How to calculate the probability of at least an item dropping. So I play this online MMORPG where a monster drops items once killed.
There are a total of 10 items that can drop. Each of them has a 1% chance of dropping. A maximum of 2 items can drop at once. What are the odd's that at least one of the item drops?
Question 2
The actual number of items is 15 different items.
7 items have a .5% chance of dropping
1 item has a .6% chance of dropping
5 items have a 1% chance of dropping
2 items have a .2% chance of dropping
Assuming 2 items can drop at a max, what are the new odds then of getting at least 1 item to drop?
There are a total of 10 items that can drop. Each of them has a 1% chance of dropping. A maximum of 2 items can drop at once. What are the odd's that at least one of the item drops?

First, note that the requirement that &quot;A maximum of 2 items can drop at once&quot; directly implies that the drops cannot be independent.  Without much more information regarding how the choices are being made, there's no way to give you a mathematically correct answer.
Here are two simplified options (I'll point out a big problem with both afterwards):

Roll a $100$-sided die ten times, once for each item.  If the roll is a $1$, then that item is added to a vector of possible drops.  After you've finished all ten rolls, check how many items are in the vector.  If it's two or fewer, the player receives all (0, 1, or 2) of them.  If it's more than two, roll a $\binom{k}{2}$-sided die to choose which two the player will receive (each side corresponds to a different pair of items).
Roll a $100$-sided die once.  If the roll result $R_1$ is greater than $10$, then nothing drops for the player, and we don't continue.  Otherwise, item #$R_1$ drops for the player and we roll a $100$-sided die again.  If the second roll $R_2$ is between $1$ and $10$ inclusive but not $R_1$ again, then item #$R_2$ drops as well.

The probability that at least one drop occurs is then either $1-(1-0.01)^{10} \approx 0.095617925$ in case (1) or $0.1$ in case (2).
The problem in both cases is that the actual drop rates are wrong.  Choosing a pair of items in the first case lowers the drop rates to less than 1% while rolling a second time in the second case raises the drop rates to above 1%.  If I were to design a system to drop at most two items out of ten, I would do the second case, as it's less work-intensive and players tend to under-estimate small probabilities more often than they over-estimate them (there's a statistical reason for this), but if we want to fix it so the final drop probabilities are 1%, we need to fix both of them by solving for an unknown probability $p$ rather than thinking of it as a $100$-sided die roll.
Doing this yields the following equations in each case:

$$(1-p)^9(1+4p) = 0.95 \implies p\approx 0.0100116 \\ \implies \text{Prob. that at least one item drops} = 1-(1-p)^{10} \approx 0.0957241$$
$$p + 9p^2 = 0.01 \implies p\approx 0.0092328 \\ \implies \text{Prob. that at least one item drops} = 10p \approx 0.092328$$


In any case, I'm not touching question 2, as it has the same lack of independence shown here, with the same problems, but with the added complication of not being able to assume the correction factors are equal...
$$1-(1-0.01)^{10}$$
We found this by modelling it as a Binomial random variable.
Question 2 answer
First off, you should try to ask your question in one go (or ask a separate question). This makes it fair to the answerer(s).
Anyway, it is easier to ask the complement question: what is the chance of nothing dropping. This is the same strategy as earlier, really. After you answer this question (getting the answer $x$), then take $1-x$ and you have your required answer.