Theorem 2 (Moivre-Laplace (local)). A in each n independent tests is equal to R n testing event A will occur once, is approximately equal (the more n, the more accurate) the value of the function
,
Where 
, . The table of function values is given in the appendix. 1.
Example 6.5. The probability of finding a porcini mushroom among others is equal. What is the probability that among 300 porcini mushrooms there will be 75?
Solution. According to the conditions of the problem 
, . We find 
. From the table we find 
.
.
Answer: 
.
Theorem 3 (Moivre-Laplace (integral)). If the probability of an event occurring A in each n independent tests is equal to R and is different from zero and one, and the number of tests is large enough, then the probability that in n tests number of successes m is between and , approximately equal (the more n, the more accurate)
,
Where R- probability of success in each test, , 
, values are given in appendix. 2.
Example 6.6. In a batch of 768 watermelons, each watermelon is unripe with probability . Find the probability that the number of ripe watermelons will be in the range from 564 to 600.
Solution. By condition By Laplace’s integral theorem
Answer:
Example 6.7. The city is visited daily by 1,000 tourists who go out for lunch during the day. Each of them chooses one of two city restaurants for lunch with equal probabilities and independently of each other. The owner of one of the restaurants wants that, with a probability of approximately 0.99, all tourists who come to his restaurant can dine there at the same time. How many seats should there be in his restaurant for this?
Solution. Let A= “tourist dined with interested owner.” Event occurrence A Let's consider it a "success" 
, . We are interested in this smallest number k, that the probability of occurrence is no less than k"successes" in a sequence of independent trials with a probability of success R= 0.5 is approximately equal to 1 – 0.99 = 0.01. This is precisely the probability of the restaurant being overcrowded. Thus, we are interested in this smallest number k, What . Let us apply the Moivre-Laplace integral theorem
Whence it follows that
.
Using the table for F(X) (Appendix 2), we find 
, Means . Therefore, the restaurant should have 537 seats.
Answer: 537 places.
From Laplace's integral theorem we can obtain the formula
.
Example 6.8. The probability of an event occurring in each of 625 independent trials is 0.8. Find the probability that the relative frequency of occurrence of an event will deviate from its probability in absolute value by no more than 0.04.
Local theorem of Moivre-Laplace(1730 Moivre and Laplace)
If the probability $p$ of occurrences of event $A$ is constant and $p\ne 0$ and $p\ne 1$, then the probability $P_n (k)$ is that event $A$ will appear $k$ times in $n $ tests, is approximately equal (the larger the $n$, the more accurate) the value of the function $y=\frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \frac ( 1 ) ( \sqrt ( 2 \pi ) ) \cdot e^ ( - ( x^2 ) / 2 ) =\frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \varphi (x)$
for $x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $. There are tables containing the values of the function $\varphi (x)=\frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \cdot e^ ( - ( x^2 ) / 2 ) $
so \begin(equation) \label ( eq2 ) P_n (k)\approx \frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \cdot \varphi (x)\,\,where\,x =\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) \qquad (2) \end(equation)
function $\varphi (x)=\varphi (( -x ))$ is even.
Example. Find the probability that event $A$ will occur exactly 80 times in 400 trials if the probability of this event occurring in each trial is $p=0.2$.
Solution. If $p=0.2$ then $q=1-p=1-0.2=0.8$.
