A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.
1. Find the probability that overbooking occurs
2. Find the probability that the flight has empty seats
+26387 A small regional carrier accepted 21 reservations for a particular flight with 18 seats. 11 reservations went to regular customers who will arrive for the flight. Each of the raining passengers will arrive for the flight with a 45% chance, independently of each other.
1. Find the probability that overbooking occurs
2. Find the probability that the flight has empty seats
My attempt:
1. Find the probability that overbooking occurs
overbooking mean that more than 7 of the other 10 will arrive for the flight:
probability of 8 arriving is \(10C_8*0.45^8*0.55^2 = 45*0.45^8*0.55^2= 0.02288958944\)
probability of 9 arriving is \(10C_9*0.45^9*0.55^1 = 10*0.45^9*0.55^1= 0.00416174353\)
probability of 10 arriving is \(10C_{10}*0.45^{10}*0.55^0 = 1*0.45^{10}*1= 0.00034050629\)
That sum is 0.02739183926.
The probability that overbooking occurs is \(\mathbf{0.0274}\)
2. Find the probability that the flight has empty seats
Just 7, all seats are full. The probability is \(10C_7*0.45^7*0.55^3 = 120*0.45^7*0.55^3= 0.07460310632\)
this plus the 0.02739183926 is 0.10199494558.
That is the probability no empty seats will occur.
The probalbility hat empty seats will occur is the compliment \((1-0.10199494558) = 0.89800505442\)
The probability that the flight has empty seats is \(\mathbf{0.8980}\)
