Alot of size n = 30 contains three nonconforming units. what is the probability that a sample of five units selected at random contains exactly one nonconforming unit? what is the probability that it contains one or more nonconformances?

Exactly 1 nonconforming unit = 0.369458128 
1 or more nonconforming units = 0.433497537  
Due to formatting issues, I'll use the notion C(n,x) for N choose X. Which would be n!/(x!(n-x)!).  
Now for the case of exactly 1, the probability will be:
 P = C(3,1)*C(30-3,4)/C(30,5)
 To explain it, you choose exactly 1 nonconforming item out of the 3 possible, then fill the sample with 4 more conforming items to complete the sample size of 5. Finally, you divide by the number of different ways you can select 5 items out of the entire group of 30. So doing the math, you get P = C(3,1)*C(30-3,4)/C(30,5) = 3*17550/142506 = 0.369458128 = 36.9458128% 
 For the case of 1 or more nonconforming units you do the sum of x ranging from 1 to 3 with the formula
 C(3,x)*C(30-3,5-x)/C(30,5)
 or you can set x to 0 and evaluate
 1 - C(3,0)*C(30-3,5-0)/C(30,5) 
 Let's do it both ways.
 C(3,1)*C(30-3,5-1)/C(30,5) +
 C(3,2)*C(30-3,5-2)/C(30,5) +
 C(3,3)*C(30-3,5-3)/C(30,5)
 = 3*17550/142506 + 3*2925/142506 + 1*351/142506
 = 0.369458128 + 0.061576355 + 0.002463054
 = 0.433497537
 
 And doing it the "simple" way
 1 - C(3,0)*C(30-3,5-0)/C(30,5) 
 = 1 - 1*80730/142506 = 1 - 0.566502463 = 0.433497537

hope it is correct if not i'm srry i tried to