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?
The above question is a GRE practice question and the answer is 11/12.
Of the $12$ people. $11$ qualify. No need of machinery. We can save ourselves the difficulty of calculating $5+6$ by noting that precisely $1$ person doesn't qualify.
If one really wants to use machinery, let $A$ be the event the chosen one is a woman, and $B$ the event the chosen one is a dessert-nonbringing man. We want $\Pr(A\cup B)$. This is $\Pr(A)+\Pr(B)-\Pr(A\cap B)$. But the events $A$ and $B$ are disjoint. Thus our probability is $\frac{6}{12}+\frac{5}{12}$.