Suppose a machine produces metal parts that contains some defective parts with probability 0.05. How many parts should be produced in order that the probability of at least one part being defective is 1/2 or more ? ( Given `log_(10)95=1.977` and `log_(10)2=0.3` )

Options

A. 11
B. 12
C. 15
D. 14

Answer

To solve the problem, we need to determine how many parts should be produced so that the probability of at least one part being defective is 1/2 or more.

### Step-by-step Solution:

1. **Define the Probability of Defectiveness**:
- Let \( p \) be the probability of a part being defective. Given \( p = 0.05 \).
- Therefore, the probability of a part being non-defective (failure) is \( q = 1 - p = 0.95 \).

2. **Set Up the Probability Condition**:
- We want the probability of at least one defective part to be at least 1/2. This can be expressed as:
\[ P(\text{at least one defective}) \geq \frac{1}{2} \]
- The complement of this event (no defective parts) is:
\[ P(\text{no defective parts}) = q^n = (0.95)^n \]
- Therefore, we can write:
\[ 1 - (0.95)^n \geq \frac{1}{2} \]

3. **Rearranging the Inequality**:
- Rearranging gives:
\[ (0.95)^n \leq \frac{1}{2} \]

4. **Taking Logarithms**:
- We take the logarithm (base 10) of both sides:
\[ \log_{10}((0.95)^n) \leq \log_{10}\left(\frac{1}{2}\right) \]
- Using the property of logarithms, this simplifies to:
\[ n \cdot \log_{10}(0.95) \leq \log_{10}(1) - \log_{10}(2) \]
- Since \( \log_{10}(1) = 0 \), we have:
\[ n \cdot \log_{10}(0.95) \leq -\log_{10}(2) \]

5. **Substituting Known Values**:
- We know \( \log_{10}(2) = 0.3 \) (given).
- Therefore, we can rewrite the inequality:
\[ n \cdot \log_{10}(0.95) \leq -0.3 \]

6. **Calculating \( \log_{10}(0.95) \)**:
- We can calculate \( \log_{10}(0.95) \) using the property of logarithms:
\[ \log_{10}(0.95) = \log_{10}\left(\frac{95}{100}\right) = \log_{10}(95) - \log_{10}(100) \]
- Given \( \log_{10}(95) = 1.977 \) and \( \log_{10}(100) = 2 \):
\[ \log_{10}(0.95) = 1.977 - 2 = -0.023 \]

7. **Substituting Back into the Inequality**:
- Now substituting \( \log_{10}(0.95) \) back into the inequality:
\[ n \cdot (-0.023) \leq -0.3 \]
- Dividing both sides by -0.023 (remember to reverse the inequality):
\[ n \geq \frac{-0.3}{-0.023} \approx 13.04 \]

8. **Conclusion**:
- Since \( n \) must be a whole number, we round up to the nearest whole number:
\[ n \geq 14 \]
- Therefore, at least **14 parts** should be produced to ensure that the probability of at least one part being defective is 1/2 or more.