1. **State the problem:** We have 6 batteries in total, 2 of which are defective. We select 2 batteries at random without replacement. We want to find the probability that both selected batteries are defective. 2. **Formula used:** The probability of an event is given by $$\text{Probability} = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}}$$ 3. **Calculate total number of ways to select 2 batteries from 6:** $$\binom{6}{2} = \frac{6!}{2!(6-2)!} = \frac{6 \times 5}{2 \times 1} = 15$$ 4. **Calculate number of favorable outcomes (both defective):** Since there are 2 defective batteries, the number of ways to select both defective is: $$\binom{2}{2} = 1$$ 5. **Calculate the probability:** $$P(\text{both defective}) = \frac{1}{15}$$ 6. **Interpretation:** There is a 1 in 15 chance that both batteries selected are defective when choosing 2 out of 6 without replacement.