Question

1. joel often flies with united airlines from grand rapids, michigan to los angeles, california, which requires him to have a layover and make a connecting flight. about 22% of united flights are delayed, including connecting flights when the first flight is not delayed. if joels first flight is delayed, there is a 34% chance his second flight is also delayed.

a. are the departure times for joels first and second flights independent? how do you know?
b. what is the probability that at least one of his two flights is delayed? show your method.
c. suppose joel takes 3 trips from grand rapids to los angeles (and back) in the month of march. what is the probability that at least one of his twelve flights is delayed? show your method.

Explanation:

Step1: Determine independence

For two events \(A\) (first - flight delay) and \(B\) (second - flight delay), if \(P(B|A)=P(B)\), the events are independent. Here, \(P(B) = 0.22\) and \(P(B|A)=0.34\). Since \(0.34
eq0.22\), the events are not independent.

Step2: Calculate probability of at least one delay (for two - flights)

Let \(P(A)\) be the probability of the first - flight delay (\(P(A)=0.22\)), \(P(B|A) = 0.34\), \(P(B|\overline{A})=0.22\) (given).
First, find \(P(\overline{A}\cap\overline{B})\):
\(P(\overline{A})=1 - 0.22=0.78\), \(P(\overline{B}|\overline{A})=1 - 0.22 = 0.78\)
By the multiplication rule \(P(\overline{A}\cap\overline{B})=P(\overline{A})P(\overline{B}|\overline{A})=0.78\times0.78 = 0.6084\)
Then \(P(\text{at least one delay})=1 - P(\overline{A}\cap\overline{B})=1-0.6084 = 0.3916\)

Step3: Calculate probability of at least one delay (for twelve - flights)

Assume each flight delay is an independent event (after part (a), we know the two - flight case is not independent, but if we assume for the twelve - flight case, using the complement rule). Let \(p = 0.22\) (probability of a single - flight delay)
The probability that a flight is not delayed is \(q=1 - p=0.78\)
The probability that all twelve flights are not delayed is \(q^{12}=(0.78)^{12}\approx0.0434\)
The probability that at least one of the twelve flights is delayed is \(1-(0.78)^{12}\approx1 - 0.0434=0.9566\)

Answer:

a. The departure times are not independent. Because \(P(B|A)=0.34
eq P(B) = 0.22\)
b. The probability that at least one of his two flights is delayed is \(0.3916\)
c. The probability that at least one of his twelve flights is delayed is approximately \(0.9566\)