A 1 hour rainfall of 10 cm has return period of 50 year. The probabili...
Return period of rainfall, T = 50 years
Probability of occurrence once in 50 years, p=1/50 =0.02
Probability of occurrence in each of 2 successive years
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A 1 hour rainfall of 10 cm has return period of 50 year. The probabili...
Given information:
- 1 hour rainfall of 10 cm has a return period of 50 years
To find:
- Probability of 1 hour or more rainfall of 10 cm occurring in each of two successive years
Solution:
Definition of return period:
The return period of an extreme event is the average time between occurrences of an event of equal or greater magnitude.
From the given information, we know that a 1 hour rainfall of 10 cm has a return period of 50 years. This means that in a given year, the probability of this extreme event occurring is 1/50 or 0.02.
Probability of 1 hour or more rainfall of 10 cm occurring in two successive years:
Since we want to find the probability of this extreme event occurring in two successive years, we can use the following formula:
P(A and B) = P(A) x P(B|A)
- P(A) = Probability of the extreme event occurring in year 1
- P(B|A) = Probability of the extreme event occurring in year 2, given that it occurred in year 1
Probability of the extreme event occurring in year 1:
Since the event has a return period of 50 years, the probability of it occurring in a given year is 1/50 or 0.02. Therefore,
P(A) = 0.02
Probability of the extreme event occurring in year 2, given that it occurred in year 1:
If the extreme event occurred in year 1, the probability of it occurring again in year 2 is still 0.02 since the events are independent of each other. Therefore,
P(B|A) = 0.02
Using the formula above, we can find the probability of the extreme event occurring in two successive years:
P(A and B) = P(A) x P(B|A)
P(A and B) = 0.02 x 0.02
P(A and B) = 0.0004
Therefore, the probability of 1 hour or more rainfall of 10 cm occurring in each of two successive years is 0.0004 or 0.04%.
Answer: Option D (0.0004)
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