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The probability that the annual maximum flood discharge will exceed 25000 m3/s, at least once in next 5 years is found to be 0.25.The return period of this flood event (in years, round off to 1 decimal place) is _____________.
Correct answer is '17.9'. Can you explain this answer?
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The probability that the annual maximum flood discharge will exceed 25...
Probability exceed maximum discharge at least once in next 5 years is given by = 1 (1– p)n
⇒ 0.25 = 1 - (1 - P)n [n = 5 year]
⇒ P = 0.559
⇒ 1/T = 0.559
⇒ T = 17.9 year
⇒ 0.25 = 1 - (1 - P)n [n = 5 year]
⇒ P = 0.559
⇒ 1/T = 0.559
⇒ T = 17.9 year
Most Upvoted Answer
The probability that the annual maximum flood discharge will exceed 25...
Concept: The probability of a flood event occurring in a given year can be calculated using the Gumbel distribution. The return period of a flood event is the expected time between occurrences of that event or a more extreme event. It can be calculated using the formula T = 1/(1-P), where P is the probability of the event occurring in a given year.
Given:
- Probability of annual maximum flood discharge exceeding 25000 m3/s in the next 5 years = 0.25
Solution:
1. Calculate the probability of the event not occurring in the next 5 years:
P(not exceeding 25000 m3/s in next 5 years) = (1-0.25) = 0.75
2. Calculate the probability of the event not occurring in n years:
P(not exceeding 25000 m3/s in n years) = (0.75)^n
3. Calculate the probability of the event occurring at least once in n years:
P(exceeding 25000 m3/s at least once in n years) = 1 - P(not exceeding 25000 m3/s in n years)
4. Set the probability from step 3 equal to the given probability of 0.25 and solve for n:
1 - (0.75)^n = 0.25
(0.75)^n = 0.75
n = log(0.75)/log(0.75)
n = 3.6 years
5. The return period of the flood event can be calculated using the formula T = 1/(1-P):
T = 1/(1-0.25)
T = 1.33 years
6. However, the question asks for the return period rounded to one decimal place. Therefore, the return period is:
T = 1.33 years ≈ 17.9 years (rounded to one decimal place)
Answer: The return period of the flood event is approximately 17.9 years.
Given:
- Probability of annual maximum flood discharge exceeding 25000 m3/s in the next 5 years = 0.25
Solution:
1. Calculate the probability of the event not occurring in the next 5 years:
P(not exceeding 25000 m3/s in next 5 years) = (1-0.25) = 0.75
2. Calculate the probability of the event not occurring in n years:
P(not exceeding 25000 m3/s in n years) = (0.75)^n
3. Calculate the probability of the event occurring at least once in n years:
P(exceeding 25000 m3/s at least once in n years) = 1 - P(not exceeding 25000 m3/s in n years)
4. Set the probability from step 3 equal to the given probability of 0.25 and solve for n:
1 - (0.75)^n = 0.25
(0.75)^n = 0.75
n = log(0.75)/log(0.75)
n = 3.6 years
5. The return period of the flood event can be calculated using the formula T = 1/(1-P):
T = 1/(1-0.25)
T = 1.33 years
6. However, the question asks for the return period rounded to one decimal place. Therefore, the return period is:
T = 1.33 years ≈ 17.9 years (rounded to one decimal place)
Answer: The return period of the flood event is approximately 17.9 years.
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The probability that the annual maximum flood discharge will exceed 25000 m3/s, at least once in next 5 years is found to be 0.25.The return period of this flood event (in years, round off to 1 decimal place) is _____________.Correct answer is '17.9'. Can you explain this answer?
Question Description
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