Friday, February 9, 2007
In a country in which people only want boys, every family continues to
In a country in which people only want boys, every family continues to have children until they have a boy. if they have a girl, they have another child. if they have a boy, they stop. what is the proportion of boys to girls in the country?
Subscribe to:
Post Comments (Atom)
If both having a boy or a girl are equal probable then the ratio should be 1.
ReplyDeleteThanks for the post
ReplyDeleteI think the answer should be 1/2.
ReplyDeleteTake this scenarios, we have x families. If we assume the having a boy or a girl is equal-probable. Then x/2 families will have boys and x/2 families will have girls. Next time x/4 family will have boys and x/4 family will have girls and so on. At the end, we get (x/2+x/4+x/8+...) boys and (x/2+x/4+x/8+...) girls and the ratio is 1. This is considering we have no twins, triples,....
ReplyDeleteThe ratio is 1 if the family population is an odd value. The ratio is 1 plus 1 boy if the family population is an even value. In fact, in the case of an even family population the last two families will have 1 boy and 1 girl and the family with a boy stops, while the family with a girl makes another baby. This baby will be a boy since that family has already had a girl.
ReplyDeleteIf the chance of having a boy and a girl is not equal....
ReplyDeleteSay P(birth is male)=p
The number of girls had by each family is distributed as a geometric random variable. The expected number of female births before a male birth is then (1-p)/p.
Thus, for every 1 boy, there are (1-p)/p girls.
We have to compute the number of girls according to the distribution:
ReplyDeletesum(P(having k girls) * k) = sum (k=0 -> infinite, k/2^(k+1)) = 0*1/2 + 1*1/4 + 2*1/8 + 3*1/16 ...
while the number of boys according to the distribution is 1.
Computing the mathematical serie for the girls gave me that the average number of girls according to the distribution is 1.
So there are 50% girls and 50% boys, but the answer is bit tricky to compute in the lands of probabilities.
If we assume certain things [no twins, no fertility issues such as no Y chromosomes, etc.,] the chance of having a boy is always 50% - regardless of the couple's previous track record. That is, couple xy's first, second, third, etc., child has a 50/50 chance of being a boy [and 50 percent chance of being a girl - assume no mutants :)]. So the proportion in this assuming country is 1:1. If there are fertility issues, the ratio skews more in favor of girls because couples that are incapable of having boys will continue manufacturing girls until the cows come home.
ReplyDeleteAssume that having a girl or a boy is a 50/50 chance.
ReplyDeletelet's take 4 families as an example
1) boy - stop
2) girl + boy - stop
2) boy - stop
4 girl + boy - stop
Total 4 boys 2 girls.
Hence the proportion is 2:1
It should be
ReplyDeleteNo of girls to no of boys =
No of girls in the village : No of families
Consider the following sample space:
Family 1: GB
Family 2: B
Family 3: GGGB
Family 4: GGGGB
Family 5: GGB
Family 6: B
Hence we have 10 girls and 6 boys (no. of families in the village is also 6)
Important: Question asked is not the probablity of a girl to the probablity of a boy, its instead the proportion of the no. of girls in the village to the no. of boys.
probablity is chance of an event likely to happen
proportion is something to do with magnitudes of 2 entities involved..
and importantly, proportions stands for something which exists... like boys to gals ratio in a class..
wheras probablity of boys to gals in a class is an invalid question..
yes it can be made valid by extending it to "probablity of a boy topping the class to a gal"
10:6 is wrong because the sample space is infinite, and the frequency of certain couples are higher than that of others. Here is an easier solution:
ReplyDeleteLet us define E(Babies)= # babies an average family has
E(Babies)= 1/2*1+ 1/2*(2+1/2(3+1/2(4...))))
The second parenthesis is like starting over again, but you have to add 1 to the expectation.
E(Babies)=1/2*1+1/2(E(Babies)+1)
E(Babies)*1/2=1
E(Babies)=2
E(Girls)=E(Babies)-E(Boys)=2-1
E(Girls):E(Boys)=1:1
This, I believe, is the how you derive the (1-p)/p (which is correct if boys and girls aren't equally likely)
b=0:1
ReplyDeletegb=1:1
ggb=2:1
gggb=3:1
.
.
.
.
.so the answer must be X:1
出逢いという奇跡を起こす真の「出逢い」系サイトの誕生です。友達、割り切り、恋愛…十人十色の目的に対応した真面目系恋人探し専用出逢い掲示板で、あなたの運命の人を探してみませんか？
ReplyDeletehey i dont think this one has an answer...its hugely probabilistic. it cant even be (1-p)/p because it is dependant on probability.
ReplyDeleteLet b= probability of a boy=0.5
ReplyDeleteg= probability of a girl=0.5
b + gb + ggb + gggb + ggggb.....
=b(1+g+gg+ggg+gggg+.....)
=b(1/(1-g))
=b/(1-g)
=0.5/0.5
=1