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The first part of this question states: Consider a very small town with 50 families with children. Let X be the number of children in a family picked at random from the 50 families with children in the town. Suppose that the family size distribution is given by $$ fX(x) = \begin{cases} 0.3 & \text{if } x = 1, \\ 0.4 & \text{if } x = 2, \\ 0.26 & \text{if } x = 3, \\ 0.04 & \text{if } x = 4. \end{cases} $$ I then had to calculate the cumulative distribution function and the Expected value and Var(X). I managed to do that all okay. But with the next part of the question I am really stuck: Now suppose that you pick a child at random from the children in this town - each child is equally likely to be picked - and ask the child how many children there are in their family (including the child you asked). Let Y be the size of the child's family. (i) How many children are there in the town?

My thoughts, at first I thought that this might involve me forming a Poisson distribution but they I realised I won't have any parameter to form the distribution with. What distribution would be best to used then? Because it can't be a Bernoulli trial or a Geometric distribution either.

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    The question may be truncated, since it has a (i) without a (ii). But if you really only want to know the answer to (i), you're thinking way too complicated. You already have the expected number of children per family; just multiply it by the number of families.2011-04-26
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    (ii) might be: what is the distribution (or the expected number) of children in the randomly picked child's family?2011-04-26
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    For instance, if you have only two families, one with one child and one with four children, then I guess this distribution is $\frac{1}{5}\delta_1+\frac{4}{5}\delta_4$. [This](http://en.wikipedia.org/wiki/Conditional_expectation) is the relevant wikipedia article.2011-04-26
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    @joriki There is a part (ii ) which asks me to calculate the expected value. Once I know what distribution to use for this question then I will know how to calculate the expected value.2011-04-26
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    Can you figure out how many children of each type of family there are? If so, that should give you the distribution you're looking for. Don't think too complicated; it has nothing to do with geometric distributions or stuff like that; it's a simple question of how many children in families with $k$ children there are if there are $m$ families with $k$ children.2011-04-26

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Hint: Call $N$ the total number of children and $K_x$ the number of families with $x$ children. Write $N$ and $P(X=x)$ as a function of $(K_y)$. Compute the number of children whose family has size $x$. Conclude.

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    This will weight the probabilities by the number of children in the family. You are more likely to get a child from a four child family than the fraction of four child families in the population, which is the point of the problem.2011-04-26
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    @Ross: Yes you are. (Is your comment addressed to me?)2011-04-26
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    No, it was addressed OP. Too late to edit now. I thought it helped explain what the problem was supposed to show. Sometimes, like this, there is a takeaway that people can miss if they just find the answer.2011-04-26
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    I will attempt what you suggested. Don't think I understand it 100% yet though.2011-04-26
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    Where are the % you do not get? Be as specific as possible. Your first step is to write $N$ the total number of children when you know there are $K_1$ families with $1$ child, $K_2$ families with $2$ children, and so on. So, $N=$... ?2011-04-26