2
$\begingroup$

Decide whether this statement is true or false: Let $(\Omega, \mathbb{F}, \mathbb{P})$ be a probability space, if for two events $A,B \in \mathbb{F}$ $\hspace{1cm}$ $\mathbb{P}(A \cup B)= \mathbb{P}(A) + \mathbb{P}(B)$ holds, then $ A \cap B= \emptyset$

In the solutions it is stated that this statement is false, however I do not really understand why? Isn't the definition that for two mutually exclusive events $\mathbb{P}(A \cup B)= \mathbb{P}(A) + \mathbb{P}(B)$ and $\mathbb{P} (A \cap B) = \emptyset$?

| cite | improve this question | |
$\endgroup$
  • 1
    $\begingroup$ Not quite. There are possible events which have probability $0$. Say you are drawing a number $X$ uniformly on $[0,1]$. Let $A$ be the event "$X≤.5$" and let $B$ be the event "$X≥.5$". Then $P(A)=P(B)=\frac 12$ and $P(A\cup B)=1$, but $A\cap B\neq \emptyset$. $\endgroup$ – lulu Jun 29 at 12:38
  • $\begingroup$ @lulu: thank you for the example $\endgroup$ – xxDianaxx Jun 29 at 13:25
4
$\begingroup$

It is always true that $P(A\cup B)+P(A \cap B)=P(A)+P(B)$. Hence $P(A\cup B)=P(A)+P(B)$ iff $P(A\cap B)=0$. But that does not imply that $A \cap B $ is the empty set. It can be any event of probability $0$.

For example take $A=[0, \frac 1 2]$ and $B=[\frac 1 2 ,1]$ on $[0,1]$ with Lebesgue mesure.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the explanation and the example, so P(A)=0.5=P(B), and $P (A \cup B)= P(A)+P(B)-P(A \cap B)$ $\rightarrow 1=0.5+0.5-P(A \cap B) \rightarrow P(A \cap B)=0$ with $P(A \cap B)=P(0.5)$ but why exactly is $P(0.5)=0$? $\endgroup$ – xxDianaxx Jun 29 at 13:09
  • 1
    $\begingroup$ That is a property of Lebesgue measure. $\endgroup$ – Kavi Rama Murthy Jun 29 at 13:12
  • $\begingroup$ thank yo, will look it up $\endgroup$ – xxDianaxx Jun 29 at 13:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.