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I have two non-mutually exclusive events with probability $P(A)$ and $P(B)$. In addition, I am given the intersection of both events: $P(A \cap B)$

Is it then valid to say:

$$ P(A' \cup B') = 1 - P(A) - P(B) + P(A \cup B) $$

Using the following identities:

$$ P(A' \cup B') = P(A') + P(B') - P(A' \cap B') $$ $$ P(A') = 1 - P(A) $$ $$ P(B') = 1 - P(B) $$ $$ P(A' \cap B') = P((A \cup B)') = 1 - P(A \cup B) $$

The big thing I'm not sure about here is the use of DeMorgan's laws to simplify the intersection. Does this all look right?

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  • $\begingroup$ Looks fine to me! $\endgroup$
    – user21436
    Commented Jan 22, 2012 at 20:40
  • $\begingroup$ Looks okay to me. You could also note that $P(A'\cup B')=1-P(A\cap B)$. $\endgroup$ Commented Jan 22, 2012 at 20:42
  • $\begingroup$ although not a proof, Venn's diagram can be helpful to convince yourself. $\endgroup$
    – fast tooth
    Commented May 3, 2014 at 1:20

1 Answer 1

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Your expression is right But why struggle this hard, Apply De-Morgan's Law straight away with the question at hand, like this below:

$$P(A'\cup B')=P((A \cap B)')=1-P(A\cap B)$$

You have struggled hard enough to get to the same point, since, $$1-P(A \cap B)=1-(P(A)+P(B)-P(A \cup B))=1-P(A)-P(B)+P(A \cup B)$$

Do note that, the expression I gave you here is better because, it uses just the information you have been given, viz, $P(A \cap B)$.

Hope this helps!

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  • $\begingroup$ Wow, missed that (I don't know how). Thanks very much. $\endgroup$
    – nopcorn
    Commented Jan 22, 2012 at 20:50

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