A box contains 5 defective and 10 non defectives lamps. 8 lamps are drawn at random in succession without replacement. What is the probability that the 8th lamp is then fifth defective.

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?

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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.

A box contains 5 defective and 10 non-defective lamps. 8 lamps are drawn at random in succession without replacement. What is the probability that the 8th lamp is the fifth defective?

Learn how to calculate the probability that the 8th lamp drawn is the fifth defective in a set of 5 defective and 10 non-defective lamps. This problem demonstrates key concepts in combinatorics and probability theory, using the hypergeometric distribution and combinations. Suitable for Grades 11-12.

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