466,038 views
45 votes
45 votes
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.

User Alex Lowe
by
2.4k points

1 Answer

19 votes
19 votes

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<9)=1-\lbrack P(x=9)+P(x=10)+P(x=11)\rbrack

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<9)=1-\lbrack P(x=9)+P(x=10)+P(x=11)\rbrack \\ P(x<9)=1-\lbrack0.0107+0.0017+0.0001\rbrack \\ P(x<9)=1-0.125 \\ P(x<9)=0.9875 \end{gathered}

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.

User Gil Fink
by
3.1k points