Suppose we pick $s$ objects independently and at each step probability that object is defective is $1/h$ then probability that each object is not defective $s$ steps is $$(1-1/h)^s$$ which as $h$ large enough is well-approximated by $$e^{-s/h}$$
My query is does this mean that as $s$ exceeds a threshold existence of at least one object is defective is certain?
When can you make this claim?
Probabilistic method seems to have these kind of claims about existence of certain objects. Does the claim follow from probabilistic method in this case. If not how to make this precise from the view of probailistic method. That is how to show existence of such a defective object from probabilistic method after $s$ exceeds a threshold?
The question seems to me to be confused. As you've computed, as long as $h > 1$ there is some positive probability that you will not draw any defective objects, so there's no way of guaranteeing that you will find defective objects.
This is not how the probabilistic method works at all. The probabilistic method applied to a situation like this gives results of the following form: if you can prove that the expected number of defective objects is positive, then defective objects exist (but may be arbitrarily hard to find).