$P_ ( 400 ) (( 80 ))\approx \frac ( 1 ) ( \sqrt ( n\cdot p\cdot q ) ) \varphi (x)\,\,where\,x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $
$ \begin(array) ( l ) x=\frac ( k-n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) =\frac ( 80-400\cdot 0.2 ) ( \sqrt ( 400 \cdot 0.2\cdot 0.8 ) ) =\frac ( 80-80 ) ( \sqrt ( 400\cdot 0.16 ) ) =0 \\ \varphi (0)=0.3989\,\,P_ ( 400 ) (( 80 ))\approx \frac ( 0.3989 ) ( 20\cdot 0.4 ) =\frac ( 0.3989 ) ( 8 ) =0.0498 \\ \end(array) $
Moivre-Laplace integral theorem
The probability P of the occurrence of event $A$ in each trial is constant and $p\ne 0$ and $p\ne 1$, then the probability $P_n (( k_1 ,k_2 ))$ that the event $A$ will occur from $k_ ( 1 ) $ up to $k_ ( 2 ) $ times in $n$ trials, equals $ P_n (( k_1 ,k_2 ))\approx \frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \int\limits_ ( x_1 ) ^ ( x_2 ) ( e^ ( - ( z^2 ) / 2 ) dz ) =\Phi (( x_2 ))-\Phi (( x_1 ))$
where $x_1 =\frac ( k_1 -n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) , x_2 =\frac ( k_2 -n\cdot p ) ( \sqrt ( n\cdot p\cdot q ) ) $ ,where
$\Phi (x)=\frac ( 1 ) ( \sqrt ( 2\cdot \pi ) ) \int ( e^ ( - ( z^2 ) / 2 ) dz ) $ -found from tables
$\Phi (( -x ))=-\Phi (x)$-odd
Odd function. The values in the table are given for $x=5$, for $x>5,\Phi (x)=0.5$
Example. It is known that 10% of products are rejected during inspection. 625 products were selected for control. What is the probability that among those selected there are at least 550 and at most 575 standard products?
Solution. If there are 10% defects, then there are 90% standard products. Then, by condition, $n=625, p=0.9, q=0.1, k_1 =550, k_2 =575$. $n\cdot p=625\cdot 0.9=562.5$. We get $ \begin(array) ( l ) P_ ( 625 ) (550.575)\approx \Phi (( \frac ( 575-562.5 ) ( \sqrt ( 625\cdot 0.9\cdot 0.1 ) ) ) )- \Phi (( \frac ( 550-562.5 ) ( \sqrt ( 626\cdot 0.9\cdot 0.1 ) )) \approx \Phi (1.67)- \Phi (-1, 67)=2 \Phi (1.67)=0.9052 \\ \end(array) $
Consider a sequence of $n$ independent experiments, in each of which event $A$ can occur with probability $p$, or not occur - with probability $q=1-p$. Let us denote by P n (k) the probability that event $A$ will occur exactly $k$ times out of $n$ possible.
In this case, the value P n(k) can be found using Bernoulli’s theorem (see lesson “ Bernoulli scheme. Examples of problem solving »):
This theorem works great, but it has a flaw. If $n$ is large enough, then find the value P n (k) becomes unrealistic due to the huge amount of computation. In this case it works Local theorem of Moivre-Laplace, which allows you to find the approximate value of the probability:
Local theorem of Moivre - Laplace. If in Bernoulli's scheme the number $n$ is large and the number $p$ is different from 0 and 1, then:
Function φ ( x) is called the Gaussian function. Its values have long been calculated and entered into a table that can be used even in tests and exams.
The Gaussian function has two properties that you should consider when working with a table of values:
- φ (− x) = φ ( x) — Gaussian function — even;
- For large values x we have: φ ( x) ≈ 0.
The local theorem of Moivre-Laplace gives an excellent approximation of Bernoulli's formula if the number of trials n big enough. Of course, the formulation “the number of tests is quite large” is very conditional, and different sources give different figures. For example:
- A common requirement is: n· p · q> 10. Perhaps this is the minimum limit;
- Others suggest working with this formula only for $n > 100$ and n· p · q > 20.
In my opinion, it is enough to simply look at the problem statement. If it is clear that the standard Bernoulli theorem does not work due to the large amount of calculations (for example, no one will count the number 58! or 45!), feel free to use the Local De Moivre-Laplace theorem.
In addition, the closer the probability values $q$ and $p$ are to 0.5, the more accurate the formula. And, conversely, at borderline values (when $p$ is close to 0 or 1), the Local Moivre-Laplace theorem gives a large error, significantly different from the real Bernoulli theorem.
However, be careful! Many higher mathematics tutors themselves make mistakes in such calculations. The fact is that a rather complex number containing an arithmetic square root and a fraction is substituted into the Gaussian function. This number must be found before substitution into the function. Let's consider everything on specific tasks:
Task. The probability of having a boy is 0.512. Find the probability that among 100 newborns there will be exactly 51 boys.
