Mathematics, 30.03.2020 17:54 gearchanger52p02zf8

An airport limousine can accomodate up to four passengers on any one trip. The company willaccept a maximum of six reservations for a trip, and a passenger must have a reservation. Fromprevious records, 20% of all those making reservations do not appear for the trip. Answer thefollowing questions, assuming independence wherever appropriate.(a)[3 pts]If six reservations are made, what is the probability that at least one individual witha reservation cannot be accomodated on the trip

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Answer from: randall10

65.53% probability that at least one individual witha reservation cannot be accomodated on the trip

Step-by-step explanation:

For each passenger, there are only two possible outcomes. Either they appear for the trip, or they do not. The probability of a passenger appearing for the trip is independent of other passengers. So we use the binomial probability distribution 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.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

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

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

n = 6, p = 0.2

If six reservations are made, what is the probability that at least one individual witha reservation cannot be accomodated on the trip

Less than two passengers do not appear for the trip. So

$P(X \leq 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{6,0}.(0.2)^{0}.(0.8)^{6} = 0.2621$

$P(X = 1) = C_{6,1}.(0.2)^{1}.(0.8)^{5} = 0.3932$

$P(X \leq 2) = P(X = 0) + P(X = 1) = 0.2621 + 0.3932 = 0.6553$

65.53% probability that at least one individual witha reservation cannot be accomodated on the trip

Answer from: 2021andrewkell

The probability that at least one individual with a reservation cannot be accommodated on the trip is 0.6553 or 65.53%

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Passengers can accommodate on any trip = Up to 4

Reservations accepted for any trip = Up to 6

Percentage of passengers that don't show up = 20%

2. Assuming independence wherever appropriate, if six reservations are made, what is the probability that at least one individual with a reservation cannot be accommodated on the trip.

This is a binomial distribution due to the independence of each trial of any previous one.

Upon saying that, we can calculate the distribution table, assuming n = 6 and p = 0.8 because this is the percentage of a passenger shows up for any trip (1 - 0.2)

Binomial distribution (n=6, p=0.8)

f(x)F(x)1 - F(x)

xPr[X = x]Pr[X ≤ x]

00.00010.0001

10.00150.0016

20.01540.0170

30.08190.0989

40.24580.3446

50.39320.7379

60.26211.0000

P (5) + P (6) = 0.3932 + 0.2621

P = 0.6553