7. 0/1 points | Previous Answers DevoreStat9 2.E.068 My Notes Ask Your Teacher A

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7. 0/1 points | Previous Answers DevoreStat9 2.E.068 My Notes Ask Your Teacher A friend who lives in Los Angeles makes frequent consulting trips to Washington, DC.; 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. 25% of the time and late into L.A. 10% of the time. For airline #2, these percentages are 25% and 20%, whereas for airline #3 the percentages are 25% and 20% 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 airline #2 airline #3 Need Help? Read It Talk to a Tutor Submit Answer Save Progress Practice Another Version

Explanation / Answer

probability that she got delayed at exactly one of two destinations

=P( airline #1 and got delayed at exactly one of two destinations+airline #1 and got delayed at exactly one of two destinations+airline #1 and got delayed at exactly one of two destinations)

=0.50*((1-0.25)*0.1+0.25*(1-0.10))+0.2*((1-0.25)*0.2+0.25*(1-0.20))+0.3*((1-0.25)*0.2+0.25*(1-0.20))=0.325

airline#1 ==0.50*((1-0.25)*0.1+0.25*(1-0.10))/0.325 =0.4615

airline#2 ==0.2*((1-0.25)*0.2+0.25*(1-0.20))/0.325 =0.2154

airline#3 ==0.3*((1-0.25)*0.2+0.25*(1-0.20))/0.325 =0.3231

2)

probability that both test are positive =P( disese and both test are positive+no disease and both test are positive)

=0.05*0.93*0.93+(1-0.05)*0.04*0.04=0.044765

therefore probability of disease given two test are positive =0.05*0.93*0.93/0.044765=0.9660

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