You've lost me here. What do you mean by "probability of samples sizes"? The probability of what?
I assume that by "sample sizes below 50%" you mean "samples of 500000 parts or less". If there are 1% defective parts, then there would be an average of .01(500000)= 5000 defective parts in a sample of 500000 parts, still a probability, of course, of 1%.
U didnt get my problem correctly. Actually total parts were 1 million which have 1% defective parts. So defective parts were 10000 in 1 million parts. Now I want to take different sample size from these 1 million parts i.e I m taking 10%, 30%, 50%,70% and 90% of these 1 million parts. Here I want to calculate probability of finding maximum 5000 defective parts from each sample size. Suppose I m taking sample size 10% of 1 million parts which is 100000 parts. Now to calculate probability of finding 5000 defective parts in this sample size is 0 which is found by using binomial distribution where success of finding defective parts p=0.01 (as we have 1% of defective parts in 1 million parts), failure q=0.99, r=5000 and n=100000. Similarly when I take any sample size whose fraction is less than 50% of 1 million parts, probability is always zero. When I take sample size 50% of 1 million parts i.e 500000, probability of finding 5000 defective parts is 0.5 and when I take any sample size above 50% of 1 million parts, probability of finding 5000 defective parts is 1. My issue is that why i m not getting any value between 0-0.5 or 0.5-1 although sample sizes are changing linearly. I think now I make my point clear so can you please answer my question now
I am going back to the beginning of this thread.
Let u = number in population.
Let d = number of defectives in population, where 0 < d < u.
Let s = number in random sample, where 0 < s < u - d.
Let c = cap on number of defectives in sample, where -1 < c < (s + 1) and c < (d + 1)
Let n(k) = probability of exactly k defectives in sample, where - 1 < k < (c + 1).
Let p(c) = probability of no more than c defectives in sample.
\(\displaystyle \displaystyle n(k) = \binom{s}{k} * \binom{d}{k} * \binom{u - d}{s - k} \div \binom{u}{s} \implies\)
\(\displaystyle n(k) = \dfrac{s!}{k! * (s - k)!} * \dfrac{d!}{k! * (d - k)!} * \dfrac{(u - d)!}{(s - k)! * (u + k - d - s)!} * \dfrac{s! * (u - s)!}{u!} > 0.\)
\(\displaystyle \displaystyle p(c) = \sum_{i=0}^cn(i) > 0.\)
Your statements that the probabilities are either 0 or 1 depending on sample size are wrong.
Now for numbers as large as yours, you either need to do some programming to calculate the probabilities or find some approximations that make sense. Probably integrals will give you a decent approximation, but I have forgotten too much of my calculus to give it a try.