# flares probability

There are 33 no of flares distributed in 2 hemisphere north and south. the no of flares in north hemisphere is 14 and no of flares in southern hemisphere is 19. then how to find the probability of distribution of dominant hemisphere (in this case southern) =0.189

for more detail please see the attachment

how the probability is counted in this attachment?

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Please share with us your thoughts on the problem, failed approaches, etc. – Sasha Oct 6 '12 at 14:44
Please state the question more clearly. In particular their seems to be no random event going on (you tell us the distribution is actually $14:19$), so where does probability come in in the first place? – Marc van Leeuwen Oct 6 '12 at 16:46

Number of ways for at least 19 flares to be in the southern hemisphere/total number of ways to be distributed is $\frac{{33\choose 19} + {33\choose 20}+...+{33 \choose 33}}{2^{33}} =0.243425$

An approximation to the answer is $\frac{1}{2}\mbox{erf} (\frac{19-\frac{33}2}{\sqrt2\log_2 33})=0.189912$

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thank you so much Angela Richardson. what is the last step of your ans?? i don't understand an approximation to the ans ??what is "erf "??? what is the use of probability=0.243425. what is the approximation formula?? please give me the detail of these formula its very important for me. please find the probability if total no. of flares are 480 in which 249 in north and 231 in southern hemisphere. – beena bhatt Oct 7 '12 at 2:10
I suspect that by wording the question "dominant hemisphere (in this case southern)" a two-sided probability is sought. – hardmath Oct 7 '12 at 17:42
@Angela Richardson what is the use of 0.243425 in this Approximation ??? – beena bhatt Oct 10 '12 at 3:16
@beena: It's somewhat ironic that you ask about a well-known abbreviation, "erf", that you could easily look up e.g. on Wikipedia, and in the same comment twice use an obscure abbreviation, "ans", that someone who doesn't happen to understand it would be hard-pressed to find a definition for. – joriki Oct 12 '12 at 12:58

The probabilities in Table $1$ of that paper seem to be flawed. The paper describes how to calculate them, but they don't actually seem to be calculated that way. Angela's answer is a nice attempt at explaining the discrepancy, but a) that value should have been rounded to $0.190$, not $0.189$, and b) the other values in the table don't match that approach. A particularly clear example is the column titled "$\gt50^\circ$", in which $3$ and $1$ flares should result in a probability of

$$\frac{\binom 41+\binom 40}{2^4}=\frac5{16}\;,$$

whereas the table gives $0.125=\frac2{16}$. I don't see a reasonable explanation for that value. It could be explained as the sum of the probabilities of events on both sides that are more extreme than the one observed (as opposed to events on one side that are more or equally extreme, which is what the text says), but that explanation doesn't work for the other columns. Angela's approximation gives $0.191$ in this case.

You might also want to check out this paper, which has a slightly more elaborate, though not much clearer explanation of the probabilities.

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@ joriki sir thank you so much for your attention. Sir I saw the Paper which you link but its not very enough to find the probability asked by me. Sir Please find the method which they used in research paper.how they calculate the probability. – beena bhatt Oct 15 '12 at 3:01
@beena: As I said, I already tried and I couldn't. – joriki Nov 4 '12 at 13:41