The question which I write is a changed format of what I have faced, in my project. I changed it to an attractive question and named it by "tall movie-watchers problem":
Suppose that a small cinema has only two rows of chairs (in the same height) and each row has 20 chairs. We have 6 tall invited-persons who must sit in the front row and 6 short and non-invited watchers who must not sit in the front. In the movie-release day, first, the non-invited watchers come and sit in the back row seats and then wait for the invited watchers to come and sit at their specified chair.
The question is if we generate the 6 chair numbers (for invitation letters) by a uniformly-distributed random generator, what is the probability that 50% or more of the non-invited watchers (3 or more) cannot see the movie?
Note: Suppose that non-invited persons sit randomly.
As the people are interchangeable, your sample space is just the $20 \choose 6$ ways to choose the occupied rear seats. To have everyone see the movie, the occupied ones can be chosen in $\binom{14}{6}$ ways. To have one person not see the movie, you choose his seat in $6 \choose 1$ ways and the rest of the seats in $14 \choose 5$ ways. Finally, the chance that at most two people cannot see the movie is $\frac{\binom{14}{6}+\binom{14}{5} \binom{6}{1}+\binom{14}{4} \binom{6}{2}}{\binom{20}{6}}$ and the chance that three or more cannot see the movie is $1-\frac{\binom{14}{6}+\binom{14}{5} \binom{6}{1}+\binom{14}{4} \binom{6}{2}}{\binom{20}{6}}=\frac {291}{1292}$