If given a set of events that are known to be 2-way universal, is there a closed-form solution for the probability of their intersection? If so, how would you go about finding it?
I know that for 2 independent events, $$\Pr(A \cap B) = \Pr(A) \Pr(B)$$
But what about for the intersection of multiple events, i.e. $$\Pr( \cap_{i=1}^n X_i)$$
If you only know that $$\text{Var}(X + Y) = \text{Var}(X) + \text{Var}(Y)$$ but $$\text{Var}(X + Y + Z) \ne \text{Var}(X) + \text{Var}(Y) + \text{Var}(Z)$$
That is, events are pair-wise independent, but not 3-way independent, 4-way indpendent, etc.
If $X_1, X_2, ..., X_n$ are independent events then
$$\Pr(\bigcap_{i=1}^n X_i) = \prod_{i=1}^{n} Pr(X_i)$$
If the events are merely pairwise independent or threewise independent or not any kind of independent at all, use the chain rule of probability:
$$\Pr(\bigcap_{i=1}^n X_i) = Pr(X_n | \bigcap_{i=1}^{n-1} X_i) Pr(X_{n-1} | \bigcap_{i=1}^{n-2} X_i) ... Pr(X_3 | \bigcap_{i=1}^{2} X_i) Pr(X_2 | X_1) P(X_1)$$
That's one way of computing it. I started with $X_n$ then I went to $X_{n-1}$ all the way up to $X_2$ and then finally $X_1$. The order doesn't matter.
About your second conern, we cannot computer $Var(X)$ if $X$ is an event. $X$ has to be a random variable. We can computer $Var(1_X)$ where $1_X$ is an indicator random variable on the event $X$.
The random variables $1_X, 1_Y and 1_Z$ are independent if and only if the events $X, Y and Z$ are independent.
As for $Var(X+Y+Z)$ where $X,YandZ$ are random variables, we have
$$Var(X+Y+Z) = Var(X) + Var(Y) + Var(Z) + 2Cov(X,Y) + 2Cov(X,Z) + 2Cov(Y,Z)$$
If the random variables are pairwise independent, then we already have
$$Var(X+Y+Z) = Var(X) + Var(Y) + Var(Z)$$
We do not need the random variables to be independent to conclude that. Them being pairwise independent is already good enough because we already have the covariances being zero.