- #1
mathmari
Gold Member
MHB
- 5,049
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Hey! 
I am looking at the following:
An airline assumes that $5 \%$ of all passengers that have booked for a flight will not appear for departure. They therefore book a flight with $50$ seats by selling $52$ tickets. It is assumed that one passenger independently of the other cancels his flight.Do we model the facts as a probability space as follows?
$\Omega=\{(x_1, \ldots , x_{52}) \mid x_i\in \{0,1\}\}$
$x_i=\left\{\begin{matrix}
1, & \text{ appears } \\
0, & \text{ does not appear }
\end{matrix}\right.$
A="Overbooking"
$A=\{(x_1, \ldots , x_{52})\in \Omega\mid \sum_{i=1}^{52}b_i\geq 51\}$
(Wondering)
In this case we have Bernoulli process of length $52$, since we are just interested whether an event occurs or not, i.e., if a passenger appers or not. For the single experiment, there are only two possible results. The single experiment is repeated $ 52 $ times independently of each other, since $52$ tickets are selled and it holds that a passenger independently of the other cancels his flight. We are also not interested which passenger doesn't appear but how many. From all these it follows that we have a Bernoulli process. Is this correct? (Wondering) We consider "Success" that a passenger doesn't appear. The propability that teh Success occurs, i.e., that a passenger doen;t appear, is always the same and equal to $5\%$. I want to calculate the probability that $k$ passengers don't appear, where $k\in \{0,1, \ldots , 52\}$.
I have done the following:
The probability that $k$ passengers don't appear, i.e. that we have exactly $k$ times Success, is equal to:
\begin{equation*}B(52;0,05;k)=\binom{52}{k}\cdot (0,05)^k\cdot (1-0,05)^{52-k}=\binom{52}{k}\cdot (0,05)^k\cdot (0,95)^{52-k}\end{equation*}
Is that correct? (Wondering) Then I want to calculate the probability that a passenger cannot take the flight although he has a ticket.
Does this mean that exacttly one passenger cannot take the flight, although it has a ticket or at least one passenger? (Wondering)
If it is meant exactly one, we want to calculate the probability that $51$ appear, i.e. that one passenger doesn't appear.
This probability is equal to
\begin{align*}B(52;0,05;1)&=\binom{52}{1}\cdot (0,05)^1\cdot (1-0,05)^{52-1}=\frac{52!}{1!\cdot (52-1)!}\cdot 0,05\cdot (0,95)^{51}=\frac{51!\cdot 52}{1!\cdot 51!}\cdot 0,05\cdot (0,95)^{51} \\ & = 52\cdot 0,05\cdot (0,95)^{51}=2,6\cdot (0,95)^{51}\approx 0,19=19\%\end{align*}
If it meant at least one, we want to calculate the probability that $52$ or $51$ passengers appear, i.e., that $0$ or $1$ passenger doesn't appear.
This probability is equal to
\begin{align*}B(52;0,05;0)+B(52;0,05;1)&=\binom{52}{0}\cdot (0,05)^0\cdot (1-0,05)^{52-0}+\binom{52}{1}\cdot (0,05)^1\cdot (1-0,05)^{52-1} \\ & =\frac{52!}{0!\cdot (52-0)!}\cdot 1\cdot (0,95)^{52}+\frac{52!}{1!\cdot (52-1)!}\cdot 0,05\cdot (0,95)^{51} \\ & =\frac{52!}{1\cdot 52!}\cdot (0,95)^{52}+\frac{51!\cdot 52}{1!\cdot 51!}\cdot 0,05\cdot (0,95)^{51} \\ & =1\cdot (0,95)^{52}+ 52\cdot 0,05\cdot (0,95)^{51} \\ & =(0,95)^{52}+ 2,6\cdot (0,95)^{51} \\ & \approx 0,07+ 0,19 \\ & =0,26=26\% \end{align*}
Are both cases correct? (Wondering) At a suggested solution there is the following:
\begin{align*}P("\text{Overbooking}")&=P(52 \text{ apprear })\cup P(51 \text{ apprear }) \\ & = 1-\binom{52}{0}\cdot (0,05)^0\cdot (0,95)^{52}+1-\binom{52}{1}\cdot (0,05)^1\cdot (0,95)^{51} \\ & = 1-(1\cdot 1\cdot 0,95^{52})+1-(52\cdot 0,05\cdot (0,95)^{51}) \\ & = 0,259496\end{align*}
Is this the same as I did? Why do we take here 1-B(52;0,05;0) and 1-B(52;0,05;1) ? (Wondering)
I am looking at the following:
An airline assumes that $5 \%$ of all passengers that have booked for a flight will not appear for departure. They therefore book a flight with $50$ seats by selling $52$ tickets. It is assumed that one passenger independently of the other cancels his flight.Do we model the facts as a probability space as follows?
