A friend who lives in Los Angeles makes frequent consulting trips to Washington,
ID: 2921373 • 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 air- line #3. For airline #1, ights 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 own on air- lines #1, #2,
Explanation / Answer
probability that she arrived late at exactly one of the two destinations =P( airline 1 and arrived lat in exactly one of the destinations+airline 2 and arrived lat in exactly one of the destinations+airline 3 and arrived lat in exactly one of the destinations)=0.5*(0.3*(1-0.1)+(1-0.3)*0.1)+0.3*(0.25*(1-0.20)+(1-0.25)*0.20)+0.2*(0.40*(1-0.25)+(1-0.4)*0.25)
=0.365
hence posterior probabilities of having own on air- lines #1 given late on exactly one flight
=0.5*(0.3*(1-0.1)+(1-0.3)*0.1)/0.365 =0.4658
posterior probabilities of having own on air- lines #2 given late on exactly one flight
=0.3*(0.25*(1-0.20)+(1-0.25)*0.20)/0.365 =0.2877
posterior probabilities of having own on air- lines #3 given late on exactly one flight
=0.2*(0.40*(1-0.25)+(1-0.4)*0.25)/0.365 =0.2466
please revert for any clarification required