GRE probability question: What is the probability that it is a woman, or a man who is not bringing a dessert? 
John invites 12 friends to a dinner party, half of which are men. Exactly one man and one woman are bringing desserts. If one person from this group is selected at random, what is the probability that it is a woman, or a man who is not bringing a dessert?

I know that the two events are mutually exclusive; so that, the answer is simply $6/12 + 5/12 = 11/12$. However, when I looked at the answer in the book that I study from, I found the same result but an approach that's really weird for me. Here is it literally:
$$P(woman) = 6/12 = 1/2$$
$$P(not\ bringing\ a\ dessert\ ) = 10/12 = 5/6$$
$$P(woman\ and\ not\ bringing\ a\ dessert) = 1/2\ *\ 5/6=5/12$$
$$P(woman\ or\ a\ man\ not\ bringing\ a\ dessert) = 1/2 + 5/6 - 5/12 = 11/12$$
May anyone explain the reasoning behind that approach? And is it technically right?
 A: Your approach is certainly right. The book's approach is right as well. It is just that they should have said $P(\text{woman OR man not bringing dessert}) = P(\text{woman OR not bringing dessert})$ (this works because there are just two genders). Then you don't have mutually exclusive events, so you have to use inclusion-exclusion. $$P(\text{woman OR not bringing dessert}) = P(\text{woman}) + P(\text{not bringing dessert}) - P(\text{woman} \cap \text{not bringing dessert}).$$
The term $P(\text{woman} \cap \text{not bringing dessert}) = 5/12$, thus they have $1/2 + 10/12 - 5/12 = 11/12$.
What you are doing is you are breaking up your space into mutually exclusive events. You are saying it is equivalent to calculating $P(\text{woman} \sqcup \left(\text{not woman AND no dessert}\right))$. That way you have $1/2 + 5/12$.
There's another way you can solve this problem. This probability is equal to $1 -
 P(\text{man bringing dessert}) = 1 - 1/12 = 11/12$.
NOTE: Edited after a misinterpretation pointed out by @Abdu Magdy
