To solve this problem, we need to calculate the probability that the 8th lamp drawn is the 5th defective lamp.

Problem Breakdown:

• We have a total of $5$ defective lamps and $10$ non-defective lamps, making a total of $15$ lamps.
• We are drawing $8$ lamps without replacement, and we want the 8th lamp to be the 5th defective.

Key Steps:

1. Distribution of the First 7 Lamps:

   • For the 8th lamp to be the 5th defective, we must have exactly 4 defective lamps and 3 non-defective lamps drawn among the first 7 lamps.
   • The number of ways to choose 4 defective lamps from the 5 available is: $\binom{5}{4} = 5$
   • The number of ways to choose 3 non-defective lamps from the 10 available is: $\binom{10}{3} = \frac{10 \times 9 \times 8}{3 \times 2 \times 1} = 120$

2. Choosing the 8th Lamp:

   • After drawing the first 7 lamps, there will be 1 defective lamp remaining and 7 non-defective lamps remaining.
   • The probability that the 8th lamp is the remaining defective lamp is: $\frac{1}{8}$

3. Total Number of Ways to Choose 8 Lamps:

   • The total number of ways to choose 8 lamps from the 15 available is: $\binom{15}{8} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9 \times 8}{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1} = 6435$

4. Final Probability:

   • The total number of favorable outcomes is the product of the combinations for the first 7 lamps and the probability for the 8th lamp: $5 \times 120 \times \frac{1}{8} = 75$
   • Therefore, the probability is: $\frac{75}{6435}$
   • Simplifying this fraction: $\frac{75}{6435} = \frac{5}{429}$

Conclusion:

The probability that the 8th lamp is the 5th defective is $\frac{5}{429}$.

Would you like any further details or clarifications?

---

Here are 5 related questions to expand on this topic:

1. What is the probability of drawing exactly 4 defective lamps out of 8?
2. How would the probability change if the selection was with replacement?
3. What if the number of defective lamps was increased to 7 while keeping the total at 15?
4. How does drawing without replacement affect the probabilities compared to drawing with replacement?
5. What is the probability of drawing at least 1 defective lamp out of the 8 drawn?

Tip: For problems involving "without replacement," combinations (not permutations) are often the key to calculating probabilities, as the order of selection does not matter.