19. How many airplanes? (11th grade, 8 points) The work of the Absent-Minded Scientist involves long business trips, and therefore he often flies on the same airline. This airline has many identical airplanes, and all of them have names. Since the Scientist does not fly every day or even every week, it can be assumed that each time he gets a random airplane. Out of curiosity and habit, the Absent-Minded Scientist records the name of the airplane he flies on each time. In his fifteenth flight, the Scientist found himself on board an airplane proudly named "Siméon Denis Poisson." After takeoff, the Scientist took out his notebook to record the name of the airplane and discovered that he had already flown on "Poisson" once before, and there were no other repetitions before that. Estimate the number of airplanes in the airline.
Solution. Consider the random variable $X$ "the ordinal number of the flight when the Scientist first gets a plane he has flown before". We will make a point estimate of the number of planes using the method of moments. For this, we need to solve the equation $\mathrm{E} X=15$ at least approximately.

Let's find $\mathrm{E} X$. Denote $n$ as the number of planes in the airline. Consider the event $A_{j}$ "there were no repetitions up to and including the $j$-th flight" $(j=1, \ldots, n)$. Let $I_{j}$ be the indicator of the event $A_{j}$. Then

$$
X=1+I_{1}+I_{2}+\ldots+I_{n}
$$

Let's find the expected values of the indicators.

$$
\mathrm{EI}_{j}=\mathrm{P}\left(A_{j}\right)=\frac{n(n-1)(n-2) \ldots(n-j+1)}{n^{j}}=\frac{n!}{(n-j)!\cdot n^{j}}
$$

Therefore,

$$
\mathrm{E} X=\mathrm{E}\left(1+I_{1}+I_{2}+\ldots+I_{n}\right)=\sum_{j=0}^{n} \frac{n!}{(n-j)!\cdot n^{j}}=\frac{n!}{n^{n}} \sum_{j=0}^{n} \frac{n^{n-j}}{(n-j)!}=\frac{n!}{n^{n}} \sum_{k=0}^{n} \frac{n^{k}}{k!}
$$

In the last transformation, we made the substitution $k=n-j$.

Consider the resulting sum separately:

$$
\sum_{k=0}^{n} \frac{n^{k}}{k!}=e^{n} \sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=e^{n} F(n)
$$

where $F(x)$ is the cumulative distribution function of the Poisson distribution with parameter $n$. The median of the Poisson distribution with an integer parameter $n$ is $n$. This means that

$$
F(n-1) \leq \frac{1}{2} \leq F(n)
$$

therefore

$$
F_{P}(n)=\sum_{k=0}^{n} \frac{n^{k} e^{-n}}{k!}=\frac{1}{2}+\theta \frac{n^{n} e^{-n}}{n!}
$$

where $0 \leq \theta \leq 1$. Setting $\theta=1 / 2$ for definiteness, we get:

$$
\mathrm{E} X=\frac{n!e^{n}}{n^{n}} F_{P}(n) \approx \frac{n!e^{n}}{n^{n}}\left(\frac{1}{2}+\frac{n^{n} e^{-n}}{2 n!}\right)=\frac{n!e^{n}}{2 n^{n}}+\frac{1}{2}
$$

Using Stirling's formula

$$
\mathrm{E} X \approx \frac{\sqrt{2 \pi n} \cdot n^{n} e^{n}}{2 n^{n} e^{n}}+\frac{1}{2}=\sqrt{\frac{\pi n}{2}}+\frac{1}{2}
$$

To find the estimate $x=\hat{n}$, solve the equation $\sqrt{\frac{\pi x}{2}}+\frac{1}{2}=15$. With rounding to the nearest integer, we get: $x \approx 134$.

Answer: approximately 134 (point estimate using the method of moments).

![](https://cdn.mathpix.com/cropped/2024_05_06_11f8b78b86004d31b3e4g-17.jpg?height=2420&width=1113&top_left_y=206&top_left_x=403)

[^0]:    ${ }^{1}$ English name - Stem-and-Leaf plot.

[^1]:    ${ }^{2}$ The mode is the number that appears in the set more times than any other.

[^2]:    ${ }^{3}$ It is believed that this problem was invented by Academician Andrei Dmitrievich Sakharov. I. F. Ginzburg. Academician A.D. Sakharov. Scientific works. Collection. - M.: AOZT "Izdatel'stvo TsentrKom", 1995.

[^3]:    ${ }^{4}$ The number $e$ is also called Euler's number.

[^4]:    ${ }^{5}$ The median of a random variable $X$ is a number $m$ such that $\mathrm{P}(X \leq m) \geq 0.5$ and $\mathrm{P}(X \geq m) \geq 0.5$. If the random variable has a finite density function $y=f(x)$, then the line $x=m$ divides the figure bounded by the x-axis and the density graph into two figures, each with an area of 0.5.