5. joel often flies with united airlines from grand rapids, michigan to los angeles, california, which…

5. 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.\na. are the departure times for joels first and second flights independent? how do you know?\nb. what is the probability that at least one of his two flights is delayed? show your method.\nc. 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.

5. 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.\na. are the departure times for joels first and second flights independent? how do you know?\nb. what is the probability that at least one of his two flights is delayed? show your method.\nc. 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.

Answer

Explanation:

Step1: Check independence for part a

Two events (A) (first - flight delay) and (B) (second - flight delay) are independent if (P(B|A)=P(B)). Given (P(\text{flight delay}) = 0.22) and (P(B|A)=0.34). Since (0.34\neq0.22), the events are not independent.

Step2: Calculate probability for part b

Let (A) be the event that the first flight is delayed and (B) be the event that the second flight is delayed. (P(A) = 0.22), (P(B|A)=0.34), (P(B|\overline{A})=0.22) By the law of total - probability, (P(B)=P(B|A)P(A)+P(B|\overline{A})P(\overline{A})) (P(\overline{A})=1 - P(A)=1 - 0.22 = 0.78) (P(B)=0.34\times0.22+0.22\times0.78=0.22\times(0.34 + 0.78)=0.22\times1.12 = 0.2464) The probability that at least one is delayed is (P(A\cup B)=P(A)+P(B)-P(A\cap B)) (P(A\cap B)=P(B|A)P(A)=0.34\times0.22 = 0.0748) (P(A\cup B)=0.22+0.2464 - 0.0748=0.3916)

Step3: Calculate probability for part c

Let (p = 0.22) be the probability that a single flight is delayed. The probability that a single flight is not delayed is (q=1 - p=1 - 0.22 = 0.78) For (n = 12) flights, the probability that none of the flights is delayed is (q^{n}=(0.78)^{12}) The probability that at least one flight is delayed is (P(X\geq1)=1 - P(X = 0)) (P(X = 0)=(0.78)^{12}\approx0.059) (P(X\geq1)=1-(0.78)^{12}\approx1 - 0.059 = 0.941)

Answer:

a. The departure times are not independent. Because (P(\text{second - flight delay}|\text{first - flight delay})\neq P(\text{second - flight delay})) b. The probability that at least one of the two flights is delayed is (0.3916) c. The probability that at least one of the twelve flights is delayed is approximately (0.941)