MHB Is Overbooking Probability Calculated Correctly?

[mathmari]

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 like Serena]

Hey mathmari! (Smile)

That looks all correct. (Nod)
The calculation of the suggested solution is wrong.
This becomes clear when we verify that calculation. The final answer does not match.
Their initial formula and final answer are correct, but the calculation in between is not. (Nerd)

 

[mathmari]

That looks all correct. (Nod)

Great! (Happy)

The calculation of the suggested solution is wrong.
This becomes clear when we verify that calculation. The final answer does not match.
Their initial formula and final answer are correct, but the calculation in between is not. (Nerd)

What exactly is wrong at the suggested solution? (Wondering)

 

[I like Serena]

Great! (Happy)

What exactly is wrong at the suggested solution? (Wondering)

It calculates $P(\lnot (52 \text{ appear})) + P(\lnot (51 \text{ appear}))$.
It's a halfway effort to calculate $P(\text{not overbooked})$.
Properly it should be:
$$P(\text{not overbooked}) = P(\lnot (52 \text{ appear}) \land \lnot (51 \text{ appear}))
= P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))$$
Note the second mistake where we have $+$ instead of $\times$.

 

[mathmari]

It calculates $P(\lnot (52 \text{ appear})) + P(\lnot (51 \text{ appear}))$.
It's a halfway effort to calculate $P(\text{not overbooked})$.
Properly it should be:
$$P(\text{not overbooked}) = P(\lnot (52 \text{ appear}) \land \lnot (51 \text{ appear}))
= P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))$$
Note the second mistake where we have $+$ instead of $\times$.

Does it hold that $P(\lnot (52 \text{ appear}) \land \lnot (51 \text{ appear}))= P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))$ because the events $\lnot (52 \text{ appear})$ and $\lnot (51 \text{ appear})$ are independent? (Wondering)

It holds that $P(\lnot (52 \text{ appear}))=1-P(52 \text{ appear})=1-B(52;0,05;0)=1-0,07=0,93$ and $P(\lnot (51 \text{ appear}))=1-P(51 \text{ appear})=1-B(52;0,05;1)=1-0,19=0,81$.

Therefore, we get $P(\text{not overbooked})=P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))=0,93 \times 0,81=0,7533$, right? (Wondering)

 

[I like Serena]

Does it hold that $P(\lnot (52 \text{ appear}) \land \lnot (51 \text{ appear}))= P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))$ because the events $\lnot (52 \text{ appear})$ and $\lnot (51 \text{ appear})$ are independent? (Wondering)

It holds that $P(\lnot (52 \text{ appear}))=1-P(52 \text{ appear})=1-B(52;0,05;0)=1-0,07=0,93$ and $P(\lnot (51 \text{ appear}))=1-P(51 \text{ appear})=1-B(52;0,05;1)=1-0,19=0,81$.

Therefore, we get $P(\text{not overbooked})=P(\lnot (52 \text{ appear})) \times P(\lnot (51 \text{ appear}))=0,93 \times 0,81=0,7533$, right? (Wondering)

Ah, we cannot assume they are independent.
So:
$$P(not\ overbooked)=1-P(overbooked)=1-P(52\ appear)-P(51\ appear)$$

 

[mathmari]

Ah, we cannot assume they are independent.
So:
$$P(not\ overbooked)=1-P(overbooked)=1-P(52\ appear)-P(51\ appear)$$

Ah ok! Thank you! (Handshake)