mathematics

please have the answer

The probability that they will make any triangle is 1 (100%), since the only way to avoid it is by making a perfectly straight line, and no line they make will be perfectly linear.

Well, they technically could all land on a straight line, couldn't they?

Throwing three points at random in a straight line is incredible. You can come close but the monte carlo will say that is will take an infinite number of trys. So the probability of throwing a triangle is a certainty.

The second problem is interesting. Try this construction. Look at the two points that make the longest leg and the two points that make the shortest leg. If the angle between those two legs is greater than a right angle (>+/-90 deg) the triangle is obtuse, less than a right angle (<+/-90 deg) acute.

This leaves this with the right angle(=+/-90 deg), but just like throwing a straight line (=0 & +/-180 deg) is incredible, so it is a certainty the triangle will always be acute or obtuse.

So now the distribution is reduced to the angle described by three points chosen at random, and if this angle is >0 and <+/-90) the triangle is acute, if >90 and <180 or >-180 and <-90, it is obtuse. This kinda describes a circular geometric figure, that is divided in half.

So my estimate of throwing a obtuse triangle is 1/2.

But it might just be easier to flip a coin, and if it lands on its edge---give me a call.

Rap

I get:
pi / [(8/3)pi - 2sqrt(3)]
which is slightly more than .639

Draw a circle with the longest leg (length = 2r) as the diameter.

Since we've already identified the long leg, the third point must be within 2r of the first two points.

Draw two intersecting circular arcs of radius 2r. Each will be centered at one of the end points of the long leg. The resulting figure will look like an eye (CBS logo).

The two intersecting arcs define the boundary for the third point.

All right triangles will have the third point on the circle.
All obtuse triangles will have the third point inside the circle.
All acute triangles will have the third point outside the circle, but inside or on the intersecting arcs.

The probability that the triangle is obtuse is the ratio of the area of the circle to the area bounded by the circular arcs.

circle area = pi*r^2
larger area = 2[(1/3)pi(2r)^2 - sqrt(3)r^2] = r^2 * [(8/3)pi - 2sqrt(3)]

markr is right. The probability of throwing a triangle is bounded by the catseye described by by the intersection of the two arcs is a certainity an obtuse triangle is bounded a circle inscribed within this catseye. The ratio of the two areas is the probability of throwing a obtuse triangle.

Area of catseye
A=2l^2(4pi-3sqrt(3))/3
Area of Circle
Ac=pi*l^2
Probability of throwing obtuse triangle
P(E)=Ac/A=3pi/(4pi-3sqrt(3))*100%=63.9%

Rap

I guess you could throw the same point if you were really unlucky, e.g. (0,0,0) (0,0,0) (0,0,0) if you random generator isn't too random or you are just very unucky, if even two of the points are the same you have your line.

Of course it also depends on the dimension of your world space. I presume you are in a 3 dimensional space, becuase if its 1 dimensional you always get a point, if its two you get a line, if its three its normally a triangle, if its 4 or higher you'd get a very unsual space, possibly with a 3D representation.

You're off by one. A point is dimensionless. One dimension is a line, two is a plane, etc.

Also, I don't know that the answer differs for dimensions greater than two. The three points will always lie in the same plane.