An airport limousine can accommodate up to four passengers on any one trip. The

Question

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 reservation. From previous records, 25% of all those making reservations do not appear for the trip. Answer the following questions, assuming independence wherever appropriate. (Round your answers to three decimal places.) (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 places when the limousine departs? places (c) Suppose the probability distribution of the number of reservations made is given in the accompanying table. Number of reservations 3,4,5,6 Probability,0.14 ,0.23 ,0.35,0.28 Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

Explanation / Answer

25% of the people having reservations do not turn up for the trip. So the probability of a person not turning up = 0.25

We can model the problem as a binominal distribution with n=6, p=0.75, q=0.25

Then we have the following table of probabilities for various values of number of persons not turning up:

(a) With 6 reservations, the probability that at least one individual with a reservation cannot be accommodated on the trip = probability that 5 persons turn up + probability that 6 persons turn up = 0.1780 + 0.3560 = 0.5340

(b) Places are available if less than 4 persons turn up. We have 1,2,3 and 4 places available when 3,4,5 and 6 persons do not turn up respectively with the probabilities mentioned in the above table.

Expected value of available places = 1 x 0.1318 + 2 x 0.033 + 3 x 0.0044 + 4 x 0.0002 = 0.2119

(c) We have the following table ofprobabilities:

X=0, Probability = 0.14 x 0.0156 + 0.23 x 0.0039 + 0.35 x 0.001 + 0.28 x 0.0002 = 0.0035

X=1, Probability = 0.14 x  0.1406 + 0.23 x 0.0469 + 0.35 x 0.0146 + 0.28 x 0.0044 = 0.0368

X=2, Probability = 0.14 x  0.4219 + 0.23 x 0.2109 + 0.35 x 0.0879 + 0.28 x 0.033 = 0.1476

X=3, Probability = 0.14 x  0.4219 + 0.23 x 0.4219 + 0.35 x 0.2637 + 0.28 x 0.1318 = 0.2853

X=4, Probability = 0.23 x 0.3164 + 0.35 x (2373+0.3955) + 0.28 x (0.178+0.356+0.2966) = 0.5268

k, persons not turning up Probability 0 0.1780 1 0.3560 2 0.2966 3 0.1318 4 0.0330 5 0.0044 6 0.0002

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.