Probability of getting a boy (explanation of answer given in textbook) Context:In the countryside, where sons traditionally have been valued, if the first child is a son, the couple may have no more children. But if the first child is a daughter, the couple may have another child. Regardless of the sex of the second child, no more are permitted. How will this policy affect the mix of males and females?
Question:To pose the question mathematically, what is the probability that a randomly selected child from the countryside is a boy?
Answer:Let B be the event that the randomly selected child from the countryside is a boy. Let E be the event that the randomly selected child is the first child of the family and F be the event that he or she is the second child of the family. Clearly, P(E) = 2/3 and P(F) = 1/3. By the law of total probability,
$$P(B)=P(B|E)P(E)*P(B|F)P(F)=1/2*2/3+1/2*1/3=1/2$$
I do not get why the probabilty of picking the first child is 2/3. The sample space is $$\{B,G,GG,GB\} $$ There are 4 first children out of 6, is that why it is 2/3? But the probability of picking a boy in a one child family is 1/2. Therefore I thought that the the probability of E should be$$ P(E)=1/2+1/2 * P(\text{picking first child among } \{G,GG,GB\})$$. Thus in my opinion, there are not enough assumptions specified in the question. Of course the probability of getting a boy or girl at any time is 1/2.
 A: The question is making an implicit assumption that all families with a first child of a daughter have a second child. Thus, the effective sample space is $\{B,GG,GB\}$. The probability of $B$ is $0.5$, with $GG$ and $GB$ having equal probability each summing to $0.5$, i.e., they are each $0.25$. Thus, among $n$ families, there will be $0.5n$ first boys and $0.5n$ first girls, but $0.5 \times 0.5n = 0.25n$ second girls and $0.25n$ second boys. Thus, there are $0.5n + 0.5n = n$ first children out of a total of $0.5n + 0.5n + 0.25n + 0.25n = 1.5n$ children, so the probability of a first child is $\frac{n}{1.5n} = \frac{2}{3}$. Note if you divide the numbers by  $n$, or equivalently set $n = 1$, you'll get the average values per family, with this being $0.5$ each of first boys and girls, and $0.25$ each of second boys and girls.
Also, in addition to the explanation given in the book, note there are $0.5n$ first children which are boys and $0.25n$ second children which are boys. Thus, as you stated, the probability of a boy would be $\frac{0.75n}{1.5n} = 0.5$.
A: Although the sample space is that $B, G, GB, GG$ a family with only one girl will never happen.  Once a girl is born the family will have a second child in a try for a boy.  (Bleagh)
So half the families will have one boy.  One quarter of the families will have a boy and a girl, and one quarter of the families will have $2$ girls.
But then the book does a complicated conditional probability.  Why? How cares if the child you pick is  first born or second.  Just count how many boys and girls there are.
If there are $n$ families there will by $\frac n2$ boys from the $\frac n2$ $1$ boy families, and $\frac n4$ boys from the $\frac n4$ boy and girl families.  That's $\frac {3n}4$ boys.  There are $\frac n4$ girls from that boy and girl families, and $2*\frac n4$ girls from the $2$ girl families.
So there are $\frac {3n}4$ boys and $\frac {3n}4$ girls and as those are equal the probability is $\frac 12$.
FWIW (which isnt much) There are $n$ first borns and $\frac n2$ second borns but I don't see why that is important.  Oh.... I guess have the first borns are boys and half the second born are boys so... again probability is $\frac 12$.
