Solution

The correct answer is \(e^{\frac{-5}{29}}\)

Key PointsThe reliability of a system in the context of reliability engineering is often modeled using the exponential distribution. The exponential reliability function is given by: \( R(t) = e^{-\lambda t}\)

where:

• \(R(t)\) is the reliability at time  t,
• \( \lambda\) is the failure rate (reciprocal of the mean time between failures).
• In this case, you are given the time between failures for the software:
  • ​\(\text{Time between failures} = [6, 4, 8, 5, 6]\)
• To find the failure rate \( \lambda\), you can use the formula:
  • ​\(\lambda = \frac{1}{\text{Mean time between failures}}\)
• Mean time between failures (MTBF) is calculated as the average of the time between failures:
  • ​\(\text{MTBF} = \frac{6 + 4 + 8 + 5 + 6}{5} = \frac{29}{5}\)
• Now, the failure rate (\( \lambda\)) is:
  • \( \lambda = \frac{1}{\text{MTBF}} = \frac{5}{29}\)
• Substitute this into the reliability function:
  • ​\(R(t) = e^{-\frac{5}{29} t}\)
• For one hour of operation ( $t = 1$ ), the reliability becomes:
  • ​\(R(1) = e^{-\frac{5}{29} \times 1} = e^{-\frac{5}{29}}\)

So, the correct answer is: \( \text{e}^{-\frac{5}{29}}\)

Therefore, the correct option is: \(\text{Option 3: } e^{-\frac{5}{29}}\)