For my problem, a game drop table that has 8 individual items with 0.2% drop rate , 6 other individual items with 0.1% drop rate, and a last item with a 97.8% drop rate. Every time a monster is killed, 5 loot bags of the same drop table as aforementioned drops. Let's say the 14 items are group A and the last item is group B, then group A has a 2.2% chance of dropping, and group B has a 97.8% chance of dropping.
Here's the question, if you kill that type of monster 100 times, is the probability of getting group A 2.2% or is it ( 1 – probability of not getting it ) which is (1 – (0.978)^100 ) or ~89.2%?
I read the closest question, but the answer didn't really answer the question, or maybe I didn't understand the answer.
Best Answer
Asking what is the probability to get group A in any of the attempts, is equal to 1 - the probability not to get any, which is way easier to calculate.
So not getting any group A means you get B every time, 100 times in a row not related one to each other, so $P(X_1=X_2=...=X_{100}=B)=P(X_1=B)\cdot...\cdot P(X_{100}=B)=0.987^{100}$ So $P(\exists n<101: X_n=A)=1-P(\forall n<101:X_n=B) = 1-0.987^{100} $