A friend who lives in Los Angeles makes equent consulting trips to Washington, D
ID: 3071204 • Letter: A
Question
A friend who lives in Los Angeles makes equent consulting trips to Washington, D.C ; 50% of the time she travels on airline #1, 30 o the time on ar ine #2, and the remaining 20% o the time on airline3. For airline #1 flights are late into D.C 40% of the time and labe into LA. 10% o the bme. For airline #2, these percentages are 25 and 10 whereas or airline #3 the percentages are 30% and 15%, we learn that on a particular trip she arr ved late at exactly one of the two destinations, what are the posterior probabilities of having flown on airlines #1, #2 and #3? Assume that the chance of a late arrival in LA. is unaffected by what happens on the flight to D Hint: From the tip of each first-generation branch on a tree diagram draw three second-generation branches labeled, respectively, 0 late, 1 late, and 2 late.1 (Round your answers to four decimal places.) airline 1 airline 3 airlinc *2 Need Help?ReaTalk to a TutorExplanation / Answer
P(late by Airline 1) = P(Airline 1)*P(late in DC and on time in LA) + P(Airline 1)*P(late in LA and on time in DC)
= 0.5*(0.4(1-0.1)) + 0.5*(0.1*(1-0.4))
= 0.18 + 0.03
= 0.2100
P(late by Airline 2) = P(Airline 2)*P(late in DC and on time in LA) + P(Airline 2)*P(late in LA and on time in DC)
= 0.3*(0.25(1-0.1)) + 0.3*(0.1*(1-0.25))
= 0.0675 + 0.0225
= 0.0900
P(late by Airline 3) = P(Airline 3)*P(late in DC and on time in LA) + P(Airline 3)*P(late in LA and on time in DC)
= 0.2*(0.3(1-0.15)) + 0.2*(0.15*(1-0.3))
= 0.051 + 0.021
= 0.0720