If $X$ is the random variable representing the number of girls out of 2 children. The probability to get a child of a given sex rahter than the other is 1/2. X equals 0 with probability 1/4, 1 with probability 1/2, and 2 with probability 1/4.
We can define a new random variable $Z$ the number of girls conditional on the older child being a girl. According to this book (p. 73), when $Z = 1$ the probability is 1/2 and 2 with probability 1/2.
However, when I compute it :
$$\frac{\overbrace{1/2}^{probability \ of \ [the \ number \ of \ girls = 1]}}{\underbrace{1/2}_{probability \ of \ the \ older \ child \ being \ a \ girl}} = 1$$
What do I get wrong ?
Best Answer
Conditional probability. $$ P(R|S) = \frac{P(R\cap S)}{P(S)} $$ Here $R$ is "both children are girls" and $S$ is "the oldest child is a girl".
Let's take sample space consisting of four items: $$ (G,G),\quad(G,B),\quad (B,G),\quad (B,B) $$ each with probability $1/4$. Then $$ R = \{(G,G)\} \\ S = \{(G,G),(G,B)\} \\ R\cap S = \{(G,G)\} $$ and therefore $$ P(R|S) = \frac{P(R\cap S)}{P(S)} = \frac{\;1/4\;}{1/2} = \frac{1}{2} . $$