Answer :
Answer:
83.29% probability that at least 2 flights 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 other flights. 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.
[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.
80 % of its flights arriving on time.
This means that 100-80 = 20% arrive late, so [tex]p = 0.2[/tex]
A test is conducted by randomly selecting 15 Southwest flights and observing whether they arrive on time.
This means that [tex]n = 15[/tex]
Find the probability that at least 2 flights arrive late.
Either less than 2 arrive late, or at least 2 does. The sum of the probabilities of these events is decimal 1. So
[tex]P(X < 2) + P(X \geq 2) = 1[/tex]
We want [tex]P(X \geq 2)[/tex]. Then
[tex]P(X \geq 2) = 1 - P(X < 2)[/tex]
In which
[tex]P(X = 2) = P(X = 0) + P(X = 1)[/tex]
So
[tex]P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}[/tex]
[tex]P(X = 0) = C_{15,0}.(0.2)^{0}.(0.8)^{15} = 0.0352[/tex]
[tex]P(X = 1) = C_{15,1}.(0.2)^{1}.(0.8)^{14} = 0.1319[/tex]
[tex]P(X = 2) = P(X = 0) + P(X = 1) = 0.0352 + 0.1319 = 0.1671[/tex]
[tex]P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1671 = 0.8329[/tex]
83.29% probability that at least 2 flights arrive late.