Question

A small regional carrier accepted 21 reservations for a particular flight with 19 seats. 10 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 44% chance, independently of each other.(Report answers accurate to 4 decimal places.)Find the probability that overbooking occurs. Find the probability that the flight has empty seats.

Answer 1

We have 19 available seats, and 21 reservations.

Of the 21 reservations, 10 are sure, so we have 10 out of the 19 seats that are surely occupied.

Then, we have 9 seats for 11 reservations, each one with 44% chance of being occupied.

We have to calculate the probability that the plane is overbooked. This means that more than 9 of the reservations arrive.

This can be modelled as a binomial distirbution with n = 11 and p = 0.44, representing the 44% chance.

Then, we have to calculate P(x > 9).

This can be calculated as:

$P(x>9)=P(x=10)+P(x=11)$

and each of the terms can be calculated as:

$\begin{gathered} P(x=10)=\dbinom{11}{10}\cdot0.44^{10}\cdot0.56^1=11\cdot0.0003\cdot0.56=0.0017 \\ P(x=11)=\dbinom{11}{11}\cdot0.44^{11}\cdot0.56^0=1\cdot0.0001\cdot1=0.0001 \end{gathered}$

Then:

$P(x>9)=P(x=10)+P(x=11)=0.0017+0.0001=0.0018$

We have a probability of 0.18% of being overbooked (P = 0.0018).

If we want to calculate the probability of having empty seats, we need to calculate P(x<9), meaning that less than 9 of the reservations arrive.

We can express this as:

$P(x$

We have to calculate P(x=9) as we already have calculated the other two terms:

$P(x=9)=\dbinom{11}{9}\cdot0.44^9\cdot0.56^2=55\cdot0.0006\cdot0.3136=0.0107$

Finally, we can calculate:

$\begin{gathered} P(x$

There is a probability of 0.9875 that there is one or more empty seats.

Answer:

There is a probability of 0.0018 of being overbooked.

There is a probability of 0.9875 of having at least one empty seat.