What is the probability of failure for a control valve tested on a three-year interval with a failure rate of once every four years?

What is the probability of failure for a control valve tested on a three-year interval with a failure rate of once every four years?

• A. 0.3
• B. 0.53
• C. 0.47
• D. 0.75

Answer: (B)

To determine the probability of failure for a control valve tested on a three-year interval, given a failure rate of once every four years, it's important to first understand the concepts involved. The failure rate provides insight into how often failures occur over a specified duration. In this case, when we say the valve has a failure rate of once every four years, we can express this as a failure rate of 0.25 failures per year (1 failure/4 years). Next, when evaluating the probability of failure within three years, we can utilize the exponential failure model. The probability of survival, which is the opposite of failure, can be expressed mathematically by the formula: \[ P(\text{failure in time } t) = 1 - e^{-\lambda t} \] where \( \lambda \) is the failure rate, and \( t \) is the time interval (in years). In this scenario, \( \lambda = 0.25 \) failures per year and \( t = 3 \) years. We can calculate: \[ P(\text{failure in 3 years}) = 1 - e^{-0.25 \times 3} \] Now, calculating \( -0.25 \times

To determine the probability of failure for a control valve tested on a three-year interval, given a failure rate of once every four years, it's important to first understand the concepts involved. The failure rate provides insight into how often failures occur over a specified duration. In this case, when we say the valve has a failure rate of once every four years, we can express this as a failure rate of 0.25 failures per year (1 failure/4 years).

Next, when evaluating the probability of failure within three years, we can utilize the exponential failure model. The probability of survival, which is the opposite of failure, can be expressed mathematically by the formula:

[ P(\text{failure in time } t) = 1 - e^{-\lambda t} ]

where ( \lambda ) is the failure rate, and ( t ) is the time interval (in years). In this scenario, ( \lambda = 0.25 ) failures per year and ( t = 3 ) years.

We can calculate:

[ P(\text{failure in 3 years}) = 1 - e^{-0.25 \times 3} ]

Now, calculating ( -0.25 \times