Flood Frequency – Exceedance in a Future Period from Return Period An annual maximum flood magnitude has a return period T = 8 years. What is the probability that this magnitude will be exceeded at least once during the next 5 years?
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A0.625
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B0.966
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C0.487
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D0.529
Answer
Correct Answer: 0.487
Explanation
Introduction / Context:Return period T relates to annual exceedance probability p by p = 1/T, assuming independence year to year. Hydrologists often need the probability of at least one exceedance over multiple future years for risk-informed design and communication.
Given Data / Assumptions:
- Return period T = 8 years ⇒ annual exceedance probability p = 1/8.
- Number of years n = 5.
- Events in different years are statistically independent.
Concept / Approach:
The probability of no exceedance in one year is (1 − p). Over n years, P(no exceedance) = (1 − p)^n. Therefore, P(at least one exceedance) = 1 − (1 − p)^n.
Step-by-Step Solution:
Compute p = 1/8 = 0.125.No exceedance in 5 years: (1 − 0.125)^5 = 0.875^5.Evaluate 0.875^5 ≈ 0.5129.Therefore, at least one exceedance = 1 − 0.5129 ≈ 0.487.Verification / Alternative check:
Binomial distribution with parameters (n = 5, p = 0.125); summing probabilities of k ≥ 1 exceedances reproduces 1 − (1 − p)^n.
Why Other Options Are Wrong:
- 0.625 and 0.529: Do not correspond to the correct computation.
- 0.966: Would imply an extremely high risk inconsistent with p = 0.125.
Common Pitfalls:
Confusing return period with a guarantee of occurrence exactly once every T years; forgetting independence assumption when applying the formula.
Final Answer:
0.487