probability – Variance of sum of indicator random variables.

I’m reading some notes and there’s part that I’m not sure about.

Lemma 6.1. Given a sequence of random variables $X_{1}, X_{2}, ldots, X_{n}$, let $X=sum_{i} X_{i} .$ Then
$$
operatorname{Var}(X)=sum_{i=1}^{n} operatorname{Var}left(X_{i}right)+sum_{i neq k} operatorname{Cov}left(X_{i}, X_{j}right)
$$

When a random variable $X$ can be written as a sum of indicator random variable of a set of events $mathcal{A} subset Sigma$, that is
$$
X=sum_{A in mathcal{A}} I_{A}
$$

then it is possible to express the variance of $X$ in a simple way as a sum. We note that $mathbb{E}left(I_{A}right)=mathbb{P}(A)$, and that $left(I_{A}right)^{2}=I_{A}$ for all $A$, and so
$$
operatorname{Var}left(I_{A}right)=mathbb{E}left(left(I_{A}right)^{2}right)-left(mathbb{E}left(I_{A}right)right)^{2}=mathbb{P}(A)-mathbb{P}(A)^{2}=mathbb{P}(A)(1-mathbb{P}(A))
$$

Also, it is a simple check that
$$
operatorname{Cov}left(I_{A}, I_{B}right)=mathbb{E}left(I_{A} I_{B}right)-mathbb{E}left(I_{A}right) mathbb{E}left(I_{B}right)=mathbb{P}(A cap B)-mathbb{P}(A) mathbb{P}(B)=mathbb{P}(A)(mathbb{P}(B mid A)-mathbb{P}(B))
$$

Therefore by Lemma $6.1$ we see that
$$
operatorname{Var}(X)=sum_{A in mathcal{A}} mathbb{P}(A)(1-mathbb{P}(A))+sum_{A neq B} mathbb{P}(A)(mathbb{P}(B mid A)-mathbb{P}(B))=sum_{A in mathcal{A}} mathbb{P}(A)left(sum_{B in mathcal{A}} mathbb{P}(B mid A)-mathbb{P}(B)right)
$$

The last inequality, when written out, would contain terms like $mathbb{P}(A)left( mathbb{P}(B|A) – mathbb{P}(B) – mathbb{P}(B cap A) + mathbb{P}(A)mathbb{P}(B) right)$, so the last two summands must cancel each other out, which is only possible if $A$ and $B$ are independent. But do we know this as a fact?

I’m working with examples like the following:

For example, in a random graph, $T$ is the sum of the indicator random variables $I_{{A text { is a triangle }}}$ over $A subset(n)$ such that $|A|=3 .$ So to evaluate $operatorname{Var}(T)$ we can use the above sum.