A friend who lives in Los Angeles makes frequent consulting trips to Washington,
ID: 3306910 • Letter: A
Question
A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; 50% of the time she travels on airline #1, 30% of the time on airline #2, and the remaining 20% of the time on airline #3. For airline #1, flights are late into D.C. 30% of the time and late into L.A. 10% of the time. For airline #2, these percentages are 25% and 20%, whereas for airline #3 the percentages are 40% and 25%. 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.]
Explanation / Answer
probability that she got delayed at exactly one of two destinations
=P( airline #1 and got delayed at exactly one of two destinations+airline #1 and got delayed at exactly one of two destinations+airline #1 and got delayed at exactly one of two destinations)
=0.50*((1-0.3)*0.1+0.3*(1-0.10))+0.3*((1-0.25)*0.2+0.25*(1-0.20))+0.2*((1-0.25)*0.4+0.25*(1-0.40))=0.365
airline#1 ===0.50*((1-0.3)*0.1+0.3*(1-0.10))/0.365 =0.4658
airline#2 ==0.3*((1-0.25)*0.2+0.25*(1-0.20))/0.365 =0.2877
airline#3 ==0.2*((1-0.25)*0.4+0.25*(1-0.40))/0.365 =0.2466