There is a country where everyone wants a boy. Every family continue to have babies till a boy is born. If the probability of having a girl or a boy is the same, what is the proportion of boys to girls after some time in that country ?
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11 comments:
2 girls to one boy.
Don't know but I want to live there.
One to one
probability versus actual statistics does not produce a definitive answer. So a guess is the best answer. Long winded explanation of "I don't know"!
Since the families keep having children until they have a boy, the possible outcomes are:
Girls Boys
0 1
1 1
2 1
3 1
4 1
5 1
6 1
7 1
8 1
9 1
10 1
Etc.
The number of girls is the sum of an arithmetic series (n/2)*(2+(n-1)). The number of boys is n+1.
Since both sums are unbounded, the probability is determined by the maximum number of children (assum)ng an equal distribution of all outcomes - which in itself is not likely).
Anyways for n = 10 (ie. up to eleven children, Prob is 1/5.
Answer is 1.
Girls Boys Probability
0 1 0.5
1 1 0.25
2 1 0.125
3 1 0.0625
4 1 0.03125
5 1 0.015625
6 1 0.0078125
7 1 0.00390625
8 1 0.001953125
9 1 0.000976563
10 1 0.000488281
11 1 0.000244141
12 1 0.00012207
13 1 6.10352E-05
14 1 3.05176E-05
Though the number of girls seems to be increasing, the likelihood of that event is getting lower.
With 50% chances a couple will have only a boy. And for the rest 50% chances atleast one girl. If you calculate #girls*probability and sum it up, it comes out to be 1 and similarly #boys*probablity and sum it up, it comes out to be 1 (both are an Infinite series with sum 1). Thus the proportion is 1.
Intuitively, Population proportion will be male biased only if events like female foeticide, etc. starts happening (Where chances of having a boy are forcefully increased, eg. Some states in India). Else the likelihood is to have a balanced mix of population.
Desiring a male child doesn't seems to be a problem but female foeticide is.
If there are n families, the total no of boys born is n because births take place until a boy is born in each family.
Considering all n families, the total no of girls born before n boys (is total no of failures before nth success) follows negative binomial distribution with parameters (n, 1/2)
Thus, expected no of girls born = mean of negative binomial distribution
= nq/p = n (since p=q=1/2)
Thus, proportion = no of boys/no of girls
= n/n =1
How does one get to know the correct answer? Will the author post the correct answer?
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