Picking from a pool of 20000 different numbers, each experiment we pick 1000 different numbers(one numbers can be picked repeatedly in different experiments but not in the same experiment), and we pick n times(n experiments).whats the probability that at least one number is picked fewer than 3 times in the entire n experiments?
From What I got, the probability of a number in this pool getting picked at least once is (1-the probability of this number never getting picked).
So In Experiment 1, the probability of this number not getting picked is 19/20.
In Experiment 2, the probability of this number not getting picked is (19/20)^2
So the probability of this number not getting picked at experiment n is (19/20)^n
And that's what I got if just at least 1, what if we change that to 2 or 3 or d?
If the counting is cumulative, then there is no point in segregating the picks between experiments. We simply make $1,000n$ choices. It is not correct to divide $1,000$ by $20,000.$ That would be right if we picked without replacement, that is, if a number can only be picked once. Then we pick $1,000$ numbers, and look to see if the designated number is among them. Then also, it make sense to take about $n$ experiments, because we would replace the $1,000$ numbers at the end of the experiment.
Now for the actual question. It seems to me like you're taking about a specific number. That is, you're asking something like, "What is the probability that $7$ gets picked fewer than $3$ times?," not "What is the probability that some number among all $20,000$ gets picked fewer than $3$ times?"
The probability that $7$ is not picked on a particular draw is $\frac{19999}{20000}=.99995.$ To make things easier to type, and top read, set $p=.9995, N=1000n.$ The probability that $7$ is never picked is $p^N.$ What about the probability that $7$ is picked exactly once? There are $N$ ways to select which draw $7$ will be picked on, and the probability that $7$ will be picked on that draw is $1-p$. Then $7$ isn't picked on any of the remaining $N-1$ draws, so the probability that $7$ is picked on exactly one draw is $N(1-p)p^{N-1}.$
To finish it off, you need to compute the probability that $7$ is picked on exactly $2$ draws, and add up the three probabilities.