Question

4. An airline sells 338 tickets for a flight to Manila which has 335 seats. It is estimated that 98% of all ticketed passengers show up for the flight. Find the probability that the flight will depart with (at least one) empty seats? (10)

Answer 1

Solution:

Total number of Tickets = 338

Total number of seats = 335

Number of extra Tickets = 338 - 335 = 3

Probability of passenger show up for the flight = 98% = 0.98x

Let X denote the number of passenger show up for the flight

Then X Bin(338,0.98)

PDF of X is given as P(X = x) = px (1 - p)n-x when x = 0,1,2,....,n

= 0 Otherwise

To depart the flight with at least one empty seats, number of passenger show up for the flight is at most 334.

Hence the probability that the flight will depart with (at least one) empty seats = P(X 334)

Using the normal approximation to the given binomial distribution,

X follow normal distribution with mean = np = 338*0.98 = 331.24 & variance = np(1-p) = 338*0.98*0.02 = 6.6248.

Therefore P(X 334) = P(X < 334.5) Using Continuity Correction

=

= P(Z < 1.27)

= 0.89796

Hence he probability that the flight will depart with at least one empty seats is 0.89796.