Probability of multiple mutually exclusive events succeeding at least once How do you calculate the probability of multiple mutually exclusive events succeeding at least once in n attempts?
I would like to know this to solve the following problem:

In a game, defeating a monster has a chance to drop only 1 item out of 4 possible items (or drop nothing). Each item has a fixed drop chance;
  A has 2% chance,
  B has 2% chance,
  C has 2.5% chance
  and D has 2.5% chance.
  What are the odds of each items having dropped at least once after defeating the monster 42 times?

Hopefully I can also understand how to apply the calculation to similar problems. Such as rolling 5 and 6 at least once in n dice rolls.
 A: Let $A_i$ denote the event that item $i$ is dropped after n defeats. Then the event in question is $\bigcap_i A_i$. $$P(\bigcap_i A_i)=1- P(\bigcup_i A_i^c)$$ where $A_i$ denotes the complement of $A_i$. 
Now use Poincaré's exclusion-inclusion principle. 
In your case $A_i^c$ is the event that item $i$ is not dropped through the 42 raids. According to the formula mentioned above
$$ P(\bigcup_{i=1}^{5} A_i^c) = \sum_{k=1}^{5} (-1)^{k-1} S_k $$
where
$$ S_k = \sum_{1\leq j_1<j_2<\ldots <j_k \leq 5} P(A_{j_1}^c \cap \ldots \cap A_{j_k}^c) .$$
For some fixed indices $j_1<j_2<\ldots< j_k$
$$ P(A_{j_1}^c\cap \ldots \cap A_{j_i}^c) = (1-p_{j_1}-p_{j_2} -... - p_{j_k})^{42} $$
where $p_j$ is the probability of the monster dropping item $j$ after one raid. In other words $S_k$ is the probability that $k$ items won't be dropped. So $S_1$ consists of the cases when 1 item is not dropped, i.e. either item $A$ is not dropped, or item $B$ is not dropped etc. But $S_1$ 
counts twice the cases when two items are missing, hence you need to subtract $S_2$. $S_2$ counts twice the cases when 3 items are missing so you need to add $S_3$ etc. 
Let's look at a simpler case: there are 3 sets and you want to calculat $P(A\cup B \cap C)$. The first thought would be to add up, i.e. $P(A)+P(B)+P(C)$, but if the sets are not disjoint, you counted some elements twice, i.e. the intersections of each 2 sets. So you subtract it from the sum. Then you realize you subtracted too the intersection of the three sets 1 to many times, hence you add it, so you end up with
$$ P(A\cup B \cup C) = P(A)+P(B)+P(C) - P(A\cap B)-P(A\cap C)-P(B\cap C) + P(A\cap B \cap C). $$
Poincaré's formula is the generalization of this for $n$ sets.
A: Let $p_i$ be the probability that item $i$ is dropped after a given defeat. Then the probability that none of the items in a set $S$ was dropped after $42$ defeats is $\left(1-\sum_{i\in S}p_i\right)^{42}$. Thus, by inclusion–exclusion, with $I$ denoting the set of all items, the probability that each item has been dropped at least once is
$$
\sum_{S\subseteq I}(-1)^{|S|}\left(1-\sum_{i\in S}p_i\right)^{42}\;.
$$
In your case this is
$$
1-2\cdot0.98^{42}-2\cdot 0.975^{42}+0.96^{42}+0.95^{42}+4\cdot0.955^{42}-2\cdot0.935^{42}-2\cdot0.93^{42}+0.91^{42}\approx0.133\;.
$$
