied by the plant to the trading base does not meet all the requirements of the standard. A batch of 250 products was delivered to the base. Find the most likely number of products that meet the requirements of the standard and calculate the probability that this batch will contain the most likely number of products.
Solution. By condition n=250,\,q=\frac(1)(15),\,p=1-\frac(1)(15)=\frac(14)(15). According to inequality (3.4) we have
250\cdot\frac(14)(15)-\frac(1)(15)\leqslant(m_0)\leqslant250\cdot\frac(14)(15)+\frac(1)(15)
where 233,\!26\leqslant(m_0)\leqslant234,\!26. Consequently, the most likely number of products that meet the requirements of the standard in a batch of 250 pcs. equals 234. Substituting the data into formula (3.5), we calculate the probability of having the most probable number of products in the batch:
P_(234,250)\approx\frac(1)(\sqrt(2\pi\cdot250\cdot\frac(14)(15)\cdot\frac(1)(15)))\approx0,\!101
Local Laplace theorem
It is very difficult to use Bernoulli's formula for large values of n. For example, if n=50,\,m=30,\,p=0,\!1, then to find the probability P_(30.50) it is necessary to calculate the value of the expression
P_(30.50)=\frac(50{30!\cdot20!}\cdot(0,\!1)^{30}\cdot(0,\!9)^{20} !}
Naturally, the question arises: is it possible to calculate the probability of interest without using Bernoulli’s formula? It turns out that it is possible. Laplace's local theorem gives an asymptotic formula that allows us to approximately find the probability of events occurring exactly m times in n trials, if the number of trials is large enough.
Theorem 3.1. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then the probability P_(m,n) that event A will appear exactly m times in n trials is approximately equal (the more accurate, the larger n) to the value of the function
Y=\frac(1)(\sqrt(npq))\frac(e^(-x^2/2))(\sqrt(2\pi))=\frac(\varphi(x))(\sqrt (npq)) at .
There are tables that contain function values \varphi(x)=\frac(1)(\sqrt(2\pi))\,e^(-x^2/2)), corresponding to positive values of the argument x. For negative values of the argument, the same tables are used, since the function \varphi(x) is even, i.e. \varphi(-x)=\varphi(x).
So, approximately the probability that event A will appear exactly m times in n trials is
P_(m,n)\approx\frac(1)(\sqrt(npq))\,\varphi(x), Where x=\frac(m-np)(\sqrt(npq)).
Example 3. Find the probability that event A will occur exactly 80 times in 400 trials if the probability of event A occurring in each trial is 0.2.
Solution. By condition n=400,\,m=80,\,p=0,\!2,\,q=0,\!8. Let us use the asymptotic Laplace formula:
P_(80,400)\approx\frac(1)(\sqrt(400\cdot0,\!2\cdot0,\!8))\,\varphi(x)=\frac(1)(8)\,\varphi (x).
Let's calculate the value x determined by the task data:
X=\frac(m-np)(\sqrt(npq))=\frac(80-400\cdot0,\!2)(8)=0.
According to the table adj. 1 we find \varphi(0)=0,\!3989. Required probability
P_(80,100)=\frac(1)(8)\cdot0,\!3989=0,\!04986.
Bernoulli's formula leads to approximately the same result (calculations are omitted due to their cumbersomeness):
P_(80,100)=0,\!0498.
Laplace's integral theorem
Suppose that n independent trials are conducted, in each of which the probability of occurrence of event A is constant and equal to p. It is necessary to calculate the probability P_((m_1,m_2),n) that event A will appear in n trials at least m_1 and at most m_2 times (for brevity we will say “from m_1 to m_2 times”). This can be done using Laplace's integral theorem.
Theorem 3.2. If the probability p of the occurrence of event A in each trial is constant and different from zero and one, then approximately the probability P_((m_1,m_2),n) that event A will appear in trials from m_1 to m_2 times,
P_((m_1,m_2),n)\approx\frac(1)(\sqrt(2\pi))\int\limits_(x")^(x"")e^(-x^2/2) \,dx, Where .
