Probability of not A or not B Probability of A = $87\%$
Probability of B = $37\%$
Probability of both A and B = $25\%$
I've determined that the probability of A or B = $97\%$,
the probability of not A and not b = $3\%$
I'm not quite sure how to proceed to determine the probably of "not A or not B". I have a feeling it's equal to $1 - P( A \text{ and } B)$ but don't really understand if this is right, and if yes, why?
 A: This is sort of a reply to the OP's comment, but I believe it addresses the confusion in the question.
IMHO the best way to understand basic logic is to think about common language situations - though always remembering that common language is not always as precise as we need for mathematics.
Suppose you are interested in taking two subjects,say MAT234 and CSC567, but they are scheduled at the same time and you can't do both.  Then you have to skip one or the other.
Now notice that in the previous paragraph we have described the same situation in two ways.  If M means "I will take MAT234" and C means "I will take CSC567", we have

not (M and C)

and

(not M) or (not C)

These mean the same thing, which is one of De Morgan's laws.  Note that it includes the possibility that you change your mind and do neither subject: this works fine with the above formulae since in mathematics we always take "or" to mean the inclusive or, unless clearly specified otherwise.
If you say as in your comment "you don't want M or C to happen", that means you won't take either of the subjects, that is,

not ( M or C )

which is the same as saying you will do neither,

(not M) and (not C)



*

*this is De Morgan's other law.  Note it is not the same as saying you won't take both subjects as we discussed above.


Hope this helps.
