Together it's a possibility

If all the the points lie in the same semicircle, they will form an obtuse triangle.

Probability that the three points form an acute triangle = $\frac {1}{π-0}(\int_{0} ^{π} \frac {\theta }{2π} d\theta )=\frac {1}{4}$

Hence, probability percentage of obtuse angled triangle = $\frac {3}{4}×100=75$

(Probability of a right angled triangle is 0)

Three points thus chosen on a unit circle are all individual events. As choosing any point doesn't affect the choice of placing any other point on the same circle.

Now let's consider the probability of choosing one point on a circumference of the circle, lying in a semi-circle . The probability of this outcome is $1$ , since any point lies on infinitely many semi-circles on the circumference of the circle

Now, let's consider the probability of any two point, chosen on the circumference, lying in a semi-circle . The probability of this turns out to be, $\dfrac{1}{2}$ , since There's only 1 way to put 2 points on the same semi-circle, out of possibilities of them being either on the same semi-circle or on 2 different ones.

Now for choosing any three points. It matters on what point you pick first. Because, out of 3 points on a circle, any 2 points must lie on the same semi-circle, with respect to the third one .

Thus, we have 3 different points, which can be singly selected to place on a semi-circle in $^3C_1 = 3$ ways.

Therefore, the probability of any 3 points on a circle being lying on the same semi-circle

$\begin{aligned} \ \ \ \ \ \ \ \ \ \ &= ^3C_1 \times 1 \times \dfrac{1}{2} \times \dfrac{1}{2} \\ &= 3 \times \dfrac{1}{4} \\ &= \dfrac{3}{4} \end{aligned}$

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In general, for n chosen points on a circle, the probability of them lying on the same semi-circle

$\begin{aligned}&= ^nC_1 \times 1 \times \dfrac{1}{2} \times \dfrac{1}{2} \times \dfrac{1}{2} \times \cdots \\ &= n \times \dfrac{1}{2^{n-1}} \\ &= \dfrac{n}{2^{n-1}} \\ \end{aligned}$