A box contains 10 screws, 3 of which are defective. Two screws are drawn at random with replacement. The probability that none of the two screws is defective will be A. 100% B. 50% C. 49% D. None of these

The probability of drawing a non-defective screw is $\frac{7}{10}$. Since the screws are drawn with replacement, the probability of drawing two non-defective screws is $\frac{7}{10} \times \frac{7}{10} = \boxed{\frac{49}{100}}$.

Option A is incorrect because it is impossible to guarantee that none of the two screws drawn

Option B is incorrect because it is the probability of drawing one defective

Option C is incorrect because it is the probability of drawing two defective screws, which is also not what the question asks.