Problem regarding the fitting cube into sphere I was solving the following problem
Suppose I have a sphere of radius 1 metre.
The sphere is colored with red and blue such that it has disconnected regions of red and blue colors.
Now I have to make a cube that fits in the sphere such that 
Each vertex of the cube touches a red-colored region.
This is possible if one of the following option is true
a) The aggregate area of red part is $11 m^2$ .
b) The aggregate area of red part is $10 m^2$ .
c) The aggregate area of red part is NOT $11 m^2$ .
Answer given was (a)
Where aggregate area is the sum of areas of all regions.
Well I approached by calculating total surface area 
 of sphere = $12.56 m^2$
So subtracting from $11m^2$ gives me the minimum residue and so the answer follows.
Is my reasoning correct or any other explanation for this the answer??
 A: I do not see any reasoning that you put forth, esp if you were not given any options.

(I will avoid any concerns about the validity of area, and assume that all your regions are closed / measurable. I do not think you are concerned about it either.)
You will need $> \frac{7}{8} $ of the surface to be red, in order to get a cube. This can be shown using the probabilistic method. 
Consider all possible arrangements of the cube. What is the expected number of vertices that are red? 
For a given vertex, as we move it across all arrangements, then the probability that the vertex is red is just the area of the red surface, which is $ > \frac{7}{8} $. 
Applying the linearity of expectation, for all 8 given vertices, the expected number of vertices that are red is $> \frac{7}{8} \times 8  = 7 $. 
Since the expected number of vertices that are red is more than 7, and the number of vertices that are red is an integer, there must be some figuration where there are strictly more than 7 red vertices, which means that this cube has 8 red vertices. Hence we are done.

(This part might not be true, though I think it is.) 
Conversely, it is easy to find a shading of the cube sphere with less that $\frac{7}{8}$ covered, and having no cube which can fit into it. 

 Take a sphere where $x, y, z, \geq 0 $ is not colored red. One of the vertices of the cube must be all non-negative, and hence is not red.

Since $4 \pi = 12.56$ (and not 12.86 like you listed), $\frac{7}{8} * 4 \pi < 11$ (just barely), so 11 suffices.
