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This question already has an answer here:

We know that the probabilty of having boy is $\frac{3}{4}$ and every family has kids until they have boy, after boy they have no children, what is the ratio between boys and girls in population?

I suppose we want to calcualte this using expected value.

Every family has 1 boy so $E(boys) = 1 $

However i am not sure how to calculate expected value of girls. The boys have exponential distribution. So we can calculate probabilty of 0..1... inf girls by calculating "what is the probability that xth child is boy" e.g

$\sum_{i = 1}^{\inf} (\frac{1}{4})^{i-1}*\frac{3}{4}$

and expected value by

$\sum_{i = 1}^{\inf} i* (\frac{1}{4})^{i-1}*\frac{3}{4}$

my question is, this formula basicly tell us expected value of of boys e.g $x *P(x)$ where x is on what attempt family had boys * probability that it was on that attempt.

This does not tell us nothing about girls. So if i modified it to

$\sum_{i = 1}^{\inf} i* (\frac{1}{4})^{i}*\frac{3}{4}$

This should tell us $x * P(x)$ where x = number of girls and P(x) = probability that family had boys on i + 1 attempt = probability that family had i girls.

And just adjusting it to

$\frac{1}{4}*\sum_{i = 1}^{\inf} i* (\frac{1}{4})^{i-1}*\frac{3}{4} = \frac{1}{4} * E(boys) = \frac{1}{4} * \frac{1}{\frac{3}{4}} = \frac{1}{4} * \frac{4}{3} =\frac{1}{3}$

So boys:girls should be $1:\frac{1}{3} = 3:1$

But i am not sure if logic behind this is correct. Can i adjust the formula this way for it to make sense?

Thanks for answers.

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marked as duplicate by Rohan, NCh, Shailesh, Dirk, Claude Leibovici Jan 5 '18 at 9:38

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ no, the question may be simmiliar but the task is and calculation is different $\endgroup$ – trolkura Dec 28 '17 at 14:09
  • $\begingroup$ I hope that you can check the calculations there and implement them in yours as well. $\endgroup$ – Rohan Dec 28 '17 at 14:10
  • $\begingroup$ calcualtions are compeltely different $\endgroup$ – trolkura Dec 28 '17 at 14:10
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I think this question could definitely use better formatting. Here is how I would approach this problem.

Let $X$ be the random variable that assigns to a family how many kids they had. Notice that $X$ takes values in $\{1,2,3,\dots\}$. We have that for each $k\geq 0$

$$\mathbb{P}(X=k+1)={\left(\frac14\right)}^k\,\frac34$$

Hence, we have

\begin{align} \mathbb E(X) &=\sum_{k\geq0}^\infty\, (k+1)\cdot\mathbb{P}(X=k+1)\\ &=\sum_{k\geq0}^\infty\, (k+1)\cdot{\left(\frac14\right)}^k\,\frac34\\ &=4\,\sum_{k\geq0}^\infty\, (k+1)\cdot{\left(\frac14\right)}^{k+1}\,\frac34\\ &=3\,\sum_{k\geq1}^\infty\, k\cdot{\left(\frac14\right)}^{k} \end{align}

This kind of sum can be found considering term by term differentiation of the

$$f(x)=\frac{1}{1-x}=\sum_{k\geq0}x^k$$

and the result is $\mathbb E(X)=3\cdot \frac49=\frac43$.

It follows that on average a family has '$4/3$' kids. Since a kid is always necessarily a boy, we have that on average a family has '$1/3$ of a girl' and $1$ boy, so the ration between boys and girls is $3$.

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Shortcut.

If $B$ denotes the number of boys then $B=1$ so that $\mathsf EB=1$ (as you state correctly).

If $G$ denotes the number of girls then: $$\mathsf EG=\frac34\cdot0+\frac14\cdot(1+\mathsf EG)=\frac14+\frac14\mathsf EG\tag1$$so that $\mathsf EG=\frac13$.

If you do not understand why $(1)$ is valid, then let me know.

Then: $$\mathsf EB:\mathsf EG=1:\frac13=3:1$$(as you state correctly).

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