prob that a plane leaves on time is .93 arrives on time.91 both arrives and departs on time is .86?

Question

find the prob that a flight departs on time, given that it arrives on time
find the prob that the flight arrives on time, given that it departs on time
please explain

Answer 1

P(depart | arrive) = P(depart AND arrive) / P(arrive)
... = 0.86 / 0.91 =0.95

P(arrive | depart) = P(depart AND arrive) / P(depart)
... = 0.86 / 0.93 = 0.92

(Based on Bayes' Rule, which gives definition of conditional probability: P(a AND b) = P(a | b) P(b.)

Answer 2

93% it leave on time, 91% it arrives on time, and 86% chance that it does both on time, What is it that the question was asking for, add more details.

Answer 3

This is an example of Bayesian probability calculations. Look it up in your textbook; it has been a few too many years since I worked with this stuff.

Answer 4

WTF are you talking about?? It makes no sense AT ALL!!