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three points are randomly chosen on a circle. what the probability that

1.triangle formed is right angled triangle.

2.triangle formed is acute angled triangle.

3.triangle formed is obtuse angled triangle.

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    @Sharat Chandra: The reason that the probability is $0$ has been very well explained by Rahul Narain. Here is a simpler situation. A **real** number is picked "at random" from the interval $[0,1]$. What is the probability the number is $1/\pi$? The probability that our chosen number lies in a specific interval of width $\epsilon$ is $\epsilon/1$, that is, $\epsilon$. Now think of smaller and smaller intervals around $1/\pi$.2012-03-03

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The question is about points on a circle, so that's the one I'm answering here. The triangle is right angled precisely if two of the three points lie on a diameter of the circle. This is Thales' theorem. As already noted, the probability that this occurs is $0$. The triangle is obtuse precisely if all three points lie on the same side of some diagonal of the circle. The probability for that is $\frac{3}{4}$ (assuming a uniform distribution on the circle).

One way of seeing this is to introduce for each vertex $v_k$ a random variable $X_k$ as follows:

$ X_k = \begin{cases} 1 & \textrm{if the other vertices lie on the half circle starting at } v_k \textrm{ in clockwise direction}\\ 0 & \textrm{otherwise} \end{cases} $

Then at most one of $X_1, X_2, X_3$ can be equal to one. Therefore

$ \mathbb{P}(\textrm{triangle is obtuse}) = \mathbb{P}(X_1+X_2+X_3 = 1) = \mathbb{E}(X_1+X_2+X_3) = 3 \mathbb{E}(X_1). $

The last equality follows from the fact that $X_1, X_2, X_3$ all have the same probability distribution. Now $\mathbb{E}(X_1) = \frac{1}{4}$ since both $v_2$ and $v_3$ have a probability of $\frac{1}{2}$ to lie on the half circle starting at $v_1$.