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Assume that each child born is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that (i) the youngest is a girl, (ii) at least one is girl?
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Let first and second girls are denoted by $\mathrm{G}_1$ and $\mathrm{G}_2$ and Boys by $B_1$ and $B_2$.
Sample space
$$
\mathrm{S}=\left\{\left(\mathrm{G}_1 \mathrm{G}_2\right),\left(\mathrm{G}_1 \mathrm{~B}_2\right),\left(\mathrm{G}_2 \mathrm{~B}_1\right),\left(\mathrm{B}_1 \mathrm{~B}_2\right)\right\}
$$
Let $\mathrm{A}=$ Both the children are girls $=\left\{\mathrm{G}_1 \mathrm{G}_2\right\}$
$\mathrm{B}=$ youngest child is a girls $=\left\{\mathrm{G}_1 \mathrm{G}_2, \mathrm{~B}_1 \mathrm{G}_2\right\}$
$\mathrm{C}=$ at least one is a girl $=\left\{\mathrm{G}_1, \mathrm{~B}_2, \mathrm{G}_1 \mathrm{G}_2, \mathrm{~B}_1 \mathrm{G}_2\right\}$
$$
\begin{aligned}
&\mathrm{A} \cap \mathrm{B}=\left\{\mathrm{G}_1 \mathrm{G}_2\right\}, \mathrm{A} \cap \mathrm{C}=\left\{\mathrm{G}_1 \mathrm{G}_2\right\} \\
&\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{4}
\end{aligned}
$$
$$
\mathrm{P}(\mathrm{B})=\frac{2}{4}, \mathrm{P}(\mathrm{C})=\frac{3}{4}
$$
(i) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{P(A \cap B)}{P(B)}=\frac{1}{4} \div \frac{2}{4}=\frac{1}{2}$
(ii) $P(A \mid C)=\frac{P(A \cap C)}{P(C)}=\frac{1}{4} \div \frac{3}{4}=\frac{1}{3}$.
Sample space
$$
\mathrm{S}=\left\{\left(\mathrm{G}_1 \mathrm{G}_2\right),\left(\mathrm{G}_1 \mathrm{~B}_2\right),\left(\mathrm{G}_2 \mathrm{~B}_1\right),\left(\mathrm{B}_1 \mathrm{~B}_2\right)\right\}
$$
Let $\mathrm{A}=$ Both the children are girls $=\left\{\mathrm{G}_1 \mathrm{G}_2\right\}$
$\mathrm{B}=$ youngest child is a girls $=\left\{\mathrm{G}_1 \mathrm{G}_2, \mathrm{~B}_1 \mathrm{G}_2\right\}$
$\mathrm{C}=$ at least one is a girl $=\left\{\mathrm{G}_1, \mathrm{~B}_2, \mathrm{G}_1 \mathrm{G}_2, \mathrm{~B}_1 \mathrm{G}_2\right\}$
$$
\begin{aligned}
&\mathrm{A} \cap \mathrm{B}=\left\{\mathrm{G}_1 \mathrm{G}_2\right\}, \mathrm{A} \cap \mathrm{C}=\left\{\mathrm{G}_1 \mathrm{G}_2\right\} \\
&\mathrm{P}(\mathrm{A} \cap \mathrm{B})=\frac{1}{4}, \mathrm{P}(\mathrm{A} \cap \mathrm{C})=\frac{1}{4}
\end{aligned}
$$
$$
\mathrm{P}(\mathrm{B})=\frac{2}{4}, \mathrm{P}(\mathrm{C})=\frac{3}{4}
$$
(i) $\mathrm{P}(\mathrm{A} \mid \mathrm{B})=\frac{P(A \cap B)}{P(B)}=\frac{1}{4} \div \frac{2}{4}=\frac{1}{2}$
(ii) $P(A \mid C)=\frac{P(A \cap C)}{P(C)}=\frac{1}{4} \div \frac{3}{4}=\frac{1}{3}$.
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