$\Omega=\{(x_1, \ldots , x_{52}) \mid x_i\in \{0,1\}\}$
$x_i=\left\{\begin{matrix}
1, & \text{ appears } \\
0, & \text{ does not appear }
\end{matrix}\right.$
A="Overbooking"
$A=\{(x_1, \ldots , x_{52})\in \Omega\mid \sum_{i=1}^{52}b_i\geq 51\}$
(Wondering)
In this case we have Bernoulli process of length $52$, since we are just interested whether an event occurs or not, i.e., if a passenger appers or not. For the single experiment, there are only two possible results. The single experiment is repeated $ 52 $ times independently of each other, since $52$ tickets are selled and it holds that a passenger independently of the other cancels his flight. We are also not interested which passenger doesn't appear but how many. From all these it follows that we have a Bernoulli process. Is this correct? (Wondering) We consider "Success" that a passenger doesn't appear. The propability that teh Success occurs, i.e., that a passenger doen;t appear, is always the same and equal to $5\%$. I want to calculate the probability that $k$ passengers don't appear, where $k\in \{0,1, \ldots , 52\}$.
I have done the following:
The probability that $k$ passengers don't appear, i.e. that we have exactly $k$ times Success, is equal to:
\begin{equation*}B(52;0,05;k)=\binom{52}{k}\cdot (0,05)^k\cdot (1-0,05)^{52-k}=\binom{52}{k}\cdot (0,05)^k\cdot (0,95)^{52-k}\end{equation*}
Is that correct? (Wondering) Then I want to calculate the probability that a passenger cannot take the flight although he has a ticket.
Does this mean that exacttly one passenger cannot take the flight, although it has a ticket or at least one passenger? (Wondering)
If it is meant exactly one, we want to calculate the probability that $51$ appear, i.e. that one passenger doesn't appear.
This probability is equal to
\begin{align*}B(52;0,05;1)&=\binom{52}{1}\cdot (0,05)^1\cdot (1-0,05)^{52-1}=\frac{52!}{1!\cdot (52-1)!}\cdot 0,05\cdot (0,95)^{51}=\frac{51!\cdot 52}{1!\cdot 51!}\cdot 0,05\cdot (0,95)^{51} \\ & = 52\cdot 0,05\cdot (0,95)^{51}=2,6\cdot (0,95)^{51}\approx 0,19=19\%\end{align*}
If it meant at least one, we want to calculate the probability that $52$ or $51$ passengers appear, i.e., that $0$ or $1$ passenger doesn't appear.
This probability is equal to
\begin{align*}B(52;0,05;0)+B(52;0,05;1)&=\binom{52}{0}\cdot (0,05)^0\cdot (1-0,05)^{52-0}+\binom{52}{1}\cdot (0,05)^1\cdot (1-0,05)^{52-1} \\ & =\frac{52!}{0!\cdot (52-0)!}\cdot 1\cdot (0,95)^{52}+\frac{52!}{1!\cdot (52-1)!}\cdot 0,05\cdot (0,95)^{51} \\ & =\frac{52!}{1\cdot 52!}\cdot (0,95)^{52}+\frac{51!\cdot 52}{1!\cdot 51!}\cdot 0,05\cdot (0,95)^{51} \\ & =1\cdot (0,95)^{52}+ 52\cdot 0,05\cdot (0,95)^{51} \\ & =(0,95)^{52}+ 2,6\cdot (0,95)^{51} \\ & \approx 0,07+ 0,19 \\ & =0,26=26\% \end{align*}
Are both cases correct? (Wondering) At a suggested solution there is the following:
\begin{align*}P("\text{Overbooking}")&=P(52 \text{ apprear })\cup P(51 \text{ apprear }) \\ & = 1-\binom{52}{0}\cdot (0,05)^0\cdot (0,95)^{52}+1-\binom{52}{1}\cdot (0,05)^1\cdot (0,95)^{51} \\ & = 1-(1\cdot 1\cdot 0,95^{52})+1-(52\cdot 0,05\cdot (0,95)^{51}) \\ & = 0,259496\end{align*}
Is this the same as I did? Why do we take here 1-B(52;0,05;0) and 1-B(52;0,05;1) ? (Wondering)