it is going to be trickier then. I sleep.simple qA 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:
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..
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value. The probability for the random variable to fall within a particular region is given by theintegral of this variable창€™s density over the region.
Taking this into consideration: (Working on this basis) (And making use of gaussian function)
Let the function be : f(x) = k* e^(-x^2/2)
This integral of this function gives us the probability at a distance "x" from the source
Hence P(No Accident after 3 kms) = Integral of k*e^(-x^2/2)dx = 1/5 (Limits 0 to 3)
From this , k= 5/3 (Had to use online calculator)
Therefore,
Prob(No accident after 1KM) = Integral of 5/3 * e^(-x^2/2)dx (Limits 0 to 1) = 5/12 (Which is more than the previous case as expected)
@anytomdickandhary Sir is there any flaw in this?? I just did it..Didn't think!!
P.S. The reason i didn't like the concept of linearity is because that would make the prob too symmetrical/ideal... Just wanted to make it (The solution) a bit more difficult :P
@catter2011I think in the light of given information I would arrive at the same answer as yours. Also, I would subscribe to your logic.Some puys have pointed out that how do we know if the probability decreases linearly or not.... which seems to be a valid point. But then the question arises that no information is given on the nature of probability distribution, so how do we select a given distribution and reject the others?Why uniform probability distribution (same as linear probability of distance) is a good choice under given scenario?if we do not chose linear distribution then it would mean that some points on the roads are more prone to accident as compared others. i.e chances of accident exactly 1m away from home is different as compared to point 1m just before office. Now to me this appears a bit more incorrect assumption as compared to assuming that each point on the road is equally probable for an accident. Hence uniform probability distribution is a better choice .... which means total probability of an accident is a linear function of distance from office.Hence (4/5)*(1/3) = 4/15 is the probability of accident=>safe landing = 1 - 4/15 = 11/15ATDH.
sir i do agree with your logic but again, a question comes that there is a chance of an other answer also. with the amount of info in hand, we can safely go with yours' and other puys' solution.
it is going to be trickier then. I sleep.simple qA 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:
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value. The probability for the random variable to fall within a particular region is given by theintegral of this variable €™s density over the region.Taking this into consideration: (Working on this basis) (And making use of gaussian function)Let the function be : f(x) = k* e^(-x^2/2)This integral of this function gives us the probability at a distance "x" from the sourceHence P(No Accident after 3 kms) = Integral of k*e^(-x^2/2)dx = 1/5 (Limits 0 to 3)From this , k= 5/3 (Had to use online calculator)Therefore,Prob(No accident after 1KM) = Integral of 5/3 * e^(-x^2/2)dx (Limits 0 to 1) = 5/12@anytomdickandhary Sir is there any flaw in this?? I just did it..Didn't think!!
i sort of like this explanation....though lets not talk about answers here..
In probability theory, a probability density function (pdf), or density of a continuous random variable, is a function that describes the relative likelihood for this random variable to take on a given value. The probability for the random variable to fall within a particular region is given by theintegral of this variable €™s density over the region.Taking this into consideration: (Working on this basis) (And making use of gaussian function)Let the function be : f(x) = k* e^(-x^2)This integral of this function gives us the probability at a distance "x" from the sourceHence P(No Accident after 3 kms) = Integral of k*e^(-x^2/2)dx = 1/5 (Limits 0 to 3)From this , k= 5/3 (Had to use online calculator)Therefore,Prob(No accident after 1KM) = Integral of 5/3 * e^(-x^2/2)dx (Limits 0 to 1) = 5/12@anytomdickandhary Sir is there any flaw in this?? I just did it..Didn't think!!
From calculations perspective there is no flaw.
However, we should be careful before using any distribution function {as you have used k*e^(-x^2) }.
Think a bit (no need to calculate!) and you would know that if you use this function then probability of accident in zone 0To me this does not sound very logical, hence I would not prefer to use e^(-x^2) as the distribution function.
Last and most interesting point: if you measure distance from office then above argument would completely reverse.
From calculations perspective there is no flaw. However, we should be careful before using any distribution function {as you have used k*e^(-x^2) }. Think a bit (no need to calculate!) and you would know that if you use this function then probability of accident in zone 0 To me this does not sound very logical, hence I would not prefer to use e^(-x^2) as the distribution function.Last and most interesting point: if you measure distance from office then above argument would completely reverse.ATDH.
Sir the function is not the prob function of accident but of non accident
Prob(No accident) = k*e^(-x^2/2)
I suppose there is nothing wrong in this?? (I guess you were talking about this only)
it is going to be trickier then. I sleep.simple qA 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:
approx 6.1km/h...????
Avg speed to go from A to B and to come back from B to A = total distance/total time=300/7.5= 40 km/h
In a certain country the average monthly income is calculated on the basis of 14 months in a calendar year while the average monthly expenditure was to be calculated on the basis of 99 months per year. this leads people having an underestimation of their savings , since there would be an underestimation of the income and an overestimation of the expenditure per month.then,1. mr.jack comes back from ussr and convinces his community comprising 273 families to start calculating the average income on the basis of 12 months per year. now if it is know that the average estimated income in his community is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a. 251.60 rs b.282.75 rs c. 312.75 rs d . cannot be determined.2. miss rose comes back from usa and convinces his community comprising 546 families to start calculating the average income and average expenditure on the basis of 12 months per year. now if it is know that the average estimated income on the island is (according to old system) 87 rs per month.then what will be the change in the average estimated savings for the country (assume that there are no other change)a.251.60 rs b. 565.5 rs c. 625.5 rs d . cannot be determined.please tell.
@anytomdickandhary As you have come out.. please answer this one too.. :P
sir i do agree with your logic but again, a question comes that there is a chance of an other answer also. with the amount of info in hand, we can safely go with yours' and other puys' solution.
please refer to chillfactor's post in the next page to get the correct solution.
approx 6.1km/h...????Avg speed to go from A to B and to come back from B to A = total distance/total time=300/7.5= 40 km/havg speed from A to B = 150/3.25=46.1 km/hDiff = 6.1 km/h(that what I understand)
Sir the function is not the prob function of accident but of non accidentProb(No accident) = k*e^(-x^2/2)I suppose there is nothing wrong in this?? (I guess you were talking about this only)
This is turning out to be very interesting here. And I have always enjoyed when people argue on probability distribution functions. (This as per me is point where they start mapping probability in real life application).
There is a certain amount of abstractness involved but I will try to explain my point of view. Please allow me some time before I am able to put the right resources in place to put my point.
I shall definitely share my ideas with you, allow me some time please.
This is turning out to be very interesting here. And I have always enjoyed when people argue on probability distribution functions. (This as per me is point where they start mapping probability in real life application).There is a certain amount of abstractness involved but I will try to explain my point of view. Please allow me some time before I am able to put the right resources in place to put my point. I shall definitely share my ideas with you, allow me some time please.ATDH.
Sir jee thoda jaldi karna 😛 IIFT hai kal..thodi der me nikalna hai ..Some distance to cover..
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..
This is turning out to be very interesting here. And I have always enjoyed when people argue on probability distribution functions. (This as per me is point where they start mapping probability in real life application).There is a certain amount of abstractness involved but I will try to explain my point of view. Please allow me some time before I am able to put the right resources in place to put my point. I shall definitely share my ideas with you, allow me some time please.ATDH.
sir do involve me also, i would definitely like to discuss this question in every possible detail.
I just did it..Didn't think!! 
..though lets not talk about answers here..