In a box containing 100 bulbs, 10 are defective. What is the probability that out of a sample of 5 bulbs, (i) none is defective and (ii) exactly 2 are defective ?

Answer

To solve the problem, we need to calculate the probabilities for two scenarios: (i) none of the 5 bulbs is defective, and (ii) exactly 2 out of the 5 bulbs are defective.

### Step-by-Step Solution:

**Total bulbs = 100**
**Defective bulbs = 10**
**Non-defective bulbs = 100 - 10 = 90**
**Sample size = 5**

#### Part (i): Probability that none of the 5 bulbs is defective

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 (total bulbs selected) is given by:
\[ P(X = r) = \binom{n}{r} P^r Q^{n-r} \]
Here, we want P(X = 0) for none defective:
\[ P(X = 0) = \binom{5}{0} \left(\frac{1}{10}\right)^0 \left(\frac{9}{10}\right)^{5} \]

3. **Calculate**:
- \(\binom{5}{0} = 1\)
- \(\left(\frac{1}{10}\right)^0 = 1\)
- \(\left(\frac{9}{10}\right)^{5} = \left(\frac{9^5}{10^5}\right) = \frac{59049}{100000}\)

4. **Final probability**:
\[ P(X = 0) = 1 \cdot 1 \cdot \frac{59049}{100000} = \frac{59049}{100000} = 0.59049 \]

#### Part (ii): Probability that exactly 2 out of the 5 bulbs are defective

1. **Use the binomial probability formula again**:
We want P(X = 2):
\[ P(X = 2) = \binom{5}{2} \left(\frac{1}{10}\right)^{2} \left(\frac{9}{10}\right)^{3} \]

2. **Calculate**:
- \(\binom{5}{2} = \frac{5!}{2!(5-2)!} = \frac{5 \times 4}{2 \times 1} = 10\)
- \(\left(\frac{1}{10}\right)^{2} = \frac{1}{100}\)
- \(\left(\frac{9}{10}\right)^{3} = \frac{729}{1000}\)

3. **Combine these values**:
\[ P(X = 2) = 10 \cdot \frac{1}{100} \cdot \frac{729}{1000} = \frac{7290}{100000} = 0.0729 \]

### Final Answers:
- (i) The probability that none of the 5 bulbs is defective is **0.59049**.
- (ii) The probability that exactly 2 of the 5 bulbs are defective is **0.0729**.