A probability problem on limit theorem (!) Each of the 300 workers of a factory takes his lunch in one of the three competing restaurants (equally likely, so with probability $1/3$). How many seats should each restaurants have so that, on average, at most one in 20 customers will remain unseated?
How can I approach this?

It is not clear to me why it is given as an excercise to the limit theorems chapter. 
 A: Hint: Let the customers be labelled $1$ to $300$. We assume that each worker chooses which restaurant to go to by flipping a fair $3$-sided coin.
Let $X_i=1$ if $i$ gets seated, and $0$ if she doesn't. Let $Y=\sum X_i$. We interpret the question as asking for the capacity $c$ that will ensure that $E(Y)\le \frac{300}{20}$.
The expectation of a sum is the sum of the expectations, so we want $300E(X_1)\le \frac{300}{20}$, and therefore $\Pr(X\gt c)\le 0.05$.
The number $W$ of customers other than $1$ going to the restaurant chosen by Customer $1$ has mean $\frac{299}{3}$. (One is probably expected to use $100$.) The random variable $W$ has variance $(299)(1/3)(2/3)$, and roughly normal distribution. We want to choose $c$ so that $c$ is at least $1.96$ standard deviation units up from $\frac{299}{3}$.
Remark: It is probably expected that one will use mean $100$, variance $300(1/3)(2/3)$ and find $c$ so that with probability $\le 0.05$, a normal with mean $100$ and variance $300(1/3)(2/3)$ is $\ge c+1$. That will not make much difference.   
A: That this is an exercise on the limit theorems chapter is actually a clue. Let $n=300$ denote the number of workers, $s$ the number of seats in each restaurant and $X$ the number of workers trying to get a seat in restaurant 1. Then $X-s$ customers remain unseated at restaurant 1 if $X\gt s$, and none if $X\leqslant s$ hence the mean number of customers who remain unseated at restaurant 1 is $E[X-s;X\gt s]$. By symmetry, the mean number of customers who remain unseated at any of the three restaurants is $m=3\cdot E[X-s;X\gt s]$ and one asks for $s$ such that $m\leqslant20$.
The distribution of $X$ is binomial $(n,\frac13)$ hence its mean is $\frac13n$ and its variance is $\frac29n$. If $n$ is large enough for the central limit theorem to apply, then 
$$
X=\tfrac13n+\tfrac13\sqrt{2n}\cdot Y,
$$ 
where $Y$ is roughly standard normal. Defining 
$$
s=\tfrac13n+\tfrac13\sqrt{2n}\cdot r,
$$ one gets $m=\sqrt{2n}\cdot E[Y-r;Y\gt r]$. Since $Y$ is approximately standard normal, for every $r$, 
$$
E[Y-r;Y\gt r]\approx\varphi(r)-r(1-\Phi(r)),
$$ 
hence one solves numerically $\varphi(r)-r(1-\Phi(r))=\frac{20}{\sqrt{2n}}=\frac13\sqrt6$. The root is $r^*\approx-0.665$, which yields $s^*=\frac13n+\frac13\sqrt{2n}\cdot r^*\approx100-\frac{20}9\sqrt6\approx94.56$. 
Finally, if the gaussian approximation applies, each restaurant should have at least $95$ seats.
