Possible mistake in book choosing Kth child Suppose that there exist N families on the earth and that the maximum number of children a family has is c. For $j=0,1,2 \ldots c$, let $\alpha_j$ be the fraction of families with $j$ children ($\sum_{j=0}^c\alpha_j=1$). 
A child is selected at random from the set of all children in the world. Let this child be the $K$th born of his or her family; then $K$ is a random variable. Find $E(K)$.
Textbook Solution:
Let $A_j$ be the event that the person belongs to a family with j children. Then$$P(K=k)=\sum_{j=0}^cP(K=k|A_j)P(A_j)=\sum_{j=k}^c\frac{1}{j}\alpha_j$$
Therefore$$E(K)=\sum_{k=1}^ckP(K=k)=\sum_{k=1}^ck\sum_{j=k}^c\frac{1}{j}\alpha_j=\sum_{k=1}^c\sum_{j=k}^c\frac{k}{j}\alpha_j$$
I feel that there is a problem with $P(A_0)$. We can't condition on this event. Am I right? My solution is $$ P(K=k)=\frac{\sum_{i=k}^c\alpha_i}{\sum_{l=1}^cl\alpha_l}$$ and $$E(K)=\frac{\sum_{k=1}^c\sum_{i=k}^ck\alpha_i}{\sum_{l=1}^cl\alpha_l}$$
 A: There are $N\alpha_j$ families with $j$ children in the world described in the question, therefore
$S = \sum_{j=0}^c j(N\alpha_j) = N\sum_{j=0}^c j\alpha_j$ children altogether.
(Note that $S = N E(X)$ where $X$ is the number of children in a randomly chosen family.)
The most reasonable interpretation of randomly selecting one of these children is that each child has $\frac 1S$ probability to be selected, regardless of which family the child belongs to.
From that it follows that 
$$P(A_j) = \frac{j\alpha_j N}{S} = \frac{j\alpha_j}{\sum_{m=0}^c m\alpha_m}.$$
We could put this into the textbook answer instead of the incorrect
$P(A_j) \stackrel?= \alpha_j.$
But I think I like your approach better. There is one $k$th-born child in each family of $k$ or more children.
Hence the total number of $k$th-born children is
$N\sum_{j=k}^c \alpha_j,$
and therefore
$$ P(K = k) = \frac{N\sum_{j=k}^c \alpha_j}{S}
 = \frac{\sum_{j=k}^c j\alpha_j}{\sum_{m=0}^c m\alpha_m},$$
which is the formula you derived.
Therefore I agree with your solution.

Here is another approach. Consider the contribution each child makes to $E(K).$
Number all the children from $1$ to $S$ and let $k_i$ be the birth order of child number $i.$ Then
$$ E(K) = \frac1S \sum_{i=1}^S k_i.$$
Separate the sum into subtotals for each size of family.
For example, for $0 \leq m \leq c$ there are $N\alpha_m$ families of size $m$ and each of those families has one child with $k_i = 1,$ one child with $k_i = 2,$ and so forth up to their one child with $k_i = m.$
So the contribution of one family of size $m$ to the sum is $1 + 2 + \cdots + k.$
Then
\begin{align}
\sum_{i=1}^S k_i
& = N\alpha_1 + N\alpha_2(1 + 2) + N\alpha_3(1 + 2 + 3) +
\cdots + N\alpha_m \sum_{k=1}^m k +
\cdots + N\alpha_c \sum_{k=1}^c k \\
& = N\alpha_1 + 3N\alpha_2 + 6N\alpha_3 +
\cdots + \frac12 m(m+1)N\alpha_m +
\cdots + \frac12 c(c+1)N\alpha_c \\
&= \frac12 N \sum_{m=1}^c  m(m+1)\alpha_m.
\end{align}
Therefore
$$ E(K) = \frac{\sum_{i=1}^S k_i}{S}
 = \frac{\sum_{m=1}^c  m(m+1)\alpha_m}{2\sum_{j=0}^c j\alpha_j}.$$
I believe this is equal to your solution as well.
A: The fraction of families with $j$ children is not the same thing as the probability that a child chosen at random belongs to a family with $j$ children. You have used $\alpha_j$ to represent these two different quantities. For example, if half the families have $0$ children, the fraction of families with $j$ children is $1\over2$, but the probability that a child chosen at random belongs to a $0$-child family is $0$. As @lulu suggests, it is not the case that $P(A_j)=\alpha_j$. You have to do more work to figure out what $P(A_j)$ is. It is not just $\alpha_j$.