When solving problems that require the application of Laplace's integral theorem, special tables are used, since indefinite integral \int(e^(-x^2/2)\,dx) is not expressed through elementary functions. Table for the integral \Phi(x)=\frac(1)(\sqrt(2\pi))\int\limits_(0)^(x)e^(-z^2/2)\,dz given in appendix. 2, where the values of the function \Phi(x) are given for positive values of x, for x<0 используют ту же таблицу (функция \Phi(x) нечетна, т. е. \Phi(-x)=-\Phi(x) ). Таблица содержит значения функции \Phi(x) лишь для x\in ; для x>5 we can take \Phi(x)=0,\!5 .
So, approximately the probability that event A will appear in n independent trials from m_1 to m_2 times is
P_((m_1,m_2),n)\approx\Phi(x"")-\Phi(x"), Where x"=\frac(m_1-np)(\sqrt(npq));~x""=\frac(m_2-np)(\sqrt(npq)).
Example 4. The probability that a part is manufactured in violation of standards is p=0,\!2. Find the probability that among 400 randomly selected non-standard parts there will be from 70 to 100 parts.
Solution. By condition p=0,\!2,\,q=0,\!8,\,n=400,\,m_1=70,\,m_2=100. Let's use Laplace's integral theorem:
P_((70,100),400)\approx\Phi(x"")-\Phi(x").
Let's calculate the limits of integration:
lower
X"=\frac(m_1-np)(\sqrt(npq))=\frac(70-400\cdot0,\!2)(\sqrt(400\cdot0,\!2\cdot0,\!8)) =-1,\!25,
upper
X""=\frac(m_2-np)(\sqrt(npq))=\frac(100-400\cdot0,\!2)(\sqrt(400\cdot0,\!2\cdot0,\!8) )=2,\!5,
Thus
P_((70,100),400)\approx\Phi(2,\!5)-\Phi(-1,\!25)=\Phi(2,\!5)+\Phi(1,\!25) .
According to the table adj. 2 we find
\Phi(2,\!5)=0,\!4938;~~~~~\Phi(1,\!25)=0,\!3944.
Required probability
P_((70,100),400)=0,\!4938+0,\!3944=0,\!8882.
Application of Laplace's integral theorem
If the number m (the number of occurrences of event A in n independent trials) changes from m_1 to m_2, then the fraction \frac(m-np)(\sqrt(npq)) will vary from \frac(m_1-np)(\sqrt(npq))=x" to \frac(m_2-np)(\sqrt(npq))=x"". Therefore, Laplace’s integral theorem can also be written as follows:
P\left\(x"\leqslant\frac(m-np)(\sqrt(npq))\leqslant(x"")\right\)=\frac(1)(\sqrt(2\pi))\ int\limits_(x")^(x"")e^(-x^2/2)\,dx.
Let us set the task of finding the probability that the deviation of the relative frequency \frac(m)(n) from the constant probability p by absolute value does not exceed the specified number \varepsilon>0 . In other words, we find the probability of the inequality \left|\frac(m)(n)-p\right|\leqslant\varepsilon, which is the same -\varepsilon\leqslant\frac(m)(n)-p\leqslant\varepsilon. We will denote this probability as follows: P\left\(\left|\frac(m)(n)-p\right|\leqslant\varepsilon\right\). Taking into account formula (3.6) for this probability we obtain
P\left\(\left|\frac(m)(n)-p\right|\leqslant\varepsilon\right\)\approx2\Phi\left(\varepsilon\,\sqrt(\frac(n)(pq ))\right).
Example 5. The probability that the part is non-standard is p=0,\!1. Find the probability that, among 400 randomly selected parts, the relative frequency of occurrence of non-standard parts will deviate from the probability p=0,\!1 in absolute value by no more than 0.03.
