Here's the problem:
Air travel is often overbooked by tour operators, because a booked trip will not be taken with a probability of 10%. In this task we assume that the decisions of the individual bookers to take the trip or not are independent of each other.
a) A tour operator books 56 flights with an airline and sells 60 flights to his customers. What is the probability that the seats are sufficient?
b) A tour operator wants to sell 60 seats to his customers. How many seats in the aircraft does he have to book so that there is at least a 95% probability that they will be enough?
c) An organizer books 43 seats on an airline. How many seats can he sell to his customers so that there is a probability of at least 95% that the seats are enough?
My thoughts:
a) $$p(x \leq 4)= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) + p(x = 4) \\
= \sum\limits_{x=0}^{4} \binom{n}{k} {0.9}^{n-k}{0.1}^{k} \\
=0.271
$$
b)
We are looking for a k for which applies:
$$1-p(x \leq k) \geq 0.95 \Rightarrow 1-p(x \leq k) \geq 0.05 $$
$$\Rightarrow \\
p(x=0) = 0.0018 \\
p(x \le1 1) = 0.0138 \\
p(x <= 2) = 0,053 \\
\Rightarrow
$$
the organiser must book at least 59 places, so that there is at least a 95% probability that the places will be sufficient.
c)
The probability that a passenger will appear is 0.9.
In order for at least 43 passengers to appear with a probability of at least 0.95, the probability of 0 to 42 passengers coming must fall to a maximum of 0.05.
So we have k=42 and p=0.9 (because in the cumulative distribution the sum of 0 to 42 is then calculated) and experiment with n until it fits.
$\binom{n}{42} 0.9^n 0.1^{n – 42} \leq 0.05$
However, I am confronted with the problem of making sure whether the solution for a) and b) is correct or not and I am not sure how to solve the subtask c).Can someone please explain it to me?
I would also appreciate other calculation possibilities.
Thanks so much 🙂
Best Answer
For a) there are not enough seats if three or less people could not make the trip $$p(x \leq 3)= p(x = 0) + p(x = 1) + p(x = 2) + p(x = 3) = \sum\limits_{k=0}^{3} \binom{n}{k} {0.9}^{n-k}{0.1}^{k} \approx 0.1374$$ Thus, the probability to have enough seats is $1-p(x \leq 3)=0.8626$
For b) your reasoning and result look good to me.
For c), look at the result from b). To achieve at least $95\%$ probability, the difference between bookings and seats cannot be greater than $1$ (considering also that we have fewer seats in our case). Thus, the organizer cannot sell more than $44$ tickets. However, $1-p(x \le 1) \approx 0.943$ which is less than required $95\%$. This tell us that the only way to guarantee sufficient seats is to sell exactly $43$ tickets (or less).