1. **Problem statement:** We have a pack of 10 watches, of which 3 are defective. We select 2 watches at random. We want to find the probability that at least one of the selected watches is defective. 2. **Understand the approach:** Instead of directly calculating the probability of "at least one defective," it is easier to find the complement: the probability that "none are defective" (i.e., both are not defective), and then subtract that from 1. 3. **Calculate total ways to select 2 watches:** The total number of ways to select 2 watches from 10 is given by the combination: $$ \binom{10}{2} = \frac{10 \times 9}{2 \times 1} = 45 $$ 4. **Calculate ways to select 2 non-defective watches:** There are 3 defective watches, so there are $10-3=7$ non-defective watches. The ways to select 2 non-defective watches are: $$ \binom{7}{2} = \frac{7 \times 6}{2 \times 1} = 21 $$ 5. **Calculate the probability none are defective:** $$ P(\text{none defective}) = \frac{21}{45} = \frac{7}{15} $$ 6. **Calculate the probability at least one is defective:** $$ P(\text{at least one defective}) = 1 - P(\text{none defective}) = 1 - \frac{7}{15} = \frac{8}{15} $$ 7. **Answer:** The probability that at least one watch is defective is $\boxed{\frac{8}{15}}$. This corresponds to option (E).