flares probability

Question

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?

Answer 1

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 $\frac{1}{2}\mbox{erf} (\frac{19-\frac{33}2}{\sqrt2\log_2 33})=0.189912$

Answer 2

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.