Solution. By condition n=400,\,p=0,\!1,\,q=0,\!9,\,\varepsilon=0,\!03. We need to find the probability P\left\(\left|\frac(m)(400)-0,\!1\right|\leqslant0,\!03\right\). Using formula (3.7), we obtain
P\left\(\left|\frac(m)(400)-0,\!1\right|\leqslant0,\!03\right\)\approx2\Phi\left(0,\!03\sqrt( \frac(400)(0,\!1\cdot0,\!9))\right)=2\Phi(2)
According to the table adj. 2 we find \Phi(2)=0,\!4772 , therefore, 2\Phi(2)=0,\!9544 . So, the desired probability is approximately 0.9544. The meaning of the result is as follows: if you take a sufficiently large number of samples of 400 parts each, then in approximately 95.44% of these samples the deviation of the relative frequency from the constant probability p=0.\!1 in absolute value will not exceed 0.03.
Poisson's formula for unlikely events
If the probability p of the occurrence of an event in a single trial is close to zero, then even with a large number of trials n, but with a small value of the product np, the probability values P_(m,n) obtained from the Laplace formula are not accurate enough and the need for another approximate formula arises.
Theorem 3.3. If the probability p of the occurrence of event A in each trial is constant but small, the number of independent trials n is sufficiently large, but the value of the product np=\lambda remains small (no more than ten), then the probability that event A will occur m times in these trials is
P_(m,n)\approx\frac(\lambda^m)(m\,e^{-\lambda}. !}
To simplify calculations using the Poisson formula, a table of Poisson function values has been compiled \frac(\lambda^m)(m\,e^{-\lambda} !}(see appendix 3).
Example 6. Let the probability of producing a non-standard part be 0.004. Find the probability that among 1000 parts there will be 5 non-standard ones.
Solution. Here n=1000,p=0.004,~\lambda=np=1000\cdot0,\!004=4. All three numbers satisfy the requirements of Theorem 3.3, therefore, to find the probability of the desired event P_(5,1000), we use the Poisson formula. From the table of values of the Poisson function (Appendix 3) with \lambda=4;m=5 we obtain P_(5,1000)\approx0,\!1563.
Let's find the probability of the same event using Laplace's formula. To do this, we first calculate the value of x corresponding to m=5:
X=\frac(5-1000\cdot0,\!004)(\sqrt(1000\cdot0,\!004\cdot0,\!996))\approx\frac(1)(1,\!996)\approx0 ,\!501.
Therefore, according to Laplace’s formula, the desired probability
P_(5,1000)\approx\frac(\varphi(0,\!501))(1,\!996)\approx\frac(0,\!3519)(1,\!996)\approx0,\ !1763
and according to Bernoulli’s formula its exact value is
P_(5,1000)=C_(1000)^(5)\cdot0,\!004^5\cdot0,\!996^(995)\approx0,\!1552.
Thus, the relative error in calculating the probabilities P_(5,1000) using the approximate Laplace formula is
\frac(0,\!1763-0,\!1552)(0,\!1552)\approx0,\!196, or 13.\!6\%
and according to the Poisson formula -
\frac(0,\!1563-0,\!1552)(0,\!1552)\approx0,\!007, or 0.\!7\%
That is, many times less.
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In this lesson we will find the probability of an event occurring in independent trials when repeating trials . Trials are called independent if the probability of one or another outcome of each trial does not depend on what outcomes other trials had. . Independent tests can be carried out both under the same conditions and under different conditions. In the first case, the probability of the occurrence of some event is the same in all trials, in the second case it varies from trial to trial.
Examples of independent retests :
- one of the device nodes or two or three nodes will fail, and the failure of each node does not depend on the other node, and the probability of failure of one node is constant in all tests;
- produced in some constant technological conditions a part, or three, four, five parts, will turn out to be non-standard, and one part may turn out to be non-standard regardless of any other part and the probability that the part will turn out to be non-standard is constant in all tests;
- out of several shots at a target, one, three or four shots hit the target regardless of the outcome of the other shots and the probability of hitting the target is constant in all trials;
- when dropping a coin, the machine will operate correctly one, two, or other number of times, regardless of the outcome of other coin drops, and the probability that the machine will operate correctly is constant across all trials.
These events can be described in one diagram. Each event occurs in each trial with the same probability, which does not change if the results of previous trials become known. Such tests are called independent, and the circuit is called Bernoulli scheme . It is assumed that such tests can be repeated as many times as desired.
