I want to solve this problem, without using calculator or table:
A passenger airline company has found from past experience that 20% of the customers who buy tickets for a flight do not show up for the journey. The company wishes to ensure that a particular flight is at least 95% full with a probability of 0.9. How many tickets should it sell if the capacity of the flight is 300?
My effort:
Let n be the number of tickets sold, and X be the number of passengers showing up. Then $P(X) = \binom {n}{x} p^x(1-p)^{n-x} $ where $p = \frac{4}{5}$.
minimum number of seats to be filled up is $ 300 * \frac{95}{100} = 285$ .
To solve for n, we need to solve the equation $P(X\ge285) = \frac{9}{10} $ $$\Rightarrow \frac{9}{10} = 1 - P(X<285) = 1- \frac{1}{5^n}\left(\binom{n}{1}4+\binom{n}{2}4^2+ \cdots + \binom{n}{284}4^{284}\right) $$ Now this expression on right seems very similar to binomial expansion, but it has an unknown n. Is there a way to easily solve this for n? Or if any better way to resolve the problem, please suggest.