# If the sum of the probabilities of events is equal to the probability of their union, does that imply that the events are disjoint?

Axiomatically, probability is a function $P$ that assigns a real number $P(A)$ to each event $A$ if it satisfies the three fundamental assumptions (Kolmogorov's assumptions):

1. $P(A) \geq 0 \ \text{for every} A$
2. $P(\Omega) = 1$
3. $\text{If} \ A_1, A_2, \cdots \text{are disjoint, then}\\ P\left(\bigcup_{i=1}^{\infty}A_i\right) = \sum\limits_{i=1}^{\infty}P(A_i)$

My question is, in the last assumption, is the converse assumed? If I show that the probabilities for a certain number of events can be added to get the probability of their union, can I directly use this axiom to claim that the events are disjoint?

• The are essentially disjoint. Dec 17, 2017 at 23:47

No, but you can conclude that the probability of any shared events is zero.

Disjoint means that $A_i \cap A_j=\emptyset$ for any $i\ne j$. You cannot conclude that, but you can conclude that $P(A_i \cap A_j)=0$ for all $i\ne j$. Any shared elements must have probability zero. Same goes for all higher-order intersections as well.

In other words, you can say, with probability 1, that none of the sets can occur together. I have seen such sets called almost disjoint or almost surely disjoint but such terminology is not standard I think.

• Late Late question here. How can I prove that $P(A_{i}\cup A_{j})=0$? Sep 1, 2021 at 12:27
• @ArghyadeepChatterjee You intended $P(A_{i}\cap A_{j})=0$ and then asked this at math.stackexchange Sep 1, 2021 at 13:27

Not really, for example, consider the uniform distribution.

Let $A_1 = [0,0.5) \cup (\mathbb{Q} \cap [0,1])$ and $A_2=[0.5,1] \cup (\mathbb{Q} \cap [0,1])$ and $A_i =\emptyset$ for $i>2$.

$P(A_1)=0.5$ and $P(A_2)=0.5$ and they sum to $1$ but they are not disjoint. $A_1 \cap A_2 \neq \emptyset$.

They can still intersect with probability measure $0$.