Question

A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.;\(60 \%\) of the time she travels on airline #1, \(20 \%\) 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. \(40 \%\) of the time and late into L.A. \(15 \%\) of the time. For airline \(\# 2\), these percentages are \(35 \%\) and \(10 \%\), whereas for airline # 3 the percentages are \(30 \%\) and \(15 \%\). 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.

Answer 1

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