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The maximum one day rainfall depth at 20 year return period of a city is 150 mm. The probability of one day rainfall equal to or greater than 150 mm in the same city occurring twice in 20 successive years is__________
Correct answer is between '0.18,0.19'. Can you explain this answer?
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The maximum one day rainfall depth at 20 year return period of a city ...
Given Information:
- Maximum one-day rainfall depth at a 20-year return period = 150 mm
To Find:
- Probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years
Assumptions:
- Rainfall events are independent and identically distributed (i.i.d) over the years
Approach:
1. Calculate the probability of one-day rainfall equal to or greater than 150 mm in a single year.
2. Since the events are i.i.d, we can use the binomial distribution to calculate the probability of the event occurring exactly twice in 20 successive years.
3. The probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years is the sum of probabilities of the event occurring exactly twice, thrice, up to 20 times.
Calculations:
Step 1: Probability of one-day rainfall equal to or greater than 150 mm in a single year
- The return period is given as 20 years. Return period is the reciprocal of the exceedance probability.
- Exceedance probability = 1 / Return period = 1 / 20 = 0.05
- The probability of one-day rainfall being less than or equal to 150 mm in a single year = 1 - Exceedance probability = 1 - 0.05 = 0.95
Step 2: Probability of the event occurring exactly twice in 20 successive years
- We can use the binomial distribution formula: P(X = k) = nCk * p^k * (1 - p)^(n - k)
where n = 20 (number of trials), k = 2 (number of successful events), and p = 0.05 (probability of success)
- P(X = 2) = 20C2 * 0.05^2 * (1 - 0.05)^(20 - 2) = (20! / (2! * (20 - 2)!)) * 0.05^2 * 0.95^18
Step 3: Probability of the event occurring exactly twice to 20 times in 20 successive years
- We need to calculate the sum of probabilities for k = 2 to k = 20
- P(X ≥ 2) = P(X = 2) + P(X = 3) + ... + P(X = 20)
Final Answer:
- The probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years is the sum of probabilities calculated in Step 3.
- The correct answer is between 0.18 and 0.19.
- Maximum one-day rainfall depth at a 20-year return period = 150 mm
To Find:
- Probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years
Assumptions:
- Rainfall events are independent and identically distributed (i.i.d) over the years
Approach:
1. Calculate the probability of one-day rainfall equal to or greater than 150 mm in a single year.
2. Since the events are i.i.d, we can use the binomial distribution to calculate the probability of the event occurring exactly twice in 20 successive years.
3. The probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years is the sum of probabilities of the event occurring exactly twice, thrice, up to 20 times.
Calculations:
Step 1: Probability of one-day rainfall equal to or greater than 150 mm in a single year
- The return period is given as 20 years. Return period is the reciprocal of the exceedance probability.
- Exceedance probability = 1 / Return period = 1 / 20 = 0.05
- The probability of one-day rainfall being less than or equal to 150 mm in a single year = 1 - Exceedance probability = 1 - 0.05 = 0.95
Step 2: Probability of the event occurring exactly twice in 20 successive years
- We can use the binomial distribution formula: P(X = k) = nCk * p^k * (1 - p)^(n - k)
where n = 20 (number of trials), k = 2 (number of successful events), and p = 0.05 (probability of success)
- P(X = 2) = 20C2 * 0.05^2 * (1 - 0.05)^(20 - 2) = (20! / (2! * (20 - 2)!)) * 0.05^2 * 0.95^18
Step 3: Probability of the event occurring exactly twice to 20 times in 20 successive years
- We need to calculate the sum of probabilities for k = 2 to k = 20
- P(X ≥ 2) = P(X = 2) + P(X = 3) + ... + P(X = 20)
Final Answer:
- The probability of one-day rainfall equal to or greater than 150 mm occurring twice in 20 successive years is the sum of probabilities calculated in Step 3.
- The correct answer is between 0.18 and 0.19.
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The maximum one day rainfall depth at 20 year return period of a city is 150 mm. The probability of one day rainfall equal to or greater than 150 mm in the same city occurring twice in 20 successive years is__________Correct answer is between '0.18,0.19'. Can you explain this answer?
Question Description
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