Question

A small regional carrier accepted 23 reservations for a particular flight with 20 seats. 14 reservations went to regular customers who will arrive for the flight. each of the remaining passengers will arrive for the flight with a 50 % chance ,independently of each other. Find the probability that overbooking occurs find the probability that the flight has empty seats

Answer 1

Answer:

P(Overbooking) = 0.0898

P(Empty seats) = 0.7461

Explanation:

The probability that overbooking occurs is the probability that arrives more than 6 passengers from the 9 that remain.

This probability can be calculated as:

$P(x)=\frac{n!}{x!(n-x)!}\cdot p^x\cdot(1-p)^{n-x}$

Where n is the total number of remaining passengers, and p is the probability that a passenger will arrive for the flight. So, the probability that x people arrive is:

$P(x)=\frac{9!}{x!(9-x)!}\cdot0.5^x\cdot(1-0.5)^{9-x}$

So, the probability that arrives 7, 8, or 9 people is:

$\begin{gathered} P(7)=\frac{9!}{7!(9-7)!}\cdot0.5^7\cdot(1-0.5)^{9-7}=0.0703 \\ P(8)=\frac{9!}{8!(9-8)!}\cdot0.5^8\cdot(1-0.5)^{9-8}=0.0176 \\ P(9)=\frac{9!}{9!(9-9)!}\cdot0.5^9\cdot(1-0.5)^{9-9}=0.002 \end{gathered}$

Therefore, the probability that overbooking occurs is:

$\begin{gathered} P(\text{Overbooking)}=P(7)+P(8)+P(9) \\ P(\text{Overbooking)}=0.0898 \end{gathered}$

On the other hand, the probability that the flight has empty seats is the probability that arrives fewer than 6 people for the flight.

So, using the same equation for P(x), we get that the probability that the flight has empty sats is:

$\begin{gathered} P(\text{Empty seats)=P(0) +P(1) + P(2) + P(3) +P(4) + P(5)} \\ P(\text{Empty seats) = 0.7461} \end{gathered}$

Therefore, the answers are:

P(Overbooking) = 0.0898

P(Empty seats) = 0.7461