Question

BlueSky Air has the best on-time arrival rate with 80% of its flights arriving on time. A test is conducted by randomly selecting 16 BlueSky Air flights and observing whether they arrive on time.
What is the probability that 2 flights of BlueSky Air arrive late? (Give your answer to the nearest three decimal places).

Answer 1

Answer:

0.211 = 21.1% probability that 2 flights of BlueSky Air arrive late

Step-by-step explanation:

For each flight, there are only two possible outcomes. Either they arrive late, or they do not. The probability of a flight arriving late is independent of any other flight. 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.

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

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.

BlueSky Air has the best on-time arrival rate with 80% of its flights arriving on time.

This means that 100 - 80 = 20% are late, which means that $p = 0.2$

Randomly selecting 16 BlueSky Air flights

This means that $n = 16$

What is the probability that 2 flights of BlueSky Air arrive late?

This is P(X = 2).

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

$P(X = 2) = C_{16,2}.(0.2)^{2}.(0.8)^{14} = 0.211$

0.211 = 21.1% probability that 2 flights of BlueSky Air arrive late