# Doubt regarding mutually exclusive events and their complements

Premise: My understanding is that if two events are mutually exclusive, it need not be that they are complementary. However I came across a question that said the following - If $P(A') = \alpha$ and $P(B') =\beta$ then $P( A\cap B)$ must be ?

The answers is greater than or equal to $1 - \alpha- \beta$. I get the idea, but what if $A$ and $B$ are mutually exclusive - Wouldn't the answer given be wrong then? Because my premise is that disjoint does not mean $\alpha + \beta = 1$, so $1 - \alpha- \beta$ would have a positive value clearly not equal to zero!.

• Are you sure you mean $P(A')$ but $P(B)$? That doesn't seem right for treating $\alpha$ and $\beta$ with the same sign later on. – hmakholm left over Monica Apr 24 '18 at 11:33
• If nothing else is specified you can just take $A$ and $B$ to be mutually exclusive and then $P(A\cap B) = 0$. What you can do with the intersection is to upper-bound it by $\mathrm{min}\left[P(A), P(B)\right]$ (when an event implies the other one), so $\mathrm{min}\left[1-\alpha, \beta\right]$ (assuming by $A^\prime$ you mean the complementary of $A$). – derpy Apr 24 '18 at 11:37
• @HenningMakholm I'm sorry, had made a mistake while typing out the question. I've made the edits now – Meera Unni Apr 24 '18 at 11:46

If $P(A^\complement)=\alpha$ and $P(B^\complement)=\beta$, then $P(A\cap B)\ge 1-\alpha-\beta$.
(Note that here $\beta$ is the probability of the complement of $B$, not of $B$ itself).
This doesn't contradict the fact that $P(A\cap B)=0$ when $A$ and $B$ are mutually exclusive, because in that case $\alpha+\beta>1$ so the bound from the claim is negative (and therefore trivially satisfied).
• I did'nt get your last line, could you please explain a bit more ?, what I felt was if A intersection B is 0 and A union B is 1 then we can safely say that 1 -$\alpha$ -$\beta$ = 0, but I don't think this could be the case if A union B is not equal to 1 – Meera Unni Apr 24 '18 at 11:51
• @madhavU: In that case you're right that $1-\alpha-\beta$ is not $0$. But then $1-\alpha-\beta$ is negative, and then the probability of $A\cap B$ is certainly greater than this negative number. – hmakholm left over Monica Apr 24 '18 at 12:08
• You may be getting confused by the fact that $\alpha$ and $\beta$ are not the probabilities of $A$ and $B$, but of their complements. So $1-\alpha-\beta = P(A)+P(B)-1$. – hmakholm left over Monica Apr 24 '18 at 12:08