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Could anyone please help with rules for complements of events- I have difficulty grasping these.

Specifically, suppose there are two events $A$ and $B$, and the corresponding complements are denoted by $A'$ and $B'$

What is $\def\P{\operatorname{\sf P}}\P(A'\cup B')$ and $\P(A'\cap B')$ when:

  • (i) $A$ and $B$ are independent, and
  • (ii) when $A$ and $B$ are not independent.

(From some previous examples, it seems $\P(A'\cup B') = P(A') + P(B') -1$ in the independent case but I am not sure if this is correct and why this is so.)

Any intuitive explanation would be very helpful.

Thanks

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  • $\begingroup$ Welcome to MSE. Please use MathJax. $\endgroup$ – José Carlos Santos Jan 18 '18 at 8:46
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    $\begingroup$ Study your "previous examples" again. They should not be telling you that at all. $\endgroup$ – Graham Kemp Jan 18 '18 at 8:55
  • $\begingroup$ Hint: de Morgan's Laws for set operations.$$A'\cup B' = (A\cap B)'\\A'\cap B'=(A\cup B)'$$ $\endgroup$ – Graham Kemp Jan 18 '18 at 8:59
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In general, the probability rule for the complement of event is the following:

Let $ A $ be any event and let $ A^c $ be the complement of $ A $. And let $ P(A) $ be probability of event $ A $ happening.

Then $ P(A^c) = 1- P(A) $ .

Now, let's solve specifically for your example:

Case (i) $A$ and $B$ are independent.

Because of independence $ P(A \cap B) = P(A)P(B) $

By De Morgan's Law $ A^c \cup B ^c = (A \cap B)^c $

Then

$ P(A^c \cup B^c ) = P((A \cap B)^c)=1-P(A \cap B) $. Since $ A $ and $ B $ are independent, $ P(A^c \cup B^c) =1- P(A)P(B) $.

By De Morgan's Law $ A^c \cap B^c = (A \cup B)^c $

Then

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

Case (ii) $A$ and $B$ are not independent.

The only difference in this case is that you need to use conditional probabilities. So, instead of $P(A \cap B)=P(A)P(B), $ we need to use $P(A \cap B) = P(A)P(B|A)=P(B)P(A|B) $.

By De Morgan's Law $ A^c \cup B ^c = (A \cap B)^c $

Then

$ P(A^c \cup B^c ) = P((A \cap B)^c)=1-P(A \cap B) $. Using conditional probabilities above $ P(A^c \cup B^c) =1- P(A)P(B|A)= 1- P(B)P(A|B) $.

By De Morgan's Law $ A^c \cap B^c = (A \cup B)^c $

Then

$ P(A^c \cap B^c) = P((A \cup B)^c) = 1-P(A \cup B)=1-(P(A)+P(B)-P(A \cap B)) = 1- (P(A) + P(B) - P(A)P(B|A)) = 1-P(A) - P(B)+P(A)P(B|A)=1-P(A) - P(B)+P(B)P(A|B) $

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    $\begingroup$ This is really helpful. Thanks very much. $\endgroup$ – Denson Jan 18 '18 at 10:38
  • $\begingroup$ @Denson No problem. Please accept my answer so I can get reputation points. :) $\endgroup$ – G.T. Jan 18 '18 at 10:54
  • $\begingroup$ Hi G.T apologies am new to this. Not quite sure how to accept- but will do so once I figure it out $\endgroup$ – Denson Jan 18 '18 at 15:02
  • $\begingroup$ @Denson Welcome. This is great place indeed and you will love it! $\endgroup$ – G.T. Jan 18 '18 at 18:29

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