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

83.29% probability that at least 2 flights arrive late.

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) = (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.