A friend Aho lives in Los Angclcs ma cs freq cnt consulting tips to Washington,

Question

A friend Aho lives in Los Angclcs ma cs freq cnt consulting tips to Washington, D.C.; 50% o c t mc she travels cn airline 41 20% cf the tr e cn airline 2, and thc rema nnq 30% ofthc time on anne3. For rlinc #1, fights are late into D.C 40% of the time and ate into L 25 at the time. For al Ine *2, these e ce tages are 30 and 259b, whereas for airline ÷3 the percentageare 40% and 159 e leam that on a particular trip she arrived late at exactly one the two destinations what are the posterior probanilities o having Hown on airlines +1 #2, ar #3 ? Assumit that he dial ce ur a ldte dmval in LA. ls una ected by hal happens ur he ligh u D.C H : Fro " ie ip ur each "st ene auon rdt ch un a u ee 'ayam did Utree secorld-generation a ches labeled respectively, 0 late 1 late, and 2 ate. Round your answers to four decimal plazes) airline #1 airline #2 airline #3 Need Help? eaTelk to Tule

Explanation / Answer

Solution:

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) = 0.40 and
P (Late in LA | Airline 1) = 0.25

P (Late in DC | Airline 2) = 0.30 and
P (Late in LA | Airline 2) = 0.25

P (Late in DC | Airline 3) = 0.40 and
P (Late in LA | Airline 3) = 0.15

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*0.40*(1-0.25) + 0.5*(1- 0.40) *0.25 + 0.2*0.30*(1-0.25) + 0.2*(1-0.30) *0.25 + 0.3*0.40*(1-0.15) + 0.3*(1-0.40) *0.15

= 0.15 + 0.075 + 0.045 + 0.035 + 0.102 + 0.027

= 0.434

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)

= [ 0.15 + 0.075] / 0.434

= 0.5184

Therefore 0.5184 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)

= [ 0.045 + 0.035] / 0.434

= 0.1843

Therefore 0.1843 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)

= [ 0.102 + 0.027] / 0.434

= 0.2972

Therefore 0.2972 is the required probability here.

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