A 6-hour rainfall of 6 cm at a station was found to have a return period of 40 years. The probability that a 6-hour rainfall of this or larger magnitude will occur at least once in 20 successive years at that station is ______ (rounded off to three decimal places).

Question

Which one of the following gases is primarily responsible for the formation of photochemical smog?

Answer 1

A neat way to see this problem is through the idea of a Bernoulli trial repeated every year. In any single year, the chance that this rainfall depth is met or exceeded is $p = 1/T = 1/40 = 0.025$, so naturally the chance that it is NOT met in that year is $1-p = 0.975$. Since successive years are independent of each other from a probability standpoint, the chance that the rainfall stays below this value in every one of the 20 years is the product of 20 identical factors, $(0.975)^{20}$. Taking logarithms, $\ln(0.975) \approx -0.02532$, so $20\ln(0.975) \approx -0.5064$, giving $(0.975)^{20} \approx e^{-0.5064} \approx 0.603$. The event we actually want, "at least one exceedance in 20 years," is simply everything except "no exceedance in all 20 years," so its probability is $P = 1-(0.975)^{20} \approx 1-0.603 = 0.397$. This value falls inside the accepted band of $0.350$ to $0.450$, confirming the computation.

\[\boxed{P \approx 0.397}\]

Answer 2

A neat way to see this problem is through the idea of a Bernoulli trial repeated every year. In any single year, the chance that this rainfall depth is met or exceeded is $p = 1/T = 1/40 = 0.025$, so naturally the chance that it is NOT met in that year is $1-p = 0.975$. Since successive years are independent of each other from a probability standpoint, the chance that the rainfall stays below this value in every one of the 20 years is the product of 20 identical factors, $(0.975)^{20}$. Taking logarithms, $\ln(0.975) \approx -0.02532$, so $20\ln(0.975) \approx -0.5064$, giving $(0.975)^{20} \approx e^{-0.5064} \approx 0.603$. The event we actually want, "at least one exceedance in 20 years," is simply everything except "no exceedance in all 20 years," so its probability is $P = 1-(0.975)^{20} \approx 1-0.603 = 0.397$. This value falls inside the accepted band of $0.350$ to $0.450$, confirming the computation.

\[\boxed{P \approx 0.397}\]

A 6-hour rainfall of 6 cm at a station was found to have a return period of 40 years. The probability that a 6-hour rainfall of this or larger magnitude will occur at least once in 20 successive years at that station is ______ (rounded off to three decimal places).

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Correct Answer: 0.397

Solution and Explanation

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