Question

A small regional carrier accepted 22 reservations for a particular flight with 19 seats. 12 reservations went to regular customers who will arrive for the flight. Each of the remaining passengers will arrive for the flight with a 51% chance, independently of each other. Hint: Use the binomial distribution with p = 0.51.

Find the probability that overbooking occurs.

Find the probability that the flight has empty seats.

Answer 1

There are total 19 seats in the small regional carrier , out of those 19 seats, 12 are confirmed.

So, (19-12)=7 seats are to be filled.

The small regional carrier accepted 22 reservations.

So there are (22-12)=10 people to come.

Let, X be the random variable denoting the number of people who shows up among those 10 people.

The 10 passengers will arrive for the flight with a 51% chance.

So, probability of success for each case, p= 51%= 0.51

X will follow a binomial distribution with p=0.51 and n= 10.

X ~ Bin (10 , 0.51)

a) The probability that overbooking occurs i.e. P(X > 7)

  

The probability that overbooking occurs is 0.0621.

b) The probability that the flight has empty seats i.e. P(X<7)

  

The probability that the flight has empty seats is 0.8112