Question: A 10-digit long number is picked randomly and each digit's pick is independent and has an equal probability of being picked (1/9 because there's digits 1 to 9). 

Let $X = \#\{\text{missing digits}\}$ (i.e. if the # is 1357768931, X = 2 because there's no 2 or 4)

What is E(X)? Var(X)?

I am so confused as to how to start this. Some guidance please?

Answer: 
We assume on the uncertain basis of hints in the wording that for some reason the digit $0$ is forbidden. In order to avoid computing probabilities, we use the method of *indicator* random variables, and exploit the linearity of expectation. 

For $i=1$ to $9$, let $X\_i=1$ if digit $i$ is missing, and let $X\_i=0$ otherwise. Then the number $X$ of missing digits is $X\_1+\cdots+X\_9$.

We have $E(X\_1+\cdots+X\_9)=E(X\_1)+\cdots +E(X\_9)=9E(X\_1)$.  

The probability that the digit $1$ is missing is $\left(\frac{8}{9}\right)^{10}$. Thus the required expectation is $9\cdot \left(\frac{8}{9}\right)^{10}$.

For the variance, it will all be easy once we know $E(X^2)$, that is, $E((X\_1+\cdots+X\_9)^2)$. 

Expand the square. We get $X\_1^2+\cdots+X\_9^2$, plus a bunch of mixed terms.

It is easy to calculate $E(X\_1^2+\cdots+X\_9^2)$, indeed we have already done so, since $X\_i^2=X\_i$. 

There are $(9)(8)$ "mixed" terms $X\_iX\_j$ where $i\ne j$. We find $E(X\_iX\_j)$. This is $\left(\frac{7}{9}\right)^{10}$, since with probability $\frac{7}{9}$ a digit in a specified position in the list of $10$ is neither $i$ nor $j$. Now put the pieces together. 

**Remark:** The method can be adapted to "real" $10$-digit numbers, where $0$ is allowed, except as the first digit. But then $X\_1$ is "special" and things get somewhat more complicated, less symmetrical.
First thing to note: the range of $X$ is $\{0,1,\ldots,8\}$. Second thing to note: there are $9^{10}$ possible 10-digit numbers. Then the expectation is given by
\begin{align}
E(X) &= \sum\_{x=0}^8xP(X=x)\\
&= \sum\_{x=0}^8 x\left(\frac{\#\mbox{ 10-digit numbers with exactly }x\mbox{ missing digits}}{9^{10}}\right).
\end{align}
Then you have to compute the numerator for all values of $x$. A few ones are straight-forward: how many with no missing digit? How many with 8 missing digits? Can you do all of them?

Finally, to compute the variance, you can do as above to compute $E(X^2)$, and then use the fact that
$$Var(X) = E(X^2) - E(X)^2.$$