A box contains 15 transistors, 5 of which are defective. An inspector takes out one transistor at random, examines it for defects, and replaces it. After it has been replaced another inspector does the same thing, and then so does a third inspector. The probability that at least one of the inspectors find a defective transistor is equal to :

Options

A. `1//27`
B. `8//27`
C. `19//27`
D. `26//27`

Answer

To solve the problem, we need to find the probability that at least one of the three inspectors finds a defective transistor when they each randomly select one from a box containing 15 transistors, of which 5 are defective.

### Step-by-Step Solution:

1. **Identify Total and Defective Transistors**:
- Total transistors = 15
- Defective transistors = 5
- Good (non-defective) transistors = 15 - 5 = 10

2. **Calculate the Probability of Selecting a Good Transistor**:
- The probability of selecting a good transistor (non-defective) in one draw is:
\[ P(\text{Good}) = \frac{\text{Number of Good Transistors}}{\text{Total Transistors}} = \frac{10}{15} = \frac{2}{3} \]

3. **Calculate the Probability That None of the Inspectors Find a Defective Transistor**:
- Since each inspector draws a transistor independently, the probability that all three inspectors select a good transistor is:
\[ P(\text{None find defective}) = P(\text{Good})^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} \]

4. **Calculate the Probability That At Least One Inspector Finds a Defective Transistor**:
- The probability that at least one inspector finds a defective transistor is the complement of the probability that none find a defective transistor:
\[ P(\text{At least one finds defective}) = 1 - P(\text{None find defective}) = 1 - \frac{8}{27} = \frac{27 - 8}{27} = \frac{19}{27} \]

5. **Final Answer**:
- Therefore, the probability that at least one of the inspectors finds a defective transistor is:
\[ \frac{19}{27} \]

### Conclusion:
The final answer is \( \frac{19}{27} \).