Probability of getting $0$ defective $ = \frac{{}^{16}C_8}{{}^{20}C_8} = \frac{16!12!}{20!8!} = \frac{9.10.11.12}{17.18.19.20} = \frac{11.3}{17.19}=0.1021$
Probability of getting $1$ defective $ = \frac{{}^{16}C_7. {}^4C_1}{{}^{20}C_8} = \frac{16!12!8.4}{20!9!} = \frac{10.11.12.8.4}{17.18.19.20} = \frac{11.12.8}{17.9.19} =0.3632 $
So, probability of not more than one defective $ = 0.102+0.3632 = 0.4654$