Question

Problem 5 Say that you operate an airport shuttle service from Will Rogers World Airport in Oklahoma City to Norman. The shuttle can accommodate up to four passengers on any one trip from the airport to Norman. Your service will accept a maximum of six reservations for a trip, and a passenger must make a reservation to take the shuttle. Based on trip history, 20% of those making reservations do not appear for the trip. Assume independence in the following as appropriate. a. If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip? b. If six reservations are made, what is the expected number of available spaces when the shuttle departs? C. Assume the following probability distribution for the number of reservations made for any given trip. And assume a random variable M denotes the number of passengers on a randomly selected trip. Calculate the probability distribution function for M. 3 4 5 Number of reservations Probability 6 0.4 0.1 0.2 0.3

Answer 1

5 a) Number of seats available =4

If more than 4 passengers turn up , then some people cannot be accommodated

Let X be the number of passengers appear for the trip out of 6 reservations

If X >4 , then there is at least one passenger who cannot be accommodated

X follow Binomial with n=6 , p= 0.80 ( as trials are finite and independent and probability of appearance is same in each trial)

To find P(X>4)

P(X>4) =P(X=5) +P(X=6)

  

= 0.3932 +0.2621

=0.6554

Probability that at least one passenger cannot be accommodated = 0.6554

b)Total number of seats available =4

Let X be the number of passengers appear for the trip out of 6 reservations

X follow Binomial with n=6 , p= 0.80

Expected number of passengers appear for the trip out of 6 reservations = E(X) =np = 6*0.80=4.8

Expected number of available spaces = 4 - 4.8 = - 0.8

c) Let number of reservations = Y

with P(Y)

Let Z be the number of passengers appear for the trip out of Y reservations

Z follow Binomial with probability n=y , p= 0.8

Given , M = number of passengers on any given trip

Probability of M is the joint probability of number of reservation and number of passengers appear

P(M) = P(ZI Y) *P(Y)

  

M can take values 0,1,2,3,4,5,6

Y can take values 3,4,5,6

The probability distribution is given below

Y	P(Y)	Z	P(Z)	P(M)
3	0.1	0	0.0080	0.0008
3	0.1	1	0.0960	0.0096
3	0.1	2	0.3840	0.0384
3	0.1	3	0.5120	0.0512
4	0.2	0	0.0016	0.0003
4	0.2	1	0.0256	0.0051
4	0.2	2	0.1536	0.0307
4	0.2	3	0.4096	0.0819
4	0.2	4	0.4096	0.0819
5	0.3	0	0.0003	0.0001
5	0.3	1	0.0064	0.0019
5	0.3	2	0.0512	0.0154
5	0.3	3	0.2048	0.0614
5	0.3	4	0.4096	0.1229
5	0.3	5	0.3277	0.0983
6	0.4	0	0.0001	0.0000
6	0.4	1	0.0015	0.0006
6	0.4	2	0.0154	0.0061
6	0.4	3	0.0819	0.0328
6	0.4	4	0.2458	0.0983
6	0.4	5	0.3932	0.1573
6	0.4	6	0.2621	0.1049