Probability of winning at least several raffle tickets My question is quite similar to this: Probability of winning a prize in a raffle, but what if I want to know the probability of at least a certain number of times?  
Let's modify that example in the aforementioned question a little bit into this:

There are 1600 Tickets and 40 Prizes. This time I'm buying 20 Tickets.

The question: What is the probability of winning at least 4 prizes?
Through simulation, I get somewhere around ≈0.0012 . Is there a systematic way I can calculate the correct probability?
It is also calculated that winning at least once is ≈0.3992 (applying the method from the aforementioned question)
 A: The probability of winning exactly $k$ of the $40$ prizes when buying $20$ of the $1600$ tickets is 
$$\frac{\dbinom{40}{k}\dbinom{1560}{20 - k}}{\dbinom{1600}{20}}$$
where $\binom{40}{k}$ is the number of ways of selecting exactly $k$ of the $40$ winning tickets, $\binom{1560}{20 - k}$ is the number of ways of selecting exactly $20 - k$ of the $1600 - 40 = 1560$ non-winning tickets, and $\binom{1600}{20}$ is the number of ways of selecting $20$ of the $1600$ available tickets.
Therefore, the probability of winning less than $4$ prizes when buying $20$ of the $1600$ tickets is 
$$\sum_{k = 0}^{3} \frac{\dbinom{40}{k}\dbinom{1560}{20 - k}}{\dbinom{1600}{20}} = \frac{\dbinom{40}{0}\dbinom{1560}{20}}{
\dbinom{1600}{20}} + \frac{\dbinom{40}{1}\dbinom{1560}{19}}{\dbinom{1600}{20}} + \frac{\dbinom{40}{2}\dbinom{1560}{18}}{\dbinom{1600}{20}} + \frac{\dbinom{40}{3}\dbinom{1560}{17}}{\dbinom{1600}{20}}$$
To find the probability of winning at least four prizes, subtract the probability of winning fewer than four prizes from $1$.
