A friend who lives in Los Angeles makes frequent consulting trips to Washington,
ID: 3365029 • Letter: A
Question
A friend who lives in Los Angeles makes frequent consulting trips to Washington, D.C.; 40% of the time she travels on airline #1, 30% 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. 25% of the time and late into L.A. 20% of the time. For airline #2, these percentages are 25% and 20%, whereas for airline #3 the percentages are 15% and 10%. 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.] (Round your answers to four decimal places.)
A friend who lives in Los Angeles m on alrline #1, 30% of the time on airline #2, and the remaining 30% of the time on airline #3 For airline #1 nights are late into D.C 25% of the Lime and late into L.A 20% o the time. For dirline 2 these percentages are 25% and 20%, whereas or airline #3 the percentages are 15% and 10% 1 we learn that on a particular trip she arrived late at exactly one of the tw destinations, what are the posterior probabilities of having flown on airlines #I, #2, and #32 Assume that the chance of a late arrival in LA. is unaffected by what happens on the flight to D.c. [lint: 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.] (Round your answers to tour decimal places.) airline #1 airline #2 airline3 akes frequent consulting trips to washington, DC.,40% of the time she travelsExplanation / Answer
Here, we are given that:
P( airline 1) = 0.4,
P( airline 2) = 0.3,
P( airline 3) = 0.3
P( late to DC | airline 1) = 0.25,
P( late to LA | airline 1) = 0.2
P( late to DC | airline 2) = 0.25,
P( late to LA | airline 2) = 0.2
P( late to DC | airline 3) = 0.15,
P( late to LA | airline 3) = 0.1
P( exactly one dest late | airline 1)P( airline 1) = 0.4*(0.25*0.8 + 0.75*0.2) = 0.14
P( exactly one dest late | airline 2)P( airline 2) = 0.3*(0.25*0.8 + 0.75*0.2) = 0.105
P( exactly one dest late | airline 3)P( airline 3) = 0.3*(0.15*0.9 + 0.85*0.1) = 0.066
Therefore, using law of total addition, we get:
P( exactly one dest late ) = 0.14 + 0.105 + 0.066 = 0.311
Therefore we get the posterior probabilities here as:
P( airline 1 | exactly one dest late) = P( exactly one dest late | airline 1)P( airline 1) / P( exactly one dest late )
P( airline 1 | exactly one dest late) = 0.14 / 0.311 = 0.4502
P( airline 2 | exactly one dest late) = 0.105 / 0.311 = 0.3376
P( airline 3 | exactly one dest late) = 0.066 / 0.311 = 0.2122