An airline claims that there is a 0.10 probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight. This outcome is independent from flight to flight. Sam is a frequent flier who always purchases coach-class tickets.

Requried:
a. What is the probability he will receive at least one upgrade during the next two weeks?
b. What is the probability that in a set of 20 flights, Sam will be upgraded 3 times or fewer?

Respuesta :

Using the binomial distribution, it is found that there is a:

a) 0.3439 = 34.39% probability he will receive at least one upgrade during the next two weeks.

b) 0.8671 = 86.71% probability that in a set of 20 flights, Sam will be upgraded 3 times or fewer.

For each flight, there are only two possible outcomes, either he receives an upgrade, or he dos not. The probability of receiving an upgrade in a flight is independent of any other flight, hence, the binomial distribution is used to solve this question.

Binomial probability distribution

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

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

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

An airline claims that there is a 0.10 probability that a coach-class ticket holder who flies frequently will be upgraded to first class on any flight, hence [tex]p = 0.1[/tex].

Item a:

He takes 4 flights, hence [tex]n = 4[/tex].

The probability is:

[tex]P(X \geq 1) = 1 - P(X = 0)[/tex]

In which:

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

[tex]P(X = 0) = C_{4,0}.(0.1)^{0}.(0.9)^{4} = 0.6561[/tex]

Then:

[tex]P(X \geq 1) = 1 - P(X = 0) = 1 - 0.6561 = 0.3439[/tex]

0.3439 = 34.39% probability he will receive at least one upgrade during the next two weeks.

Item b:

20 flights, hence [tex]n = 20[/tex].

The probability is:

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3)[/tex]

Then:

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

[tex]P(X = 0) = C_{20,0}.(0.1)^{0}.(0.9)^{20} = 0.1216[/tex]

[tex]P(X = 1) = C_{20,1}.(0.1)^{1}.(0.9)^{19} = 0.2702[/tex]

[tex]P(X = 2) = C_{20,2}.(0.1)^{2}.(0.9)^{18} = 0.2852[/tex]

[tex]P(X = 3) = C_{20,3}.(0.1)^{3}.(0.9)^{17} = 0.1901[/tex]

[tex]P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.1216 + 0.2702 + 0.2852 + 0.1901 = 0.8671[/tex]

0.8671 = 86.71% probability that in a set of 20 flights, Sam will be upgraded 3 times or fewer.

To learn more about the binomial distribution, you can take a look at https://brainly.com/question/24863377