We have 300 days, 150 of them are rainy and 150 sunny days. What is the probability to have for example 6, 7 and 8 rainy days successively during that 300 days? And that is the probability to get such a rows 1, 2, 3, 4, 5 times. I’d be very interested to see how to calculate this task
There is WAY not enough information. Your idea may presume independence of days. This is quite a foolish assumption for weather.
If you want to talk about days selected randomly, that would be a different conversation. Consecutive days? No.
By the way, the "probability of weather" is 100%!
What information do you need?
That rainy or sunny days is only example, not real situation. I need some explanation how to calculate probability of rainy weeks (6 days or 7) number during that 300 days. And the answer should be like, it is 99% probability to have rainy week once, 75% twice and so on... .
Well, if you insist on making calculation based on nonsense, feel free to do so.
According to you 150/150, each day as a 50% chance of rainfall or clear.
Two days 25% Both Rain 50% one ran and the other sunny 25% both sunny.
And on and on and on.
However, like I said, this assumes independence of days. Do you REALLY believe that rain today has NOTHING to do with rain tomorrow?
Why do you think it is based on nonsense?
First of all it is just a task, it can be the toss of a coin, or something with probability of 50%/50%.
Of course it is related if we have only 300 days (or something). But if it is ~, not.
It is every day (toss of a coin) the same probability of rainy, sunny days, even if we have 10 rainy days in a row, 11 day is 50% probability of rain again
and again about 50%, 25% and on and on, actually it should be 150/300*149/300*148/300...
Your idea may presume independence of days. This is quite a foolish assumption for weather.
______________________
The Leverage Season 5 DVD crew helps a veteran hockey player who thinks team management!
If you have a two state Markov chain model of the rain state, and tommorows rain state depends only on today's. Then, if the long term probability of rain is 0.5, the rain state tommorow is independent of the rain state today.Originally Posted by TKHunny
.
Perfectly reasonable question. On any day there is a 50% chance of rain. The probability of 3 days in succession or not is the probability of any three days being rainy. What is the probability of getting three heads, in succession or not? .5x.5x.5
You are assuming that this is a binomial distribution, with the probability of rain on any one day being 1/2 independent of the probability of rain the previous day. That is the simplest assumption you can make based on the information given but it is an assumption, not given in the problem. And, I suspect it is not true of rain in "the real world".
This has nothing to do with binomial distribution: out of N trials what is the probablity of n successes if there is a 50% chance of success for each trial.
The premise is obvious from the given information. If someone reports that an experiment is successful 50% of the time with no other information, one can logically only conclude that the probability of success in any one experiment is 50%, and that the probability of three successes in any three experiments, sequential or not, is .5x.5x.5.
The question is quite natural and logical: what can one conclude from the given information? If you were given more information about the distribution of rainy days it would be a different problem.
If I know that an experiment has a 50% chance of success, and I conduct 10 experiments in a row, the probability of three successes (rainy days) out of ten (binomial distribution) is not the same as the probability of the first, second, and fifth experiment, or any three specified experiments, sequential or not, being successful (rainy), which is .5x.5x.5, or.5^n for n successes (rainy days).
EDIT: The probability that it will rain three days in the week (binomial distribution) is not the same as the probability it will rain Mon, Tues, and Thu.
Yes, that's true (and uses the binomial distribution- I don't know why you said that "This has nothing to do with binomial distribution" before). But, as I said before that involves assuming there is a 50% chance of rain on any given day. You say "one can logically only conclude that the probability of success in any one experiment is 50%". No, that is NOT the only thing one can conclude- but it certainly is the simplest thing to do.
I have a bowl with 150 red and 150 white balls. What is the probability of picking 3, or n, red balls in succession or on my first, third, fifth,...try (with replacement)? .5^n
Why isn't that a complete, simple, problem? Does one have to make assumptions? The balls are mixed? but that's obvious. This is a routine problem in probability.
See post #7 above, which says with a reasonable-ish model of the correlation of rain state from day-to-day that it rains on about half of all days (which is the data given in the question) implies that rain tomorrow is independent of rain today (or yesterday, ...)
.
Actually,
for Markov model of two-state weather (wiki, examples of Markov chains):
p(n+1)=p(n)P. In the limit, p = pP.
If you assume lim p(n) = p = |1/2, 1/2 | and solve p=pP for P you get P = |x,1-x ; 1-x,x |
So no matter what x is, the long term PREDICTION is for 50/50 chance of rain or shine.
So you can’t conclude from long term PREDICTION what short term will be. Which means you can’t conclude that on any particular day the chance of rain or shine is 50/50. But the problem as stated requires you to know the probability on a given day which is only possible if x=1/2 in which case the probability of rain or shine on any particular day is 50/50 and you don’t have to assume it.
  |
1Thanks
LinkBack URL
About LinkBacks