So, total tests according to the Bernoulli scheme n= 100. In addition, p = 0.512, q= 1 − p = 0.488.
Because the n= 100 is a fairly large number, we will work according to the Local Theorem of Moivre - Laplace. notice, that n· p · q= 100 · 0.512 · 0.488 ≈ 25 > 20. We have:
Since we rounded the value n· p · q to a whole number, the answer can also be rounded: 0.07972 ≈ 0.08. There is simply no point in taking other numbers into account.
Task. The telephone exchange serves 200 subscribers. For each subscriber, the probability that he will call the station within one hour is 0.02. Find the probability that exactly 5 subscribers will call within an hour.
According to Bernoulli's scheme, n= 200, p = 0.02, q= 1 − p = 0.98. notice, that n= 200 is not a weak number, so we use the Local Theorem of Moivre - Laplace. First, let's find n· p · q= 200 · 0.02 · 0.98 ≈ 4. Of course, 4 is too small, so the results will be inaccurate. However, we have:
Let's round the answer to the second decimal place: 0.17605 ≈ 0.18. It doesn't make sense to take more characters into account anyway, since we rounded n· p · q= 3.92 ≈ 4 (up to exact square).
Task. The store received 1000 bottles of vodka. The probability that a bottle will break during transportation is 0.003. Find the probability that the store receives exactly two broken bottles.
According to Bernoulli's scheme we have: n= 1000, p = 0.003, q= 0.997. From here n· p · q= 2.991 ≈ 1.73 2 (we selected the nearest exact square). Since the number n= 1000 is large enough, we substitute all the numbers into the formula of the Local Theorem of Moivre - Laplace:
We deliberately leave only one decimal place (in fact, it will turn out to be 0.1949...), since we initially used rather rough estimates. In particular: 2.991 ≈ 1.73 2. The three in the numerator inside the Gaussian function arose from the expression n· p = 1000 · 0.003 = 3.
The probability that in n independent trials, in each of which the probability of an event occurring is p(0< p < 1), событие наступит ровно k раз, приближенно равна
Table of function values φ(x); for negative values of x, use the same table (function φ (x) is even: φ(-x) = φ(x)).
Example No. 1. In each of 700 independent trials, event A occurs with a constant probability of 0.35. Find the probability that event A occurs: a) exactly 270 times; b) less than 270 and more than 230 times; c) more than 270 times.
Solution. Since the number of experiments n = 700 is quite large, we use Laplace’s formulas.
a) Given: n = 700, p = 0.35, k = 270.
Let's find P 700 (270). We use Laplace's local theorem.
We find:
We find the value of the function φ(x) from the table:
b) Given: n = 700, p = 0.35, a = 230, b = 270.
Let's find P 700 (230< k < 270).
We use Laplace's integral theorem (23), (24). We find: 
We find the value of the function Ф(x) from the table:
c) Given: n = 700, p = 0.35, a = 270, b = 700.
Let's find P 700 (k > 270).
We have:
Example No. 2. At a steady-state technological process at a weaving mill, 10 thread breaks occur per 100 spindles per hour. Determine: a) the probability that 7 thread breaks will occur on 80 spindles within an hour; b) the most likely number of thread breaks on 80 spindles within an hour.
Solution. The statistical probability of a thread breaking within an hour is p = 10/100 = 0.1 and, therefore, q = 1 – 0.1 = 0.9; n = 80; k = 7.
Since n is large, the local Laplace theorem (23) is used. We calculate: 
Let's use the property φ(-x) = φ(x), find φ(0.37) ≈ 0.3726, and then calculate the desired probability: 
Thus, the probability that 7 thread breaks will occur on 80 spindles within an hour is approximately 0.139.
The most probable number k 0 of occurrences of an event during repeated tests will be determined by formula (14). We find: 7.1< k 0 < 8,1. Поскольку k 0 может быть только целым числом, то k 0 = 8.
Example No. 3. The probability that a part is first grade is 0.4. 150 parts made. Find the probability that there are 68 first-class parts among them.
Example No. 4. The probability of an event occurring in each of the independent trials is p.
Find the probability that the event will occur n times if m tests are carried out.
Provide your answer to three significant figures.
р=0.75, n=87, m=120