Dice: Rolling at least N successes where number of succeses vary by dice value Rules
I have four different types of dice: six-, eight-, ten- and twelve-sided (d6, d8, d10 & d12, respectively).
The number of successes vary by the value rolled (and thus indirectly by dice type).

*

*One success is gained by rolling 6 or 7.

*Two successes are gained by rolling 8 or 9.

*Three successes are gained by rolling 10 or 11.

*Four successes are gained by rolling 12.

This means that a 1d6 can result in at most 1 success, 1d8 1-2 successes, 1d10 1-3, and 1d12 1-4.
Successes are added together after the roll, so rolling 6 dice and getting [12, 3, 8, 7, 10, 1] will result in 4 + 2 + 1 + 3 = 10 successes.
Input is the number of dice and how many sides they have, and the minimum amount of successes I want to achieve.
Question
My main question is this:

Given that I roll a known combination of d6s, d8s, d10s and d12s, how do I calculate the probability of rolling N or more successes? Q1

(though feel free to answer any other questions in this post as well, indexed Q$n$ for your convenience)
Context
I know how to calculate the probability of rolling at least $N$ successes for an arbitrary number of d6's, since they can only yield one success at most.
I am stuck, however, when it comes to calculating at least $N$ successes when rolling a mix of differently sided dice, where some of them can yield more than one success.
For example, with $5$d6, $1$d8, $1$d12, how likely am I to roll $\geq$ 4 successes? Q2

EDIT: It's been brought to my attention that there is no closed form solution to this question.
That is fine; any solution or clever approximation that's more efficient than running 100k simulated rolls is a sufficient answer.
Can the problem be split into separate probabilities that can later be combined? E.g., given 5d6 & 1d12 and that I'm looking for the probability of at least $k$ successes, can I calculate the probabilities for each die type separately and later combine them somehow? Q3
Also, how would I go about calculating $\geq k$ successes for 1d12? For 2d12? For $n$d12? Q4
Currently, I can 'solve' the problem by running a simulation, but it irks me that I am not able come up with anything better.
 A: A straightforward combinatorial answer.
I assume that all dices are fair, that is any side of any $d_i$ has a probability $1/i$ to be dropped after a roll.
Let for any $i$ and any non-negative integer $k$, $P_i(k)$ be a probability to have exactly $k$ successes. For instance $P_8(0)=5/8$, $P_8(1)=1/4$, $P_8(2)=1/8$, and $P_8(k)=0$ otherwise.
It follows that if we have $i$ fixed and have $n$ instances of a dice $d_i$ then for each non-negative integer $k$ a probability $P_i(k,n)$  to have exactly $k$ successes is
$$\sum_{k_1+k_2+\dots k_{n}=k\hskip5pt} \prod_{j=1}^{n} P_i(k_j).$$
In particular, $P_i(k,n)=0$ iff
$$(i=6 \wedge k>n) \vee (i=8 \wedge k>2n) \vee (i=10 \wedge k>3n) \vee (i=12 \wedge k>4n).$$
In particular, if $n=0$ then $P_i(0,0)=1$ and $P_i(k,0)=0$ for each $k>0$.
If $n>1$ then probability $P_i(k,n)$ can also be calculated recurrently by a formula
$$P_i(k,n)=\sum_{k_1+k_2=k} P_i(k_1)P_i(k_2,n-1).$$
In special cases an expression for $P_i(k,n)$ can be simplified. For instance, $P_6(k,n)={n\choose k} 5^{n-k}6^{-n}$.
Finally, if we have $i$ fixed and have $n_i$ instances of a dice $d_i$ for each $i$, for each non-negative integer $k$ a probability $P(k)$  to have at least $k$ successes is
$$\sum_{k_1+k_2+k_3+k_4\ge k} P_6(k_1,n_1) P_8(k_2,n_2) P_{10}(k_3,n_3)P_{12}(k_4,n_4).$$
In particular, $P(k)=0$ iff $k>n(6)+2n(8)+3n(10)+4n(12)$.
