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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?

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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

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  • $\begingroup$ But P(man not bringing a dessert) should have equalled $5/12$ not $10/12$, Consequently, P(woman ∩ man not bringing dessert) should have equalled $5/12 \ *\ 1/2 = 5/24$ $\endgroup$ – Abdu Magdy Oct 21 '17 at 18:52
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    $\begingroup$ I interpret $P(\text{man not bringing a dessert})$ as $P(\text{not bringing a dessert} | \text{man}) = \frac{P(\text{not bringing dessert} \cap \text{man})}{P(\text{man})} = \frac{\frac{5}{12}}{\frac{1}{2}} = \frac{5}{6}$, but YMMV. In my experience with GRE problems, sometimes the language is indeed ambiguous. $\endgroup$ – Abhiram Natarajan Oct 21 '17 at 19:02
  • $\begingroup$ I'm sorry but I really need to understand... Based on the definition of mutually inclusive events; if we only are to choose one person, May we choose a woman and a man not bringing a dessert at the same time? $\endgroup$ – Abdu Magdy Oct 21 '17 at 19:23
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    $\begingroup$ >> May we choose a man ... same time? You certainly cannot choose that way. It's more like when you have "not bringing dessert", it can be either man or woman. Let's try this way. They should have said $P(\text{woman OR not bringing dessert})$. That way, you don't have mutually exclusive events. So by inclusion exclusion, you have $1/2 + 10/12 - 5/12$. I said this wrong in my answer. I shall edit it right away. $\endgroup$ – Abhiram Natarajan Oct 21 '17 at 20:41
  • $\begingroup$ Sorry about the confusion. $\endgroup$ – Abhiram Natarajan Oct 21 '17 at 21:00

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