I was given the next question:
I am given a graph $G$ with 6 nodes and a probability $p$ of each edge to be created.
I need to prove/disprove that for each $p$, we have a case where either 3 nodes are all connected to each other or none of them are connected to each other.
Any help will be appreciated. I tried to solve it by negation but couldn't find a right solution.
As explained by Henry, this is not really a probabilistic question. Are you familiar with Ramsey theory and Ramsey number?
The following result is well known (and easy to prove, see Henry's link) : consider $K_6$ the complete graph on 6 vertices. (all vertices are pairwise connected).
For any any coloring of $K_6$ with two colors (red and blue for instance), you must have a blue triangle or a red triangle.
This relate to your problem by saying that the blue edges are present, while the red ones are not. Hence for any graph on 6 vertices, you must have three vertices forming a triangle, are 3 vertices with no edges.
This is because the diagonal ramsey number $R(3)$ is 6.
Therefore, for any random graph model, (or any way to generate a (simple) graph on 6 vertices), the result holds.