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TheSodesa
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Homework Statement
A travel agency knows from experience, that 5% of passengers who have booked a ticket will not show up for the flight. Therefore the company sells 260 tickets for a plane that can only take 255 passengers.
What is the probability that all of the passengers arriving for the flight get a seat, assuming they arrive independently of each other?
Correct answer: ##0.997##
Homework Equations
The binomial distribution mass function:
\begin{equation}
f(x) = {n \choose k}p^{k}(1-p)^{n-k}
\end{equation}
The Attempt at a Solution
The passengers arrive independently of each other, and if ##x## passengers show up, ##n-x## passengers don't. Therefore we can use the binomial distribution to model the situation.
Now the company overbooks by 5, therefore there are 5 seats that may have two people, leaving the other without a seat. Therefore ##n=10##. Every person will have a seat if out of these 10 at most 5 show up.
Let ##X = \text{ the number of people that show up out of the 10}## and ##A = \text{ a person will show up}##. Then ##p = P(A) = 0.95 \Rightarrow (1-p) = P(\overline{A}) = 0.05##
, and
\begin{align*}
P(X \leq 5)
&= f(0) + f(1) + f(2) + f(3) + f(4) + f(5)\\
&= {10 \choose 0}(0.95)^{0}(0.05)^{10} + {10 \choose 1}(0.95)^{1}(0.05)^{9}\\
&+ {10 \choose 2}(0.95)^{2}(0.05)^{8} + {10 \choose 3}(0.95)^{3}(0.05)^{7}\\
&+ {10 \choose 4}(0.95)^{4}(0.05)^{6} + {10 \choose 5}(0.95)^{5}(0.05)^{5}\\
&= 0.0000636898314453125
\end{align*}
Not quite what I was expecting. What am I missing? Do I need to take into account the other passengers on the plane somehow?
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