# How many seats should be available such that the probability.. is less than $0.01$?

Two railway companies respectively deploy one train (to get from city X to city Y). In total, $1000$ people randomly choose the train, respectively with probability $\frac{1}{2}$.

How many seats should make one railway company available in the train, such that the probability, that at least one of their passengers needs to stand, is less than $0.01$?

I think for these problem I need to use theorem of De Moivre Laplace.

I call total amount of people $n = 1000$

Probability for choose train is $p = \frac{1}{2}$

But formula for it is strange and I'm not sure how use it correct:

$$\lim_{n \rightarrow \infty}\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}}\leq x\right) = \Phi(x)$$

When insert everything in formula we have

$$\lim_{n \rightarrow \infty} \left(\frac{1000-0.5n}{\sqrt{0.5n(1-0.5)}}\right) = \lim_{n \rightarrow \infty} \left(\frac{1000-0.5n}{\sqrt{0.25n}}\right) = \lim_{n \rightarrow \infty}\left(\frac{1000}{\sqrt{0.25n}} - \sqrt{n}\right) = 0-\infty = -\infty$$

But I see from solution something went wrong :(

It was also hard find the correct formula on internet because in script there is strange thing. How solve this correct because I think similar question can asked in lesson and I want do it correct in test.

• Since you should consider a fixed $n$ CLT resp. Moivre Laplace in this Bernoulli case is not the right choice. – Konstantin Jan 17 '18 at 20:56

$X\sim \text{Bin}(n,p)$. If $k$ are the available seats we are interested in the event $\{X \geq k+1 \}$ and we want $\Bbb P (X \geq k+1) \leq 0.01$
By $$\lim_{n \rightarrow \infty}\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}}\leq x\right) = \Phi(x)$$ we could say $n=1000$ is large enough and we can approximate $$\Bbb P (X \geq k+1) = 1-\Bbb P (X \leq k) =1- \Bbb P \left(\frac{X-1000 \cdot \frac 1 2}{\sqrt{1000 \cdot \frac 1 2(1-\frac 1 2)}}\leq \frac {k - 1000\cdot \frac 1 2}{\sqrt{1000 \cdot \frac 1 2(1-\frac 1 2)}} \right)\\ = 1-\Bbb P \left(\frac{X-np}{\sqrt{np(1-p)}}\leq \frac {k - 500}{\sqrt{250}} \right) \approx 1- \Phi\left(\frac {k - 500}{\sqrt{250}}\right) \overset{!}{\leq} 0.01$$ This leads to $$\Phi\left(\frac {k - 500}{\sqrt{250}}\right) \geq 0.9 \Leftrightarrow \frac {k - 500}{\sqrt{250}}\geq 1.28155 \Leftrightarrow k \geq \sqrt{250}\cdot 1.28255 +500 = 520.278896065$$ The company should make 521 seats available.
Let's denote $X$ the number of people joining company A's railway. Furthermore use $s$ for the number of seats we search for.
You want to achieve: $0.01 > \mathbb{P}(X-s>0) = \mathbb{P}(X>s)$, where $X = \sum_{i_1}^n X_i$ and all $X_i$ admit the given Bernoulli distribution.