P(UD) be Probability of non-defective screws=7/10
P(D) be Probability of Defective screws=3/10
D1 First defective screw drawn P(D1) = 3/10.
D2 second defective screw drawn P(D2) = 3/10.
probability that none of two screws is defective will be $P(\overline{D1}\cap \overline{D2})=P(\overline{D1})*P(\overline{D2})$ { as both events are independent }
=$\frac{7}{10} * \frac{7}{10}=\frac{49}{100}= 49%$ %
As only one two outcomes are possible so D1 or D2 complement will be non-defective screw UD1 OR UD2