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
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) $