Probability of Selecting Jars I have 20 jars, and 15 jars have 2 balls each while 5 jars have 1 ball each. If, from the 35 balls, I select 30 at random, what's the probability that I have sampled x (15, 16, etc.) number of jars (each jar being the same except that 15 held 2 balls)?
I've tried a combinatorics approach, but I've put myself at a dead end.
 A: It helps organize our thoughts if we temporarily assume that every ball is labeled differently.  Say, for example, with its urn number, and it whether it is ball1 or ball2 (in the case of multiple balls in same urn).
For exactly 15 jars to have been used:

 the jars must have been specifically the ones with two balls in them each.

This occurs with probability:

 $1/\binom{35}{30}\approx 3\cdot 10^{-6}$

For exactly 16 jars to have been used, one of two things will be true:

 You picked two balls each from each of 14 of the jars with two balls available, one from the remaining jar with two balls available, and one from a jar with only one available,  OR two balls each from each of 14 jars with two balls available and one from each of two jars with one available.

We count how many ways we can accomplish this:

 In the first case, choose which of the jars with two balls had only one of the balls used.  Then, which of the balls.  Finally, choose which of the jars with only one ball was used.  There are then $15\cdot 2\cdot 5=150$ possibilities for this case.  The second case, pick which jar with two balls we didn't take any of, and then pick which of the jars with only one available were used.  There are then $15\cdot\binom{5}{2}=150$ possibilities.  The probability of occurrence is then $300/\binom{35}{30}$.

Equivalently worded, we may think of where the locations of the unused balls are kept.  Doing so will arrive at the same conclusions as above.  You may continue similarly for the remaining cases.

For 18 jars to have been used, keeping track of cases becomes messier than before, so I will introduce additional notation.  Consider the triple $(A,a,b)$.  Let $A$ be the number of jars which have two available that had both used.  Let $a$ be the number of jars which have two available that had only one used.  Let $b$ be the number of jars used which had only one available.
For example, in the case of $15$ jars used, we had $(15,0,0)$.  In the case of $16$ jars used, we had $(14,1,1)$ or $(14,0,2)$.  Note that the total number of balls taken will be $2A+a+b$.  The total number of jars used will be $A+a+b$.  Also, $A+a$ cannot exceed 15, and $b$ cannot exceed five.
Note also that to have $k$ jars used, $A$ must be $30-k$.  This is because we want $A+a+b=k$ and $2A+a+b=30$.  Subtracting the first equation from the second yields the result.

For $18$ jars we have the following possibilities:
$\begin{array}{|c|}\hline
(12,1,5)\\
\hline (12,2,4)\\
\hline (12,3,3)\\
\hline\end{array}$
$(12,0,6)$ is not possible since there are not six available jars with only one available.  Similarly, $(12,4,2)$ is not possible since there are not enough jars with two available for that to happen.
Counting each case:  Pick the jars used for $A$.  Pick the jars used for $a$.  Pick which of the balls was used for each jar in $a$.  Pick the jars used for $b$.
$\begin{array}{|c|c|}\hline
(12,1,5)&\binom{15}{12}\binom{3}{1}\cdot 2\cdot \binom{5}{5}\\
\hline (12,2,4)&\binom{15}{12}\binom{3}{2}\cdot 2^2\cdot \binom{5}{4}\\
\hline (12,3,3)&\binom{15}{12}\binom{3}{3}\cdot 2^3\cdot \binom{5}{3}\\
\hline\end{array}$
Summing over all cases gives $66430$ possible outcomes.  Dividing by the sample space size, $\binom{35}{30}$ yields the probability $\frac{4745}{23188}\approx 0.2046$
