There are *** 10 points at the midpoint of the four vertices and sides of a regular tetrahedron. If we take four points, the probability that these four points are not * * * faces is

A.5/7

B.7/10

C.24/35

d .47/70

I thought the same thing.

Take three points at random and determine the * * * face, then find the face composed of the fourth point and these three points.

According to "Resources", 14 1 seed ball must be divided by something, and 14 1 is to be a molecule. So the numerator in the four answers is 80% of the divisor of 14 1. Test results, only d is.

So I chose D. ...

Wait for the teacher to supplement the arrangement and combination. ...

I don't understand ... Hee hee ...

Take four points of 10 as C( 10)(4)=2 10.

Just find the * * * surface.

* * * surface is divided into three situations:

1, four points are all on a certain surface of the tetrahedron, each surface has six points, and there are

C(6)(4)= 15, and 4* 15=60 kinds of four sides.

2. Three of them are * * * lines, and the other point and these three points are not on a certain surface of the tetrahedron, but on the midpoint of a straight line different from the straight line where these three points are located. Obviously, there are only six cases (because the tetrahedron has only six sides).

3. The straight line where two points are located is parallel to the straight line where the other two points are located, and these four points are not on a certain surface of the tetrahedron. Drawing shows that there are only three situations.

So the four points that are not * * * are taken differently: 210-60-6-3 =141.