Question

A small regional carrier accepted 23 reservations for a particular flight with 19 seats. 18 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 45% chance, independently of each other.
Find the probability that overbooking occurs.     
Find the probability that the flight has empty seats.    

Answer 1

here this is binomial with parameter n=5 and p=0.45

1)

probability that overbooking occurs =P(2 or more will arrive) =1-P(X<=1)

=1-(P(X=0)+P(X=1))

=1-((5C0)*(0.45)^0*(0.55)^5+(5C1)*(0.45)^1*(0.55)^4)

=1-0.2562 =0.7438

2) probability that the flight has empty seats =P(none of 5 will arrive) =(1-0.45)^5=0.0503