Suppose a die is rolled $14$ times, and $X$ be the number of faces that appear exactly 2 times. How can I find $E(X)$? It seems that this question is relatively straightforward as it resembles the binomial. However, I cannot complete the entire argument if I try to do it in accordance with $X=0$, $X=1$, and so on and then compute the probability of each. Does anyone have any ideas?
 A: Just multiply the probability that some face occurs exactly twice (which can be calculated with the binomial distribution) with $6$, the number of faces. Here, we use that expectations are additive : $E(X+Y)=E(X)+E(Y)$
A: I am using bionomial distribution to solve this question. 
$p(\text{particular face}) = \frac{1}{6}$ 
$q(\text{Not particular face}) = 1 - \frac{1}{6} = \frac{5}{6}$
Now according to bionamial distribution
$P(X=r) = C(n,r) (q)^{n-r} (p)^r$
n is the number of dice. And X is particular result.
Here X = 2 and n = 14.
$P(X=2) = C(14,2) * \left(\frac{5}{6}\right)^{12} * \left(\frac{1}{6}\right)^2$
And exactly 6 pairs. So,
= $6 * C(14,2) * \left(\frac{5}{6}\right)^{12} * \left(\frac{1}{6}\right)^2$
A: The probability $p$ that any given value, e.g., $6$, appears exactly $2$ times is given by 
$$p={14\choose 2}\cdot\left({1\over6}\right)^2\cdot\left({5\over6}\right)^{12}\ .$$
The expected number $E$ of  values appearing exactly two times is therefore given by
$$E=6p={2685546875\over2176782336}\doteq1.23372\ .$$
