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. 
 A: Hint: The random variable takes values $0,1,2, \cdots 5$. Find the probability of each event. Also, $X^2$ takes values $0,1,4,16,25$ with the same probabilities, as computed above, and $Var(X)=E(X^2)-E(X)^2$
A: 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 $$Var(X)=np(1-p)=5\cdot\left(\frac{1}{91}\right)\cdot\left(1-\left(\frac{1}{91}\right)\right)$$
A: Finding the distribution of $X$ is feasible, but not pleasant. We show a quick way of finding the mean. The variance could also be computed in this way, but there are complications, and the resulting work may be roughly as complicated as finding the distribution. 
For $i=1$ to $13$, define random variable $U_i$ by $U_i=1$ if positions $i$, $i+1$, and $i+2$ are occupied by members of the same family. Let $U_i=0$ otherwise.
Then $X=U_1+U_2+\cdots+U_{13}$, and therefore by the linearity of expectation we have $E(X)=E(U_1)+E(U_2)+\cdots+E(U_{13}$.
The probability that $U_i=1$ is $\frac{2}{14}\cdot\frac{1}{13}$. Thus $E(U_i)=\frac{2}{(14)(13)}$, and therefore 
$E(X)=13\cdot \frac{2}{(14)(13)}=\frac{1}{7}$. 
