@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge@Estallar12 ..can u all please elaborate the concept behind it...i don't have an OA to it..
You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?
@nick_baba said:You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge@Estallar12 ..can u all please elaborate the concept behind it...i don't have an OA to it..
4/15 ??
@nick_baba said:You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge ..can u all please elaborate the concept behind it...i don't have an OA to it..
11 out of 15 ?
@kingsleyx said:4/15 ??
@catter2011 said:11 out of 15 ?
how can i tell u that it is right or wrong when i don't have an OA to this..justify your approaches..
@nick_baba said:how can i tell u that it is right or wrong when i don't have an OA to this..justify your approaches..
4/5 mrngs in 3 Km's have a close call => evry mrng, every Km is having a close call of 1/3* 4/5 => 4/15, hence safety call for each KM is 11 out of 15
@catter2011 said:4/5 mrngs in 3 Km's have a close call => evry mrng, every Km is having a close call of 1/3* 4/5 => 4/15, hence safety call for each KM is 11 out of 15
@nick_baba same approach used by me..what is the answer in your opinion?
@catter2011 said:4/5 mrngs in 3 Km's have a close call => evry Km is having a close call of 1/3* 4/5 => 4/15, hence safety call for each KM is 11 out of 15
in this case it seems that u have taken 1 km as a unit...but what if the probability is changing lets say every 500 m or 100 m?? what i mean to say is that how can u take a discrete value...??
the prob of being safe is max(i.e 1) when he is at home and it is reducing in a particular fashion till the time he reaches his office..now what is the trend in which this prob reduces??...and how have you quantified it?
@catter2011 said:
4/5 mrngs in 3 Km's have a close call => evry mrng, every Km is having a close call of 1/3* 4/5 => 4/15, hence safety call for each KM is 11 out of 15
You have assumed the unit to be 1 KM... I would say.. Answer changes when unit distances assumed changes..
Suppose..take an element dx at a distance "x" away from home.. then the probability changes with "x"..But as far as your solution goes..it is constant for a KM..I am unable to comply to this
- no close calls --> must be 11/15 !!@nick_baba said:in this case it seems that u have taken 1 km as a unit...but what if the probability is changing lets say every 500 m or 100 m?? what i mean to say is that how can u take a discrete value...??the prob of being safe is max(i.e 1) when he is at home and it is reducing in a particular fashion till the time he reaches his office..now what is the trend in which this prob reduces??...and how have you quantified it?
yes it changes.. when distance reduces.. and question is targeting for a particular distance..(1 KM) and according to u.. he is 100% safe when distance is 0, or when home is office.. i assumed linearity...
@soumitrabengeri said:@nick_babasame approach used by me..what is the answer in your opinion?
though i won't deny that it could be one of the ans..but my point is u have quantified the prob in unit..now what if that unit changes?? i hope u understand what i'm trying to convey..
@catter2011 said:yes it changes.. when distance reduces.. and questions is targeting for a particular distance.. and according to u.. he is 100% safe when distance is 0, or when home is office.. i assumed linearity...
Can't assume linearity..
@nick_baba said:You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge@Estallar12 ..can u all please elaborate the concept behind it...i don't have an OA to it..
I think,
11/15.
P = 1/3*4/5 = 4/15
P' = 1-4/15 = 11/15
Do you have options ? I somehow think it might be 1/5 also..
@catter2011 said:yes it changes.. when distance reduces.. and question is targeting for a particular distance..(1 KM) and according to u.. he is 100% safe when distance is 0, or when home is office.. i assumed linearity...
see what i'm looking for here is not an answer but i just want to discuss it..u have assumed prob to be a linear function but in this case as far as i can comprehend, prob is a distributed function. so we have to go by the basics of prob which is area under the curve of that pdf.
@sujamait said:I think,11/15.P = 1/3*4/5 = 4/15 P' = 1-4/15 = 11/15Do you have options ? I somehow think it might be 1/5 also..
sir i don't have any options to this..it was just a question of discussion i was going through...but one thing for which i'm sure is that the prob cannot be 1/5..because the more closer we are to a destination, more is the prob for achieving the target..so in my opinion i don't think(may be i'm going wrong somewhere) it should be 1/5.
@nick_baba said:sir i don't have any options to this..it was just a question of discussion i was going through...but one thing for which i'm sure is that the prob cannot be 1/5..because the more closer we are to a destination, more is the prob for achieving the target..so in my opinion i don't think(may be i'm going wrong somewhere) it should be 1/5.
yup thts fine..but is that 4/5 thing dependent on distance ? I'm nt sure..
@sujamait said:yup thts fine..but is that 4/5 thing dependent on distance ? I'm nt sure..
though it is not clearly mentioned in the question, but we can safely assume that it should depend on it, because the longer the path of travel is, more are the chances of having an accident(but it doesn't mean that prob of accident is not 4/5, what my point is that this prob is now distributed in a wider domain)
@nick_baba said:You live 3km from the office, and you take an auto rickshaw to work every morning. Given how busy Mumbai Road is in Hyderabad, you have observed that 4 out of every 5 mornings, your auto rickshaw driver gets quite close to making an accident. If you move closer to the office, and now only live 1km away, how often do you expect to have a clear run with no close calls to the office?@anytomdickandhary @sujamait @Angadbir @Brooklyn @naga25french @krum @scrabbler @deedeedudu @techsurge@Estallar12 ..can u all please elaborate the concept behind it...i don't have an OA to it..
i think chillfactor has given the desired solution later in the posts and has put end to the arguments.
My earlier arguments stand withdrawn.
ATDH.
@nick_baba said:though it is not clearly mentioned in the question, but we can safely assume that it should depend on it, because the longer the path of travel is, more are the chances of having an accident(but it doesn't mean that prob of accident is not 4/5, what my point is that this prob is now distributed in a wider domain)
it is going to be trickier then. 
I sleep.
I sleep.simple q
A man drives 150 km from A to B in 3 hours and 20 minutes and returns to the place in 4 hours and 10 minutes. If A1 is the average speed of the entire trip, then the average speed of the trip from A to B, exceeds A1 by: