Question

The rates of on-time flights for commercial jets are continuously tracked by the U.S. Department of Transportation. Recently, Southwest Air had the best rate with 80 % of its flights arriving on time. A test is conducted by randomly selecting 15 Southwest flights and observing whether they arrive on time. (a) Find the probability that at least 2 flights arrive late.

Answer 1

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.

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

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.

80 % of its flights arriving on time.

So 100 - 80 = 20% arrive late, which means that $p = 0.2$

15 Southwest flights

This means that $n = 15$

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

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$

Then

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

So

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

$P(X = 0) = C_{15,0}.(0.2)^{0}.(0.8)^{15} = 0.0352$

$P(X = 1) = C_{15,1}.(0.2)^{1}.(0.8)^{14} = 0.1319$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0352 + 0.1319 = 0.1671$

Then

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.1671 = 0.8329$

83.29% probability that at least 2 flights arrive late.