What's the probability that at least one passenger gets no seat?

Question

Two airplane companies respectively use one airplane (to go from country A to country B). There are in total $1000$ people randomly choose the airplane, respectively with probability $\frac{1}{2}$.

The company offers planes with $510$ seats for passengers. What's the probability that at least one passenger gets no seat?

I think theorem of De Moivre Laplace is good to use here:

First of we have $P(X \geq 510 +1) = p$

Then

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

$$P(X \geq 510+1) = 1-P(X \leq 510) = 1-\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}} \leq \frac{510}{\sqrt{1000 \cdot \frac{1}{2}(1-\frac{1}{2})}}\right) \\ = 1-\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}} \leq \frac{510}{5\sqrt{10}}\right) \approx 1-\Phi\left(\frac{510}{5\sqrt{10}}\right) \leq p$$

But from here I don't know how continue and if correct till here? I need read value from table but I don't know how and find it for this high value?

Answer 1

There is a error in your calculation:

$$P(X \geq 510+1) = 1-P(X \leq 510) = 1-\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}} \leq \frac{\color{red}{\mathbf{10}}}{\sqrt{1000 \cdot \frac{1}{2}(1-\frac{1}{2})}}\right) \\ = 1-\mathbb{P}\left(\frac{X-np}{\sqrt{np(1-p)}} \leq \frac{10}{5\sqrt{10}}\right) \approx 1-\Phi\left(\frac{10}{5\sqrt{10}}\right) \approx 1-\Phi\left(0.632 \right) = 1- 0.7357 = 0.2643$$

Answer 2

The exact answer is:

$$P = 1 - \frac{\sum_{j=490}^{510}{1000 \choose j}}{2^{1000}}$$

Pick one plane. There needs to be no fewer than $490$ and no more than $510$ passengers on that plane. The sum takes care of those cases; you want all the other cases.