On choosing three vertex of a cube at random, what's the probability that the three are in a same face?
The counting principle, of course.
The first pick if free, we may pick any vertex.
There are 8 vertices on the cube, we've picked one leaving 7. Of these only 6 share a face with the one we've picked. So we have
$\displaystyle 1 \cdot \frac{6}{7}$
so far.
Now, we've picked two vertices.
1) If the two vertices are on the same edge, there are 4 possible of 6 vertices left to pick, so the probability will be
$\displaystyle 1 \cdot \frac{6}{7} \cdot \frac{4}{6} = \frac{4}{7}$
2) If the two vertices are not on the same edge, they are on a diagonal across one face. In this case there is only one possible face that these could be on, so there are only two possible vertices out of 6, so the probability will be
$\displaystyle 1 \cdot \frac{6}{7} \cdot \frac{2}{6} = \frac{2}{7}$
Only one of these choices is possible, so we add the probabilities:
$\displaystyle P = \frac{4}{7} + \frac{2}{7} = \frac{6}{7}$
(Which seems entirely too large a chance for my intuition. Did I goof somewhere?)
-Dan
Am I missing something in this problem? Because it seems easy.
The first vertex can be anywhere. The second vertex needs to be on the same face. There are 7 points and 3 favorable points. So (3/7). The next one is 2 favorable points out of 6 so it is (2/6). So (3/7)(2/6)=1/7
Well, Well. I seems that we don’t agree, any of us.
There are $\displaystyle \left( {\begin{array}{c} 8 \\ 3 \\ \end{array}} \right) =56 $ possible sets of three vertices on a cube.
There are $\displaystyle \left( {\begin{array}{c} 4 \\ 3 \\ \end{array}} \right) =4 $ sets of three on any face.
Having six faces that means that there are 24 sets of three that share a face.
The probability of sharing a face ought to be: $\displaystyle \frac{{24}}{{56}} = \frac{3}{7}$