Conditional probability. A family has 3 children

Question

A couple has $3$ children.

1) What is the probability that three are girls if the eldest is a girl?

2) Solve the same problem, provided that it is known that one of the children is a girl.

1) Let $A$ be the event that the $3$ children are girls. And let $B$ be the event that the eldest is a girl. The sample space is $GGG$ $BGG$ $BBG$ $BBB$.

Since it is given that the eldest is a girl, then $P(B)=\dfrac34$.

$P(A$ and $B)=\dfrac14$

$P(A|B)=\dfrac{(1/4)}{(3/4)}=\dfrac13$

2) I am confused about the second part. Give me a hint, please.

Answer 1

Let $C_1,C_2,C_3$ take values in $\{G,B\}$ such that e.g. $\{C_1=G\}$ denotes the event that the firstborn child is a girl. To keep things more general let $p$ denote the probability on a girl.

1) To be found is: $$P(C_1=C_2=C_3=G\mid C_1=G)=P(C_2=C_3=G\mid C_1=G)=P(C_2=G\wedge C_3=G)=P(C_2=G)P(C_3=G)=p^2$$

The second and third equality both rest on independence.

2) To be found is:$$P(C_1=C_2=C_3=G\mid G\in\{C_1,C_2,C_3\})=$$$$P(C_1=C_2=C_3=G\wedge G\in\{C_1,C_2,C_3\})/P(G\in\{C_1,C_2,C_3\})=$$$$P(C_1=C_2=C_3=G)/[1-P(G\notin\{C_1,C_2,C_3\})]=\frac{p^3}{1-(1-p)^3}$$

Answer 2

Here, the (conditional) sample spaces with equally likely outcomes are easily listed.

So let $\Omega$ denote the original sample space of all possible gender combinations of siblings.

Let $\Omega_{C}$ denote the restricted sample space induced by a given condition $C$. (Actually, $\Omega_{C} = C$).

1) Here $C$ is "first born is a girl". So, let $G$ denote the first born girl and letters $g$,$b$ may denote the other siblings: $\Omega_{C} = \{Ggg, Ggb, Gbg, Gbb\}$ $$\rightarrow P(Ggg | C)= \frac{1}{4}$$

2) Here $C$ is the condition that "(at least) one kid is a girl". Similarly as in 1) we get: $\Omega_{C} = \Omega \setminus \{bbb\} = \{ggg, ggb, gbg, gbb, bgg, bgb, bbg\}$ $$\rightarrow P(ggg | C)= \frac{1}{7}$$