Sum of probabilities of Non-disjoint sets can be greater than one? I just want to confirm whether I am understanding this correctly: The requirements of a probability measure require the probabilities of pairwise disjoint sets to sum to one. Non-disjoint probabilities can sum to greater than one, then (I'm not sure if we would ever care about this, though).
An example I am thinking of is if we had 10 coin flips. The probability of the first coin flip being heads is $\frac{1}{2}$, the probability of the second coin flip being heads is $\frac{1}{2}$, the probability of the third coin flip being heads is $\frac{1}{2}$, and so on...
 A: Yes, the sum can be greater than $1$. (You're also right that we never really care about this - at least, as far as I know . . .)
An even simpler example:


*

*What's the probability that $8=8$?

*What's the probability that $1\not=17$?

*What's the sum of these probabilities?
A: 
The requirements of a probability measure require the probabilities of pairwise disjoint sets to sum to one.

No. Given pairwise disjoint events $A_1, A_2, ...$ a probability space $(\Omega, \mathscr F, \mathbb P)$, we have by a Kolmogorov axiom:
$$P(\bigcup_n A_n) = \sum_n P(A_n)$$
Consider $(\Omega, \mathscr F, \mathbb P) = ([0,1], \mathscr B([0,1]), \lambda)$ and $A_1 = (0.4,0.6), A_2 = (0.7,0.8), A_m = \emptyset$ for $m \ge 3$
Then
$$P(\bigcup_n A_n) = \sum_n P(A_n) = 0.3 \ne 1$$
However, if we have pairwise disjoint $A_n$ s.t. $P(\bigcup_n A_n) = 1$, then
$$P(\bigcup_n A_n) = \sum_n P(A_n) = 1$$

Non-disjoint probabilities can be greater than 1 eg
$$\sum_{n=1}^{\infty} P([0,1/n]) = \sum_{n=1}^{\infty} 1/n = \infty$$
Consider the Borel-Cantelli Lemmas.
If we have independent events $A_1, A_2, ...$, then
$$\sum_n P(A_n) = \infty \to P(\limsup A_n) = 1$$
$$\sum_n P(A_n) < \infty \to P(\limsup A_n) = 0$$
As to why we would care? Look at cases 2,3,4,5 here.
We can have $1 = \sum_n P(A_n)$, $1 < \sum_n P(A_n) < \infty$ or $\sum_n P(A_n) = \infty$
