Question

AMERICAN AIRLINES HAS A POLICY OF BOOKING AS MANY AS 15 PEOPLE ON AN AIRPLANE THAT CAN SEAT ONLY 14. (PAST STUDIES HAVE REVEALED THAT ONLY
85% OF THE BOOKED PASSENGERS ACTUALLY ARRIVE FOR THE FLIGHTS.) FIND THE PROBABILITY THAT IF AMERICAN AIRLINES BOOKS 15 PEOPLE, NOT ENOUGH SEATS
WILL BE AVAILABLE

Answer 1

Using the binomial distribution, there is a 0.0874 = 8.74% probability that not enough seats will be available.

What is the binomial distribution formula?

The formula is:

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

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

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.

For this problem, the values of the parameters are given by:

n = 15, p = 0.85.

The probability that not enough seats will be available is P(X = 15), as the only outcome in which not enough seats will be available is when all 15 people who bought the ticket show up, hence:

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

$P(X = 15) = C_{15,15}.(0.85)^{15}.(0.15)^{0} = 0.0874$

0.0874 = 8.74% probability that not enough seats will be available.

More can be learned about the binomial distribution at https://brainly.com/question/24863377

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