From this question and answer, I've obtained the formula for binomial distribution
$$ \binom n kp^k(1-p)^{n-k} $$
and used it to obtain probabilities of a hit, where a hit is a die roll at X or lower, with probability $p$ derived from X.
For example, rolling 3 dice ($k$) hitting on a 2 or lower ($p$ = $\frac{2}6$), the probability of getting 1 hit ($n$) exactly is 44.44%, or if I want at least 1 hit it's 70.37% by adding all probabilities from $\sum_{n=1}^k$
6 ($k$) dice at 1 ($p$ = $\frac{1}6$), hoping for at least 1 hit have the probability of 66.51%
These are two separate dice rolls, one with 3@2, another with 6@1, each having independent probabilities for "at least" 1 hit, or "at least" 2 hits, etc.
How do I sum up these two separate probabilities? I would like to know, after rolling 3@2 and 6@1, what is the probability of getting at least 1 hit, and also exactly 1 hit.
Thank you for your time.
Edit 09/09/2016
I have found a near answer to what I was asking here, but I'm still looking for a way to simplify the formula, or get an answer without having to iterate so much.
For example, we'll have 3 dice hitting on 1 and 2 dice hitting on a 2 or lower. We are looking for the probability of at least one hit (and the max is 5, because we have 5 dice)
k = hits
y = 3 dice @ 1
z = 2 dice @ 2
$P(k) = \binom n kp^k(1-p)^{n-k}$
The formula ends up being
$P_y(\ge1) = P_y(1) + P_y(2) + P_y(3)$
$P_z(\ge1) = P_z(1) + P_z(2)$
$P_{y+z}(2) = P_y(1) * P_z(1)$
$P_{y+z}(3) = (P_y(2) * P_z(1)) + (P_y(1) * P_z(2))$
$P_{y+z}(4) = (P_y(3) * P_z(1)) + (P_y(2) * P_z(2))$
$P_{y+z}(5) = P_y(3) * P_z(2)$
$$P_{y+z}(\ge1) = P_y(\ge1) + P_z(\ge1) + P_{y+z}(2) + P_{y+z}(3) + P_{y+z}(4) +P_{y+z}(5) = 74.28\%$$
This works, but it includes a lot of iteration. I'm wondering if there is some way to simplify the formula, so that fewer calculations need to be made.