8. 1 points My Notes Ask Your Teach A friend who lives in Lus An eles makes requ

Question

8. 1 points My Notes Ask Your Teach A friend who lives in Lus An eles makes requen consulting lips o Washington, D C i 50% of he line she ravels un airline 1, 30% of he Line on airline 2 and lhe rerna" " ng 20% o the lime on airline #3 hor airline # 1 flights are late into l).C. 40% of the time and late into l A 3()?0 the time tor airline #2, these percentages are 30% and 10 i, whereas or airline #3 the percentages are 41 and 35% 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 LA. is unaffected by what happens on the flight to D.C. [int: Trom the tip of cach first-generation branch on a tree diagram, draw three second-generation branches labele, respectively, late, 1 late, and 2 late.] (Round your answers to four declmal places.) airline #1 airline #2 airline #3

Explanation / Answer

Use of the term "at exactly one" denotes the use of P("exactly one")= P(E n F') - P(E' n F)
E={late to DC} F={late to LA}
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B0= Both Not Late B1= "exactly one" Late B2= Both Late

P(A1)=.50 | P(A2)=.30 | P(A3)=.20
Ldc=.40 | Ldc=.30 | Ldc=.40
Lla=.30 | Lla=.10 | Lla=.35
NLdc=.60 | NLdc=.70 | NLdc=.60
NLla=.70 | NLla=.90 | NLla=.65

Reference back: P("exactly one" flight late on Ai)= P(Ai)*P(E n F') + P(Ai)*P(E' n F)
E={late to DC} F={late to LA}

A1 options:
Flight A1 AND Late to DC AND Not late to LA (.5)(.40)(.70) = 0.14
OR
Flight A1 AND Late to LA AND Not late to DC (.5)(.30)(.60) = 0.09
SO: = P(A1 n B1) = 0.14 + 0.09 = 0.23

A2 options:
Flight A2 AND Late to DC AND Not late to LA (.3)(.30)(.90) = 0.081
OR
Flight A2 AND Late to LA AND Not late to DC (.3)(.10)(.70) = 0.021
SO: = P(A2 n B1) = 0.102

A3 options:
Flight A3 AND Late to DC AND Not late to LA (.2)(.40)(.65) = 0.052
OR
Flight A3 AND Late to LA AND Not late to DC (.2)(.35)(.60) = 0.042
SO: = P(A3 n B1) = 0.094

P(B1) = 0.23 + 0.102 + 0.094 = 0.426

Airline #1 = P(A1 n B1)/ P(B1) = 0.23/0.426 = 0.54

Airline #2 = P(A2 n B1)/ P(B1) = 0.102/0.426 = 0.2394

Airline #3 = P(A3 n B1)/ P(B1) = 0.094/0.426 = 0.221

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