If `3%` of electric bulbs manufactured by a company are defective , then the probability that a sample of 100 bulbs has no defective is

Options

A. 0
B. `e^(-3)`
C. `1-e^(-3)`
D. `3-e^(-3)`

Answer

To solve the problem of finding the probability that a sample of 100 electric bulbs has no defective bulbs when 3% of the bulbs are defective, we can use the Poisson distribution. Here’s a step-by-step solution:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- The percentage of defective bulbs is given as 3%.
- The total number of bulbs in the sample (n) is 100.

2. **Calculate the Expected Number of Defective Bulbs (λ)**:
- Since 3% of 100 bulbs are defective, we can calculate the expected number of defective bulbs (λ) as follows:
\[ \lambda = \frac{3}{100} \times 100 = 3 \]

3. **Set Up the Poisson Distribution**:
- The Poisson distribution formula is given by:
\[ P(X = x) = \frac{e^{-\lambda} \lambda^x}{x!} \]
- In this case, we want to find the probability of having no defective bulbs (x = 0).

4. **Substitute Values into the Poisson Formula**:
- Substitute λ = 3 and x = 0 into the formula:
\[ P(X = 0) = \frac{e^{-3} \cdot 3^0}{0!} \]

5. **Simplify the Expression**:
- Since \(3^0 = 1\) and \(0! = 1\), we can simplify:
\[ P(X = 0) = e^{-3} \cdot \frac{1}{1} = e^{-3} \]

6. **Final Probability**:
- The probability that a sample of 100 bulbs has no defective bulbs is:
\[ P(X = 0) = e^{-3} \]

### Conclusion:
The final answer is \( P(X = 0) = e^{-3} \).