There are 10 defective bulbs in a group of 100 bulbs. If a sample of 8 bulbs are selected at random then find the probability that:
(i) in this sample 3 bulbs are defective and 5 are non-defective.
(ii) in this sample at least one bulb is defective.

Explanation

To solve the problem step-by-step, we will break it down into two parts as described in the question.

**Given:**
- Total bulbs = 100
- Defective bulbs = 10
- Non-defective bulbs = 100 - 10 = 90
- Sample size = 8

### Part (i): Probability of selecting 3 defective bulbs and 5 non-defective bulbs

1. **Identify the probabilities:**
- Probability of selecting a defective bulb (p) = Number of defective bulbs / Total bulbs = 10/100 = 1/10
- Probability of selecting a non-defective bulb (q) = 1 - p = 1 - (1/10) = 9/10

2. **Use the binomial probability formula:**
The probability of getting exactly r successes (defective bulbs) in n trials (sample size) is given by:
\[ P(X = r) = \binom{n}{r} p^r q^{n-r} \]
Where:
- \( n = 8 \) (total bulbs selected)
- \( r = 3 \) (defective bulbs)
- \( n - r = 5 \) (non-defective bulbs)

3. **Calculate the binomial coefficient:**
\[ \binom{8}{3} = \frac{8!}{3!(8-3)!} = \frac{8 \times 7 \times 6}{3 \times 2 \times 1} = 56 \]

4. **Calculate the probability:**
\[ P(X = 3) = \binom{8}{3} \left(\frac{1}{10}\right)^3 \left(\frac{9}{10}\right)^5 \]
\[ = 56 \times \left(\frac{1}{10}\right)^3 \times \left(\frac{9}{10}\right)^5 \]
\[ = 56 \times \frac{1}{1000} \times \frac{59049}{100000} \]
\[ = 56 \times \frac{59049}{100000000} \]
\[ = \frac{3308448}{100000000} = 0.03308448 \]

### Part (ii): Probability of selecting at least one defective bulb

1. **Calculate the probability of selecting 0 defective bulbs:**
Using the binomial distribution:
\[ P(X = 0) = \binom{8}{0} p^0 q^8 \]
\[ = 1 \times \left(\frac{1}{10}\right)^0 \times \left(\frac{9}{10}\right)^8 \]
\[ = 1 \times 1 \times \left(\frac{9}{10}\right)^8 \]
\[ = \left(\frac{9}{10}\right)^8 = \frac{43046721}{100000000} \]

2. **Calculate the probability of at least one defective bulb:**
\[ P(\text{at least 1 defective}) = 1 - P(X = 0) \]
\[ = 1 - \left(\frac{9}{10}\right)^8 \]
\[ = 1 - \frac{43046721}{100000000} \]
\[ = \frac{100000000 - 43046721}{100000000} = \frac{56953279}{100000000} = 0.56953279 \]

### Final Answers:
(i) The probability of selecting 3 defective bulbs and 5 non-defective bulbs is approximately **0.0331**.
(ii) The probability of selecting at least one defective bulb is approximately **0.5695**.