A hendecagon ($11$-sided regular polygon) is drawn with all sides and diagonals. Each of these segments is then painted (independently) yellow with probability $\frac{1}{4}$ or red with probability $\frac{3}{4}$.
Let $T$ denote the number of monochromatic triangles that were obtained - by monochromatic triangle we mean three different vertices of the initial polygon connected by segments of the of the same color.
Determine the expected value and variance of the variable $T$.
We can determine a potentially monochromatic triangle by choosing $3$ from $11$ vertices as on the picture:
So there are $ \ {{11}\choose{3}} = \frac{11!}{3! \cdot 8!} = \frac{11 \cdot 10 \cdot 9 \cdot 8!}{6 \cdot 8!} = 11 \cdot 5 \cdot 3 = 165 \ $ potentially monochromatic triangles.
$$T = \sum_{i=1}^{165} t_i \ \text{ , where: } t_i = 1 \text{ if triangle i is monochromatic, } t_i = 0 \text{ otherwise. }$$
We have then that for expected value:
$$\mathbb{E} T = \mathbb{E} \sum_{i=1}^{165} t_i = \sum_{i=1}^{165} \mathbb{E} t_i = 165 \mathbb{P}(t_i = 1)$$
We can have $3$ yellow or $3$ red segments, therefore:
$\mathbb{P}(t_i = 1) = \left( \frac{1}{4} \right)^3 + \left( \frac{3}{4} \right)^3 = \frac{7}{16}$
And we get that: $\mathbb{E} T = 165 \cdot \frac{7}{16} = \frac{1155}{16}$
For the variance I though of using the formula: $Var(T) = \mathbb{E} T^2 - \left( \mathbb{E}T \right)^2 $.
I can get $\left( \mathbb{E}T \right)^2 $ seasly, so it's time to think about $\mathbb{E} T^2$:
We know that: $\mathbb{E} T^2 = \mathbb{E} \left( \sum_{i=1}^{165} t_i \right)^2 = \mathbb{E} \sum_{i=1}^{165} t_i^2 + \mathbb{E} 2 \sum_{j=1}^{165} \sum_{i=1}^{j-1} \ t_i \ t_j = \sum_{i=1}^{165} \mathbb{E} t_i^2 + \sum_{j=1}^{165} \sum_{i=1}^{j-1} \mathbb{E} \ t_i \ t_j$
So we need $\mathbb{E} \ t_i \ t_j$, but that seems to be a bit complicated since we have take into account $2$ possibilities:
- triangles $i$ and $j$ don't have no common vertives: $\mathbb{E} \ t_i \ t_j = \mathbb{E} t_i \cdot \mathbb{E} t_j = \left( \mathbb{E} t_i \right)^2 = \left( \frac{1155}{16} \right)^2$
- triangles $i$ and $j$ have a common vertice (therefore, they have no common segments): we have the same situation as in point $1$
- triangles $i$ and $j$ have 2 common vertives (therefore they have a common segmen) - therefore, we need 5 segments in the same colour: $\mathbb{E} \ t_i \ t_j = \left( \frac{1}{4} \right)^5 + \left( \frac{3}{4} \right)^5$
And now I dont know how to put those $2$ possibilities into my formula for $\mathbb{E} T^2$. There is probably some smart way to diferentiate in the sum of $\mathbb{E} \ t_i \ t_j$ those pairs of triangles which are of type $1/2$ and those of type $3$. Any help would be much appreciated.
Rather than decomposing $\ T^2\ $ as $\ \sum_\limits{i=1}^{165}t_i^2+2\sum_\limits{j=1}^{165}\sum_\limits{i=1}^{j-1}t_it_j\ $ and looking at common vertices, I'd suggest doing it this way $$ T^2=\sum_\limits{i=1}^{165}t_i^2+\sum_\limits{j=1}^{165}\sum_\limits{i=1\\i\ne j}^{165}t_jt_i\ , $$ and looking at common sides, as Calvin Lin has suggested in his comment.
In the sum $\ \sum_\limits{i=1\\i\ne j}^{165}t_jt_i\ $ there are $3\times8=24$ values of $\ i\ $for which the $\ i^\text{th}\ $ triangle shares exactly one side with the $\ j^\text{th}\ $, and the remaining $\ 164-24=$$\,140\ $ values of $\ i\ $ correspond to triangles that share no sides with the the $\ j^\text{th}\ $. You've already obtained $\ \Bbb{E}t_jt_t=\left(\frac{1}{4}\right)^5+\left(\frac{3}{4}\right)^5\ $ for the first lot, and for the second lot you have $\ \Bbb{E}t_jt_t=\left(\frac{7}{16}\right)^2\ $ (not your $\ \left(\frac{1115}{16}\right)^2\ $, which I'm guessing was just a slip of the keyboard). Therefore $$ \Bbb{E}\sum_\limits{i=1\\i\ne j}^{165}t_jt_i=24\left(\left(\frac{1}{4}\right)^5+\left(\frac{3}{4}\right)^5\right)+140\left(\frac{7}{16}\right)^2\ . $$ I'm sure you can finish it off from there.