A couple plans on having 2 children. Given that at least one of them is a boy, the probability that both are boys is
$$
\frac{P(both~boys)}{P(at~least~one~is~a~boy)} = \frac{0.25}{0.75} =\frac{1}{3}
$$
Furthermore, a textbook I am reading claims that the probability of both children being boys given that at least one is a boy born in spring is $\frac{7}{15}$. It doesn't explain its solution.
Why? What does the season have anything to do with whether both children are boys?
Best Answer
Specifying one of the boys was born in spring, increases the probability the second child is also a boy, because parents with two boys are more likely to have one born in spring than parents with just one. You are basically calculating:
$$ \frac{P_A}{P_B}=\frac{\frac{1}{4}(P_D+P_E)}{1-P_C}=\frac{\frac{1}{4}(\frac{6}{16}+\frac{1}{16})}{1-(\frac{7}{8})^2}=\frac{7}{15} $$ With $P_A$ as the probability of having two boys, at least one of which born in spring; $P_B$ of having at least one boy born in spring; $P_C$ of having no boys born in spring; $P_D$ of exactly one out of two children being born in spring; and $P_E$ of two out of two children being born in spring.