problem of probability and distribution Suppose there are 1 million parts which have $1\%$ defective parts i.e 1 million parts have $10000$ defective parts. Now suppose we are taking different sample sizes from 1 million like $10\%$, $30\%$, $50\%$, $70\%$, $90\%$ of 1 million parts and we need to calculate the probability of finding maximum $5000$ defective parts from these sample sizes. As 1 million parts has $1\%$ defective parts so value of success $p$ is $0.01$ and failure $q$ is $0.99$. 
Now suppose we are adding $100,000$ parts in one million parts which makes total $1,000,000$ parts but this newly added $100,000$ parts do not have any defective parts. So now what will be the value of success $p$ in total $1,000,000$ parts to find $5000$ defective parts? Please also give justification for choosing value of $p$?

There was an earlier problem of which this is a variant. In the solution to that problem I did a great many computations. For this problem, the computations are in the same style, with a different value of $p$, the probability that any one item is defective.
Some of the computations in the earlier answer were done for "general" (small) $p$, so they can be repeated with small modification. 
We had $10000$ defectives in a population of $1000000$, and added $100000$ non-defectives. So the new probability of a defective is $p=\frac{10000}{1100000}\approx 0.0090909$.  We are taking a large sample, presumably without replacement. So the distribution of the number $X$ of defectives in a sample of size $n$ is hypergeometric, not binomial. However, that does not make a significant differnce for the sample sizes of interest.
As was the case in the earlier problem, the probability of $\le 5000$ defectives in a sample of size $n$ is nearly $1$ up to a certain $n_0$, the n climbs very rapidly to (nearly) $0.5$ at a certain $n_0$, and then falls very rapidly to $0$ as $n$ increases further. 
In the earlier problem, we had $n_0=500000$. In our new situation, the appropriate $n_0$ is obtained by solving the equation 
$$n_0p=5000$.  
Solve. We get $n_0=550000$. 
If $n$ is significantly below $550000$, we will have that $\Pr(X\le 5000)$ will be nearly $1$. For example, that is the case already at $n=500000$. However, for $n$ quite close to $550000$, such as $546000$, the probability is not close to $1$. Similrly, on the other side of $550000$ but close, like $554000$, the proability that $X\le 5000$ will not be close to $0$.
In the erlier answer, you were supplied with all the formulas  to do any needed calculations if you want to explore the "near $550000$" region in detail.
