This is a probability question that I don't quite understand

**Schuyla** · April 2nd 2012, 04:10 AM

In an attempt to reduce the growth of its population, China instituted a policy limiting a family to one child. Rural Chinese suggested revising the policy to limit families to one son. Assuming the suggested policy is adopted and that any birth is as likely to produce a boy as a girl, explain how to use simulation to answer the following:

What would be the average family size?
What would be the ratio of newborn boys to newborn girls?

So if anyone can help explain this one, I would truly be grateful.

**MathoMan** · April 4th 2012, 03:35 AM

A family that would count n members would look like: n= 2parents + number of girls born + 1 son, or more precisely $n=k+3$ , where k is the number of girls born prior to the birth of a son.

First you set up a random variable X that would count the members of the family. Its values would be $x_0=3=2+0+1$ , $x_1=4=2+1+1$ , $x_2=5=2+2+1$ , $x_3=6=2+3+1$ , .... You get the pattern, right? $x_k=k+3$ where $k$ is the number of girls born before the first son.

Then you should assign probabilities to those values.
$p_k=P(X=k+3)=(0.5)^{k} \cdot 0.5 = (0.5)^{k+1}$ , where $(0.5)^{k}$ means that the $k$ members are girls born prior to a son being born, and he is born with the probability $0.5$ . So you can
So now you have

$X\sim\left(\begin{array}{ccccc}3 & 4 & 5 & 6 & \cdots \\ 0.5 & (0.5)^{2} & (0.5)^{3} & (0.5)^{4} & \cdots \end{array}\right)$

Next, find the expected value of that random variable:
$\sum\limits_{k=0}^{\infty}x_k\cdot p_k & =\sum\limits_{k=0}^{\infty}(k+3)(0.5)^{k+1}= \frac{1}{2} \sum\limits_{k=0}^{\infty}(k+3)(0.5)^{k}=$
$=\frac{1}{2} \left(\sum\limits_{k=0}^{\infty}k(0.5)^{k}+3\cdot \sum \limits_{k=0}^{\infty}(0.5)^{k}\right)= \frac{1}{2}\left(2+3\cdot 2\right)=4.$

I think that would be the size of the average family, unless I'm wrong, eh?!

**Schuyla** · April 6th 2012, 07:57 AM

Thank you for your explanation! I went a round about way of doing this and came up with my answer.

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