Probability that I am not selected in any of 2000 samples?

Question

The population contains 100 million adults, which includes myself. Simple random sampling is used to choose a sample of 1000 adults, 2000 times, independently.

I need to find the probability that I am not selected for any of these 2000 samples.

I have this so far:

P(Not selected for any of 2000 samples)=P(not in sample 1 $\cap$ not in sample 2 $\cap$ ... $\cap$ not in sample 2000)=1-P(sample 1 $\cup$ sample 2 $\cup$ ... $\cup$ sample 2000)

To find the latter probability, do I need to use the generalized form of $P(A\cup B)=P(A)+P(B)-P(A\cap B)$?

Or more specifically:

$1-[2000P(in 1 sample)-c(2000,2)P(in 2 samples)+c(2000,3)P(in 3 samples)-...+c(2000,1999)P(in 1999 samples)-P(in 2000 samples)]$

I know that the probability that I am selected for any ONE sample is: $\frac{1,000}{100,000,000}=\frac{1}{100,000}$. Then, the probability that I am selected for any two samples is $(\frac{1}{100,000})^2$ and so on.

Am I on the right path? Is there an easier way to solve this problem than computing 2000 values and then summing them?

Answer 1

You are in a sample with probability $$\frac{1000}{100000000} = 10^{-5}$$ Then, $1-10^{-}5$ is the probability that you are not in a generic sample. Since your presence in the sample $i$ is independent from your presence in the sample $j$, for all $i \neq j$, then you have to multiply all these probability. Hence $$(1-10^{-5})^{2000} = 0.980198575286329....$$ is what you are looking for.

Answer 2

Those are big numbers. Let's suppose there are 100 people, one is selected and we do it 2 times. Then probability that you are not selected at a first attempt is 99% and same for a second time. So, probability that both of those events happen is $0.99^{2}=0.9801$, thus probability that you would be selected in one (or both) attempts is $1-0.9801$ or 1.99%. I'll let you work that through with bigger numbers...