---
title: "Gamma: water supply inadequate"
subject: "Probability and Statistics"
chapter: 3
key: "semester_5/probability_and_statistics:gamma-water-supply-9-million"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/gamma-water-supply-9-million
site: BCT Engineering
---

# Gamma: water supply inadequate

Probability and Statistics · Chapter 3: Continuous Probability Distributions

- Asked in: 2073 Chaitra, 5 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/gamma-water-supply-9-million
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-3

## Question

The daily consumption of water in a certain place follow a gamma distribution with parameters α = 2 and β = 3. If the daily capacity of this city is 9 million gallon of water, what is the probability that on any given day the water supply is inadequate?

## Short answer

Daily consumption X ~ Gamma(α=2, β=3) million gallons with pdf f(x)=(x/9)e^(−x/3). Supply is inadequate if X>9. P(X>9)=∫_9^∞ f(x) dx = 4e^(−3) ≈ 0.1991 (19.91%).

## Answer

The daily consumption *X* (million gallons) follows a gamma distribution with shape α = 2 and scale β = 3, whose pdf is

f(x) = x^(α−1) e^(−x/β) / (β^α Γ(α)) = x e^(−x/3) / 9, x > 0

(Mean = αβ = 6 million gallons.) The supply is inadequate when consumption exceeds the capacity of 9 million gallons.

P(X > 9) = ∫₉^∞ (x/9) e^(−x/3) dx

Integrate by parts: ∫ x e^(−x/3) dx = −3x e^(−x/3) − 9 e^(−x/3)

```text
∫₉^∞ x e^(−x/3) dx = [−3x e^(−x/3) − 9 e^(−x/3)] from 9 to ∞
                   = 0 − (−27 e^(−3) − 9 e^(−3))
                   = 36 e^(−3)

P(X > 9) = 36 e^(−3) / 9 = 4 e^(−3)
         = 4 × 0.04979 = 0.1991
```

**Answer: the probability that the water supply is inadequate on a given day is about 0.1991 (19.91%).**

## Check yourself

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

1. The water supply is inadequate when:
   A. X < 9
   B. X > 9
   C. X = 6
   D. X < 6
   Answer: B. Supply fails when consumption exceeds the capacity of 9 million gallons, so we need P(X > 9). (Key point: Inadequate supply event)

2. With α = 2 and β = 3 the pdf simplifies to:
   A. x e^(−x/3)/9
   B. x e^(−3x)/2
   C. x^2 e^(−x/3)/27
   D. e^(−x/3)/3
   Answer: A. f(x) = x^(α−1) e^(−x/β)/(β^α Γ(α)) = x e^(−x/3)/(9 × 1). (Key point: Gamma pdf with α = 2, β = 3)

3. How is ∫ x e^(−x/3) dx evaluated?
   A. Direct power rule
   B. Integration by parts
   C. Substitution only, no parts needed
   D. Partial fractions
   Answer: B. It is a product of x and an exponential, solved by parts giving −3x e^(−x/3) − 9 e^(−x/3). (Key point: Method of integration)

4. The probability P(X > 9) equals:
   A. 4e^(−3) ≈ 0.1991
   B. 36e^(−3) ≈ 1.79
   C. e^(−3) ≈ 0.0498
   D. 1 − 4e^(−3) ≈ 0.8009
   Answer: A. The integral is 36e^(−3), divided by 9 gives 4e^(−3) = 0.1991. (Key point: Final result)

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