Conditional Probability Question EDIT!!!

a)  ￼The probability that any child in a certain family will have blue eyes is 1/4, and this feature is inherited independently by different children in the family.  If there are five children in the family and it is known that at least one of these children has blue eyes, what is the probability that at least three of the children have blue eyes?
b)  Consider a family with the five children just described.  If it is known that the youngest child in the family has blue eyes, what is the probability that at least three of the children have blue eyes?

Hello!  I'm pretty sure I understand part (a), but I'm not sure about part b.  Since the child is actually distinguished in this case, does it change the denominator?
So since its distinguished who actually has the blue eyes, namely, the young child, does the denominator just become 1/4?
Actually it turns out that the denominator is actually 1.  Can someone please explain?
Please note that the answers in (a) and (b) are actually different.  They are NOT the same.  It turns out that knowing who the child is actually simplifies the problem and removes the necessity of using conditionals.
Thanks in advance!
Part a:
Let:
A= event that at least 3 children have blue eyes
B= event that at least 1 child has blue eyes
$\therefore A \subset B$
$\Pr(A \mid B)=\cfrac{\Pr(A \cap B)}{\Pr(B)}=\cfrac{\Pr(A)}{\Pr(B)}=\cfrac{ \sum\limits_{i=3}^5 \binom{5}{i} \cdot (0.25)^i \cdot 0.75^{5-i}}{1-0.75^5}=0.1357\tag{1}$
Part b:
A=event youngest child has blue eyes
B=event at least 3 children have blue eyes
$\Pr(A \mid B)=\cfrac{ \sum\limits_{i=2}^4 \binom{4}{i} \cdot (0.25)^i \cdot 0.75^{4-i}}{1/4??}\tag{2}$

Under you notations, 
the answer of the second part should be rather
$$
Pr(B|\,A)=\frac{Pr(A\cap B)}{Pr(A)}=\frac{Pr(C)}{\frac{1}{4}}
$$
where $C=$ at least 2 children among the 4 children left have blue eyes and "the youngest child has blue eye" .
