| I'll do case two, with P(girl) = p, P(boy) = q.
With probability p a couple will have zero boys and one
girl.
With probability qp a couple will have one boy and one
girl.
With probability q^2 p a couple will have two boys and
one girl.
In general, with probability q^n p a couple will have n
boys and one girl, n = 0, 1, 2, ....
So each couple has an expectation of one girl child and
0 p + 1 q p + 2 q^2 p + 3 q^3 p + ... boy child(ren).
This series sums to (pq / (1-q)^2), or, as 1-q = p, to
(pq / p^2) or q/p boys.
Now, I'm not sure how far off of 50/50 the population can
get, if one assumes that only mixed gender couples are
having children. But N couples would have an expected
value of Nq/p boys and N girls.
Dan
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| The ratio of male babes to female babes is independent of the
"culture". If P(male) = p, and P(female) = q, and p+q = 1, then the
ratio will be p:q.
Think of it from the stork's perspective. Each baby he "delivers"
will have exactly prob p of being male. Which family that baby goes to,
and how many babes are already there is entirely irrelevant.
Regards,
Andrew.
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On first consideration it seems that the couples are favoring girls,
so there should be more of them than boys. However, as .2 points
out, you can't fool "mother" nature, so the ratio is unchanged, no
matter what the strategy is. .1 showed it for this case.
That was too easy. I'm still trying to find a challenging problem
to post that doesn't require a PhD just to comprehend.
Avery
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