Hint and Help Easy Statistics question

A lot of 30 parts is shipped to a company. A sampling plan dictates that n parts are to be taken at random without replacement. The lot will be accepted if no more than one of these n parts is defective. What is the minimum value of n such that the probability of accepting the lot is at least 80% if the lot contains 3 defective parts and the probability of accepting the lot is less than 25% if the lot contains 9 defective parts?

So I have gotten the question set up I am stuck on finding a n, as I am I just supposed to randomly plug in numbers for n that would work?

for 80% its (27)(30)+(27)(3)
n 0 n-1 1
---------------------
( 30)
n

for 25% (21) (9)+(21)(9)
n 0 n-1 1
---------------------
(30)
n


then I am to use the answer I obtain do to a simulation for how many times I would accept the lot if the lot contains 3 defective parts and 25 good parts? and for 9 defective and 21 good parts? I am stuck as to how to start it


Thank you