Question

The probability (in %) that a storm having return period of 15 years may occur in the next 10 years is ______ (rounded off to two decimal places).

Hint:

Use the hydrologic risk formula \( P = 1-(1-1/T)^n \) with T=15 and n=10.

Answer 1

Correct Answer: 49.84

A different way to frame this: the return period T tells you the annual chance the storm is equalled or exceeded is 1/T. Over n independent years, think of no storm in year 1 AND no storm in year 2 AND ... AND no storm in year n as a chain of n independent events, then subtract that chain's probability from 1 to get storm happens at least once.

1. Annual non-exceedance probability: With $T = 15$ years, the chance the storm does NOT happen in any single year is $q = 1 - \dfrac{1}{15} = \dfrac{14}{15} = 0.93333$.
2. Chain probability over 10 years: Since the years are independent, the chance of no storm in all 10 years is $q^{10}$. Using logarithms, $\ln(0.93333) = -0.068993$, so $10\ln(0.93333) = -0.68993$, and $q^{10} = e^{-0.68993} = 0.50162$.
3. Complement gives the risk: The chance of at least one storm in 10 years is the complement of no storm at all, so $P = 1 - q^{10} = 1 - 0.50162 = 0.49838$.

As a percentage this is $0.49838 \times 100 = 49.838\%$, which rounds to $49.84\%$.

Let's summarize:

• The event of exceedance in any one year is independent with probability 1/T.
• Chaining no exceedance over n years and subtracting from 1 gives the risk of at least one exceedance.

So the probability that the 15 year storm occurs at least once in the next 10 years is about $49.84\%$.