Probability of 4 specific numbers (1-3000) occuring in a sample of 400 How to calculate the probability that four specific, distinct numbers from the range 1 - 3000 occur at least once in a fixed sample of 400 random numbers from the range 1-3000? The numbers in the sample can repeat as they were randomly generated. 
My intuition would be that it is basically a set of four "scans" of the 400 numbers, so the probability of hitting the 1/3000 searched number in each of the scans is roughly 400/3000 = 2/15. This would give the total probability count as (2/15)x(2/15)x(2/15)x(2/15) = 16/50625 = 0,000316. However, I'm not sure if this accounts (and if it should account) for the fact that it is a fixed sample so it's not "re-rolled" for each of the four scans.
Thanks for any advice.   
 A: Use inclusion/exclusion principle:


*

*Include the number of combinations with at least $\color\red0$ missing values: $\binom{4}{\color\red0}\cdot(3000-\color\red0)^{400}$

*Exclude the number of combinations with at least $\color\red1$ missing values: $\binom{4}{\color\red1}\cdot(3000-\color\red1)^{400}$

*Include the number of combinations with at least $\color\red2$ missing values: $\binom{4}{\color\red2}\cdot(3000-\color\red2)^{400}$

*Exclude the number of combinations with at least $\color\red3$ missing values: $\binom{4}{\color\red3}\cdot(3000-\color\red3)^{400}$

*Include the number of combinations with at least $\color\red4$ missing values: $\binom{4}{\color\red4}\cdot(3000-\color\red4)^{400}$


Finally, divide by the total number of combinations, which is $3000^{400}$:
$$\frac{\sum\limits_{n=0}^{4}(-1)^n\cdot\binom{4}{n}\cdot(3000-n)^{400}}{3000^{400}}\approx0.000239$$
A: Label the $4$ numbers and let $E_{i}$ denote the event that number
with label $i\in\left\{ 1,2,3,4\right\} $ does not occur in the sample.
Then you are looking for $\Pr\left(E_{1}^{c}\cap E_{2}^{c}\cap E_{3}^{c}\cap E_{4}^{c}\right)=1-\Pr\left(E_{1}\cup E_{2}\cup E_{3}\cup E_{4}\right)$.
With inclusion/exclusion and symmetry we find that this equals:
$$1-4\Pr\left(E_{1}\right)+6\Pr\left(E_{1}\cap E_{2}\right)-4\Pr\left(E_{1}\cap E_{2}\cap E_{3}\right)+\Pr\left(E_{1}\cap E_{2}\cap E_{3}\cap E_{4}\right)$$
Can you take it from here?
