On the occasion of the start of winter holidays, all the boys from 8 "V" went to the shooting range. It is known that there are $n$ boys in 8 "V". In the shooting range where the boys came, there are $n$ targets. Each boy randomly chooses a target for himself, and some boys could choose 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 $n / 2$?
a) Let $p$ be the probability that a given target is hit. By symmetry, the expected number of hit targets is $n p$.

Note that the probability that a given boy will hit a given target is $1 / n$. Therefore, the probability that a given boy will not hit a given target is $1-\frac{1}{n}$. Hence (due to the independence of shots), the probability that no boy will hit a given target is $(1-1 / n)^{n}$. Thus, the probability that at least one boy will hit a given target is $1-(1-1 / n)^{n}$.

Therefore, the expected number of hit targets is $n\left(1-(1-1 / n)^{n}\right)$.

b) We will show that the expected number of hit targets is greater than $\frac{n}{2}$. For this, it is sufficient to show that $(1-1 / n)^{n}<1 / 2$.

Note that the sequence $a_{n}=\left(1-\frac{1}{n}\right)^{n}=\left(\frac{n-1}{n}\right)^{n}$ is increasing. Indeed, the sequence $b_{n}$ of the reciprocals $\left(\frac{n}{n-1}\right)^{n}$ is decreasing (see the solution to problem $\underline{61394}$).

Moreover, $\lim _{n \rightarrow \infty} b_{n}=\lim _{n \rightarrow \infty}\left(\frac{n+1}{n}\right)^{n+1}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)=e$, therefore,

$\lim _{n \rightarrow \infty} a_{n}=\frac{1}{e}<\frac{1}{2}$ (see problem $\underline{60873}$).

## Answer

a) $n\left(1-(1-1 / n)^{n}\right);$ b) cannot.