Question

Assume the likelihood that any flight on Delta Airlines arrives within 15 minutes of the scheduled time is 0.90. We randomly selected a Delta flight on four different days. What is the probability that all four flights arrived within 15 minutes of the scheduled time?

1) 0.6561
2) 0.729
3) 0.81
4) 0.9

Answer 1

The probability that all four flights arrive within 15 minutes of schedule, given each has a 0.90 chance, is 0.6561 (0.90 raised to the fourth power).

Step-by-step explanation:

The question revolves around calculating the probability that all four flights selected randomly arrive within 15 minutes of the scheduled time. Given that the probability of one flight arriving on time is 0.90, and assuming the flights are independent events, the probability of all four flights arriving on time is found by raising the single event probability to the power of the number of events (the number of flights), which is four in this case.

To calculate this, we use the formula for the probability of independent events occurring together (P(A) × P(B) × P(C) × P(D)), where P(x) is the probability of each flight arriving on time. Thus:

P(all four flights on time) = P(flight 1 on time) × P(flight 2 on time) × P(flight 3 on time) × P(flight 4 on time) = 0.90 × 0.90 × 0.90 × 0.90 = 0.6561

Therefore, the correct answer is 0.6561, which corresponds to option 1).

1