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 it arrives late, or it does not arrive late. 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.
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
In which
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)!}](https://tex.z-dn.net/?f=C_%7Bn%2Cx%7D%20%3D%20%5Cfrac%7Bn%21%7D%7Bx%21%28n-x%29%21%7D)
And p is the probability of X happening.
80 % of its flights arriving on time.
So 100 - 80 = 20% arrive late, which means that ![p = 0.2](https://tex.z-dn.net/?f=p%20%3D%200.2)
15 Southwest flights
This means that ![n = 15](https://tex.z-dn.net/?f=n%20%3D%2015)
Find the probability that at least 2 flights arrive late.
Either less than two arrive late, or at least 2 do. The sum of the probabilities of these outcomes is 1. So
![P(X < 2) + P(X \geq 2) = 1](https://tex.z-dn.net/?f=P%28X%20%3C%202%29%20%2B%20P%28X%20%5Cgeq%202%29%20%3D%201)
We want ![P(X \geq 2)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29)
Then
![P(X \geq 2) = 1 - P(X < 2)](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20P%28X%20%3C%202%29)
In which
![P(X < 2) = P(X = 0) + P(X = 1)](https://tex.z-dn.net/?f=P%28X%20%3C%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29)
So
![P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}](https://tex.z-dn.net/?f=P%28X%20%3D%20x%29%20%3D%20C_%7Bn%2Cx%7D.p%5E%7Bx%7D.%281-p%29%5E%7Bn-x%7D)
![P(X = 0) = C_{15,0}.(0.2)^{0}.(0.8)^{15} = 0.0352](https://tex.z-dn.net/?f=P%28X%20%3D%200%29%20%3D%20C_%7B15%2C0%7D.%280.2%29%5E%7B0%7D.%280.8%29%5E%7B15%7D%20%3D%200.0352)
![P(X = 1) = C_{15,1}.(0.2)^{1}.(0.8)^{14} = 0.1319](https://tex.z-dn.net/?f=P%28X%20%3D%201%29%20%3D%20C_%7B15%2C1%7D.%280.2%29%5E%7B1%7D.%280.8%29%5E%7B14%7D%20%3D%200.1319)
![P(X < 2) = P(X = 0) + P(X = 1) = 0.0352 + 0.1319 = 0.1671](https://tex.z-dn.net/?f=P%28X%20%3C%202%29%20%3D%20P%28X%20%3D%200%29%20%2B%20P%28X%20%3D%201%29%20%3D%200.0352%20%2B%200.1319%20%3D%200.1671)
Then
![P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1671 = 0.8329](https://tex.z-dn.net/?f=P%28X%20%5Cgeq%202%29%20%3D%201%20-%20P%28X%20%3C%202%29%20%3D%201%20-%200.1671%20%3D%200.8329)
83.29% probability that at least 2 flights arrive late.