Help understand the solution of a problem about conditional probability

Question

Sheldon Ross's A First Course in Probability, Chapter 3, Problem 3.75:

Problem. In a certain village, it is traditional for the eldest son (or the older son in a two-son family) and his wife to be responsible for taking care of his parents as they age. In recent years, however, the women of this village, not wanting that responsibility, have not looked favorably upon marrying an eldest son.

(a) If every family in the village has two children, what proportion of all sons are older sons?

(b) If every family in the village has three children, what proportion of all sons are eldest sons? Assume that each child is, independently, equally likely to be either a boy or a girl.

Solution: (a) Choose a child at random, letting $E$ be the event that the child is an eldest son, letting $S$ be the event that it is a son, and letting $A$ be the event that the child's family has at least one son.

$$ P(E \mid S) = \frac {P(ES)} {P(S)} = 2P(E).$$

Why $\frac {P(ES)}{P(S)}=2P(E)$?,then 2P(E)=$[P(E|A)\cdot P(A)+P(E|A')\cdot P(A')]$= 2[$0.5\times 0.75+0 \times 0.25$]=0.75,why P(E|A)=0.5?

(b):P(E|S)=$\frac {P(ES)}{P(S)}=2P(E)$=$[P(E|A)\cdot P(A)+P(E|A')\cdot P(A')]$=$2[\frac 1 3\times\frac 7 8+0\times\frac 1 8]=\frac 7 {12} $,why P(E|A)=$\frac1 3?$

Answer 1

Firstly, since it is given that a child is equally likely to be a boy or a girl, $$P(S)=\frac{1}{2}$$ $$\implies\frac{P(ES)}{P(S)}=2P(ES)$$ Secondly, since a child can only be an "eldest son" if they are, in fact, a son and not a daughter, $$2P(ES)=2P(E)$$

Answer 2

Here is an alternate approach for part $(b)$ you may find helpful.

Suppose you have $100$ families in your village.

The number of eldest sons in your village is precisely the number of families that have at least one boy which is $100\cdot \left(1-\frac{1}{8}\right)$

Meanwhile, the number of boys in your village is $\frac{1}{2}\cdot 3\cdot 100=150$

So, the proportion of sons that are eldest sons is $\frac{100\cdot \left(1-\frac{1}{8}\right)}{150}=\frac{7}{12}$.

It's not hard to see using this approach that if each family had $n$ children then the souhght after proportion would be $\frac{2\left(1-(0.5)^n\right)}{n}$.

Answer 3

Every family that has at least one son has exactly one eldest son.

If all families have $2$ children, then given that you chose a child at random and the child happened to be from a family with at least one son, you have chosen from a subpopulation of children in which half the children are eldest sons and half are not eldest sons. The child you chose has a $\frac12$ chance of being one of the eldest sons.

If all families have $3$ children, then given that you chose a child at random and the child happened to be from a family with at least one son, one-third of the children in that subpopulation are eldest sons and two-thirds are not.

I think it is a bit of a leap to just say that $P(E\mid A) = \frac12$ or $P(E\mid A) = \frac13$ without any explanation, but perhaps there were earlier problems with analogous conditional probabilities that were more thoroughly explained.