Step 1: Understanding the Question: The question asks for the 'risk' of a hydrological event. Risk is defined as the probability that an event of a certain magnitude (or greater) will occur at least once in a specified period (the design life).
Step 2: Key Formula or Approach: The probability of an event with a return period \(T\) occurring in any given year is \(P = 1/T\).
The probability of the event
not occurring in any given year is \(q = 1 - P\).
The probability of the event not occurring for \(n\) consecutive years is \(q^n = (1 - P)^n\).
The risk (\(R\)) is the probability of the event occurring at least once in \(n\) years, which is 1 minus the probability of it never occurring. \[ R = 1 - q^n = 1 - (1 - P)^n = 1 - \left(1 - \frac{1}{T}\right)^n \]
Step 3: Detailed Explanation: We are given the following values:
- Design life, \(n = 50\) years
- Return period, \(T = 100\) years
First, calculate the probability of the design flood being exceeded in a single year: \[ P = \frac{1}{T} = \frac{1}{100} = 0.01 \] Now, use the risk formula: \[ R = 1 - \left(1 - \frac{1}{100}\right)^{50} \] \[ R = 1 - (1 - 0.01)^{50} \] \[ R = 1 - (0.99)^{50} \] Calculating the value of \((0.99)^{50}\): \[ (0.99)^{50} \approx 0.605006 \] Now, calculate the risk: \[ R = 1 - 0.605006 = 0.394994 \] The question asks to correct the answer up to three decimal places. \[ R \approx 0.395 \]
Step 4: Final Answer: The risk that the design flood will be equalled or exceeded during the bridge's design life is 0.395.