Question 1161341
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Your question is rather unclear.  In the first place, what happened to the tenth grandchild?


So do you want the odds for 8 boys followed by 1 girl as opposed to any other configuration? Or do you want the odds against that occurrence?  Or do you want the odds that the next one will be one sex or the other?


There are *[tex \Large 2^9\ =\ 512] different configurations of boys and girls in a series of nine children. There is exactly one way to obtain the exact configuration 8B/1G, so the probability of that exact configuration is 1/512 which means the odds in favor of such an occurrence are 1 to 511 (odds against would be 511:1).


All of that presumes that the probability for any given pregnancy that the child will be a girl is exactly 1/2 -- just like flipping a fair coin.  Unfortunately for your question, human biology is anything but a fair coin. The particular physiology of any given egg provider, sperm donor, or a combination thereof, may skew the probability per birth one way or the other so that results such as you describe are not actually that uncommon.


The upshot is that the answer to your question is "I have no idea".  It is certainly possible that a team of researchers including experts in urology, gynecology, and genetics could, for the particular set of parents that produced the set of grandchildren you describe, determine a probability number for one sex or the other for any given birth.  If we let *[tex \Large p] represent the probability that a given birth will result in a boy, then the probability of 8 boys and 1 girl in 9 births would be given by:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(8,9,p)\ =\ {{9}\choose{8}}\,(p)^8(1-p)]


This is the probability of 8 boys and 1 girl with the girl occurring anywhere in the sequence.  For your specific situation, you would have to divide by 9.


For example, if the boy/girl ratio for the particular set of parents was 60/40, then


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ P(8,9,.6)\ =\ {{9}\choose{8}}\,(.6)^8(.4)\ \approx\ 0.0604]


Then, divided by 9 it is approximately *[tex \Large 0.0067]


Not very likely, but not surprising that it occurs once in a while, like almost 7 times in 1000.  Hence, the odds in favor would be about 7 to 993.
								
								
John
*[tex \LARGE e^{i\pi}\ +\ 1\ =\ 0]
My calculator said it, I believe it, that settles it
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