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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.

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  • $\begingroup$ Is it a typo that your $P(X>8) = 1 - P(X=6)-P(X=7) - P(X = 8)$? Shouldn't it be $P(X>8) = 1 - P(X=10)-P(X=9)$? $\endgroup$ – Lee David Chung Lin May 22 '18 at 10:48
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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*}

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  • $\begingroup$ Argh, you're right. apologies, my mistake. $\endgroup$ – Paul May 22 '18 at 12:12

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