Why is the probability of having 2 boys 7/15? 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?
 A: 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.
A: If the probability that a child is a boy is $\frac 12$ and the probability that it is born in spring is $\frac 14$, the probability that it is a boy born in spring is $\frac 18$.  The probability that at least one of two children is a boy born in spring is therefore $1 - (1 - \frac 18)^2$ = $\frac{15}{64}$.
The probability that both children are boys born in spring is $(\frac 18)^2 = \frac{1}{64}$, so the probability that exactly one boy is born in spring is $\frac{15-1}{64}$. If exactly one boy is born in spring, the probability that the other child is a boy (not a boy born in spring, so either a boy born in one of the other three seasons or a girl born in any of the four seasons) is $\frac{3}{3+4}=\frac{3}{7}$. The probability that one child is a boy born in spring and the other is a boy born in one of the other three seasons is therefore $\frac{15-1}{64}$ x $ \frac{3}{7}$ = $\frac{6}{64}$. This makes a total joint probability of two boys and at least one boy born in spring of $\frac{1+6}{64}=\frac{7}{64}$.
The desired conditional probability is therefore $\frac{7}{64}$ / $\frac{15}{64}$ = $\frac{7}{15}$.
Addendum 24/9/2021
Addressing questions in comments by koss:
Q1 “Why does Pr(exactly 1 boy born in spring) = Pr(At least 1 of the 2 children is a boy born in spring) – Pr(both children are boys born in spring)?”
A1 The following three possibilities are exhaustive: a) No boys born in spring; b) Exactly 1 boy born in spring; c) 2 boys born in spring.  Pr(At least 1 is a boy born in spring) = Pr(b) + Pr(c).  Therefore Pr(b) = Pr(At least 1 is a boy born in spring) – Pr(c).
Q2 “Why must you add 1/64 to 6/64?”
A2 6/64 is the probability that one child is a boy born in spring and the other is a boy born in one of the other three seasons.  Thus it excludes the probability that both boys were born in spring.  But we are also interested in the latter probability because the question asks for the probability of both children being boys given that at least one is a boy born in spring.  So we must add that probability which is 1/64.
Q3 “Why isn’t Pr(one child is a boy born in spring and the other is a boy born in one of the other three seasons) = Pr(two boys and at least one boy born in spring)?
A3 The former excludes, but the latter includes, the case of two boys both born in spring.
A: A couple plans on having 2 children. Judging from the probabilities 0.25 of both boys and 0.75 of at least 1 boy, I can surely say that P(boy) = 0.5 and P(girl)=0.5.
I presume that you are asking how is the probability of both boys given that at least 1 boy is born in spring = $\frac{7}{15}$
Now, since the word "spring" is somehow introduced, lets assume that it is possible for a child to be born in any of the 4 seasons with equal probability. 
Lets now have a look at the sample space:
A = All events in Set(Both Boys | One is born in Spring) = 2*(Boys in Autumn, Winter, Summer) + (Both boys in Spring)
B = All events in Set (One Boy born in Spring) = 2*(Boys in Autumn, Winter, Summer) + (Both boys in Spring) + 2*(Girl in Autumn, Winter, Summer, Spring)
Hence, Probability of (Both Boys|One boy is born in spring) = $\frac{P(A)}{P(B)}$
We assume no chance of twins.
Hope that solves the question.
A: A simple way is to keep track of the older child.
If the older child is a boy born in spring, the younger child could be a girl or boy born in any season ($8$ cases, $4$ of them favorable)
If the older child is not a boy born in spring, the younger child is. The older child could then be a girl born in any season, or a boy born in non-spring
($7$ cases, $3$ of them favorable)
Thus $Pr = \Large{\frac{4+3}{{8+7}} = \frac{7}{15}}$
