The triangle will be obtuse iff the three points lie on the same half of the circle. This is because the arc that the obtuse angle intersects is greater than half of the circle.

The first point can be anywhere, the second point can be anywhere (since given two points on a circle they must be on some similar half circle, except for the case when they are exactly opposite each other, but the probability for that is negligible since the circle is made up of infinite points), and the third point has to be on that 1/2 of the circle that both points are, so the probability is \(\frac{1}{2}\)

Now for point 2: If this point is placed opposite to point 1 (or 'almost opposite'), Michael is right (or in fact 'almost' right): Points 1 and 2 'almost' fix half of the circle as point 3 possible landing spot.

Now what if point 2 would not be opposite to point 1 but quite a bit closer: Lets say we move it half-way towards point 1 so that points 1 and 2 are a quarter circle away from each other.

Michael states that point 3 should be on 'that 1/2 of the circle that both points are on'. Please note that this is not correct: there is no exact HALF of the circle on which points 1 and 2 are... Point 3 is allowed to be between points 1 and 2 (prob. 0.25), but also allowed to deviate away from points 1 and 2 in both directions (almost prob 0.25 + almost 0.25). In this particular scenario, chances for points 1, 2 and 3 ending in the same half of the circle increased to almost 0.75.

Taking it one step further: Suppose point 2 would be extremely close to point 1. Now point 3 can be placed basically everywhere on the circle without violating the '3 points on one circle half' criterion, providing us a probability of almost 1 that points 1, 2 and 3 are on the same half.

Basically: After placing point 1, probability calculations should go over both points 2 and 3.
Probabilities run from 0.50 to 1, depending on where point 2 ends up compared to point 1.

Lets stick with 0.75 for the picture as a whole as points are distributed uniformly.

there won't be only 1 semicircle na for which they will lie on the same half there will be many such possibilities of semicircles...actually the 3rd point can lie anywhere on the larger arc formed if the former 2 points are joined to the circumference via the centre

First of all,the probability of a right triangle is 0 because the points are truly random. Pick any point on the circle and draw a diameter through the point. The remaining two points must be on the same side of the diameter. The probability is 1 X 0.5=1/2 because the second point can land on any place but the third point has to be on the same side.

Shouldn't it be infinite as a circle has infinite points and there can be any combination of points until and unless the line joining two points form a diameter

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TopNewestThe triangle will be obtuse iff the three points lie on the same half of the circle. This is because the arc that the obtuse angle intersects is greater than half of the circle.

The first point can be anywhere, the second point can be anywhere (since given two points on a circle they must be on some similar half circle, except for the case when they are exactly opposite each other, but the probability for that is negligible since the circle is made up of infinite points), and the third point has to be on that 1/2 of the circle that both points are, so the probability is \(\frac{1}{2}\)

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Agree that point 1 can be placed anywhere.

Now for point 2: If this point is placed opposite to point 1 (or 'almost opposite'), Michael is right (or in fact 'almost' right): Points 1 and 2 'almost' fix half of the circle as point 3 possible landing spot.

Now what if point 2 would not be opposite to point 1 but quite a bit closer: Lets say we move it half-way towards point 1 so that points 1 and 2 are a quarter circle away from each other.

Michael states that point 3 should be on 'that 1/2 of the circle that both points are on'. Please note that this is not correct: there is no exact HALF of the circle on which points 1 and 2 are... Point 3 is allowed to be between points 1 and 2 (prob. 0.25), but also allowed to deviate away from points 1 and 2 in both directions (almost prob 0.25 + almost 0.25). In this particular scenario, chances for points 1, 2 and 3 ending in the same half of the circle increased to almost 0.75.

Taking it one step further: Suppose point 2 would be extremely close to point 1. Now point 3 can be placed basically everywhere on the circle without violating the '3 points on one circle half' criterion, providing us a probability of almost 1 that points 1, 2 and 3 are on the same half.

Basically: After placing point 1, probability calculations should go over both points 2 and 3. Probabilities run from 0.50 to 1, depending on where point 2 ends up compared to point 1.

Lets stick with 0.75 for the picture as a whole as points are distributed uniformly.

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there won't be only 1 semicircle na for which they will lie on the same half there will be many such possibilities of semicircles...actually the 3rd point can lie anywhere on the larger arc formed if the former 2 points are joined to the circumference via the centre

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First of all,the probability of a right triangle is 0 because the points are truly random. Pick any point on the circle and draw a diameter through the point. The remaining two points must be on the same side of the diameter. The probability is 1 X 0.5=1/2 because the second point can land on any place but the third point has to be on the same side.

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0.25

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Shouldn't it be infinite as a circle has infinite points and there can be any combination of points until and unless the line joining two points form a diameter

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No one for this?

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