If the probability p occurrence of an event A is constant in each trial, then the probability that in n independent testing event A will come m times, is located by Bernoulli's formula :
(Where q= 1 – p- the probability that the event will not occur)
Let us set the task - to find the probability that an event of this type in n independent tests will come m once.
Bernoulli's formula: examples of problem solving
Example 1. Find the probability that among five parts taken at random, two are standard, if the probability that each part turns out to be standard is 0.9.
Solution. Probability of event A, consisting in the fact that a part taken at random is standard, there is p=0.9 , and there is a probability that it is non-standard q=1–p=0.1 . The event designated in the problem statement (we denote it by IN) will occur if, for example, the first two parts turn out to be standard, and the next three are non-standard. But the event IN will also occur if the first and third parts turn out to be standard and the rest are non-standard, or if the second and fifth parts are standard and the rest are non-standard. There are other possibilities for the event to occur IN. Any of them is characterized by the fact that out of five parts taken, two, occupying any places out of five, will turn out to be standard. Hence, total number various possibilities for the occurrence of an event IN is equal to the number of possibilities for placing two standard parts in five places, i.e. is equal to the number of combinations of five elements by two, and .
The probability of each possibility according to the probability multiplication theorem is equal to the product of five factors, of which two, corresponding to the appearance of standard parts, are equal to 0.9, and the remaining three, corresponding to the appearance of non-standard parts, are equal to 0.1, i.e. this probability is . Since these ten possibilities are incompatible events, by the addition theorem the probability of an event IN, which we denote
Example 2. The probability that the machine will require the attention of a worker within an hour is 0.6. Assuming that the problems on the machines are independent, find the probability that within an hour a worker’s attention will require any one machine out of the four he operates.
Solution. Using Bernoulli's formula at n=4 , m=1 , p=0.6 and q=1–p=0.4, we get
Example 3. For normal operation of the carpool, there must be at least eight vehicles on the line, and there are ten of them. The probability of each vehicle not entering the line is 0.1. Find the probability of normal operation of the car depot in the next day.
Solution. The carpool will work normally (event F), if eight or eight come on line (event A), or nine (event IN), or all ten cars event (event C). According to the theorem of addition of probabilities,
We find each term according to Bernoulli's formula. Here n=10 , m=8; 10 and p=1-0.1=0.9, since p should indicate the probability of the vehicle entering the line; Then q=0.1 . As a result we get
Example 4. Let the probability that the buyer needs men's shoes Size 41, equal to 0.25. Find the probability that out of six buyers, at least two need shoes of size 41.
Before presenting the third question of the lecture, the teacher identifies a problem that necessitates the consideration of the theorem on the repetition of experiments, noting that in the probability theory course being studied, only a particular theorem related to the repetition of independent experiments, in each of which event A appears with a constant probability, will be considered.
After which the teacher shows the proof of this theorem (derivation of Bernoulli’s formula).
To explain the physical essence of the theorem under consideration, the teacher uses an overhead projector and prepared slides.
At the end of the lecture, the teacher explains why the probability distribution of the occurrence of event A in a series of n tests, in conditions where they are inconsistent and form a complete group of events, is called binomial and draws attention to the importance of knowing this distribution for solving applied problems.
Until now, we have considered combinations of a relatively small number of events, when the direct application of the rules of addition and multiplication of probabilities did not cause great computational difficulties. However, as the number of events or the number of trials in which the event of interest may appear increases, the learned calculation method becomes very cumbersome.
Moreover, the problem was solved quite simply only if the experiments were independent.
Several experiments are called independent, if the probability of one or another outcome of each experiment does not depend on what outcomes other experiments had.
In practice, there are cases when the probability of an event occurring A in all independent experiments it can be either the same or vary from experiment to experiment. For example, if you adjust your fire after each shot, the probability of hitting the target will change with each shot.
In the case when in independent experiments the probability of the occurrence of an event changes from experiment to experiment, the general theorem on the repetition of experiments is used, and when in independent experiments the probability of the occurrence of an event does not change from experiment to experiment, a particular theorem on the repetition of experiments is used.
