Say I guess 4 numbers from 1 to 30, then 4 numbers are draw randomly from the list, without replacement, what are the chances that I will get at least 3 correct?

My idea so far:

So to get all correct, P(all 4 correct) = $\displaystyle 4! \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {1}{27} $

and to have 3 correct, it is P(3 correct) = $\displaystyle \frac {4!}{3!} \frac {1}{30} \frac {1}{29} \frac {1}{28} \frac {26}{27} $

So do I just add them up together?