Question

A person who lives in LA makes trips to D.C.; 40% of the time she goes on airline #1, 20% of the time on airline #2, and 40% of the time on airline #3. For airline #1, flights are late into D.C. 35% of the time and late into L.A. 20% of the time. For airline #2, these percentages are 40% and 15%, whereas for airline #3 the percentages are 35% and 30%. If we learn that on a particular trip she arrived 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 L.A. is unaffected by what happens on the flight to D.C. 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. Round your answers to four decimal places. #1 airline #1 airline #2 airline #3

Answer 1

	a	b	a*b	ab/Σab
x	P(x)	P(late on 1d |x)	P(x and late on 1d)	P(x|late on 1d)
airline #1	0.4	0.4100	0.1640	0.3850
airline #2	0.2	0.4300	0.0860	0.2019
airline #3	0.4	0.4400	0.1760	0.4131
		Σab =	0.4260

P(airline #1 given late on 1 d)=P(airline #1 given late on 1 d)/P(late on 1d)=0.3850

P(airline #2 given late on 1 d)=P(airline #2 given late on 1 d)/P(late on 1d)=0.2019

P(airline #3 given late on 1 d)=P(airline #3 given late on 1 d)/P(late on 1d)=0.4131