I'm stuck with a question: Two points are selected at random on a straight line segment of length 1. What is the probability that a triangle can be constructed from the results line segments.
So far all I can come up with is:
- Points p1 and p2 have equal probability of 1.
- Lengths a, b and c can be formed from two cases; if p2 > p1 then a = p1, b = p2 - p1 and c = 1 - p2. For the other case, p1 and p2 are reversed.
- The inequality must be satisfied. I can substitute p1 and p2 in, and imagine some double integral over p1 and p2 for each case, but I don't know how to deal with the inequality.
This is a standard problem. A good reference is Jim Pitman's Probability~.
He calls this the the uniform triangle space.
In general, "the probability that the 2 points will be in same "spot"" has probability zero for continuous distributions.
Another way to approach this problem is like this:
1. Assume the first point selected in the interval (0,1) is called point A. Consider the case A < 1/2. In order to have a triangle the second point to be positioned (call it B) must be placed somewhere in the interval between (1/2,1/2 + A). If B falls outside that interval then the line segments can't form a triangle. For example: if point A = 0.3 then point B must lie between 0.5 and 0.8. The width of that interval is A, and hence the probability that B will be positioned in that interval is also A.
2. Now consider if A>1/2. In this case B must land in the interval between A-1/2 and 1/2. The probability of that happeng is 1-A.
3. So you have a continuous function P(triangle) = A for A between 0 and 1. The resulting integral is:
Thanks for all the help folks. The way I (hopefully) solved it is as follows (probably a bit long-winded):
Two points, x1 and x2. Two (symmetric) cases, so considering :
To make a triangle from a unit length, all sides must be < 1/2, i.e. and
where and so the final integral is:
For , the situation is the same with the variables switched, so the final probability is twice the above, i.e. 1/2
I think you're off by a factor of 2. Your last step was to double the value you calculated, so as to cover the case x1< x2 and x1> x2. But in doing that you need to multiply the first by the probability that x1<x2 and the second by the probability that x1>x2, both of which = 1/2.