Three points have been chosen randomly from the vertices of a cube.
What is the probability that they form (a) an acute triangle; (b) a right triangle?

The first step in solving this problem is to figure out how many different triangles can be made using the vertices of a cube. Since there are 8 vertices, the first point can be picked from one of 8 vertices, the second point can be picked from the 7 remaining vertices, and the last point of the triangle can be picked from the 6 remaining vertices. There are, therefore, 8𡥚 = 336 ways to draw a triangle on (or inside) a cube, when you disregard the sequence of choosing points for the triangles. Taking the point sequence into account, we can divide 336 by 6 to come up with 56 unique triangles.

Of those 56 triangles, only 8 are regular triangles. To draw a regular triangle in a cube, the first two points must fall on opposing diagonals of a single face. The last point is taken from one of the two points on the opposite face on a corner that does not share an edge with the first two points.

The remaining triangles are all right triangles. Therefore the probability of getting a regular triangle is 8/56 = 1/7 = 0.14286..., and the probability of getting a right triangle is 48/56 = 6/7 = 0.85714...

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