Probability Calculation: Non-Defective Screws

This problem involves calculating the probability of selecting two non-defective screws sequentially without replacement from a given set.

• Total number of screws = 10
• Number of defective screws = 3
• Number of non-defective screws = 10 - 3 = 7

Step-by-Step Probability Calculation

1. First Draw Probability: Calculate the probability that the first screw selected is non-defective.

   There are 7 non-defective screws out of a total of 10 screws.

   $ P(\text{1st screw is non-defective}) = \frac{\text{Number of non-defective screws}}{\text{Total screws}} = \frac{7}{10} $

2. Second Draw Probability: Calculate the probability that the second screw selected is also non-defective, given the first was non-defective. Since the draw is without replacement, there are now 9 screws remaining, and 6 of them are non-defective.

   $ P(\text{2nd screw is non-defective} | \text{1st was non-defective}) = \frac{\text{Remaining non-defective screws}}{\text{Remaining total screws}} = \frac{6}{9} $

3. Combined Probability: The probability that both screws are non-defective is the product of the probabilities calculated in the previous steps.

   $ P(\text{Neither screw is defective}) = P(\text{1st non-defective}) \times P(\text{2nd non-defective} | \text{1st non-defective}) $

   $ P(\text{Neither screw is defective}) = \frac{7}{10} \times \frac{6}{9} = \frac{42}{90} $

4. Simplification: Simplify the final fraction. The greatest common divisor of 42 and 90 is 6.

   $ \frac{42}{90} = \frac{42 \div 6}{90 \div 6} = \frac{7}{15} $

   The fraction $\frac{7}{15}$ is equivalent to $\frac{14}{30}$.

The probability that neither of the two screws drawn is defective is $\frac{7}{15}$, which matches the option $\frac{14}{30}$.