"Expected Probability": Withdrawing a white ball from a bucket with a random variable X = number of whites I have a problem with the subject of "Expected Probability" (I don't really know what is the right name for it).
This is an example of a question: (I am not looking for the specific answer, just for the idea)
What is the probability to withdraw 2 white balls (without returning) from a bucket, that contain X white balls and 120-X black balls. when X is a random variable.
from what I understand, I should represent the probability as function of X and then to get the expected value of it. 
        P(two white)     →    X(X-1)/(120∙199)   →   E(X(X-1)/(120∙199))

is this correct? what am I really doing here? maybe you can forward me to a relevant place that I can read about this subject ?
thanks
 A: For any fixed value of $X$, the probability you seek is
$$
\frac{\binom{X}{2}}{\binom{120}{2}} = \frac{X (X-1)}{120 \cdot 199}
$$
as you write.
However now that $X$ is a random variable, it has some distribution. It is clear that $X$ is an integer in $[0,120]$, let's say that $\mathbb{P}[X=k] = p_k$ for $k \in [0,120]$. In case $k=0,k=1$ it's not possible to draw two white balls... Then,
$$
\mathbb{P}[\text{draw 2 white balls}]
 = \sum_{k=2}^{120} p_k \frac{k (k-1)}{120 \cdot 199}
 = \frac{1}{199 \cdot 120} \sum_{k=2}^{120} p_k k (k-1),
$$
which depends on the distribution of $X$.
EDIT Further simplification is possible.
Note that $\mathbb{E}[X^n] = \sum_{k=0}^{120}k^n p_k$.
Then,
$$
\begin{split}
\mathbb{P}[\text{draw 2 white balls}]
 &= \frac{1}{199 \cdot 120} \sum_{k=2}^{120} p_k k (k-1) \\
 &= \frac{1}{199 \cdot 120}
    \left( \sum_{k=0}^{120} p_k k (k-1)
           - p_0 \cdot 0 \cdot (-1) - p_1 \cdot 1 \cdot 0 \right) \\
  &= \frac{1}{199 \cdot 120} \sum_{k=0}^{120} p_k k (k-1) \\
  &= \frac{1}{199 \cdot 120}
     \left(
           \sum_{k=0}^{120} k^2 p_k - \sum_{k=0}^{120} k p_k
     \right) \\
  &= \frac{\mathbb{E}[X^2] - \mathbb{E}[X]}{199 \cdot 120}
\end{split}
$$
A: You wrote:
        P(two white)     →    X(X-1)/(120∙199)   →   E(X(X-1)/(120∙199))

It seems hard to talk students out of using arrows in this way.  I don't know where they learn to do that.
One could write
$$
\Pr(\text{two white}\mid X) = \frac{X(X-1)}{120\cdot199},
$$
i.e. the conditional probability of getting two white balls given the value of the random variable $X$ is $X(X-1)/(200\cdot199)$.  Then we'd have
$$
\Pr(\text{two white}) = \mathbb E(\Pr(\text{two white}\mid X)) = \mathbb E\left(\frac{X(X-1)}{120\cdot199}\right).
$$
That last quantity depends on the probability distribution of the random variable $X$.
