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

Step 1: Recall the risk (probability of occurrence) formula in hydrology.
For a hydrologic event with return period T years, the probability that it occurs at least once in n successive years is given by the risk formula:
\[ P(\text{at least once in } n \text{ years}) = 1 - \left(1 - \frac{1}{T}\right)^{n} \]
Here the annual exceedance probability is \( p = 1/T \), so (1-p) is the probability the event does not occur in one year, and \( (1-p)^n \) is the probability it does not occur in any of the n years, assuming each year is an independent event. Subtracting from 1 gives the probability that it occurs at least once.

Step 2: Substitute the given values.
Here T = 15 years and n = 10 years, so
\[ p = \frac{1}{15} = 0.06667 \]
\[ 1-p = \frac{14}{15} = 0.93333 \]

Step 3: Raise to the power 10.
\[ (0.93333)^{10} = 0.50162 \]
This is found by multiplying 0.93333 by itself ten times, or equivalently by taking \( 10 \ln(0.93333) = -0.68993 \) and then \( e^{-0.68993} = 0.50162 \).

Step 4: Find the probability of at least one occurrence.
\[ P = 1 - 0.50162 = 0.49838 \]
Converting to a percentage:
\[ P = 49.838\% \approx 49.84\% \]

Final Answer:
The probability that a 15 year return period storm occurs at least once in the next 10 years is about 49.84%.
\[ \boxed{49.84\%} \]