Lets say I have 8 objects. From those 8 objects lets say, 2 of those objects are "optimal". I am allowed to pick an item at random- my objective is to pick atleast one of the "optimal objects". I have four tries/attemps to pick one. What's the probability of you picking atleast one "optimal" object from those 4 tries? Note that picking an item doesn't remove it from the choices the next try - the number of total objects to pick from remains the same. Picking an object does not remove it.
I approached this seemingly very simple problem in two ways, but each way gives a different probability:
Method 1: There's a $\frac{2}{8}$ chance of you picking any "optimal" object at each try. Repeating this $4$ times gives the probability: $P =\frac{2}{8} * 4 = 1$
Method 2: There's are two possibilities of outcome: either I pick an optimal object by the end or I don't. Although these outcomes obviously don't have equal probability, the existence of the possibility of not picking an "optimal" object and the fact that's the probability of not picking an optimal object isn't zero, then it would be impossible for the probability of picking an optimal object to be 1! Both outcomes can't be true and since the probability of not picking an optimal object is greater than 0, then the probability of picking an object has to be less than 1, which is a contradiction with Method 1.
What's the fallacy in my logic?