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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 12:00:48 IST
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A doctor is to visit a patient. From the past experience, it is known that the probabilities that he will come by train, bus, scooter or by any other means of transport are respectively 3 ,1, 1 and2 10 5 10 5 . The probabilities that he will be late are 1,1and 1 4 3 12 if he comes by train, bus and scooter respectively, but if he comes by other means of transport, then he will not be late. When he arrives, he is late. What is the probability that he comes by train. ans is 5/7????
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alphawoman1 |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 12:24:57 IST
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i m getting 1/2 are these values correct?? if yes, plz give the values again..as they are not properly typed...
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Na pine ka shouk tha na pilane ka shonk tha. Hume to sirf NAZREIN milane ka shonk tha. Par kya karen hum nazre hi unse mila baithe jinhe nazron se PILANE ka shonk tha. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 12:29:18 IST
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3/10, 1/5, 1/10, 2/5 resp fortrain bus scooetr and other prob that hell be late is 1/4, 1/3, 1/12 and 0 resp
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alphawoman1 |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 13:06:01 IST
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answer is 1/2 only.. simple application of baye's theorem solves it..
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JEE and OLYMPIA INFINATUM
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 13:17:22 IST
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agree with @hash_include
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Na pine ka shouk tha na pilane ka shonk tha. Hume to sirf NAZREIN milane ka shonk tha. Par kya karen hum nazre hi unse mila baithe jinhe nazron se PILANE ka shonk tha. |
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![[Post New]](/templates/default/images/icon_minipost_new.gif) 18 Mar 2008 13:34:30 IST
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HEY, IT IS 1/2... A READER SSUM... P(T)=3/10,P(B)=1/5,P(S)=1/10 ANS P(OTH)=2/5 P(L/T)=1/4,P(L/B)=1/3,P(L/S)=1/12,P-(L/OTH)=0 P(T/L)= P(T)P(L/T)/[P(B)P(L/B)+P(T)P(L/T)+P(S)P(L/S)+ P(OTH)P(L/OTH)] =1/2.
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