Assume on average $7\%$ of passengers holding a valid ticket don't show up or are too late to their flight. Now, we want to sell more tickets than there are seats. How many tickets can be sold for a plane with $240$ seats to assure with $99\%$ probability that all punctual passengers with a ticket also get a seat?
Suppose $X$ is the number of people who show up. Then $X \sim Bin(n, 0.93)$.
We now want to find $n$ where $\mathbb{P}(X_n \leq 240) \geq 0.99.$
We should probably use the normal approximation to the binomial distribution.
From the Limit Theorem of de Moivre and Laplace we know that:
$\mathbb{P}(X_n \leq 240) \approx \Phi(\frac{240-0.93 \cdot n}{\sqrt{n\cdot 0.93 \cdot 0.07}})$.
How can I now find $n$?