Why independent events are never mutually exclusive?

I am a novice at probability and I came across this statement in my maths book. Two independent events are never mutually exclusive. I am not getting why it is so. Can anyone explain using a ven diagram?

Let $A, B$ are mutually exclusive, and $P(A)>0,P(B)>0$. Then $P(A\cap B)=0$. But if they are independent then $P(A\cap B)=P(A)P(B)>0$. Contradicion.

• In other words the claim is justified if both events $A$ and $B$ have positive probability. But if not (as in Ross's example) the claim may be false. Commented Jul 29, 2017 at 4:32
• @hardmath yes, indeed! For example degenerate at a single point distribution, this will not hold. Commented Jul 29, 2017 at 4:33

It is not strictly true. If you have two events and A always happens and B never does(or the reverse), they are mutually exclusive because you never have both happen together. The probability of A does not depend on whether B happens because it is $1$ and similarly the probability of $B$ does not depend on whether A happens because it is $0$. In a Venn diagram all the events are in the $A \cap \lnot B$ region.

Outside this corner case the statement is true. If you know A happened you know B did not, so independence is violated.

• Sir, what is $A \cap \lnot B$ (not familiar with the notation!) Commented Jul 29, 2017 at 4:29
• It is the intersection of $A$ and $\lnot B$. In the standard 2 statement Venn diagram it is the full circle of $A$ less the lens of $A \cap B$. The answer you accepted assumes $P(A) \gt 0, P(B) \gt 0$. I am pointing out that if that assumption (which is not part of the question) is violated, the statement is false. Commented Jul 29, 2017 at 4:32
• Also $\neg B$ is the complement of $B$. Commented Jul 29, 2017 at 4:35
• @RossMillikan thanks.... that is why I put that conditions, as I said in the comment section this will not hold for degenerate at a single point distribution! Commented Jul 29, 2017 at 4:39
• @GrahamKemp thanksssss, I see here people use that notation but did not know the meaning. Every time I see this notation, I skip that post... so at last I had to ask it. Btw why people use that notation... $B^c$ or $B'$ are enough,isn't!!! Is this became useful because of logic gate or something like that... Commented Jul 29, 2017 at 4:46