Q:

Conditional probability for the first kid being female

I've got the following question:
We have a family with 3 children and every child has the same probability for male or female. What is the chance for the youngest child being female, if we know
a) the oldest one is female
b) we have a least one female child
c) we have exactly one female
I tried to solve it we conditional probability:
a) $P(X_1=M|X_3=M)=\frac{P(X_1=M,X_3=M)}{P(X_3=M)}=\frac{\frac{1}{2} \cdot \frac{1}{2}}{\frac{1}{2}}=\frac{1}{2}$
b) $P(X_1=M \vee X_2=M \vee X_3=M|X_3=M)=\frac{X_1=M \vee X_2=M \vee X_3=M}{X_1=M}=\frac{\frac{7}{8} \cdot \frac{1}{2}}{\frac{1}{2}}=\frac{7}{8}$
c) $P(X_i=M|X_1=M)=\frac{\frac{3}{8} \cdot \frac{1}{2}}{\frac{1}{2}}=\frac{3}{8}$
Are my ideas correct and also my solutions?
Thanks a lot in advance.

A:

Let $F$ be the event that the first child is female and $E_a$, $E_b$, $E_c$ be the evidence in each case. We will use Bayes' theorem.
(a) You state that "every child has the same probability for male or female". Hence, information about any individual child does not give you any more information. Thus,
$$
P(F|E_a)=\frac{P(E_a|F)P(F)}{P(E_a)}=\frac{1/2\times 1/2}{1/2} = 1/2.
$$
(b) There is at least one female child. There are $2^3$ possible configurations of children. However, one of them (all male) is not possible. Thus $P(E_b)=7/8$. We have that $P(E_b|F)=1$ because $E_b$ is true if the first child is female. $P(F)$ remains $1/2$. Hence,
$$
P(F|E_b)=\frac{P(E_b|F)P(F)}{P(E_b)}=\frac{1\times 1/2}{7/8}=\frac{4}{7}.
$$
(c) There is exactly one female child. Again, there are $2^3$ possible configurations of children. However, only three are allowed (first child female, second child female, third child female). Thus $P(E_c)=3/8$. The probability that $E_c$ occurs given that the first child is female is $P(E_c|F)=1/2\times 1/2=1/4$ because we require the second and third child to be both boys. $P(F)$ remains $1/2$. Hence,
$$
P(F|E_c)=\frac{P(E_c|F)P(F)}{P(E_c)}=\frac{1/4\times 1/2}{3/8}=\frac{1}{3}.
$$

