We have $20$ students and $15$ lessons. In every lesson, one student is randomly picked and asked a question. Find expected value and variance. There are $20$ students and $15$ lessons. In each lesson, one student is randomly picked and asked a question by the teacher. Find expected value of amount of students asked a question during the 15 lessons and find its variance.
If $X$ is random variable representing number of students asked, then I guess to find $\mathbf{E}X$ we have to write something like $X=X_1+X_2+\dots$, but I cannot find some smart way to define those $X_i$, so a hint would be greatly appreciated.
 A: The trick to a lot of problems about discrete random variables is to write them as a sum of indicator random variables and exploit symmetries.  This is an example of such a problem.  Since you're interested in second moments, we just need to understand what happens to any two students at a time.
There's nothing special about the numbers 20 and 15 here.  Let's say there are $s$ students and $\ell$ lessons.  Later on we'll plug in $s = 20, \ell = 15$.
Let $X_i$ be the number of times that student $i$ is asked a question.  Let $Y_i = 1_{X_i \ge 1}$ be 1 if student $i$ is asked a question at least once, and 0 otherwise.  Let $Y = Y_1 + \cdots + Y_s$ be the total number of students asked a question; we're trying to find $E(Y)$.
Now $E(Y) = E(Y_1) + \cdots + E(Y_s)$ since expectation is linear, and then $E(Y) = s E(Y_1)$ by symmetry - all the students are the same.  $E(Y_1)$ is the probability that student 1 is asked at least one question; this is $1-(1-1/s)^\ell$.  Therefore $E(Y) = s (1-(1-1/s)^\ell)$.  With the particular numbers you asked about, this is $20 \times (1 - (19/20)^{15}) \approx 10.73$.
To find the variance is a bit trickier.  Of course $Var(Y) = E(Y^2) - E(Y)^2$.  We know $E(Y)$ and thus $E(Y)^2$.  Now the thing to do here is to write $Y = Y_1 + \cdots + Y_s$, and thus we get
$$E(Y^2) = E( (Y_1 + \cdots + Y_s)^2 ) = E(Y_1^2 + \cdots + Y_s^2) + E(Y_1 Y_2 + Y_1 Y_3 + \cdots + Y_{s-1} Y_s)$$
where the first term has the squares of each of $Y_1$ through $Y_s$, and the second term has all the terms of the form $Y_j Y_k$ where $j \not = k$.  Now we can rewrite this as
$$ s E(Y_1^2) + s(s-1) E(Y_1 Y_2) $$
by symmetry.  Every student is the same so $E(Y_j^2)$ doesn't depend on $j$, and every pair of students is the same so $E(Y_j Y_k)$ doesn't depend on $j, k$ as long as $j$ and $k$ aren't the same.  
Let's turn to $E(Y_1^2)$.  Since $Y_1$ is an indicator random variable it's always 0 or 1, so $Y_1 = Y_1^2$; thus $E(Y_1^2) = E(Y_1) = (1-(1-1/s)^\ell).$
Now, what is $E(Y_1 Y_2)$?  $Y_1 Y_2$ is 1 if both students 1 and 2 get asked a question, and zero otherwise.  This is the tricky part.  We need the principle of inclusion-exclusion:
$$P(Y_1 = 1 \hbox{ or } Y_2 = 1) = P(Y_1 = 1) + P(Y_2 = 1) - P(Y_1 = 1 \hbox{ and } Y_2 = 1)$$
which we can rearrange to get
$$P(Y_1 = 1 \hbox{ and } Y_2 = 1) =  P(Y_1 = 1) + P(Y_2 = 1) - P(Y_1 = 1 \hbox{ or } Y_2 = 1)$$
Of course $P(Y_1 = 1) = E(Y_1) = 1-(1-1/s)^\ell$ and $P(Y_2 = 1) = E(Y_2) = 1-(1-1/s)^\ell$.  And we can see that $P(Y_1 = 1 \hbox{ or } Y_2 = 1) = 1-(1-2/s)^\ell$, since $(1-2/s)^\ell$ is the probability that neither $1$ nor $2$ ever gets asked a question.  Thus we have
$$E(Y_1 Y_2) = P(Y_1 = 1 \hbox{ and } Y_2 = 1) = 2 (1-(1-1/s)^\ell) - (1-(1-2/s)^\ell)$$
and with the particular numbers you asked about, this is $2 (1 - (19/20)^{15}) - (1 - (18/20)%{15}) \approx 0.2793$.
So we can finally put everything together to get
$$E(Y^2) = s (1-(1-1/s)^\ell) + s(s-1) [2 (1-(1-1/s)^\ell) - (1-(1-2/s)^\ell)] $$
and thus 
$$Var(Y) = s (1-(1-1/s)^\ell) + s(s-1) [2 (1-(1-1/s)^\ell) - (1-(1-2/s)^\ell)] - (s (1-(1-1/s)^\ell))^2 $$
(which can probably be simplified a bit).  If you plug in $s = 20, \ell = 15$ you get about $1.64895$ ,the same numerical answer as Yoel.
A: Concerning how to define $X_i$, I can see two "bad" ways to do it:
Bad way 1
\begin{cases}
X_i=1 & \text{if at least $i$ different students are asked a question}\\
X_i=0 & \text{otherwise}
\end{cases}
then you'd have to compute the expected value of a sum of non independant random variables, and I don't recommend it.
Bad way 2
\begin{cases}
X_i=1 & \text{if the $i$-th student was asked a question}\\
X_i=0 & \text{if the $i$-th student was not asked a question}
\end{cases}
and once again, you end up with non independant random variables.
Right now I don't see other ways, but I'm not convinced there's a good way to do this with $X=\sum X_i$.
Instead consider the following.
Another way (detailed because it turns out I'm bad at this)
If we can compute the probabilities $P(X=k)$,
then $\mathbb EX=\sum_{k=1}^{15} kP(X=k)$.
To get $P(X=k)$, go see the edit section of Yoël's answer.
EDIT:
Corrected the "alternative way" computation after SekstusEmpiryk's remark
New version after Yoël pointed out another error
Removed old and false computation of $P(X=k)$
