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, 35% of all those making reservations do not appear for the trip.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?C) Suppose the probability distribution of the number of reservations made is given in the accompanying table.Number of reservations 3 4 5 6Probability 0.09 0.25 0.32 0.34Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

Answer:

a) 0.3191 = 31.91% probability that at least one individual with a reservation cannot be accommodated on the trip.

b) The expected number of available places when the limousine departs is 0.1.

c) [tex]P(X = 3) = 0.09[/tex]

[tex]P(X = 4) = 0.25[/tex]

[tex]P(X = 5) = 0.32[/tex]

[tex]P(X = 6) = 0.34[/tex]

Step-by-step explanation:

For each reservation, there are only two possible outcomes. Either the person appears for the trip, or the person does not. The probability of a person appearing for the trip is independent of any other person. This means that the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

In which [tex]C_{n,x}[/tex] is the number of different combinations of x objects from a set of n elements, given by the following formula.

[tex]C_{n,x} = \frac{n!}{x!(n-x)!}[/tex]

And p is the probability of X happening.

The expected value of the binomial distribution is given by:

[tex]E(X) = np[/tex]

The company will accept a maximum of six reservations for a trip, and a passenger must have a reservation.

This means that [tex]n = 6[/tex]

35% of all those making reservations do not appear for the trip.

This means that 100 - 35 = 65% appear, so [tex]p = 0.65[/tex]

A) If six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip?

This will happen if more than four appear, as the limousine can accommodate up to four passengers on any one trip. So

[tex]P(X > 4) = P(X = 5) + P(X = 6)[/tex]

In which

[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]

[tex]P(X = 5) = C_{6,5}.(0.65)^{5}.(0.35)^{1} = 0.2437[/tex]

[tex]P(X = 6) = C_{6,6}.(0.65)^{6}.(0.35)^{0} = 0.0754[/tex]

[tex]P(X > 4) = P(X = 5) + P(X = 6) = 0.2437 + 0.0754 = 0.3191[/tex]

0.3191 = 31.91% 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?

The expected number of arrivals is given by:

[tex]E(X) = np = 6*0.65 = 3.9[/tex]

Since the are four places:

4 - 3.9 = 0.1

The expected number of available places when the limousine departs is 0.1.

C) Suppose the probability distribution of the number of reservations made is given in the accompanying table.Number of reservations 3 4 5 6Probability 0.09 0.25 0.32 0.34Let X denote the number of passengers on a randomly selected trip. Obtain the probability mass function of X.

This is the probability of each outcome. So

[tex]P(X = 3) = 0.09[/tex]

[tex]P(X = 4) = 0.25[/tex]

[tex]P(X = 5) = 0.32[/tex]

[tex]P(X = 6) = 0.34[/tex]

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Yes (group 1) 40 (32.5) 20 (27.5) 60
No (group 2) 6 (13.5) 19 (11.5) 25
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b. The expected cell counts are not large enough to apply the chi-square test for homogeneity.
c. The total sample size is not large enough to apply the chi-square test for homogeneity.
d. The total sample size is too large to apply the chi-square test for homogeneity.
e. All conditions necessary to apply the chi-square test for homogeneity are satisfied here.

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