The question states that an airline has flights of 8 passengers, but sells at most 10 tickets per flight. The chance that a passenger does not show up for a flight is $0.1$. The chance that $k$ amount of tickets are sold is given as follows:
$$k=6, p=0.3$$
$$k=7, p=0.3$$
$$k=8, p=0.25$$
$$k=9, p=0.1$$
$$k=10, p=0.05$$
It asks to find the chance that the amount of passengers who show up for the flight is greater than the amount of available seats
I started by modeling the people who show up by $X \sim \mathsf{Binom}(n=10,\, p=.9)$. I found $P(X > 8)$ by finding $1 – P(X=6)-P(X=7) – P(X = 8)$, but this answer was not correct. I was thinking to solve this by finding $P(X > 8 | k > 8)$, but I'm not sure if this is correct, nor how to go about doing it. Any help is appreciated.
Best Answer
If $9$ tickets are sold, the conditional probability of overbooking is $$\left(\frac{9}{10}\right)^9$$ If $10$ tickets are sold, the conditional probability of overbooking is $$10\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^9+\left(\frac{9}{10}\right)^{10}$$ hence, the probability of overbooking is \begin{align*} &\left(\frac{1}{10}\right) \left(\frac{9}{10}\right)^9 + \left(\frac{1}{20}\right) \left( 10\left(\frac{1}{10}\right)\left(\frac{9}{10}\right)^9+\left(\frac{9}{10}\right)^{10} \right)\\[4pt] &=\frac{15109399071}{2\cdot 10^{11}}\\[4pt] &\approx.076 \end{align*}