Probability that neither friend watches favourite movie This is a question I encountered in an interview:
"There are five movie CDs and two friends A and B. The friends have an independent preference for each movie -- they rank the movies independently, and each movie has a uniform chance of being assigned any given rank. They want to watch a movie, but they can watch only one. So, friend A choosed two of his least favourite movies and removes them from the pile. Friend B then chooses his two least favourite movies from the pile and discards them. They then watch the remaining movie. What is the probability that neither gets to watch his favourite movie?"
First things first -- the number of arrangements is $5! = 120$.
I now have to find the set where 1) The B's favourite movie is ranked in the bottom 2 by A.
2) A's favourite movie is ranked in the bottom two by B.
However, I'm having a hard time trying to figure this one out. How do I proceed?
 A: The only ways that they end up watching somebodies favourite movie is 
1 - Friend B's favourite is in A's top 3 
( prob = 3/5)
2 - Friend A's favourite is also B's favourite out of A's top 3
(prob = 1/3)
the intersection between these cases is when A and B have the same favourite
(prob = 1/5)
So prob that they do watch someone's favourite is 
$$P = 3/5 + 1/3 - 1/5 = 11/15 $$
A: As you stated, we must find the number of orderings in which B's favorite movie is ranked in the bottom $2$ by A. The probability that this happens is simply $2/5$, as we can consider WLOG that B's favorite movie is movie X, and then it is just the probability movie X is one of the two out of five chosen.
Then we have $3$ movies remaining and both A and B now have their own rankings of these from $1$ to $3$. Given that B's favorite movie is now gone, we must now find the probability that B discards A's favorite movie, which will occur with probability $2/3$ for the same reasons as above.
Since our second probability is conditional on the first probability, and if the first case does not happen, then at least one of them will get their top choice, we get that the chance that nobody gets their top choice is simply $(2/5) \cdot (2/3)=\boxed{4/15} $
