"by definition A and B R.V are independent means that: $p(A∪B)=p(A)+p(B)$ right?" No, absolutely not right.

Question

Can someone please explain why?

Isn't $p(a,b)=p(a)*p(b) $ equivalent to $p(A∪B)=p(A)+p(B)$?

If not can you please give a counterexample or something?

Thanks a lot!

Answer 1

You are mixing up notation; $P(A,B)$ is the same as $P(A\cap B)$ or $P(AB)$.

Now, $A$ and $B$ are said to be independent events exactly when $P(A\cap B) = P(A)P(B)$ and since $$P(A\cup B) = P(A)+P(B)-P(A\cap B)$$ always holds regardless of whether $A$ and $B$ are independent or dependent events, you can see immediately that $P(A\cup B)$ equals $P(A)+P(B)$ exactly when $P(A\cap B) = 0$. But, when $A$ and $B$ are independent, you know that $P(A\cap B) = P(A)P(B)$ and so $P(A\cap B) = 0$ exactly when at least one of $P(A)$ and $P(B)$ is $0$.

For independent events $A$ and $B$, $P(A\cup B) = P(A)+P(B)$ holds only when at least one of $P(A)$ and $P(B)$ is $0$.

Answer 2

Suppose $p(A)=p(B)=1$. They are independent. We have $p(A\cup B)=1$ but $p(A)+p(B)=2$.