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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!.

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  • $\begingroup$ 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. $\endgroup$ – hmakholm left over Monica Apr 24 '18 at 11:33
  • $\begingroup$ 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 $). $\endgroup$ – derpy Apr 24 '18 at 11:37
  • $\begingroup$ @HenningMakholm I'm sorry, had made a mistake while typing out the question. I've made the edits now $\endgroup$ – Meera Unni Apr 24 '18 at 11:46
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The claim you're qouting is either misquoted or wrong. What is right is

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).

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  • $\begingroup$ 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 $\endgroup$ – Meera Unni Apr 24 '18 at 11:51
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    $\begingroup$ @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. $\endgroup$ – hmakholm left over Monica Apr 24 '18 at 12:08
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    $\begingroup$ 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 $1-\alpha-\beta = P(A)+P(B)-1$. $\endgroup$ – hmakholm left over Monica Apr 24 '18 at 12:08
  • $\begingroup$ Thank you so much I get it now!! $\endgroup$ – Meera Unni Apr 24 '18 at 13:01

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