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If I choose 2 points on a line XY, each point being chosen according to the uniform distribution on XY, and the choices are being made independently. The line XY is now divided into 3 parts. What is the probability that they may be made into a triangle?

=============================================================== X---P1----P2----Y

So they are asking to find the probability of the existence of a triangle with sides of length XP1, P1P2 and P2Y.

I read somewhere that it can only happen if and only if XP1 + P2Y > P1P2. Could someone explain why?

What does it mean by chosen according at the uniform distribution?

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You can find a few approaches to answering this question at… – svenkatr Oct 10 '11 at 23:54
Indeed, svenkatr's reference seems to have everything one could want. Unfortunately, this does not meet any of the acceptable criteria for closing a question. – Gerry Myerson Oct 11 '11 at 0:59
thanks, appreciate it! – John Southall Oct 11 '11 at 12:07
up vote 1 down vote accepted

A triangle has the feature that each of the three sides is less than the sum of the other two sides. Otherwise you cannot construct the triangle (try it to see why, or consider what is the shortest distance between two points).

Choosing a point with uniform probability means informally that each point is as likely to be chosen as any other. Alternatively, regard the probability of being a fraction $p$ or less of the way along the line segment as being $p$.

So if both points $P_1$ and $P_2$ are in the same half of the line segment (what probability?) then you cannot construct the triangle. Similarly if the points are in different halves but more than half the length of the line segment apart (what probability?) then you cannot construct the triangle.

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