# Overbooked plane probability

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1. Sep 24, 2016

### TheSodesa

1. The problem statement, all variables and given/known data
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: $0.997$
2. Relevant equations
The binomial distribution mass function:

f(x) = {n \choose k}p^{k}(1-p)^{n-k}

3. 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?

Last edited: Sep 24, 2016
2. Sep 24, 2016

### TheSodesa

I solved it.

I did have to take all of the other passengers into account. If $X = \text{ the number of passengers on the plane}$ and $0 \leq X \leq 255$, all passengers will get a seat. You could brute force the calculation by writing a script with a for-loop that sums $f(0), f(1), ... , f(255)$ togethter. The easier way, if you need to do this by hand, is to calculate the probability for $256 \leq X \leq 260$ and take its complement.

I'd already written the Matlab script, so I chose that route.

Last edited: Sep 24, 2016
3. Sep 24, 2016

### Ray Vickson

Your answer is way too big! If the number who do not show up is $X$, then $X \sim \text{Bin}(260,0.05)$, because any of the 260 ticketed passengers could be one of those who do not show.

4. Sep 24, 2016

### TheSodesa

Yeah, I kind of realized that just now. I would have had to calculate 256 probabilities by hand...

I replied to my own post above with the solution.

5. Sep 24, 2016

### Ray Vickson

You would not need to compute 256 probabilities by hand or any other way. The probability of not having enough seats is $P(X \leq 4)$, where $X \sim \text{Bin}(260,0.05)$.

6. Sep 24, 2016

### TheSodesa

I do see what you mean.

Let $X \sim \text{Bin}(260,0.05)$. Then

\begin{align*}
P(X \leq 4)
&= f(0)+f(1)+f(2)+f(3)+f(4)\\
&= {260 \choose 0}(0.05)^{0} (0.095)^{260} + {260 \choose 1}(0.05)^{1} (0.095)^{259}\\
&+ {260 \choose 2}(0.05)^{2} (0.095)^{258} + {260 \choose 3}(0.05)^{3} (0.095)^{257}\\
&+ {260 \choose 4}(0.05)^{4} (0.095)^{256}\\
&= 0.00316142309\\
&\approx 0.003
\end{align*}

Now the correct answer listed in the assignment paper is the complement of this: $0.997$.
Thanks for the little correction. I see my mistake now.