What is the probability of a chord inside a circle being greater than D?

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that day my friend asked me a question. what is the probability that the chord inside a circle (with radius 1) will be greater than D, where D is in the interval [0, 2].

i have came up with the answer
{\pi - 2\sin^{-1}(D/2)} \over {\pi}

is it ok? or did i do anything wrong?
 
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Am I missing something? A circle of radius 1 has diameter of length 2 and all other chords are shorter. It is impossible for a chord to have length greater than 2. The probability is 0.
 
Am I missing something? A circle of radius 1 has diameter of length 2 and all other chords are shorter. It is impossible for a chord to have length greater than 2. The probability is 0.
If you put D=2 in the expression, you will get 0. The question is what is the probability for a specific D, where 0<D<2. The answer will depend on what sort of distribution function defines the placing of the chord - there is more than 1 way to do it.
Examples:
(1) Uniform in distance from the center.
(2) Pick a point on the circumference, other end point is uniform around the circumference.
 
how will the 2 distributions affect the answer? can you be a bit more elaborate?
 
Ah, yes, I simply misread the quesition!
 
how will the 2 distributions affect the answer? can you be a bit more elaborate?
You had to make some sort of assumption about the chords to get the answer you did. I haven't worked out what the rersults would be for the 2 examples I gave, but I can make up possibilities which I know would give different results, although they might look strange. For example, uniform in the square of the distance from the center.
 
I took the time to work out the various possibilities I mentioned. To simplify notation, let s=D/2. The probabiliites for these cases are:

Case..........Prob.
uniform in arc length.......murshid islam result
uniform in distance from center.....(1-s2)1/2
uniform in distance squared......1-s2
uniform in chord length.....1-s

As you can see, there is no "right" answer.
 

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