An airport limousine can accommodate up to four passengers o

An airport limousine can accommodate up to four passengers on any one trip. The company will accept a maximum of six reservations for a trip, and a passenger must have a reservations. From experience, 80% of all those making reservations will not show up. In the following questions assume independence wherever appropriate.

(1)The random variable Y denotes the number of passengers with reservations who show up. If 3 reservations are made, then Y has a Binomial distribution. What are the parameters n and p?

(2)The random variable Y denotes the number of passengers with reservations who show up. If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated?

(3)If six reservations are made what is the expected number of available spaces when the limousine departs?

(4)

Suppose the probability distribution of Z= the number of reservations made is

The random variable X denotes the number of passengers on a randomly selected trip. What is the probability P( X = 0)?

Solution

(1)The random variable Y denotes the number of passengers with reservations who show up.

n = Number of reservations made = 3

From experience, 80% of all those making reservations will not show up.

So p = Probability that passenger made reservation and who show up = 1 - 0.80 = 0.20

So Y follows binomial distribution with n= 3 and p=0.2.

(2)The random variable Y denotes the number of passengers with reservations who show up.

Here n =6 and p = 0.2

At least one individual with a reservation cannot be accommodated on the trip is equivalent to 5 or 6 individuals turning up. We now have to find, P[Y = 5] + P[Y = 6] = ⁶C₅ * 0.2⁵ *(1-0.2)^6-5 + ⁶C₆ * 0.2⁶*(1-0.2)^6-6

= 0.0015 + 0.0001

= 0.0016

( 3)

The available no. of places Z = max[4 ? Y, 0]. Therefore, P[Z = k] = P[Y = 4 ? k], k = 1, . . . , 4, and P[Z = 0] = P[Y = 4] + P[Y = 5] + P[Y = 6]. The expected value is therefore, 1*P[Y = 3] + 2*P[Y = 2] + 3*P[Y = 1] + 4*P[Y = 0]

= 1*0.0819 + 2*0.2458 + 3*0.3932 + 4*0.2621

= 2.8017

(4)

Define the r.v. R to be the no. of reservations. First consider the case, X = 0, this can happen even if R = 3, 4, 5, 6.

Therefore P[X = 0] = P[X = 0, R = 3]+P[X = 0, R = 4]+P[X = 0, R = 5]+P[X = 0, R = 6]

= 0.15*(³C₀ * 0.2⁰*(1-0.2)^3-0) +  0.2*(⁴C₀ * 0.2⁰*(1-0.2)^4-0?) +  0.35*(⁵C₀ * 0.2⁰*(1-0.2)^5-0?) +  0.3*(⁶C₀ * 0.2⁰*(1-0.2)^6-0?)

P[X = 0]= 0.3521