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A doctor is to visit a patient. From past experience, it is known that the probabilities that he will come by train, bus, scooter or by car are repectively 3/10,1/5,1/10and2/5. The probabilities that he will be late are 1/4,1/3 and 1/12, if he comes by train,bus and scooter respectively, but if he comes by car, he will not be late. When he arrives, he is late. What is the probability that he has come by train ? |
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Answer» Solution :Let `E_1,E_2,E_3and E_4` be the events that the DOCTOR comes by train, bus, SCOOTER and car RESPECTIVELY. Then, `P(E_1)=3/10,P(E_2)=1/5,P(E_3)=1/10andP(E_4)=2/5`. Let E be the event that the doctor is late. Then, `P(E//E_1)` = probability that the doctor is late, given that he comes by train `=1/4`. `P(E//E-2)`= probability that the doctor is late, given thathe comes by bus `=1/3`. `P(E//E_3)`= probability that the doctor is late, given that the comes by scooter `=1/12`. `P(E//E_4)`= probability that the doctor is late, given that that he comes by car =0. Probability that he comes by train, given that he is late `P(E_1//E)` `(P(E_1).P(E//E_1))/(P(E_1).P(E//E_1)+P(E_2).P(E//E_2)+P(E_3).P(E//E_3)+P(E_4).P(E//E_4))`[by Bayes's theorem] `((3/10xx1/4))/((3/10xx1/4)+(1/5xx1/3)+(1/10xx1/12)+(2/5xx0))` `=(3/40xx120/18)=1/2` Hence, the required probability is `1/2`. |
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