In the probability theory course we are studying, we will consider only the particular topic of repeating experiments when it is necessary to determine the probability of an event occurring A in a series of independent experiments, in each of which event A appears with equal probability.
For example, it is necessary to calculate the probability that with five shots from a gun at constant settings, exactly two hits on the target will be obtained if the shots are independent and with each shot the probability of hitting the target is known and does not change.
If we compose possible combinations of the occurrence of the event we are interested in A 1, we get:
There will be 10 possible combinations in which the event A=(get 2 hits with five shots) occurs.
Applying the theorem about the sum and product of independent events, we have:
An increase in the number of events or the number of tests we are interested in will lead to an even greater increase in the volume of computational operations, so the task arises of finding less labor-intensive calculation methods.
Problem statement:
Let us assume, under identical conditions, to carry out n independent tests, the result of each of which may be the occurrence of either event A, or its opposite 
.
Let us denote by A 1 occurrence of an event A on the first test, A 2 - on the second test, A n- at the last test.
Due to the constancy of the test conditions:
P(A 1 ) = P(A 2 ) = … P(A n ) = p
We are interested in the probability that event A will occur exactly m times in n trials, but will not occur in the remaining n-m trials (i.e., the opposite event to event A will occur - 
).
Let us assume that the event we are interested in A occurs consecutively m times, starting from the first, i.e. an event takes place - E.
E= A 1
A 2
… A m -1
A m
(1)
m n- m
According to the condition of repetition of tests, the events included in this combination are independent, while the probabilities of the occurrence of events A 1, A 2
,… A m -1
, A m same and equal p: P(A 1
) = P(A 2
) =…= P(A m ) = p, and the probabilities of events not occurring 
same and equal q=1-р:.
Applying the rule of multiplying probabilities for independent events to expression 1, we obtain:
P(E) = P(A 1
) P(A 2
) … P(A m -1
) P(A m ) P( 
= p m (1-r) n
-
m
= p m q n -
m
Due to the constancy of test conditions, we assumed that the event of interest to us A occurs in a row m times, starting from the first. But the event A V n trials may come exactly m times in different sequences or combinations. In this case, we are indifferent to the exact sequence in which event A appears exactly m once.
The number of such combinations is equal to the number of combinations 
of n elements by m.
Since these combinations of events (similar to combination E) are incompatible and we are not interested in the sequence of occurrence of the event A in the test exactly m times, then denoting the probability we are interested in through R m, we get:
R m
=
r m (1-r) n
-
m
=
=
Where 
- number of combinations of n elements by m.
This formula is called Bernoulli's formula.
Bernoulli's formula allows us to get an answer to the question: what is the probability that when n independent tests are repeated, some event A comes exactly m times, if in each of these trials the probability of the event occurring A is constant and equal P(A) = p.
The above Bernoulli formula is extremely important in probability theory for the reason that it is associated with the repetition of tests under the same conditions, i.e. with such conditions in which the laws of probability theory manifest themselves.
Conclusion of the lecture:
In the lecture, we examined the fundamental issues of probability theory in relation to random variables, introduced the basic conceptual apparatus necessary for further study of the discipline: definition random variable, their classification; concept of the distribution law and its form for various types random variable.
In preparation for subsequent lectures and practical exercises, you must independently supplement your lecture notes while studying the recommended literature in depth and solving the proposed problems.
In addition, in subsequent lessons we will study theorems and dependencies that allow us to determine the probability of a random variable appearing the required number of times or at a certain interval, for example, the probability of hitting a target.
Explore:
Ventzel E.S. Probability theory. Textbook. Eighth edition, stereotypical. – M.: Higher School, 2002 - 575 p. – pp. 67-78, 80-84
Ventzel E.S., Ovcharov L.A.. Probability theory and its engineering applications. Study guide. Third edition, revised and expanded. – M.: “Academy”, 2003 – 464 p. – pp. 73-93
Gmurman V.E. Probability theory and mathematical statistics. Tutorial. Tenth edition, stereotypical. - M.: Higher School", 2004 - 480 p. Page 64-73
