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StackExchange has been amazing in the past, and I want to thank the collective hive-mind in advance. I have a pretty basic probability problem that I have no idea how to solve. I'm not looking for the answer, but rather, how to tackle it.

"Suppose that an experiment leads to events A and B with the following probabilities: $P(A) = 0.6$ and $P(B) = 0.7$. Show that $P(A \cap B) \geq 0.3$."

I suppose we know the events cannot be mutually exclusive, since $P(A) + P(B) > 1$. If the events are independent, then $P(A \cap B) = P(A)*P(B) = 0.42$, which I suppose is a (lower or upper?) bound for the joint probability.

Am I on the right track? Assuming that $A$ and $B$ are not completely independent, but also not mutually exclusive, what would my next step be?

Thanks,

Adam

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2 Answers 2

up vote 0 down vote accepted

Independence was a nice idea, since it gives you a way to relate $P(A)$ and $P(B)$, but in this situation we know nothing about the dependence relationship between $A$ and $B$, so there's nothing we can say using independence. Fortunately, the explanation you're looking for turns out to be even simpler.

The key here is to draw out a Venn Diagram. If $P(A\cap B$)<0.3, then $P(A\cap\neg B)>0.3$, since $P(A)=0.6$. Similarly, $P(B\cap\neg A)>0.4$. Now add the probabilities of these regions up, and you'll get that the probability of having either $A$ or $B$ but not both is more than 1, a contradiction. Hence $P(A\cap B)\geq 0.3$.

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Thanks so much @BrettFrankel! I really appreciate it! –  Adam_G Apr 29 '12 at 1:36

Hint: you need to remember that $$ P(A\cup B)=P(A)+P(B)-P(A\cap B)\leq1 $$

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@BrettFrankel: yes, thank you. –  Artem Apr 29 '12 at 1:42

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