We have the following probabilites:
[tex]\begin{gathered} P(\text{had a son)=P(s)}=0.32 \\ P(\text{had a daughter)}=P(d)=0.3 \\ P(\text{had both son and daughter)}=P(d\cap s)=0.11 \end{gathered}[/tex]Following the definition of conditional probability:
[tex]P(A|B)=\frac{P(A\cap B)}{P(A)}[/tex]In this case, we want to calculate the conditional probability that a person has a daughter given that he/she already has a son. Then, the probability is:
[tex]P(d|s)=\frac{P(d\cap s)}{P(s)})=\frac{0.11}{0.32}=0.34[/tex]therefore, the conditional probability that a person who has a son also has a daughter is 34%
