2 years, 1 month ago
Suppose the annual probability to rain in some region is p...
What is the probability to be drought (no rain) for n years?
Caveat and the second part of the problem: suppose you think like this "since the probability to rain in one year is p, the probability not to rain is 1-p, so for n years it's n(1-p)" Why is this method incorrect and is it possible that the real probability and the one given by incorrect method to *coincide*? In other words, is it possible to get the right answer with incorrect method? If so, for which n and p?
Caveat and the second part of the problem: suppose you think like this "since the probability to rain in one year is p, the probability not to rain is 1-p, so for n years it's n(1-p)" Why is this method incorrect and is it possible that the real probability and the one given by incorrect method to *coincide*? In other words, is it possible to get the right answer with incorrect method? If so, for which n and p?
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M$2 Answers
(1-p)^n
in other words: (1-p) to the nth power.
The reason is that if the probability of a drought for any year is (1-p, then the probablity for drought for n years is the probability of drought in year one AND drought in year two AND drought in year three.... AND drought in year n.
To find the probability of one event and another independent event, you multiply their probabilities. "And" generally means multiplication of probabilities.
So for part two of your question, n(1-p) is wrong. It would also mean that a 10-year drought is MORE probable than, say, a 2-year drought, which makes no sense.
Is it possible for n(1-p)to work anyway? Let's try setting n(1-p) equal to (1-p)^n, and then checking if that is possible.
n(1-p) = (1-p)^n
Then n(1-p)/(1-p) = n = (1-p)^(n-1)
But n >= 1. But (1-p) <= 1, since p is a probability and therefore between 0 and 1, and so (1-p)^(n-1) <= 1. So the only possible answer is if n=1.
So n=1 is the only situation where n(1-p) would work.
in other words: (1-p) to the nth power.
The reason is that if the probability of a drought for any year is (1-p, then the probablity for drought for n years is the probability of drought in year one AND drought in year two AND drought in year three.... AND drought in year n.
To find the probability of one event and another independent event, you multiply their probabilities. "And" generally means multiplication of probabilities.
So for part two of your question, n(1-p) is wrong. It would also mean that a 10-year drought is MORE probable than, say, a 2-year drought, which makes no sense.
Is it possible for n(1-p)to work anyway? Let's try setting n(1-p) equal to (1-p)^n, and then checking if that is possible.
n(1-p) = (1-p)^n
Then n(1-p)/(1-p) = n = (1-p)^(n-1)
But n >= 1. But (1-p) <= 1, since p is a probability and therefore between 0 and 1, and so (1-p)^(n-1) <= 1. So the only possible answer is if n=1.
So n=1 is the only situation where n(1-p) would work.
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M$
Assuming the probability of rain in any year is independent of the probability in any different year then the answer (as provided by a previous answerer) would be (1-p)^n. Also as stated by yourself in a comment, n * (1-p) is not a reasonable answer, since once n becomes larger than 1/(1-p), the probability calculated that way would exceed 1.
Asserting n * (1-p) = (1-p)^n we can see if there is any value of n for which the two methods (the correct and incorrect ones) would accidentally provide the same answer.
First, if n = 1, then the two sides are the same for any p: 1 * (1-p) = (1-p)^1 = (1-p)
Next, assuming n is not equal to 1:
Either (1-p) = 0 or we can divide by (1-p) in which case n = (1-p)^(n-1)
For the first case, p = 1, the answer would be 0 in both cases for any value of n.
For the second case, we can take the logarithm:
log (n) = (n-1) * log (1-p)
If n = 0 there is no sense talking about the probabilities, so assume n is not zero. Thus:
log (n^(1/n-1)) = log (1-p)
n^(1/(n-1)) = (1-p)
If we assume n is an integer, and we've ruled out n=0, and have dealt with n=1, then n=2, 3, 4, etc. However, for any n>1, n^(1/(n-1)) > 1 > (1-p) (since 0 <= p <= 1).
However, if p is the long-term average probability, and there is a climatological pattern ongoing which is decreasing the likelihood of rain in that region, then (1-p)^n would be incorrect, and would under-estimate the true probability for no rain over an n-year period. In such a case one would need to accurately model the climate of the region to get a more accurate answer.
The bottom line then is that if p is an accurate multi-year average which is not changing, and the probability of rain in any given year is independent of the probability of rain in any other year, the correct method would be (1-p)^n and this would accidentally be duplicated by the erroneous method n*(1-p) by any value of p if n = 1, and by any value of n if p=1.
Asserting n * (1-p) = (1-p)^n we can see if there is any value of n for which the two methods (the correct and incorrect ones) would accidentally provide the same answer.
First, if n = 1, then the two sides are the same for any p: 1 * (1-p) = (1-p)^1 = (1-p)
Next, assuming n is not equal to 1:
Either (1-p) = 0 or we can divide by (1-p) in which case n = (1-p)^(n-1)
For the first case, p = 1, the answer would be 0 in both cases for any value of n.
For the second case, we can take the logarithm:
log (n) = (n-1) * log (1-p)
If n = 0 there is no sense talking about the probabilities, so assume n is not zero. Thus:
log (n^(1/n-1)) = log (1-p)
n^(1/(n-1)) = (1-p)
If we assume n is an integer, and we've ruled out n=0, and have dealt with n=1, then n=2, 3, 4, etc. However, for any n>1, n^(1/(n-1)) > 1 > (1-p) (since 0 <= p <= 1).
However, if p is the long-term average probability, and there is a climatological pattern ongoing which is decreasing the likelihood of rain in that region, then (1-p)^n would be incorrect, and would under-estimate the true probability for no rain over an n-year period. In such a case one would need to accurately model the climate of the region to get a more accurate answer.
The bottom line then is that if p is an accurate multi-year average which is not changing, and the probability of rain in any given year is independent of the probability of rain in any other year, the correct method would be (1-p)^n and this would accidentally be duplicated by the erroneous method n*(1-p) by any value of p if n = 1, and by any value of n if p=1.
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M$
that's good
This answer is incomplete. See my answer for the complete response. The three items this answer missed are:
1. If the multi-year average is changing, you cannot use this approach.
2. If the probability of rain in one year is not independent of the probability of rain in the next or previous year this approach will not work.
3. The wrong method of n(1-p) will accidentally work not only for n=1 and any value of p, but also for p=1, for any value of n.
Yes. However, even if the value of p is valid, I'm not sure that subsequent years' probability is independent of this year's. Furthermore, from a purely mathematical viewpoint, the "Best Answer" missed that for p=1 the "wrong method" would also accidentally get you the right answer for any value of n. That answer would always be 0 of course.
"1. If the multi-year average is changing, you cannot use this approach. "
Do you mean if the annual probability is changing?
another thing is that if we keep adding, n(1-p) becomes bigger than 1
I agree with you @opher but she answered first and it's a pretty good answer ; I was satisfied by it