Calculating Overbooking Probability for a Plane with Independent Passengers

[TheSodesa]

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?

 

[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.

 

[Ray Vickson]

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?

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.

 

[TheSodesa]

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.

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.

 

[Ray Vickson]

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.

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

 

[TheSodesa]

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

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.