Ok then. Your sack of $12$ items has $1$ item of type 'X', $2$ items of type 'Y', and $9$ other items. Suppose you draw an item from the sack $n$ times.
To not have at least $2$ X and at least $3$ Y, you must have picked either:
- $0$ X and any number of $Y$ (ways: $11^n$),
- $1$ X and any number of $Y$ (ways: ${n \choose 1}11^{n-1}$),
- $0$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: $10^n - 9^n - {n \choose 1}9^{n-1}$),
- $1$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: ${n \choose 1}10^{n-1} - {n \choose 1}9^{n-1} - {n \vphantom{-1} \choose 1}{n-1 \choose 1}9^{n-2}$), or
- $2$ Y and any number of $X$ (the case of $0$ or $1$ X is already counted, so ways: ${n \choose 2}2^210^{n-2} - {n \choose 2}2^29^{n-2} - {n\vphantom{-2}\choose 2}{n-2\choose 1}2^29^{n-3}$).
So putting all these together, the probability of getting at least two X and at least three Y is:
$1 - (11^n + n11^{n-1} + (10^n - 9^n -n9^{n-1}) + (n10^{n-1} - n9^{n-1} -n(n-1)9^{n-2}) + (\frac{n(n-1)}{2}2^210^{n-2} - \frac{n(n-1)}{2}2^n9^{n-2} - \frac{n(n-1)(n-2)}{2}2^29^{n-3}))/{12^n}$
To understand the behaviour of this number for large $n$ (of course it gets very close to $1$, since the probability of picking so few X or so few Y is close to $0$), you may want to collect exponential terms with the same base together:
$\frac{12^n - 11^n(1 + \frac{n}{11}) - 10^n(1 + \frac{n}{10} + \frac{n(n-1)}{50}) + 9^n(1 + \frac{2n}{9} + \frac{3n(n-1)}{81} + \frac{2n(n-1)(n-2)}{729})}{12^n}$
In particular, for $n = 100$, the probability that you will have at least two X and at least three Y is about $0.998318$.