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

P(Overbooking) = 0.0898

P(Empty seats) = 0.7461

Step-by-step 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)=(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)=(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)=(9!)/(7!(9-7)!)\cdot0.5^7\cdot(1-0.5)^(9-7)=0.0703 \\ P(8)=(9!)/(8!(9-8)!)\cdot0.5^8\cdot(1-0.5)^(9-8)=0.0176 \\ P(9)=(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