---
title: "Gumbel exceedance probability and 0.1 flood"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-exceed-10000-3000"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-exceed-10000-3000
site: BCT Engineering
---

# Gumbel exceedance probability and 0.1 flood

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2073 Shrawan, 8 marks
- Video: [Gumbel's method in Hydrology for flood estimation | Flood Routing | GATE 2022 |](https://www.youtube.com/watch?v=gtUiLfn_E8U) — TechnoCivil
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-exceed-10000-3000
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

The annual peak discharge of a river follows the Gumbel's extreme value distribution with a mean of 10000 m$^3$/s and a standard deviation of 3000 m$^3$/s. What is the probability that the annual peak discharge is more than 15000 m$^3$/s? What is the magnitude of the peak discharge with an exceedance probability of 0.1? [Hint: $\alpha = \frac{1.28255}{\sigma}$; $\beta = \mu - 0.48\sigma$]

## Short answer

P(X>15000) =1 - exp[-e^{-α(15000-β)}] with α=1.28255/3000, β=10000-0.48×3000, giving 0.0617. Dischrge with exceedance probability 0.1: Q_0.1 = β - (1/α)ln(-ln0.9) = 13824 m³/s.

## Answer

Gumbel's cumulative distribution (probability of non-exceedance):

$$
F(x)=P(X\le x)=\exp\left[-e^{-\alpha(x-\beta)}\right]
$$

Using the hint:

$$
\alpha=\frac{1.28255}{3000}=4.2752\times10^{-4}\ \text{per m}^3/\text{s},\qquad \beta=10000-0.48\times3000=8560 \text{ m}^3/\text{s}
$$

#### Probability that the annual peak exceeds 15000 m³/s
$$
\alpha(x-\beta)=4.2752\times10^{-4}\times(15000-8560)=2.753
$$

$$
F(15000)=\exp(-e^{-2.753})=0.9383
$$

$$
P(X>15000)=1-F=0.0617
$$

**Answer: probability ≈ 0.0617 (about 6.17%), a return period of ≈ 16.2 years.**

#### Discharge with exceedance probability 0.1
Exceedance 0.1 means $F=0.9$:

$$
-e^{-\alpha(x-\beta)}=\ln 0.9\ \Rightarrow\ x=\beta-\frac{1}{\alpha}\ln(-\ln0.9)
$$

$$
x=8560-\frac{\ln(-\ln0.9)}{4.2752\times10^{-4}}=8560-\frac{(-2.2504)}{4.2752\times10^{-4}}=13824 \text{ m}^3/\text{s}
$$

**Answer: $Q_{0.1}$ ≈ 13824 m³/s** (the 10-year flood).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. The non-exceedance probability is:
   A. F(x) = exp[-e^(-alpha(x-beta))]
   B. F = 1 - e^(-x)
   C. F = e^(-alpha x)
   D. F = alpha(x-beta)
   Answer: A. This is the Gumbel CDF. (Key point: Gumbel CDF)

2. Using the hint, beta equals:
   A. mean + 0.48 sigma
   B. mean - 0.48 sigma = 8560 m3/s
   C. mean = 10000
   D. sigma/1.28255
   Answer: B. beta = 10000 - 0.48 x 3000 = 8560. (Key point: Parameters)

3. P(X > 15000) is about:
   A. 0.9383
   B. 0.5
   C. 0.0617
   D. 0.1
   Answer: C. F = 0.9383, so exceedance = 1 - F. (Key point: Exceedance probability)

4. An exceedance probability of 0.1 uses:
   A. F = 0.1
   B. F = 0.99
   C. F = 0.5
   D. F = 0.9, giving about 13824 m3/s
   Answer: D. This is the 10-year flood. (Key point: Exceedance 0.1)

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