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I've been working through some statistics problems and could use a little guidance on how to arrive at the right answer (and more importantly, why that is the right answer.)

I've detailed the problem below

"4. You are tasked with assessing the risk of damage due to hurricanes at a coastal city whose total exposed assets (e.g. building stock) are valued at 100 billion USD. The city is hit by two storms per year on average. For simplicity, you assume that each storm is independent and at landfall may have an intensity of Vmax = {30, 40, 60} m/s with probabilities of 0.5, 0.3, 0.2, respectively, and a translation speed of Vtrans = {1, 5, 10} m/s with probabilities of 0.1, 0.7, 0.2, respectively. Assume that Vmax and Vtrans are independent.

(a) Suppose storm surge depth, hsurge, depends simply on the maximum wind speed according to the following equation: hsurge = (hsurge,0)*(Vmax/30 m/s)^1.5 [m], where hsurge,0 = 1 m. However, the city is protected by a 2 m tall sea wall, and thus there is no surge at all unless hsurge exceeds this height. If the wall is breached, though, then water flows over and the wall does not reduce the surge at all. Calculate the probability that the sea wall will be breached at least once in a given year."

What I have done so far is calculated probabilities for each of the possible hsurge values using the Vmax values (30, 40, and 60).


hsurge,30 = 1m --> Probability = .5
hsurge,40 = 1.54m --> Probability = .3
hsurge,60 = 2.83m --> Probability = .2

Given that hsurge,60 is the only case in which there is any storm surge in the city, due to the surge wall, it is the only case that counts right? (I'm fairly confident that all other points are not being considered since we were given the sample space for Vmax)

What do I do with this now? There are two storms per year on average but I'm fairly certain I can't just multiply .2 by 2. Any guidance would be greatly appreciated.

1 Answers 1

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I think your work thus far is correct, and as you intuited, you can't just multiply .2 by 2. Otherwise, if there were, say, 6 storms per year, then you would have .2 * 6 = 1.2 > 1, which is a probability greater than 100%, which is not a reasonable outcome.

I will assume that there are always exactly two storms each year. Since we care only that we had at least one breach, the 20% chance occurrence of a breach during Storm 2 only counts if Storm 1 did not already breach. So, what we have is a 20% chance that the wall gets breached during Storm 1, and if it doesn't happen (80%), then, in that case only, another 20% chance (relative to the 80%) that it gets breached during Storm 2. Visualizing a decision tree of your problem might help. So we need to add the probability of the breach happening during Storm 1 (20%) with the probability of the breach happening during Storm 2 but not Storm 1 (80% * 20%). So 20% + 80% * 20% = 36%.

(TL;DR) For a much simpler alternative solution, you could also ask, what is the probability of getting no breaches, and then subtract that result from 1. So, 1 - 0.8 * 0.8 = 1 - 0.64 = 0.36.

Restated more algebraically, let B1 be the event that there is a breach during Storm 1, and similarly B2 the event that there is a breach during Storm 2. Then: P(at least 1 breach) = 1 - P(no breaches) = 1 - (P(not B1) * P(not B2)) = 1 - (0.8 * 0.8) = 1 - 0.64 = 0.36, where B1 and B2 are assumed to be independent events (as allowed in the problem description).

To be more accurate in real life, though, one would need to consider the probability distribution of having n storms in a given year (e.g., 10% chance of 0 storms, 20% chance of 1 storm, 50% chance of 2 storms, 20% chance of 3 storms, or such). However, it seems that the problem source is not interested in that since those details are not given. So I've assumed that 2 storms occur every year.