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mzilplik · Oct 13, 2009, 04:59 AM

What are the chances of getting 15 days of rain during the next 30 days? Assuming that there's a 70% of rain everyday and that what happens on one day (rain or shine) was not in any way dependent on what happened the day before. (Do I need to use BAYES' Theorem? Please help!)

mzilplik · Oct 13, 2009, 05:10 AM

Originally Posted by galactus

$(.7)^{15}$

No Bayes theorem needed.

Someone sent me this answer...

If you mean exactly 15 days (and no more), then it's
30 C 15 ways to choose the days times .70^15 * .30^15
probabilities of rain on 15 days and not rain on 15 days

155117520 * 0.00475 * 0.0000000143 = 0.0106
In other words, about 1%

mzilplik · Oct 13, 2009, 05:23 AM

Originally Posted by mzilplik

Someone sent me this answer...

If you mean exactly 15 days (and no more), then it's
30 C 15 ways to choose the days times .70^15 * .30^15
probabilities of rain on 15 days and not rain on 15 days

155117520 * 0.00475 * 0.0000000143 = 0.0106
In other words, about 1%

How did he come up with this solution?

galactus · Oct 13, 2009, 05:46 AM

This is a binomial probability.

$C(30,15)(.70)^{15}(.30)^{15}$

That is where they got it from.

Sorry, I was in error on my first post. I was thinking 15 consecutive days.

mzilplik · Oct 13, 2009, 06:26 AM

Originally Posted by galactus

This is a binomial probability.

$C(30,15)(.70)^{15}(.30)^{15}$

What is C(30,15)?
How did he arrive with 155,+++,++ ?

ebaines · Oct 13, 2009, 09:48 AM

Mzilplik:

What Galactus gave you is the probability that in the next 30 days there will be precisely 15 days of rain and 15 days of no rain. The notation C(30,15) means "the combination of 30 items taken 15 at a time," and is calculated from:

$<br /> \frac { 30)!} {(30-15)! 15!} <br />$

It might be easier to explain this if you consider a simpler version of this problem. Suppose you are interested in knowing the probability of it raining on 2 of the next 4 days. There are 6 possible combinations of rainy (R) and sunny (S) days, as follows:

RRSS
RSRS
RSSR
SRSR
SRRS
SSRR

It is assumed that each of these 6 combinations is equally likely. This value of 6 can be calculated using:
$<br /> C(4,2) = \frac {4!} {(4-2)!2!} = \frac {4*3*2*1} {(2*1)*(2*1)} = 6<br />$

So to do the math it is convenient to calculate the probability that the first combination will occur (RRSS), then multiply that by 6. Hence the math is: $C(4,2)*(.7)^2(.3)^2$

Now we extend this to the case of 15 days of rain out of 30, and you get Galactus's answer.

yr74msn · Oct 31, 2009, 08:57 AM

Can some explain the quantitative reasoning behind the answer? I just need to find a better way of wording it.

morgaine300 · Nov 1, 2009, 08:16 PM

There's a .70 probability of rain. So each day has a .70 probability that it will rain, times itself 15 times to get 15 total days. In other words:

$(.70)(.70)(.70)\text\ 15\ times\ for\ the \ 15\ days.\ Hence\ .70^{15}$

That leaves a .30 probability that it won't rain. So you need 15 days that it won't rain, at a .30 probability for each one. So

$(.30)(.30)(.30) \text\ 15\ times,\ hence\ (.30)^{15}$

Hence, the

$(.70)^{15}(.30)^{15}$

Which gives us the equation:

$p^{x}(1-p)^{n-x}$

where x is how many you're looking for (15 days) and n is the total.

You're multiplying out the probability of "success" (as defined by what you're looking for, in this case, rain) the number of times you're looking for it to happen. Then multiplying out the probability of failure times the number of times it will happen.

But that's only if the first 15 days are the ones during which it rains and the last 15 the ones when it doesn't rain. So we add the combination in front of it to account for the number of ways we could combine rain with no rain, as ebaines pretty much already went over. So we have the total equation:

$C(n,x)p^x(1-p)^{n-x}$