Probability, random line up Five distinct families arrive to a party. Each family consists of 3 people. The 15 participants of the party are arranged randomly in a line. 
Let X be the number of families that their members sit next to each other. Find E[X] and Var(x).

My attempt: Just go straight to find out the pmf of X, P(X=1), P(X=2)... up to P(x=5). Does this question ask for all members of at least 2 of the family members? If it is the second case, I have no idea how to do it. 

Suppose $$X \sim B\left(5,\left(\frac{1}{91}\right)\right)$$ n = 5 because you can have up to 5 families sitting next to each other, also in order for either of them to sit next to each other, each person must sit next to a relative, there are 3 persons in each family and 15 people overall so P(Person sitting to one relative) = 2/14 and P(Person is sitting next to two relatives) is:$$p_x(k)=\left(\frac{2}{14}\cdot\frac{1}{13}\right)=\frac{1}{91}$$, the rest is just plug and chug. $$E[X] = \sum_{k=0}^n k \cdot {n \choose k} p^k (1-p)^{n-k} = np = 5\cdot \left(\frac{1}{91}\right)$$ and $$Var(X)=np(1-p)=5\cdot\left(\frac{1}{91}\right)\cdot\left(1-\left(\frac{1}{91}\right)\right)$$
