9. 0/1 points | Previous Answers DevoreStat9 2.E.068 My Notes Ask Your Tea A fri
ID: 1885359 • Letter: 9
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9. 0/1 points | Previous Answers DevoreStat9 2.E.068 My Notes Ask Your Tea 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 25% and 10%, whereas for airline #3 the percentages are 20% 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.) airline #1 Enter a number airline #2 airline #3 Need Helpa L 1 Read It Talk to a TutorExplanation / Answer
P(1) = 0.6
P(2) = 0.2
P(3) = 0.2
Probability that flights are late into DC and LA on airline 1 P(L/1) = arrive late in DC not in LA + using #1 arrive late in LA not in DC
= 0.4*(1-0.15) + (1-0.4)*0.15
= 0.43
Probability that flights are late into DC and LA on airline2 P(L/2) = arrive late in DC not in LA + arrive late in LA not in DC
= 0.25*(1-0.1) + (1-0.25)*0.1
= 0.3
Probabilitythat flights are late into DC and LA on airline3 P(L/3)= arrive late in DC not in LA + arrive late in LA not in DC
= 0.2*(1-0.1) + (1-0.2)*0.1
= 0.26
P(L) = P(1)P(L|1) + P(1)P(L|1) + P(1)P(L|1)
= 0.6*0.43 + 0.2*0.3 + 0.2*0.26
= 0.37
Posterior probabilities are
P(1|L) = P(1)*P(L|1)/P(L)
= 0.6*0.43/0.37
= 0.69729729
P(2|L) = P(2)*P(L|2)/P(L)
= 0.2*0.3/0.37
= 0.1621621
P(3|L) = P(3)*P(L|3)/P(L)
= 0.2*0.26/0.37
= 0.14054
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