"A small regional carrier accepted 14 reservations for a particular flight with 12 seats. 8 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 56% chance, independently of each other."

a. Find the probability that overbooking occurs.
b. Find the probability that the flight has empty seats.

If 8 of the 14 reservations are regulars customers, there are just 6 reservations with the possibility to arrive or not for the flight. Additionally, there are just 4 free seats for the 6 reservations.

Now, the probability that x customers arrive for the flight follows a Binomial distribution, so it is calculated as:

[tex]p(x)=\frac{n!}{x!(n-x)!}*p^{x}*(1-p)^{n-x}[/tex]

Where p is probability that a customer arrive, so it is equal to 0.56 and n is the total number of customers with the possibility to arrive or not, in this case it is 6. Replacing values we get:

[tex]p(x)=\frac{6!}{x!(6-x)!}*0.56^{x}*(1-0.56)^{6-x}[/tex]

So, the probability that a overbooking occurs is the probability that 5 or more customers from the 6 arrive for the flight and the probability that the flight has empty seats is the probability that 3 or less customers from the 6 arrive for the flight.

Then, the probability that overbooking occurs is:

[tex]p(x\geq5 )=p(5) + p(6)\\p(x\geq5 )=\frac{6!}{5!(6-5)!}*0.56^{5}*(1-0.56)^{6-5}+\frac{6!}{6!(6-6)!}*0.56^{6}*(1-0.56)^{6-6}\\p(x\geq5 )=0.1454+0.0308\\p(x\geq5 )=0.1762[/tex]

At the same way, the probability that the flight has empty seats is:

[tex]p(x\leq 3)=p(3)+p(2)+p(1)+p(0)\\p(x\leq 3)=0.2992+0.1763+0.0554+0.0073\\p(x\leq 3)=0.5381[/tex]