karthickgururaj

This is a result of a fallout from a previous puzzle.. You choose a point "randomly" on the positive side of origin on x-axis. This point is called A. Choose an another point again on x-axis (again, positive). Call this point B. What is the probability that B is closer to the origin than A? Is it 0? Or 0.5? Or something else? If the question is not well-formed, please feel free to qualify it with "reasonable" statements.

Well, something must be wrong, since you got two different answers IMHO, as I have already stated, your initial assumption that A and B are the closest (or farthest) points seems to be wrong. I'll try to justify my point in a subsequent post, please give me some time. And on the solution above, you are solving a different problem How can you be sure that the solution to both the problems are the same? I'm going to attempt what bononova suggested, perhaps you can try that too. Solve first for three points in a circle of radius R, instead of an infinite plane.

I guess you meant to say the probability is "1" My approach is very similar to above.. but there are couple of issues you might need to address.. "since Q is also INFINITY , you can't be doing INFINITY-Q" Q is not an INFINITY but number which varies from 0 to INFINITY having a fixed value in each test point. Of course I had in mind the probability = 1. Sorry. I didn't understand you then.. I thought Q was the area between the two perpendiculars, in which case Q is infinite..

I agree that the a-priori probability that the second chosen point is closer to first is 0.5. But the posteriori probability is zero.

Ok, here is my solution, similar to how koren approached it..

I guess you meant to say the probability is "1" My approach is very similar to above.. but there are couple of issues you might need to address..

I think there is a subtle problem with the highlighted text above.

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions. Further on in the same manner... No, that would be an incorrect observation.

I agree we need to assume the points to be contained in a finite circle, and then apply the limit as the radius approaches infinity. The question can be considered suitable re-worded. [spoiler=Why circle, and not any other shape?.. ] ..because a circle is "fair" Fairness was implicit in the question. It is like asking: a point is chosen at random on x-axis, what is the probability that it is positive? "Intuitively", it looks to be 0.5. But we can get to that result only if we assume a line segment centered at origin, and then apply the limit as the segment length approaches infinity. A line segment centered at origin is "fair" to both positive and negative directions. I solved the puzzle in a slightly different way (however..), without starting with a bounding circle. I'll post that solution later, not sure if it is entirely correct.

Pick three points on a plane at random (assume uniform probability density across the plane). What is the probability that the triangle so formed is obtuse angled? Source: general internet, with some slight changes. I got a surprising answer to this one..

Ah, ok.. I misunderstood. As you have stated, it can certainly be solved with 3 questions.

And of course, if there is no "C",

Show that there is a set of 2002 consecutive positive integers containing exactly 150 primes. (You may use the fact that there are 168 primes less than 1000) (original source: from a training set for IMO).

I don't know how to solve this, but I may be able to show that it is not solvable with three questions..

Not quite ... a mistake solving for sin(a). Right approach. Eeek. Sorry.. This is why we still need pen and paper.

Assuming R = 1 meter