19. On the occasion of the start of winter holidays, all the boys from 8th grade class "V" went to the shooting range. It is known that there are $n$ boys in 8th grade class "V". At the shooting range where the boys went, there are $n$ targets. Each boy randomly selects a target for himself, and some boys could have chosen the same target. After that, all of them simultaneously fire at their targets. It is known that each boy hit his target. A target is considered hit if at least one boy hit it.

a) Find the average number of hit targets.

b) Can the average number of hit targets be less than $\frac{n}{2}$?
# Solution.

a) Let $p$ be the probability that a given target is hit. Then, by symmetry, the expected number of hit targets is $np$. Thus, it remains to find $p$. Note that the probability that a given boy will hit a given target is $\frac{1}{n}$. Therefore, the probability that a given boy will not hit a given target is $\left(1-\frac{1}{n}\right)$. Hence, due to the independence of shots, the probability that no boy will hit a given target is $\left(1-\frac{1}{n}\right)^{n}$. Therefore, the probability that at least one boy will hit a given target is $1-\left(1-\frac{1}{n}\right)^{n}$. This is the probability $p$. Thus, the expected number of hit targets is $p\left(1-\left(1-\frac{1}{n}\right)^{n}\right)$.

Answer: $p\left(1-\left(1-\frac{1}{n}\right)^{n}\right)$.

b) We will show that the expected number of hit targets is always greater than $\frac{n}{2}$. For this, it is sufficient to show that:[^0]

$$
1-\left(1-\frac{1}{n}\right)^{n}>\frac{1}{2} ;\left(1-\frac{1}{n}\right)^{n}\left(1+\frac{1}{n}\right) \cdot \frac{n}{n+1}=1
$$

The first inequality in the chain follows from the well-known Bernoulli's inequality: $(1+a)^{n} \geq 1+n a$ for $n \in \mathbb{N}$ and $a>-1$. This means that the sequence $a_{n}$ is monotonically increasing.

Now we will show that $\lim _{n \rightarrow \infty} a_{n}=\frac{1}{e}$. For this, it is sufficient to show that $\lim _{n \rightarrow \infty} b_{n}=e$. We have:

$$
\begin{gathered}
\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{-n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n-1}\right)^{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n+1}= \\
=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \cdot \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=e \cdot 1=e
\end{gathered}
$$

The last thing to prove is the inequality $\frac{1}{e} < \frac{1}{2}$, which is true since $e \approx 2.7$. The statement is fully proven.

Answer: cannot.

[^0]:    ${ }^{1}$ The sequence $1,1,2,3,5,8, \ldots$, where each term, starting from the third, is the sum of the two preceding ones, is called the Fibonacci sequence, named after the Italian mathematician Leonardo Fibonacci of Pisa, who published a study of this sequence in 1202.

[^1]:    ${ }^{2}$ The irrational number $e=2.718281828459 \ldots$ is the base of the natural logarithm. The graph of the logarithmic function with this base intersects the x-axis at an angle of $45^{\circ}$. The equality $\lim _{n \rightarrow \infty}\left(1-\frac{1}{n}\right)^{n}=\frac{1}{e}$ is one of the remarkable and widely known facts.