A friend who lives in Los Angeles makes frequent consulting trips to Washington,
ID: 3271008 • 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, 20% of the time on airline #2, and the remaining 30% of the time on airline #3. For airline #1, flights are late into D.C. 15% of the time and late into L.A. 10% of the time. For airline #2, these percentages are 25% and 15%, whereas for airline #3 the percentages are 40% 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. (Round your answers to four decimal places.) airline #1 times airline #2 times airline #3 timesExplanation / Answer
Here we are given the following probabilities for the airlines used by the friend to travel to DC from LA:
P( Airline 1) = 0.5, P( Airline 2) = 0.2, P( Airline 3) = 0.3
Also we are given that:
P( Late in DC | Airline 1) = 0.15 and
P( Late in LA | Airline 1) = 0.10
P( Late in DC | Airline 2) = 0.25 and
P( Late in LA | Airline 2) = 0.15
P( Late in DC | Airline 3) = 0.40 and
P( Late in LA | Airline 3) = 0.30
Now we will compute the probability that she arrived late on exactly one of the 2 destinations:
P( late exactly 1 destination ) = P( airline 1 and late to DC and on time to LA) + P( airline 1 and late to LA and on time to DC) + P( airline 2 and late to DC and on time to LA) + P( airline 2 and late to LA and on time to DC) + P( airline 3 and late to DC and on time to LA) + P( airline 3 and late to LA and on time to DC)
= 0.5*0.15*(1-0.1) + 0.5*(1-0.15)*0.1 + 0.2*0.25*(1-0.15) + 0.2*(1-0.25)*0.15 + 0.3*0.4*(1-0.3) + 0.3*(1-0.4)*0.3
= 0.0675 + 0.0425 + 0.0425 + 0.0225 + 0.084 + 0.054
= 0.313
Now given that the flight was late in exactly 1 destination, probability that she took airline 1 would be computed as:
= [ P( airline 1 and late to DC and on time to LA) + P( airline 1 and late to LA and on time to DC)] divided by P( late exactly 1 destination )
= [ 0.0675 + 0.0425 ] / 0.313
= 0.3514
Therefore 0.3514 is the required probability here.
Now given that the flight was late in exactly 1 destination, probability that she took airline 2 would be computed as:
= [ P( airline 2 and late to DC and on time to LA) + P( airline 2 and late to LA and on time to DC)] divided by P( late exactly 1 destination )
= [ 0.0425 + 0.0225] / 0.313
= 0.2077
Therefore 0.2077 is the required probability here.
Now given that the flight was late in exactly 1 destination, probability that she took airline 3 would be computed as:
= [ P( airline 3 and late to DC and on time to LA) + P( airline 3 and late to LA and on time to DC)] divided by P( late exactly 1 destination )
= [ 0.084 + 0.054] / 0.313
= 0.4409
Therefore 0.4409 is the required probability here.