# If $P(A \ \cup \ B) = P(A) + P(B)$, is it the case that $A$ and $B$ are disjoint?

I know that if $A$ and $B$ are disjoint events, then $P(A \cup \ B) = P(A) + P(B)$. However, is the converse true as well? Thanks.

• Disjoint events in the sense of probability simply means $P(A \cap B) = 0$, not necessarily that as sets $A\cap B = \emptyset$. For this reason (avoiding confusion) it is sometimes stated instead that $A,B$ are mutually exclusive events. Jan 29, 2016 at 15:13
• @hardmath, I don't agree with your definition of disjoint. Can you cite any references? Jan 29, 2016 at 15:18
• @hardmath I don't agree as well. If they are disjoint, then $P(A \cap B) = 0$, but the converse doesn't necessarily hold. Jan 29, 2016 at 15:20
• @TonyK: I'm drawing a distinction between the use of "disjoint" in probabilities and in sets, and suggesting to avoid confusion by using "mutually exclusive" instead of "disjoint". The terminology that events $A,B$ are disjoint (while not ideal) is certainly tolerable where the probabilistic context is clear (esp. when there is no ascription of sets to the events). Jan 29, 2016 at 15:23
• @hardmath: I am asking whether you have any evidence for your claim that this is the default meaning of "disjoint" in probability theory. Jan 29, 2016 at 15:26

No, all you can deduce is that $P(A \cap B) = 0$, because $$P(A \cup B) = P(A) + P(B) - P(A \cap B)$$ It doesn't mean that $A \cap B$ is empty.

No for example suppose you have a uniform distribution on $[0,1]$. Let $A=[0,1/2]$ and $B=[1/2,1]$. Then $P(A\cap B)=0$ but $A\cap B=\{1/2\}\not=\emptyset$.

• Thank you for coming up with an excellent counterexample! (+1) Jan 29, 2016 at 15:25
• Why is $P(A\cap B)=0$? Jan 29, 2016 at 15:29
• Because the uniform distribution on $[0,1]$ is continuous: the probability of any single particular point is $0$. Jan 29, 2016 at 15:31
• @Inuyaki If you have any continuous random variable X, $$\mathbb{P}(X = c) = \int_{c}^{c}f_{X}(x)\text{ d}x = 0$$ Jan 29, 2016 at 15:32
• @Inuyaki Another way to see that $P(A \cap B) = 0$ is to note that $P(A \cup B) = P(A) + P(B) - P(A \cap B)$ for any subsets $A$ and $B$ of $[0,1]$ . Now $P(A \cup B) = 1$, $P(A) = 1/2$, and $P(B) = 1/2$. Plugging in these values reveals $P(A \cap B) = 0$.
– Joel
Jan 29, 2016 at 15:42

EDIT: This is incorrect, disregard. The comments have valuable discussion and counterexamples.

Yes. Assuming $A$ and $B$ are independent, by definition, $P(A \cup B) = P(A) + P(B) - P(A \cap B)$. If we have $P(A \cup B) = P(A) + P(B)$, then $P(A \cap B) = 0$, which is the definition of disjoint events.

• Why do you need independence? Jan 29, 2016 at 15:10
• Hmm, is that true? Doesn't it just mean that $A\cap B$ has probability zero, not that it is empty. (Unless only empty sets have probability zero.)
– MPW
Jan 29, 2016 at 15:10
• Why do we need to assume that $A$ and $B$ are independent? What do you mean "by definition"? Jan 29, 2016 at 15:11
• Is that really the definition of disjoint events? What if $X$ is uniformly distributed on $[0,1]$, $A$ is $(X \le \frac12)$, and $B$ is $(X \ge \frac12)$? Jan 29, 2016 at 15:11
• A counterexample can even be given on a finite probability space, e.g. if $X$ is a random variable with values in $\{1,2\}$, with $\mathbb{P}(X=1)=1$ and $\mathbb{P}(X=2) = 0$. Letting $A = \{ 1, 2 \}$ and $B = \{ 2 \}$, we see $A$ and $B$ are independent, non-disjoint events such that $\mathbb{P}(A \cup B) = \mathbb{P}(A) + \mathbb{P}(B)$. Jan 29, 2016 at 15:18